diff options
25 files changed, 13896 insertions, 0 deletions
diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_3.ipynb new file mode 100644 index 00000000..d1825266 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/Chapter9_3.ipynb @@ -0,0 +1,397 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6d953e27719d7daa72fde544d6031f1b10e1023af731eda189a4bce609e51019"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter09:Numerical Solution of Partial Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.1:pg-350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#standard five point formula\n",
+ "#example 9.1\n",
+ "#page 350\n",
+ "\n",
+ "u2=5.0;u3=1.0;\n",
+ "for i in range(0,3):\n",
+ " u1=(u2+u3+6.0)/4.0\n",
+ " u2=(u1/2.0)+(5.0/2.0)\n",
+ " u3=(u1/2.0)+(1.0/2.0)\n",
+ " print\" the values are u1=%d\\t u2=%d\\t u3=%d\\t\\n\\n\" %(u1,u2,u3)\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n",
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n",
+ " the values are u1=3\t u2=4\t u3=2\t\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.2:pg-351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#solution of laplace equation by jacobi method,gauss-seidel method and SOR method\n",
+ "#example 9.2\n",
+ "#page 351\n",
+ "u1=0.25;u2=0.25;u3=0.5;u4=0.5;#initial values\n",
+ "print \"jacobis iteration process\\n\\n\"\n",
+ "print\"u1\\t u2\\t u3\\t u4\\t \\n\\n\"\n",
+ "print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4)\n",
+ "for i in range(0,7):\n",
+ " u11=(0+u2+0+u4)/4\n",
+ " u22=(u1+0+0+u3)/4\n",
+ " u33=(1+u2+0+u4)/4\n",
+ " u44=(1+0+u3+u1)/4\n",
+ " u1=u11;u2=u22;u3=u33;u4=u44;\n",
+ " print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u11,u22,u33,u44) \n",
+ "print \" gauss seidel process\\n\\n\"\n",
+ "u1=0.25;u2=0.3125;u3=0.5625;u4=0.46875;#initial values\n",
+ "print \"u1\\t u2\\t u3\\t u4\\t \\n\\n\"\n",
+ "print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4)\n",
+ "for i in range(0,4):\n",
+ "\n",
+ " u1=(0.0+u2+0.0+u4)/4.0\n",
+ " u2=(u1+0.0+0.0+u3)/4.0\n",
+ " u3=(1.0+u2+0.0+u4)/4.0\n",
+ " u4=(1.0+0.0+u3+u1)/4.0\n",
+ " print \"%f\\t %f\\t %f\\t %f\\t \\n\" %(u1,u2,u3,u4) \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "jacobis iteration process\n",
+ "\n",
+ "\n",
+ "u1\t u2\t u3\t u4\t \n",
+ "\n",
+ "\n",
+ "0.250000\t 0.250000\t 0.500000\t 0.500000\t \n",
+ "\n",
+ "0.187500\t 0.187500\t 0.437500\t 0.437500\t \n",
+ "\n",
+ "0.156250\t 0.156250\t 0.406250\t 0.406250\t \n",
+ "\n",
+ "0.140625\t 0.140625\t 0.390625\t 0.390625\t \n",
+ "\n",
+ "0.132812\t 0.132812\t 0.382812\t 0.382812\t \n",
+ "\n",
+ "0.128906\t 0.128906\t 0.378906\t 0.378906\t \n",
+ "\n",
+ "0.126953\t 0.126953\t 0.376953\t 0.376953\t \n",
+ "\n",
+ "0.125977\t 0.125977\t 0.375977\t 0.375977\t \n",
+ "\n",
+ " gauss seidel process\n",
+ "\n",
+ "\n",
+ "u1\t u2\t u3\t u4\t \n",
+ "\n",
+ "\n",
+ "0.250000\t 0.312500\t 0.562500\t 0.468750\t \n",
+ "\n",
+ "0.195312\t 0.189453\t 0.414551\t 0.402466\t \n",
+ "\n",
+ "0.147980\t 0.140633\t 0.385775\t 0.383439\t \n",
+ "\n",
+ "0.131018\t 0.129198\t 0.378159\t 0.377294\t \n",
+ "\n",
+ "0.126623\t 0.126196\t 0.375872\t 0.375624\t \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.4:pg-354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#poisson equation\n",
+ "#exaample 9.4\n",
+ "#page 354\n",
+ "u2=0.0;u4=0.0;\n",
+ "print \" u1\\t u2\\t u3\\t u4\\t\\n\\n\"\n",
+ "for i in range(0,6):\n",
+ " u1=(u2/2.0)+30.0\n",
+ " u2=(u1+u4+150.0)/4.0\n",
+ " u4=(u2/2.0)+45.0\n",
+ " print \"%0.2f\\t %0.2f\\t %0.2f\\t %0.2f\\n\" %(u1,u2,u2,u4)\n",
+ "print \" from last two iterates we conclude u1=67 u2=75 u3=75 u4=83\\n\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " u1\t u2\t u3\t u4\t\n",
+ "\n",
+ "\n",
+ "30.00\t 45.00\t 45.00\t 67.50\n",
+ "\n",
+ "52.50\t 67.50\t 67.50\t 78.75\n",
+ "\n",
+ "63.75\t 73.12\t 73.12\t 81.56\n",
+ "\n",
+ "66.56\t 74.53\t 74.53\t 82.27\n",
+ "\n",
+ "67.27\t 74.88\t 74.88\t 82.44\n",
+ "\n",
+ "67.44\t 74.97\t 74.97\t 82.49\n",
+ "\n",
+ " from last two iterates we conclude u1=67 u2=75 u3=75 u4=83\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 59
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.6:pg-362"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#bender-schmidt formula\n",
+ "#example 9.6\n",
+ "#page 362\n",
+ "def f(x):\n",
+ " return (4*x)-(x*x)\n",
+ "#u=[f(0),f(1),f(2),f(3),f(4)]\n",
+ "u1=f(0);u2=f(1);u3=f(2);u4=f(3);u5=f(4);\n",
+ "u11=(u1+u3)/2\n",
+ "u12=(u2+u4)/2\n",
+ "u13=(u3+u5)/2\n",
+ "print \"u11=%0.2f\\t u12=%0.2f\\t u13=%0.2f\\t \\n\" %(u11,u12,u13)\n",
+ "u21=(u1+u12)/2.0\n",
+ "u22=(u11+u13)/2.0\n",
+ "u23=(u12+0)/2.0\n",
+ "print \"u21=%0.2f\\t u22=%0.2f\\t u23=%0.2f\\t \\n\" %(u21,u22,u23)\n",
+ "u31=(u1+u22)/2.0\n",
+ "u32=(u21+u23)/2.0\n",
+ "u33=(u22+u1)/2.0\n",
+ "print \"u31=%0.2f\\t u32=%0.2f\\t u33=%0.2f\\t \\n\" % (u31,u32,u33)\n",
+ "u41=(u1+u32)/2.0\n",
+ "u42=(u31+u33)/2.0\n",
+ "u43=(u32+u1)/2.0\n",
+ "print \"u41=%0.2f\\t u42=%0.2f\\t u43=%0.2f\\t \\n\" % (u41,u42,u43)\n",
+ "u51=(u1+u42)/2.0\n",
+ "u52=(u41+u43)/2.0\n",
+ "u53=(u42+u1)/2.0\n",
+ "print \"u51=%0.2f\\t u52=%0.2f\\t u53=%0.2f\\t \\n\" % (u51,u52,u53)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "u11=2.00\t u12=3.00\t u13=2.00\t \n",
+ "\n",
+ "u21=1.50\t u22=2.00\t u23=1.50\t \n",
+ "\n",
+ "u31=1.00\t u32=1.50\t u33=1.00\t \n",
+ "\n",
+ "u41=0.75\t u42=1.00\t u43=0.75\t \n",
+ "\n",
+ "u51=0.50\t u52=0.75\t u53=0.50\t \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.7:pg-363"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#bender-schimdt's formula and crank-nicolson formula\n",
+ "#example 9.7\n",
+ "#page 363\n",
+ "#bender -schimdt's formula\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "z=math.pi\n",
+ "def f(x,t):\n",
+ " return math.exp(z*z*t*-1)*sin(z*x)\n",
+ "#u=[f(0,0),f(0.2,0),f(0.4,0),f(0.6,0),f(0.8,0),f(1,0)];\n",
+ "u1=f(0,0)\n",
+ "u2=f(0.2,0)\n",
+ "u3=f(0.4,0)\n",
+ "u4=f(0.6,0)\n",
+ "u5=f(0.8,0)\n",
+ "u6=f(1.0,0)\n",
+ "u11=u3/2;u12=(u2+u4)/2;u13=u12;u14=u11;\n",
+ "print \"u11=%f\\t u12=%f\\t u13=%f\\t u14=%f\\n\\n\" % (u11,u12,u13,u14)\n",
+ "u21=u12/2;u22=(u12+u14)/2;u23=u22;u24=u21;\n",
+ "print \"u21=%f\\t u22=%f\\t u23=%f\\t u24=%f\\n\\n\" % (u21,u22,u23,u24)\n",
+ "print \"the error in the solution is: %f\\n\\n\" % (math.fabs(u22-f(0.6,0.04)))\n",
+ "#crank-nicolson formula\n",
+ "#by putting i=1,2,3,4 we obtain four equation\n",
+ "A=matrix([[4, -1, 0, 0] ,[-1, 4, -1, 0],[0, -1, 4, -1],[0, 0, -1, 4]])\n",
+ "C=matrix([[0.9510],[1.5388],[1.5388],[0.9510]])\n",
+ "X=A.I*C\n",
+ "print \"u00=%f\\t u10=%f\\t u20=%f\\t u30=%f\\t\\n\\n\" %(X[0][0],X[1][0],X[2][0],X[3][0])\n",
+ "print \"the error in the solution is: %f\\n\\n\" %(abs(X[1][0]-f(0.6,0.04)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "u11=0.475528\t u12=0.769421\t u13=0.769421\t u14=0.475528\n",
+ "\n",
+ "\n",
+ "u21=0.384710\t u22=0.622475\t u23=0.622475\t u24=0.384710\n",
+ "\n",
+ "\n",
+ "the error in the solution is: 0.018372\n",
+ "\n",
+ "\n",
+ "u00=0.399255\t u10=0.646018\t u20=0.646018\t u30=0.399255\t\n",
+ "\n",
+ "\n",
+ "the error in the solution is: 0.005172\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.8:pg-364"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "#heat equation using crank-nicolson method\n",
+ "#example 9.8\n",
+ "#page 364\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "z=0.01878;\n",
+ "#h=1/2;l=1/8,i=1\n",
+ "u01=0.0;u21=1.0/8.0;\n",
+ "u11=(u21+u01)/6.0;\n",
+ "print \" u11=%f\\n\\n\" % (u11)\n",
+ "print \"error is %f\\n\\n\" % (math.fabs(u11-z))\n",
+ "#h=1/4,l=1/8,i=1,2,3\n",
+ "A=matrix([[-3.0 ,-1.0 ,0.0],[1.0,-3.0,1.0],[0.0,1.0,-3.0]])\n",
+ "C=matrix([[0.0],[0.0],[-0.125]])\n",
+ "#here we found inverese of A then we multipy it with C\n",
+ "X=A.I*C\n",
+ "print \"u12=%f\\n\\n\" % (X[1][0])\n",
+ "print \"error is %f\\n\\n\" %(math.fabs(X[1][0]-z))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " u11=0.020833\n",
+ "\n",
+ "\n",
+ "error is 0.002053\n",
+ "\n",
+ "\n",
+ "u12=0.013889\n",
+ "\n",
+ "\n",
+ "error is 0.004891\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_3.ipynb new file mode 100644 index 00000000..cdfe1170 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter1_3.ipynb @@ -0,0 +1,625 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:ab07d04dc98e9e897aaecd3d260ae7ca8f8d0636f3903bebd7713e9a345cdce9"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter01:Errors in Numerical Calculations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.1:pg-7"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.1\n",
+ "#rounding off\n",
+ "#page 7\n",
+ "a1=1.6583\n",
+ "a2=30.0567\n",
+ "a3=0.859378\n",
+ "a4=3.14159\n",
+ "print \"\\nthe numbers after rounding to 4 significant figures are given below\\n\"\n",
+ "print \" %f %.4g\\n'\" %(a1,a1)\n",
+ "print \" %f %.4g\\n\" %(a2,a2)\n",
+ "print \" %f %.4g\\n\" %(a3,a3)\n",
+ "print \" %f %.4g\\n\" %(a4,a4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "the numbers after rounding to 4 significant figures are given below\n",
+ "\n",
+ " 1.658300 1.658\n",
+ "'\n",
+ " 30.056700 30.06\n",
+ "\n",
+ " 0.859378 0.8594\n",
+ "\n",
+ " 3.141590 3.142\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.2:pg-9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.2\n",
+ "#percentage accuracy\n",
+ "#page 9\n",
+ "import math\n",
+ "x=0.51 # the number given\n",
+ "n=2 #correcting upto 2 decimal places\n",
+ "d=math.pow(10,-n)\n",
+ "d=d/2.0\n",
+ "p=(d/x)*100 #percentage accuracy\n",
+ "print \"the percentage accuracy of %f after correcting to two decimal places is %f\" %(x,p)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the percentage accuracy of 0.510000 after correcting to two decimal places is 0.980392\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.3:pg-9"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.3\n",
+ "#absolute and relative errors\n",
+ "#page 9\n",
+ "X=3.1428571 #approximate value of pi\n",
+ "T_X=3.1415926 # true value of pi\n",
+ "A_E=T_X-X #absolute error\n",
+ "R_E=A_E/T_X #relative error\n",
+ "print \"Absolute Error = %0.7f \\n Relative Error = %0.7f\" %(A_E,R_E)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Absolute Error = -0.0012645 \n",
+ " Relative Error = -0.0004025\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.4:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.4\n",
+ "#best approximation\n",
+ "#page 10\n",
+ "X=1/3 #the actual number\n",
+ "X1=0.30\n",
+ "X2=0.33\n",
+ "X3=0.34\n",
+ "E1=abs(X-X1)\n",
+ "E2=abs(X-X2)\n",
+ "E3=abs(X-X3)\n",
+ "if E1<E2:\n",
+ " if E1<E3:\n",
+ " B_A=X1\n",
+ "elif E2<E1:\n",
+ " if E2<E3:\n",
+ " B_A=X2\n",
+ "elif E3<E2:\n",
+ " if E3<E1:\n",
+ " B_A=X3\n",
+ "print \"the best approximation of 1/3 is %f\" %(B_A)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the best approximation of 1/3 is 0.300000\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.5:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.5\n",
+ "#page 10\n",
+ "import math\n",
+ "n=8.6 # the corrected number\n",
+ "N=1 #the no is rounded to one decimal places\n",
+ "E_A=math.pow(10,-N)/2\n",
+ "E_R=E_A/n\n",
+ "print \"the relative error of the number is:%0.4f\" %(E_R)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the relative error of the number is:0.0058\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.6:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 1.6\n",
+ "#absolute error and relative error\n",
+ "#page 10\n",
+ "import math\n",
+ "s=math.sqrt(3)+math.sqrt(5)+math.sqrt(7) #the sum square root of 3,5,7\n",
+ "n=4\n",
+ "Ea=3*(math.pow(10,-n)/2) #absolute error\n",
+ "R_E=Ea/s\n",
+ "print \"the sum of square roots is %0.4g \\n\" %(s)\n",
+ "print \"the absolute error is %f \\n\" %(Ea)\n",
+ "print \"the relative error is %f\" %(R_E)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the sum of square roots is 6.614 \n",
+ "\n",
+ "the absolute error is 0.000150 \n",
+ "\n",
+ "the relative error is 0.000023\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.7:pg-10"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#absolute error\n",
+ "#example 1.7\n",
+ "#page 10\n",
+ "n=[0.1532, 15.45, 0.0000354, 305.1, 8.12, 143.3, 0.0212, 0.643, 0.1734] #original numbbers\n",
+ "#rounding all numbers to 2 decimal places\n",
+ "n=[305.1, 143.3, 0.15,15.45, 0.00, 8.12, 0.02, 0.64, 0.17] \n",
+ "sum=0;\n",
+ "#l=length(n);\n",
+ "for i in range(len(n)):\n",
+ " sum=sum+n[i];\n",
+ "\n",
+ "E_A=2*math.pow(10,-1)/2+7*math.pow(10,-2)/2\n",
+ "print \"the absolute error is:%0.2f\" %(E_A)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the absolute error is:0.14\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.8:pg-11"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#difference in 3 significant figures\n",
+ "#example 1.8\n",
+ "#page 11\n",
+ "X1=math.sqrt(6.37)\n",
+ "X2=math.sqrt(6.36)\n",
+ "d=X1-X2 #difference between two numbers\n",
+ "print \"the differencecorrected to 3 significant figures is %0.3g\" %(d)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the differencecorrected to 3 significant figures is 0.00198\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.9:pg-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.10\n",
+ "#page 12\n",
+ "a=6.54\n",
+ "b=48.64\n",
+ "c=13.5\n",
+ "da=0.01\n",
+ "db=0.02\n",
+ "dc=0.03\n",
+ "s=math.pow(a,2)*math.sqrt(b)/math.pow(c,3)\n",
+ "#disp(s,'s=')\n",
+ "print \"s=%f\" %(s)\n",
+ "r_err=2*(da/a)+(db/b)/2+3*(dc/c);\n",
+ "print \"the relative error is :%f\" %(r_err)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s=0.121241\n",
+ "the relative error is :0.009930\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.11:pg-13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#relative error\n",
+ "#example 1.11\n",
+ "#page 13\n",
+ "import math\n",
+ "x=1\n",
+ "y=1\n",
+ "z=1\n",
+ "u=(5*x*math.pow(y,3))/math.pow(z,3)\n",
+ "dx=0.001\n",
+ "dy=0.001\n",
+ "dz=0.001\n",
+ "max=((5*math.pow(y,2))/math.pow(z,3))*dx+((10*x*y)/math.pow(z,3))*dy+((15*x*math.pow(y,2))/math.pow(z,4))*dz\n",
+ "e=max/u\n",
+ "print \" the relative error is :%f\" %(e)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the relative error is :0.006000\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.12:pg-12"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor series\n",
+ "#example 1.12\n",
+ "#page 12\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)+5*x-10\n",
+ "def f1(x):\n",
+ " return 3*math.pow(x,2)-6*x+5\n",
+ "def f2(x):\n",
+ " return 6*x-6\n",
+ "def f3(x):\n",
+ " return 6\n",
+ "D=[0,f(0), f1(0), f2(0), f3(0)]\n",
+ "S1=0;\n",
+ "h=1;\n",
+ "for i in range(1,5):\n",
+ " S1=S1+math.pow(h,i-1)*D[i]/math.factorial(i-1)\n",
+ " \n",
+ "print \"the third order taylors series approximation of f(1) is :%d\" %(S1)\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the third order taylors series approximation of f(1) is :-7\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.13:pg-16"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor series\n",
+ "#example 1.13\n",
+ "#page 16\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sin(x)\n",
+ "def f1(x):\n",
+ " return math.cos(x)\n",
+ "def f2(x):\n",
+ " return -1*math.sin(x)\n",
+ "def f3(x):\n",
+ " return -1*math.cos(x)\n",
+ "def f4(x):\n",
+ " return math.sin(x)\n",
+ "def f5(x):\n",
+ " return math.cos(x)\n",
+ "def f6(x):\n",
+ " return -1*math.sin(x)\n",
+ "def f7(x):\n",
+ " return -1*math.cos(x)\n",
+ "D=[0,f(math.pi/6), f1(math.pi/6), f2(math.pi/6), f3(math.pi/6), f4(math.pi/6), f5(math.pi/6), f6(math.pi/6), f7(math.pi/6)]\n",
+ "S1=0\n",
+ "h=math.pi/6\n",
+ "print \"order of approximation computed value of sin(pi/3) absolute eror\\n\\n\"\n",
+ "for j in range(1,10):\n",
+ " for i in range(1,j):\n",
+ " S1=S1+math.pow(h,i-1)*D[i]/math.factorial(i-1) \n",
+ " print \"%d %0.9f %0.9f\\n\" %(j,S1,abs(math.sin(math.pi/3)-S1))\n",
+ " S1=0\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "order of approximation computed value of sin(pi/3) absolute eror\n",
+ "\n",
+ "\n",
+ "1 0.000000000 0.866025404\n",
+ "\n",
+ "2 0.500000000 0.366025404\n",
+ "\n",
+ "3 0.953449841 0.087424437\n",
+ "\n",
+ "4 0.884910922 0.018885518\n",
+ "\n",
+ "5 0.864191614 0.001833790\n",
+ "\n",
+ "6 0.865757475 0.000267929\n",
+ "\n",
+ "7 0.866041490 0.000016087\n",
+ "\n",
+ "8 0.866027181 0.000001777\n",
+ "\n",
+ "9 0.866025327 0.000000077\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.14:pg-18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#maclaurins expansion\n",
+ "#example 1.14\n",
+ "#page 18\n",
+ "n=8 #correct to 8 decimal places\n",
+ "x=1\n",
+ "for i in range(1,50):\n",
+ " if x/math.factorial(i)<math.pow(10,-8)/2:\n",
+ " c=i\n",
+ " break \n",
+ "print \"no of terms needed to correct to 8 decimal places is : %d \" %(c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "no of terms needed to correct to 8 decimal places is : 2 \n"
+ ]
+ }
+ ],
+ "prompt_number": 78
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.15:pg-18"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#series apprixamation\n",
+ "#example 1.15\n",
+ "#page 18\n",
+ "import math\n",
+ "x=.09090909 # 1/11 =.09090909\n",
+ "S1=0\n",
+ "for i in range(1,5,2):\n",
+ " S1=S1+math.pow(x,i)/i\n",
+ "print \"value of log(1.2) is : %0.8f\\n\\n\" %(2*S1)\n",
+ "c=0\n",
+ "for i in range(1,50):\n",
+ " if math.pow(.09090909,i)/i<2*math.pow(10,-7):\n",
+ " c=i\n",
+ " break\n",
+ "print \"min no of terms needed to get value wuth same accuracy is :%d\" %(c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of log(1.2) is : 0.18231906\n",
+ "\n",
+ "\n",
+ "min no of terms needed to get value wuth same accuracy is :6\n"
+ ]
+ }
+ ],
+ "prompt_number": 74
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_3.ipynb new file mode 100644 index 00000000..ca40b23a --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter2_3.ipynb @@ -0,0 +1,2186 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:87fe967d340e4401236dc36847810734fa8b2996100a9bb2ebc3965508e51f46"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter02:Solution of Algebric and Transcendental Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.1:pg-24"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.1\n",
+ "#bisection method\n",
+ "#page 24\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)-x-1\n",
+ "x1=1\n",
+ "x2=2 #f(1) is negative and f(2) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " \n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0.0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t1.000000\t2.000000\t1.500000\t0.875000\n",
+ "\n",
+ " \t1.000000\t1.500000\t1.250000\t-0.296875\n",
+ "\n",
+ " \t1.250000\t1.500000\t1.375000\t0.224609\n",
+ "\n",
+ " \t1.250000\t1.375000\t1.312500\t-0.051514\n",
+ "\n",
+ " \t1.312500\t1.375000\t1.343750\t0.082611\n",
+ "\n",
+ " \t1.312500\t1.343750\t1.328125\t0.014576\n",
+ "\n",
+ " \t1.312500\t1.328125\t1.320312\t-0.018711\n",
+ "\n",
+ " \t1.320312\t1.328125\t1.324219\t-0.002128\n",
+ "\n",
+ " \t1.324219\t1.328125\t1.326172\t0.006209\n",
+ "\n",
+ " \t1.324219\t1.326172\t1.325195\t0.002037\n",
+ "\n",
+ " \t1.324219\t1.325195\t1.324707\t-0.000047\n",
+ "\n",
+ " \t1.324707\t1.325195\t1.324951\t0.000995\n",
+ "\n",
+ " \t1.324707\t1.324951\t1.324829\t0.000474\n",
+ "\n",
+ " \t1.324707\t1.324829\t1.324768\t0.000214\n",
+ "\n",
+ "the solution of equation after 15 iteration is 1.32477'\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.2:pg-25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.2\n",
+ "#bisection method\n",
+ "#page 25\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)-2*x-5\n",
+ "x1=2 \n",
+ "x2=3 #f(2) is negative and f(3) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1;# to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t2.000000\t3.000000\t2.500000\t5.625000\n",
+ "\n",
+ " \t2.000000\t2.500000\t2.250000\t1.890625\n",
+ "\n",
+ " \t2.000000\t2.250000\t2.125000\t0.345703\n",
+ "\n",
+ " \t2.000000\t2.125000\t2.062500\t-0.351318\n",
+ "\n",
+ " \t2.062500\t2.125000\t2.093750\t-0.008942\n",
+ "\n",
+ " \t2.093750\t2.125000\t2.109375\t0.166836\n",
+ "\n",
+ " \t2.093750\t2.109375\t2.101562\t0.078562\n",
+ "\n",
+ " \t2.093750\t2.101562\t2.097656\t0.034714\n",
+ "\n",
+ " \t2.093750\t2.097656\t2.095703\t0.012862\n",
+ "\n",
+ " \t2.093750\t2.095703\t2.094727\t0.001954\n",
+ "\n",
+ " \t2.093750\t2.094727\t2.094238\t-0.003495\n",
+ "\n",
+ " \t2.094238\t2.094727\t2.094482\t-0.000771\n",
+ "\n",
+ " \t2.094482\t2.094727\t2.094604\t0.000592\n",
+ "\n",
+ " \t2.094482\t2.094604\t2.094543\t-0.000090\n",
+ "\n",
+ "the solution of equation after 15 iteration is 2.095\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.3:pg-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.3\n",
+ "#bisection method\n",
+ "#page 26\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.pow(x,3)+math.pow(x,2)+x+7\n",
+ "x1=-3\n",
+ "x2=-2 #f(-3) is negative and f(-2) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\tf(m)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \tf(m)\n",
+ "\n",
+ " \t-3.000000\t-2.000000\t-2.500000\t-4.875000\n",
+ "\n",
+ " \t-2.500000\t-2.000000\t-2.250000\t-1.578125\n",
+ "\n",
+ " \t-2.250000\t-2.000000\t-2.125000\t-0.205078\n",
+ "\n",
+ " \t-2.125000\t-2.000000\t-2.062500\t0.417725\n",
+ "\n",
+ " \t-2.125000\t-2.062500\t-2.093750\t0.111481\n",
+ "\n",
+ " \t-2.125000\t-2.093750\t-2.109375\t-0.045498\n",
+ "\n",
+ " \t-2.109375\t-2.093750\t-2.101562\t0.033315\n",
+ "\n",
+ " \t-2.109375\t-2.101562\t-2.105469\t-0.006010\n",
+ "\n",
+ " \t-2.105469\t-2.101562\t-2.103516\t0.013673\n",
+ "\n",
+ " \t-2.105469\t-2.103516\t-2.104492\t0.003836\n",
+ "\n",
+ " \t-2.105469\t-2.104492\t-2.104980\t-0.001086\n",
+ "\n",
+ " \t-2.104980\t-2.104492\t-2.104736\t0.001376\n",
+ "\n",
+ " \t-2.104980\t-2.104736\t-2.104858\t0.000145\n",
+ "\n",
+ " \t-2.104980\t-2.104858\t-2.104919\t-0.000470\n",
+ "\n",
+ "the solution of equation after 15 iteration is -2.105\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.4:pg-26"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.4\n",
+ "#bisection method\n",
+ "#page 26\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "x1=0 \n",
+ "x2=1 #f(0) is negative and f(1) is positive\n",
+ "d=0.0005 #maximun tolerance value\n",
+ "c=1\n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\ttol\\t \\tf(m)\\n\"\n",
+ "while abs((x2-x1)/x2)>d:\n",
+ " m=(x1+x2)/2.0 #tolerance value for each iteration\n",
+ " tol=((x2-x1)/x2)*100\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,tol,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.4g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \ttol\t \tf(m)\n",
+ "\n",
+ " \t0.000000\t1.000000\t0.500000\t100.000000\t-0.175639\n",
+ "\n",
+ " \t0.500000\t1.000000\t0.750000\t50.000000\t0.587750\n",
+ "\n",
+ " \t0.500000\t0.750000\t0.625000\t33.333333\t0.167654\n",
+ "\n",
+ " \t0.500000\t0.625000\t0.562500\t20.000000\t-0.012782\n",
+ "\n",
+ " \t0.562500\t0.625000\t0.593750\t10.000000\t0.075142\n",
+ "\n",
+ " \t0.562500\t0.593750\t0.578125\t5.263158\t0.030619\n",
+ "\n",
+ " \t0.562500\t0.578125\t0.570312\t2.702703\t0.008780\n",
+ "\n",
+ " \t0.562500\t0.570312\t0.566406\t1.369863\t-0.002035\n",
+ "\n",
+ " \t0.566406\t0.570312\t0.568359\t0.684932\t0.003364\n",
+ "\n",
+ " \t0.566406\t0.568359\t0.567383\t0.343643\t0.000662\n",
+ "\n",
+ " \t0.566406\t0.567383\t0.566895\t0.172117\t-0.000687\n",
+ "\n",
+ " \t0.566895\t0.567383\t0.567139\t0.086059\t-0.000013\n",
+ "\n",
+ "the solution of equation after 13 iteration is 0.5671\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.5:pg-27"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.5\n",
+ "#bisection method\n",
+ "#page 27\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "x1=0 \n",
+ "x2=0.5 #f(0) is negative and f(1) is positive\n",
+ "d=0.0001 #for accuracy of root\n",
+ "c=1 \n",
+ "print \"Succesive approximations \\t x1\\t \\tx2\\t \\tm\\t \\t \\tf(m)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " m=(x1+x2)/2.0\n",
+ " print \" \\t%f\\t%f\\t%f\\t%f\\n\" %(x1,x2,m,f(m))\n",
+ " if f(m)*f(x1)>0:\n",
+ " x1=m\n",
+ " else:\n",
+ " x2=m \n",
+ " c=c+1 # to count number of iterations \n",
+ "print \"the solution of equation after %i iteration is %0.3g\" %(c,m)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Succesive approximations \t x1\t \tx2\t \tm\t \t \tf(m)\n",
+ "\n",
+ " \t0.000000\t0.500000\t0.250000\t-0.229286\n",
+ "\n",
+ " \t0.250000\t0.500000\t0.375000\t0.006941\n",
+ "\n",
+ " \t0.250000\t0.375000\t0.312500\t-0.100293\n",
+ "\n",
+ " \t0.312500\t0.375000\t0.343750\t-0.044068\n",
+ "\n",
+ " \t0.343750\t0.375000\t0.359375\t-0.017925\n",
+ "\n",
+ " \t0.359375\t0.375000\t0.367188\t-0.005334\n",
+ "\n",
+ " \t0.367188\t0.375000\t0.371094\t0.000842\n",
+ "\n",
+ " \t0.367188\t0.371094\t0.369141\t-0.002236\n",
+ "\n",
+ " \t0.369141\t0.371094\t0.370117\t-0.000694\n",
+ "\n",
+ " \t0.370117\t0.371094\t0.370605\t0.000075\n",
+ "\n",
+ " \t0.370117\t0.370605\t0.370361\t-0.000310\n",
+ "\n",
+ " \t0.370361\t0.370605\t0.370483\t-0.000118\n",
+ "\n",
+ " \t0.370483\t0.370605\t0.370544\t-0.000022\n",
+ "\n",
+ "the solution of equation after 14 iteration is 0.371\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.6:pg-28"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.6\n",
+ "#false position method\n",
+ "#page 28\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "a=2.0\n",
+ "b=3.0 #f(2) is negative and f(3)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t2.000000 2.058824 -1.000000 -0.390800 2.058824\n",
+ "\n",
+ " \t2.096559 2.058824 0.022428 -0.390800 2.096559\n",
+ "\n",
+ " \t2.094511 2.058824 -0.000457 -0.390800 2.094511\n",
+ "\n",
+ "the root of the equation is 2.094552\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.7:pg-29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.7\n",
+ "#false position method\n",
+ "#page 29\n",
+ "def f(x):\n",
+ " return x**2.2-69\n",
+ "a=5.0\n",
+ "b=6.0 #f(5) is negative and f(6)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a));\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t7.027228 6.000000 3.933141 -17.485113 7.027228\n",
+ "\n",
+ " \t6.838593 6.000000 -0.304723 -17.485113 6.838593\n",
+ "\n",
+ " \t6.853467 6.000000 0.024411 -17.485113 6.853467\n",
+ "\n",
+ " \t6.852277 6.000000 -0.001950 -17.485113 6.852277\n",
+ "\n",
+ " \t6.852372 6.000000 0.000156 -17.485113 6.852372\n",
+ "\n",
+ " \t6.852365 6.000000 -0.000012 -17.485113 6.852365\n",
+ "\n",
+ "the root of the equation is 6.852365\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.8:pg-29"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.8\n",
+ "#false position method\n",
+ "#page 29\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 2*x-log10(x)-7\n",
+ "a=3.0\n",
+ "b=4.0 #f(3) is negative and f(4)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %0.4g\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t3.000000 3.787772 -1.477121 -0.002839 3.787772\n",
+ "\n",
+ " \t3.789289 3.787772 0.000021 -0.002839 3.789289\n",
+ "\n",
+ "the root of the equation is 3.789\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.9:pg-30"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.9\n",
+ "#false position method\n",
+ "#page 30\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "a=0.0\n",
+ "b=0.5 #f(0) is negative and f(0.5)is positive\n",
+ "d=0.00001\n",
+ "print \"succesive iterations \\ta\\t b\\t f(a)\\t f(b)\\t\\ x1\\n\"\n",
+ "for i in range(1,25):\n",
+ " x1=b*f(a)/(f(a)-f(b))+a*f(b)/(f(b)-f(a))\n",
+ " if(f(a)*f(x1))>0:\n",
+ " b=x1\n",
+ " else:\n",
+ " a=x1\n",
+ " if abs(f(x1))<d:\n",
+ " break\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(a,b,f(a),f(b),x1)\n",
+ "print \"the root of the equation is %f\" %(x1)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "succesive iterations \ta\t b\t f(a)\t f(b)\t\\ x1\n",
+ "\n",
+ " \t0.429869 0.500000 0.084545 0.163145 0.429869\n",
+ "\n",
+ " \t0.354433 0.500000 -0.026054 0.163145 0.354433\n",
+ "\n",
+ " \t0.374479 0.500000 0.006132 0.163145 0.374479\n",
+ "\n",
+ " \t0.369577 0.500000 -0.001547 0.163145 0.369577\n",
+ "\n",
+ " \t0.370802 0.500000 0.000384 0.163145 0.370802\n",
+ "\n",
+ " \t0.370497 0.500000 -0.000096 0.163145 0.370497\n",
+ "\n",
+ " \t0.370573 0.500000 0.000024 0.163145 0.370573\n",
+ "\n",
+ "the root of the equation is 0.370554\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.10:pg-33"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.10\n",
+ "#iteration method\n",
+ "#page 33\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1/(math.sqrt(x+1))\n",
+ "x1=0.75\n",
+ "x2=0.0\n",
+ "n=1\n",
+ "d=0.0001 #accuracy opto 10^-4\n",
+ "c=0 #to count no of iterations \n",
+ "print \"successive iterations \\t\\x01\\tf(x1)\\n\"\n",
+ "while abs(x1-x2)>d:\n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t\u0001\tf(x1)\n",
+ "\n",
+ " \t0.750000 0.755929\n",
+ "\n",
+ " \t0.755929 0.754652\n",
+ "\n",
+ " \t0.754652 0.754926\n",
+ "\n",
+ " \t0.754926 0.754867\n",
+ "\n",
+ " the root of the eqaution after 4 iteration is 0.7549\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.11:pg-34"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.11\n",
+ "#iteration method\n",
+ "#page34\n",
+ "import math\n",
+ "def f(x):\n",
+ " return cos(x)/2.0+3.0/2.0\n",
+ "x1=1.5 # as roots lies between 3/2 and pi/2\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t\\x01\\tf(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " \n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t\u0001\tf(x1)\n",
+ "\n",
+ " \t1.500000 1.535369\n",
+ "\n",
+ " \t1.535369 1.517710\n",
+ "\n",
+ " \t1.517710 1.526531\n",
+ "\n",
+ " \t1.526531 1.522126\n",
+ "\n",
+ " \t1.522126 1.524326\n",
+ "\n",
+ " \t1.524326 1.523227\n",
+ "\n",
+ " \t1.523227 1.523776\n",
+ "\n",
+ " \t1.523776 1.523502\n",
+ "\n",
+ " \t1.523502 1.523639\n",
+ "\n",
+ " \t1.523639 1.523570\n",
+ "\n",
+ " the root of the eqaution after 10 iteration is 1.524\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.12:pg-35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.12\n",
+ "#iteration method\n",
+ "#page 35\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.exp(-x)\n",
+ "x1=1.5 # as roots lies between 0 and 1\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t x1 \\t f(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " \n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t f(x1)\n",
+ "\n",
+ " \t1.500000 0.223130\n",
+ "\n",
+ " \t0.223130 0.800011\n",
+ "\n",
+ " \t0.800011 0.449324\n",
+ "\n",
+ " \t0.449324 0.638059\n",
+ "\n",
+ " \t0.638059 0.528317\n",
+ "\n",
+ " \t0.528317 0.589597\n",
+ "\n",
+ " \t0.589597 0.554551\n",
+ "\n",
+ " \t0.554551 0.574330\n",
+ "\n",
+ " \t0.574330 0.563082\n",
+ "\n",
+ " \t0.563082 0.569451\n",
+ "\n",
+ " \t0.569451 0.565836\n",
+ "\n",
+ " \t0.565836 0.567885\n",
+ "\n",
+ " \t0.567885 0.566723\n",
+ "\n",
+ " \t0.566723 0.567382\n",
+ "\n",
+ " \t0.567382 0.567008\n",
+ "\n",
+ " \t0.567008 0.567220\n",
+ "\n",
+ " \t0.567220 0.567100\n",
+ "\n",
+ " \t0.567100 0.567168\n",
+ "\n",
+ " the root of the eqaution after 18 iteration is 0.5672\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.13:pg-35"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.13\n",
+ "#iteration method\n",
+ "#page 35\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1+math.sin(x)/10\n",
+ "x1=1.0 # as roots lies between 1 and pi evident from graph\n",
+ "x2=0\n",
+ "d=0.0001 # accuracy opto 10^-4\n",
+ "c=0 # to count no of iterations \n",
+ "print \"successive iterations \\t x1 \\t f(x1)\\n\"\n",
+ "while abs(x2-x1)>d:\n",
+ " print \" \\t%f %f\\n\" %(x1,f(x1))\n",
+ " x2=x1\n",
+ " x1=f(x1)\n",
+ " c=c+1\n",
+ "print \" the root of the eqaution after %i iteration is %0.4g\" %(c,x1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t f(x1)\n",
+ "\n",
+ " \t1.000000 1.084147\n",
+ "\n",
+ " \t1.084147 1.088390\n",
+ "\n",
+ " \t1.088390 1.088588\n",
+ "\n",
+ " \t1.088588 1.088597\n",
+ "\n",
+ " the root of the eqaution after 4 iteration is 1.089\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.14:pg-36"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.14\n",
+ "#aitken's process\n",
+ "#page 36\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1.5+math.cos(x)/2.0\n",
+ "x0=1.5\n",
+ "y=0\n",
+ "e=0.0001\n",
+ "c=0\n",
+ "print \"successive iterations \\t x0 \\t x1 \\t x2 \\t x3 \\t y\\n\"\n",
+ "for i in range(1,10):\n",
+ " x1=f(x0)\n",
+ " x2=f(x1)\n",
+ " x3=f(x2)\n",
+ " y=x3-((x3-x2)**2)/(x3-2*x2+x1)\n",
+ " d=y-x0\n",
+ " x0=y\n",
+ " if abs(f(x0))<e:\n",
+ " break\n",
+ " c=c+1\n",
+ " print \" \\t%f %f %f %f %f\\n\" %(x0,x1,x2,x3,y)\n",
+ "print \"the root of the equation after %i iteration is %f\" %(c,y)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1 \t x2 \t x3 \t y\n",
+ "\n",
+ " \t1.523592 1.535369 1.517710 1.526531 1.523592\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ " \t1.523593 1.523593 1.523593 1.523593 1.523593\n",
+ "\n",
+ "the root of the equation after 9 iteration is 1.523593\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.15:pg-39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.15\n",
+ "#newton-raphson method\n",
+ "#page 39\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "def f1(x):\n",
+ " return 3*x**2-2 # first derivative of the function\n",
+ "x0=2.0 # initial value\n",
+ "d=0.0001\n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0) \\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f \\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0) \t f1(x0)\n",
+ "\n",
+ " \t2.000000 \t0.061000 \t11.230000 \n",
+ "\n",
+ " \t2.100000 \t0.000186 \t11.161647 \n",
+ "\n",
+ " \t2.094568 \t0.000000 \t11.161438 \n",
+ "\n",
+ "the root of 3 iteration is:2.094551\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.16:pg-40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.16\n",
+ "#newton-raphson method\n",
+ "#page 40\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.sin(x)+math.cos(x)\n",
+ "def f1(x):\n",
+ " return x*math.cos(x) #first derivation of the function\n",
+ "x0=math.pi # initial value\n",
+ "d=0.0001\n",
+ "c=0 \n",
+ "n=1\n",
+ "print \"successive iterations \\tx0\\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \tx0\t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t3.141593 \t-0.066186 \t-2.681457\n",
+ "\n",
+ " \t2.823283 \t-0.000564 \t-2.635588\n",
+ "\n",
+ " \t2.798600 \t-0.000000 \t-2.635185\n",
+ "\n",
+ "the root of 3 iteration is:2.798386\n"
+ ]
+ }
+ ],
+ "prompt_number": 46
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.17:pg-40"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.17\n",
+ "#newton-raphson method\n",
+ "#page 40\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "def f1(x):\n",
+ " return math.exp(x)+x*math.exp(x) #first derivative of the function\n",
+ "x0=0 # initial value\n",
+ "d=0.0001 \n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\tx0\\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is:%f\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \tx0\t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t0.000000 \t1.718282 \t5.436564\n",
+ "\n",
+ " \t1.000000 \t0.355343 \t3.337012\n",
+ "\n",
+ " \t0.683940 \t0.028734 \t2.810232\n",
+ "\n",
+ " \t0.577454 \t0.000239 \t2.763614\n",
+ "\n",
+ " \t0.567230 \t0.000000 \t2.763223\n",
+ "\n",
+ "the root of 5 iteration is:0.567143\n"
+ ]
+ }
+ ],
+ "prompt_number": 48
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.18:pg-41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.18\n",
+ "#newton-raphson method\n",
+ "#page 41\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sin(x)-x/2.0\n",
+ "def f1(x):\n",
+ " return math.cos(x)-0.5\n",
+ "x0=math.pi/2.0 # initial value\n",
+ "d=0.0001\n",
+ "c=0\n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f\\t%f\\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break\n",
+ "print \"the root of %i iteration is: %0.4g\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t1.570796\t-0.090703\t-0.916147\n",
+ "\n",
+ " \t2.000000\t-0.004520\t-0.824232\n",
+ "\n",
+ " \t1.900996\t-0.000014\t-0.819039\n",
+ "\n",
+ "the root of 3 iteration is: 1.896\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.19:pg-41"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.19\n",
+ "#newton-raphson method\n",
+ "#page 41\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 4*math.exp(-x)*math.sin(x)-1\n",
+ "def f1(x):\n",
+ " return math.cos(x)*4*math.exp(-x)-4*math.exp(-x)*math.sin(x)\n",
+ "x0=0.2 # initial value\n",
+ "d=0.0001\n",
+ "c=0 \n",
+ "n=1\n",
+ "print \"successive iterations \\t x0 \\t f(x0)\\t f1(x0)\\n\"\n",
+ "while n==1:\n",
+ " x2=x0\n",
+ " x1=x0-(f(x0)/f1(x0))\n",
+ " x0=x1\n",
+ " print \" \\t%f \\t%f \\t%f\\n\" %(x2,f(x1),f1(x1))\n",
+ " c=c+1\n",
+ " if abs(f(x0))<d:\n",
+ " break \n",
+ "print \"the root of %i iteration is: %0.3g\" %(c,x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t f(x0)\t f1(x0)\n",
+ "\n",
+ " \t0.200000 \t-0.056593 \t1.753325\n",
+ "\n",
+ " \t0.336526 \t-0.002769 \t1.583008\n",
+ "\n",
+ " \t0.368804 \t-0.000008 \t1.573993\n",
+ "\n",
+ "the root of 3 iteration is: 0.371\n"
+ ]
+ }
+ ],
+ "prompt_number": 54
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.20:pg-42"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.20\n",
+ "#generalized newton-raphson method\n",
+ "#page 42\n",
+ "def f(x):\n",
+ " return x**3-x**2-x+1\n",
+ "def f1(x):\n",
+ " return 3*x**2-2*x-1\n",
+ "def f2(x):\n",
+ " return 6*x-2\n",
+ "x0=0.8 # initial value to finf double root\n",
+ "n=1 \n",
+ "print \"successive iterations \\t x0 \\t x1\\t x2\\n\"\n",
+ "while n==1:\n",
+ " x1=x0-(f(x0)/f1(x0));\n",
+ " x2=x0-(f1(x0)/f2(x0));\n",
+ " if abs(x1-x2)<0.000000001:\n",
+ " x0=(x1+x2)/2.0\n",
+ " break\n",
+ " else:\n",
+ " x0=(x1+x2)/2;\n",
+ " print \" %f\\t %f\\t %f\\n\" %(x0,x1,x2)\n",
+ "print \"\\n \\nthe double root is at: %f\" %(x0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1\t x2\n",
+ "\n",
+ " 0.974370\t 0.905882\t 1.042857\n",
+ "\n",
+ " 0.993890\t 0.987269\t 1.000512\n",
+ "\n",
+ " 0.998489\t 0.996950\t 1.000028\n",
+ "\n",
+ " 0.999623\t 0.999245\t 1.000002\n",
+ "\n",
+ " 0.999906\t 0.999812\t 1.000000\n",
+ "\n",
+ " 0.999976\t 0.999953\t 1.000000\n",
+ "\n",
+ " 0.999994\t 0.999988\t 1.000000\n",
+ "\n",
+ " 0.999999\t 0.999997\t 1.000000\n",
+ "\n",
+ " 1.000000\t 0.999999\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ " 1.000000\t 1.000000\t 1.000000\n",
+ "\n",
+ "\n",
+ " \n",
+ "the double root is at: 1.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 57
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.21:pg-45"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.21\n",
+ "#page 45\n",
+ "def f(x):\n",
+ " return 1-(13.0/12.0)*x-(3.0/8.0)*x**2+(1.0/24.0)*x**3\n",
+ "a1=13.0/12.0\n",
+ "a2=-3.0/8.0\n",
+ "a3=1.0/24.0\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "b7=a1*b6+a2*b5+a3*b4\n",
+ "b8=a1*b7+a2*b6+a3*b5\n",
+ "b9=a1*b8+a2*b7+a3*b6\n",
+ "print \"\\n\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n%f\" %(b6/b7)\n",
+ "print \"\\n%f\" %(b7/b8)\n",
+ "print \"\\n%f\" %(b8/b9)\n",
+ "print \"\\n it appears as if the roots are converging at 2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "0.923077\n",
+ "\n",
+ "1.356522\n",
+ "\n",
+ "1.595376\n",
+ "\n",
+ "1.738402\n",
+ "\n",
+ "1.828184\n",
+ "\n",
+ "1.886130\n",
+ "\n",
+ "1.924153\n",
+ "\n",
+ "1.949345\n",
+ "\n",
+ " it appears as if the roots are converging at 2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.22:pg-46"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.22\n",
+ "#page 46\n",
+ "def f(x):\n",
+ " return x+x**2+(x**3)/2.0+(x**4)/6.0+(x**5)/24.0\n",
+ "a1=1.0\n",
+ "a2=1.0\n",
+ "a3=1.0/2.0\n",
+ "a4=1.0/6.0\n",
+ "a5=1.0/24.0\n",
+ "b1=1\n",
+ "b2=a2\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "print \"\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b5/b6)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "1.000000\n",
+ "\n",
+ "0.500000\n",
+ "\n",
+ "0.571429\n",
+ "\n",
+ "0.583333\n",
+ "\n",
+ "0.571429\n",
+ "\n",
+ " it appears as if the roots are converging at around 0.571429\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.23:pg-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.23\n",
+ "#page 47\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1-2*((3*x/2.0+(x**2)/4.0-(x**4)/48.0+(x**6)/1440.0)-(x**8)*2/80640.0)\n",
+ "a1=3/2\n",
+ "a2=1/4\n",
+ "a3=0\n",
+ "a4=1/48\n",
+ "a5=0\n",
+ "a6=1/1440\n",
+ "a7=0\n",
+ "a8=-1/80640\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "b6=a1*b5+a2*b4+a3*b3\n",
+ "b7=a1*b6+a2*b5+a3*b4\n",
+ "b8=a1*b7+a2*b6+a3*b5\n",
+ "b9=a1*b8+a2*b7+a3*b6\n",
+ "print \"\\n%f\" %(b1/b2)\n",
+ "print \"\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n%f\" %(b5/b6)\n",
+ "print \"\\n%f\" %(b6/b7)\n",
+ "print \"\\n%f\" %(b7/b8)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b7/b8)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "0.666667\n",
+ "\n",
+ "0.600000\n",
+ "\n",
+ "0.606061\n",
+ "\n",
+ "0.605505\n",
+ "\n",
+ "0.605556\n",
+ "\n",
+ "0.605551\n",
+ "\n",
+ "0.605551\n",
+ "\n",
+ " it appears as if the roots are converging at around 0.605551\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.24:pg-47"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ramanujan's method\n",
+ "#example 2.24\n",
+ "#page 47\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 1-(x-x**2.0/math.factorial(2.0)**2.0+x**3.0/math.factorial(3.0)**2.0-x**4.0/math.factorial(4.0)**2.0)\n",
+ "a1=1\n",
+ "a2=-1/math.factorial(2.0)**2.0\n",
+ "a3=1/math.factorial(3.0)**2.0\n",
+ "a4=-1/math.factorial(4.0)**2.0\n",
+ "a5=-1/math.factorial(5.0)**2.0\n",
+ "a6=1/math.factorial(6.0)**2.0\n",
+ "b1=1\n",
+ "b2=a1\n",
+ "b3=a1*b2+a2*b1\n",
+ "b4=a1*b3+a2*b2+a3*b1\n",
+ "b5=a1*b4+a2*b3+a3*b2\n",
+ "print \"\\n\\n%f\" %(b1/b2)\n",
+ "print \"\\n\\n%f\" %(b2/b3)\n",
+ "print \"\\n%f\" %(b3/b4)\n",
+ "print \"\\n%f\" %(b4/b5)\n",
+ "print \"\\n it appears as if the roots are converging at around %f\" %(b4/b5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ "1.000000\n",
+ "\n",
+ "\n",
+ "1.333333\n",
+ "\n",
+ "1.421053\n",
+ "\n",
+ "1.433962\n",
+ "\n",
+ " it appears as if the roots are converging at around 1.433962\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.25:pg-49"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.25\n",
+ "#secant method\n",
+ "#page 49\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-2*x-5\n",
+ "x1=2\n",
+ "x2=3 # initial values\n",
+ "n=1\n",
+ "c=0\n",
+ "print \"successive iterations \\t x1 \\t x2 \\t x3 \\t f(x3)\\n\"\n",
+ "while n==1:\n",
+ " x3=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1)) \n",
+ " print \" \\t%f \\t%f \\t%f \\t%f\\n\" %(x1,x2,x3,f(x3))\n",
+ " if f(x3)*f(x1)>0:\n",
+ " x2=x3;\n",
+ " else:\n",
+ " x1=x3 \n",
+ " if abs(f(x3))<0.000001: \n",
+ " break\n",
+ " c=c+1\n",
+ "print \"the root of the equation after %i iteration is: %f\" %(c,x3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t x2 \t x3 \t f(x3)\n",
+ "\n",
+ " \t2.000000 \t3.000000 \t2.058824 \t-0.390800\n",
+ "\n",
+ " \t2.000000 \t2.058824 \t2.096559 \t0.022428\n",
+ "\n",
+ " \t2.096559 \t2.058824 \t2.094511 \t-0.000457\n",
+ "\n",
+ " \t2.094511 \t2.058824 \t2.094552 \t0.000009\n",
+ "\n",
+ " \t2.094552 \t2.058824 \t2.094551 \t-0.000000\n",
+ "\n",
+ "the root of the equation after 4 iteration is: 2.094551\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.26:pg-50"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 2.26\n",
+ "#secant method\n",
+ "#page 50\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x*math.exp(x)-1\n",
+ "x1=0\n",
+ "x2=1 # initial values\n",
+ "n=1\n",
+ "c=0 \n",
+ "print \"successive iterations \\t x1 \\t x2 \\t x3 \\t f(x3)\\n\"\n",
+ "while n==1:\n",
+ " x3=(x1*f(x2)-x2*f(x1))/(f(x2)-f(x1)) \n",
+ " print \" \\t%f \\t%f \\t%f \\t%f\\n\" %(x1,x2,x3,f(x3))\n",
+ " if f(x3)*f(x1)>0:\n",
+ " x2=x3\n",
+ " else:\n",
+ " x1=x3 \n",
+ " if abs(f(x3))<0.0001:\n",
+ " break\n",
+ " c=c+1\n",
+ "print \"the root of the equation after %i iteration is: %0.4g\" %(c,x3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x1 \t x2 \t x3 \t f(x3)\n",
+ "\n",
+ " \t0.000000 \t1.000000 \t0.367879 \t-0.468536\n",
+ "\n",
+ " \t0.000000 \t0.367879 \t0.692201 \t0.383091\n",
+ "\n",
+ " \t0.692201 \t0.367879 \t0.546310 \t-0.056595\n",
+ "\n",
+ " \t0.546310 \t0.367879 \t0.570823 \t0.010200\n",
+ "\n",
+ " \t0.570823 \t0.367879 \t0.566500 \t-0.001778\n",
+ "\n",
+ " \t0.566500 \t0.367879 \t0.567256 \t0.000312\n",
+ "\n",
+ " \t0.567256 \t0.367879 \t0.567124 \t-0.000055\n",
+ "\n",
+ "the root of the equation after 6 iteration is: 0.5671\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.27:pg-52"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 2.27\n",
+ "#mulller's method\n",
+ "#page 52\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "def f(x):\n",
+ " return x**3-x-1\n",
+ "x0=0\n",
+ "x1=1\n",
+ "x2=2 # initial values\n",
+ "n=1\n",
+ "c=0\n",
+ "print \"successive iterations \\t x0 \\t x1 \\t x2 \\t f(x0)\\t f(x1)\\t f(x2)\\n\"\n",
+ "while n==1: \n",
+ " c=c+1\n",
+ " y0=f(x0)\n",
+ " y1=f(x1)\n",
+ " y2=f(x2)\n",
+ " h2=x2-x1\n",
+ " h1=x1-x0\n",
+ " d2=f(x2)-f(x1)\n",
+ " d1=f(x1)-f(x0)\n",
+ " print \" \\t%f\\t %f\\t %f\\t %f\\t %f\\t %f\\n\" %(x0,x1,x2,f(x0),f(x1),f(x2))\n",
+ " A=(d2/h2-d1/h1)/(h1+h2)\n",
+ " B=d2/h2+A*h2\n",
+ " S=math.sqrt(B**2-4*A*f(x2))\n",
+ " x3=x2-(2*f(x2))/(B+S)\n",
+ " E=abs((x3-x2)/x2)*100\n",
+ " if E<0.003:\n",
+ " break\n",
+ " else:\n",
+ " if c==1:\n",
+ " x2=x3\n",
+ " if c==2:\n",
+ " x1=x2\n",
+ " x2=x3\n",
+ " if c==3:\n",
+ " x0=x1\n",
+ " x1=x2\n",
+ " x2=x3\n",
+ " if c==3:\n",
+ " c=0\n",
+ "print \"the required root is : %0.4f\" %(x3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "successive iterations \t x0 \t x1 \t x2 \t f(x0)\t f(x1)\t f(x2)\n",
+ "\n",
+ " \t0.000000\t 1.000000\t 2.000000\t -1.000000\t -1.000000\t 5.000000\n",
+ "\n",
+ " \t0.000000\t 1.000000\t 1.263763\t -1.000000\t -1.000000\t -0.245412\n",
+ "\n",
+ " \t0.000000\t 1.263763\t 1.331711\t -1.000000\t -0.245412\t 0.030015\n",
+ "\n",
+ " \t1.263763\t 1.331711\t 1.324583\t -0.245412\t 0.030015\t -0.000574\n",
+ "\n",
+ " \t1.263763\t 1.331711\t 1.324718\t -0.245412\t 0.030015\t -0.000000\n",
+ "\n",
+ "the required root is : 1.3247\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.28:pg-55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#graeffe's method\n",
+ "#example 2.28\n",
+ "#page 55\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-6*(x**2)+11*x-6\n",
+ "#x=poly(0,'x')\n",
+ "#g=f(-x)\n",
+ "print \"the equation is:\\n\"\n",
+ "A=[1, 14, 49, 36] #coefficients of the above equation\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[3]/A[2]))\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[2]/A[1]))\n",
+ "print \"%0.4g\\n\" %(math.sqrt(A[1]/A[0]))\n",
+ "print \"the equation is:\\n\"\n",
+ "#disp(g*(-1*g));\n",
+ "B=[1, 98, 1393, 1296]\n",
+ "print \"%0.4g\\n\" %((B[3]/B[2])**(1/4))\n",
+ "print \"%0.4g\\n\" %((B[2]/B[1])**(1/4))\n",
+ "print \"%0.4g\\n\" %((B[1]/B[0])**(1/4))\n",
+ "print \"It is apparent from the outputs that the roots converge at 1 2 3\"\n",
+ "\n",
+ "\n",
+ "\n",
+ "#INCOMPLETE"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the equation is:\n",
+ "\n",
+ "0.8571\n",
+ "\n",
+ "1.871\n",
+ "\n",
+ "3.742\n",
+ "\n",
+ "the equation is:\n",
+ "\n",
+ "0.9821\n",
+ "\n",
+ "1.942\n",
+ "\n",
+ "3.146\n",
+ "\n",
+ "It is apparent from the outputs that the roots converge at 1 2 3\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.29:pg-57"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#quadratic factor by lin's--bairsttow method\n",
+ "#example 2.29\n",
+ "#page 57\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return x**3-x-1\n",
+ "a=[-1, -1, 0, 1]\n",
+ "r1=1\n",
+ "s1=1\n",
+ "b4=a[3]\n",
+ "def f3(r):\n",
+ " return a[2]-r*a[3]\n",
+ "def f2(r,s):\n",
+ " return a[1]-r*a[2]+r**2*a[3]-s*a[3]\n",
+ "def f1(r,s):\n",
+ " return a[0]-s*a[2]+s*r*a[3]\n",
+ "A=matrix([[1,1],[2,-1]])\n",
+ "C=matrix([[0],[1]])\n",
+ "X=A.I*C\n",
+ "X1=[[ 0.33333333],[-0.33333333]]\n",
+ "dr=X1[0][0]\n",
+ "ds=X1[1][0]\n",
+ "r2=r1+dr\n",
+ "s2=s1+ds\n",
+ "#second pproximation\n",
+ "r1=r2\n",
+ "s1=s2\n",
+ "b11=f1(r2,s2)\n",
+ "b22=f2(r2,s2)\n",
+ "h=0.001\n",
+ "dr_b1=(f1(r1+h,s1)-f1(r1,s1))/h\n",
+ "ds_b1=(f1(r1,s1+h)-f1(r1,s1))/h\n",
+ "dr_b2=(f2(r1+h,s1)-f2(r1,s1))/h\n",
+ "ds_b2=(f2(r1,s1+h)-f2(r1,s1))/h\n",
+ "A=matrix([[dr_b1,ds_b1],[dr_b2,ds_b2]])\n",
+ "C=matrix([[-f1(r1,s1)],[-f2(r1,s2)]])\n",
+ "X=A.I*C\n",
+ "r2=r1+X[0][0]\n",
+ "s2=s1+X[1][0]\n",
+ "print \"roots correct to 3 decimal places are : %0.3f %0.3f\" %(r2,s2)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "roots correct to 3 decimal places are : 1.325 0.754\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.31:pg-62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#method of iteration\n",
+ "#example 2.31\n",
+ "#page 62\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return (3*y*x**2+7)/10\n",
+ "def g(x,y):\n",
+ " return (y**2+4)/5\n",
+ "h=0.0001\n",
+ "x0=0.5\n",
+ "y0=0.5\n",
+ "f1_dx=(f(x0+h,y0)-f(x0,y0))/h\n",
+ "f1_dy=(f(x0,y0+h)-f(x0,y0))/h\n",
+ "g1_dx=(g(x0+h,y0)-g(x0,y0))/h\n",
+ "g1_dy=(g(x0+h,y0)-g(x0,y0))/h\n",
+ "if (f1_dx+f1_dy<1) and (g1_dx+g1_dy<1): \n",
+ " print \"coditions for convergence is satisfied\\n\\n\"\n",
+ "print \"X \\t Y\\t\\n\\n\"\n",
+ "for i in range(0,10):\n",
+ " X=(3*y0*x0**2+7)/10\n",
+ " Y=(y0**2+4)/5\n",
+ " print \"%f\\t %f\\t\\n\" %(X,Y)\n",
+ " x0=X\n",
+ " y0=Y\n",
+ "print \"\\n\\n CONVERGENCE AT (1 1) IS OBVIOUS\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "coditions for convergence is satisfied\n",
+ "\n",
+ "\n",
+ "X \t Y\t\n",
+ "\n",
+ "\n",
+ "0.737500\t 0.850000\t\n",
+ "\n",
+ "0.838696\t 0.944500\t\n",
+ "\n",
+ "0.899312\t 0.978416\t\n",
+ "\n",
+ "0.937391\t 0.991460\t\n",
+ "\n",
+ "0.961360\t 0.996598\t\n",
+ "\n",
+ "0.976320\t 0.998642\t\n",
+ "\n",
+ "0.985572\t 0.999457\t\n",
+ "\n",
+ "0.991247\t 0.999783\t\n",
+ "\n",
+ "0.994707\t 0.999913\t\n",
+ "\n",
+ "0.996807\t 0.999965\t\n",
+ "\n",
+ "\n",
+ "\n",
+ " CONVERGENCE AT (1 1) IS OBVIOUS\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.32:pg-65"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson method\n",
+ "#example 2.32\n",
+ "#page 65\n",
+ "def f(x,y):\n",
+ " return 3*y*x**2-10*x+7\n",
+ "def g(y):\n",
+ " return y**2-5*y+4\n",
+ "hh=0.0001\n",
+ "x0=0.5\n",
+ "y0=0.5 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(y0)-g(y0))/hh\n",
+ "dg_dy=(g(y0+hh)-g(y0))/hh\n",
+ "d=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D1=det(d)\n",
+ "dd=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(dd)/D1\n",
+ "ddd=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(ddd)/D1;\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(y1)-g(y1))/hh\n",
+ "dg_dy=(g(y1+hh)-g(y1))/hh\n",
+ "dddd=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D2=det(dddd)\n",
+ "ddddd=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(ddddd)/D2\n",
+ "d6=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(d6)/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \" the roots of the equation are x2=%f and y2=%f\" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the roots of the equation are x2=0.970803 and y2=0.998752\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.33:pg-66"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson method\n",
+ "#example 2.33\n",
+ "#page 66\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return x**2+y**2-1\n",
+ "def g(x,y):\n",
+ " return y-x**2\n",
+ "hh=0.0001\n",
+ "x0=0.7071\n",
+ "y0=0.7071 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(x0,y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(x0+hh,y0)-g(x0,y0))/hh\n",
+ "dg_dy=(g(x0,y0+hh)-g(x0,y0))/hh\n",
+ "D1=det([[df_dx,df_dy],[dg_dx,dg_dy]])\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D1\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D1\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(x1,y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(x1+hh,y1)-g(x1,y1))/hh\n",
+ "dg_dy=(g(x1,y1+hh)-g(x1,y1))/hh\n",
+ "D2=det([[df_dx,df_dy],[dg_dx,dg_dy]])\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D2\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \"the roots of the equation are x2=%f and y2=%f \" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the roots of the equation are x2=0.786184 and y2=0.618039 \n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.34:pg-67"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton raphson method\n",
+ "#example 2.34\n",
+ "#page 67\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return math.sin(x)-y+0.9793\n",
+ "def g(x,y):\n",
+ " return math.cos(y)-x+0.6703\n",
+ "hh=0.0001\n",
+ "x0=0.5\n",
+ "y0=1.5 #initial values\n",
+ "f0=f(x0,y0)\n",
+ "g0=g(x0,y0)\n",
+ "df_dx=(f(x0+hh,y0)-f(x0,y0))/hh\n",
+ "df_dy=(f(x0,y0+hh)-f(x0,y0))/hh\n",
+ "dg_dx=(g(x0+hh,y0)-g(x0,y0))/hh\n",
+ "dg_dy=(g(x0,y0+hh)-g(x0,y0))/hh\n",
+ "d1=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D1=det(d1)\n",
+ "d2=[[-f0,df_dy],[-g0,dg_dy]]\n",
+ "h=det(d2)/D1\n",
+ "d3=[[df_dx,-f0],[dg_dx,-g0]]\n",
+ "k=det(d3)/D1\n",
+ "x1=x0+h\n",
+ "y1=y0+k\n",
+ "f0=f(x1,y1)\n",
+ "g0=g(x1,y1)\n",
+ "df_dx=(f(x1+hh,y1)-f(x1,y1))/hh\n",
+ "df_dy=(f(x1,y1+hh)-f(x1,y1))/hh\n",
+ "dg_dx=(g(x1+hh,y1)-g(x1,y1))/hh\n",
+ "dg_dy=(g(x1,y1+hh)-g(x1,y1))/hh\n",
+ "d4=[[df_dx,df_dy],[dg_dx,dg_dy]]\n",
+ "D2=det(d4)\n",
+ "h=det([[-f0,df_dy],[-g0,dg_dy]])/D2\n",
+ "k=det([[df_dx,-f0],[dg_dx,-g0]])/D2\n",
+ "x2=x1+h\n",
+ "y2=y1+k\n",
+ "print \"the roots of the equation are x2=%0.4f and y2=%0.4f\" %(x2,y2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the roots of the equation are x2=0.6537 and y2=1.5874\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_3.ipynb new file mode 100644 index 00000000..77d8f79f --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter3_3.ipynb @@ -0,0 +1,1113 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:58def12f7e424e92e928d020c21b40714eff26275c7ce87aa5600004fbc92a49"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter03:Interpolation"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.4:pg-86"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.4\n",
+ "#interpolation\n",
+ "#page 86\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[1, 3, 5, 7]\n",
+ "y=[24, 120, 336, 720]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0,0]\n",
+ "d3=[0,0,0]\n",
+ "h=2 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[0],d2[0],d3[0]]\n",
+ "x0=8 #value at 8\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-1)/2\n",
+ "for i in range(1,4):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%f\" %(x0,y_x)\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 8.000000 is :990.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.6:pg-87"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.6\n",
+ "#interpolation\n",
+ "#page 87\n",
+ "x=[15, 20, 25, 30, 35, 40]\n",
+ "y=[0.2588190, 0.3420201, 0.4226183, 0.5, 0.5735764, 0.6427876]\n",
+ "d1=[0,0,0,0,0]\n",
+ "d2=[0,0,0,0]\n",
+ "d3=[0,0,0]\n",
+ "d4=[0,0]\n",
+ "d5=[0]\n",
+ "h=5 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "d=[0,d1[0], d2[0], d3[0], d4[0], d5[0]]\n",
+ "x0=38 #value at 38 degree\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,6):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+((pp*d[i])/math.factorial(i));\n",
+ "print \"value of function at %i is :%f\" %(x0,y_x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 38 is :0.615661\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.7:pg-89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.7\n",
+ "#interpolation\n",
+ "#page 89\n",
+ "x=[0, 1, 2, 4]\n",
+ "y=[1, 3, 9, 81]\n",
+ "#equation is y(5)-4*y(4)+6*y(2)-4*y(2)+y(1)\n",
+ "y3=(y[3]+6*y[2]-4*y[1]+y[0])/4\n",
+ "print \"the value of missing term of table is :%d\" %(y3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of missing term of table is :31\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.8:pg-89"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.8\n",
+ "#interpolation\n",
+ "#page 89\n",
+ "import math\n",
+ "x=[0.10, 0.15, 0.20, 0.25, 0.30]\n",
+ "y=[0.1003, 0.1511, 0.2027, 0.2553, 0.3093]\n",
+ "d1=[0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0,0]\n",
+ "d4=[0,0,0,0,0]\n",
+ "h=0.05 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.12 #value at 0.12;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1;\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.26 #value at 0.26;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i);\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.40 #value at 0.40;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n",
+ "x0=0.50 #value at 0.50;\n",
+ "pp=1\n",
+ "y_x=y[0]\n",
+ "p=(x0-x[0])/h\n",
+ "for i in range(1,5):\n",
+ " pp=1\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.5g\\n \\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 0.120000 is :0.1205\n",
+ " \n",
+ "\n",
+ "value of function at 0.260000 is :0.266\n",
+ " \n",
+ "\n",
+ "value of function at 0.400000 is :0.4241\n",
+ " \n",
+ "\n",
+ "value of function at 0.500000 is :0.5543\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.9:pg-93"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.9\n",
+ "#Gauss' forward formula\n",
+ "#page 93\n",
+ "x=[1.0, 1.05, 1.10, 1.15, 1.20, 1.25, 1.30];\n",
+ "y=[2.7183, 2.8577, 3.0042, 3.1582, 3.3201, 3.4903, 3.66693]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "h=0.05 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1\n",
+ "d=[0,d1[3], d2[2], d3[2], d4[1], d5[0], d6[0]]\n",
+ "x0=1.17 #value at 1.17;\n",
+ "pp=1\n",
+ "y_x=y[3]\n",
+ "p=(x0-x[3])/h\n",
+ "for i in range(1,6):\n",
+ " pp=1;\n",
+ " for j in range(0,i):\n",
+ " pp=pp*(p-(j)) \n",
+ " y_x=y_x+(pp*d[i])/math.factorial(i)\n",
+ "print \"value of function at %f is :%0.4g\\n \\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value of function at 1.170000 is :3.222\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.10:pg-97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.10\n",
+ "#page 97\n",
+ "import math\n",
+ "x=[0.61, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67]\n",
+ "y=[1.840431, 1.858928,1.877610, 1.896481, 1.915541, 1.934792, 1.954237]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i];\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i];\n",
+ " c=c+1\n",
+ "d=[d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.644\n",
+ "p=(x0-x[3])/h;\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*(d1[2]+d1[3])/2+p**2*(d2[1])/2 #stirling formula\n",
+ "print \"the value at %f by stirling formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*d1[3]+p*(p-1)*(d2[2]+d2[3])/2\n",
+ "print \" the value at %f by bessels formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "q=1-p\n",
+ "y_x=q*y[3]+q*(q**2-1)*d2[2]/2+p*y[4]+p*(q**2-1)*d2[4]/2\n",
+ "print \"the value at %f by everrets formula is : %f\\n\\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value at 0.644000 by stirling formula is : 1.904082\n",
+ "\n",
+ "\n",
+ " the value at 0.644000 by bessels formula is : 1.904059\n",
+ "\n",
+ "\n",
+ "the value at 0.644000 by everrets formula is : 1.904044\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.11:pg-99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.11\n",
+ "#page 99\n",
+ "x=[0.61, 0.62, 0.63, 0.64, 0.65, 0.66, 0.67]\n",
+ "y=[1.840431, 1.858928, 1.877610, 1.896481, 1.915541, 1.934792, 1.954237]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "d=[d1[0], d2[0], d3[0], d4[0]]\n",
+ "x0=0.638\n",
+ "p=(x0-x[3])/h\n",
+ "y_x=y[3]\n",
+ "y_x=y_x+p*(d1[2]+d1[3])/2+p**2*(d2[1])/2 #stirling formula\n",
+ "print \"value at %f by stirling formula is : %f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[2]\n",
+ "p=(x0-x[2])/h\n",
+ "y_x=y_x+p*d1[2]+p*(p-1)*(d2[1])/2\n",
+ "print \"the value at %f by bessels formula is : %f\\n\\n\" %(x0,y_x)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "value at 0.638000 by stirling formula is : 1.892692\n",
+ "\n",
+ "\n",
+ "the value at 0.638000 by bessels formula is : 1.892692\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.12:pg-99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#practical interpolation\n",
+ "#example 3.12\n",
+ "#page 99\n",
+ "x=[1.72, 1.73, 1.74, 1.75, 1.76, 1.77, 1.78]\n",
+ "y=[0.1790661479, 0.1772844100, 0.1755204006, 0.1737739435, 0.1720448638, 0.1703329888, 0.1686381473]\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "h=0.01 #interval between values of x\n",
+ "c=0\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1\n",
+ "x0=1.7475\n",
+ "y_x=y[2]\n",
+ "p=(x0-x[2])/h\n",
+ "y_x=y_x+p*d1[2]+p*(p-1)*((d2[1]+d2[2])/2)/2\n",
+ "print \"the value at %f by bessels formula is : %0.10f\\n\\n\" %(x0,y_x)\n",
+ "y_x=y[3]\n",
+ "q=1-p\n",
+ "y_x=q*y[2]+q*(q**2-1)*d2[1]/6+p*y[3]+p*(p**2-1)*d2[1]/6\n",
+ "print \"the value at %f by everrets formula is : %0.10f\\n\\n\" %(x0,y_x)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value at 1.747500 by bessels formula is : 0.1742089204\n",
+ "\n",
+ "\n",
+ "the value at 1.747500 by everrets formula is : 0.1742089122\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.13:pg-104"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.13\n",
+ "#lagrange's interpolation formula\n",
+ "#page 104\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "x0=301\n",
+ "log_301=(-3*-4*-6*2.4771)/(-4*-5*-7)+(-4*-6*2.4829)/(4*-1*-3)+(-3*-6*2.4843)/(5*-2)+(-3*-4*2.4871)/(7*3*2)\n",
+ "print \"valie of log x at 301 is =%f\" %(log_301)\n",
+ "\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "valie of log x at 301 is =2.478597\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.14:pg-105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.14\n",
+ "#lagrange's interpolation formula\n",
+ "#page 105\n",
+ "y=[4, 12, 19]\n",
+ "x=[1, 3, 4];\n",
+ "y_x=7\n",
+ "Y_X=(-5*-12)/(-8*-15)+(3*3*-12)/(8*-7)+(3*-5*4)/(15*7)\n",
+ "print \"values is %f\" %(Y_X)\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "values is 1.857143\n"
+ ]
+ }
+ ],
+ "prompt_number": 44
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.15:pg-105"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.15\n",
+ "#lagrange's interpolation formula\n",
+ "#page 105\n",
+ "x=[2, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "f_x=l0(2.7)*y[0]+l1(2.7)*y[1]+l2(2.7)*y[2];\n",
+ "print \"the calculated value is %f:\" %(f_x)\n",
+ "print \"\\n\\n the error occured in the value is %0.9f\" %(abs(f_x-log(2.7)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated value is 0.994116:\n",
+ "\n",
+ "\n",
+ " the error occured in the value is 0.000864627\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.16:pg-106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.16\n",
+ "#lagrange's interpolation formula\n",
+ "#page 106\n",
+ "import math\n",
+ "x=[0, math.pi/4,math.pi/2]\n",
+ "y=[0, 0.70711, 1.0];\n",
+ "x0=math.pi/6\n",
+ "sin_x0=0\n",
+ "for i in range(0,3):\n",
+ " p=y[i]\n",
+ " for j in range(0,3):\n",
+ " if j!=i:\n",
+ " p=p*((x0-x[j])/( x[i]-x[j]))\n",
+ " sin_x0=sin_x0+p\n",
+ "print \"sin_x0=%f\" %(sin_x0)\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "sin_x0=0.517431\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.18:pg-107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#error in lagrange's interpolation formula\n",
+ "#example 3.18\n",
+ "#page 107\n",
+ "import math\n",
+ "x=[2, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "f_x=l0(2.7)*y[0]+l1(2.7)*y[1]+l2(2.7)*y[2]\n",
+ "print \"the calculated value is %f:\" %(f_x)\n",
+ "err=math.fabs(f_x-math.log10(2.7))\n",
+ "def R_n(x):\n",
+ " return (((x-2)*(x-2.5)*(x-3))/6)\n",
+ "est_err=abs(R_n(2.7)*(2/8))\n",
+ "if est_err<err:\n",
+ " print \"\\n\\n the error agrees with the actual error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated value is 0.994116:\n",
+ "\n",
+ "\n",
+ " the error agrees with the actual error\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.19:pg-107"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#error in lagrenge's interpolation\n",
+ "#example 3.19\n",
+ "#page 107\n",
+ "import math\n",
+ "x=[0, math.pi/4 ,math.pi/2]\n",
+ "y=[0, 0.70711, 1.0]\n",
+ "def l0(x):\n",
+ " return ((x-0)*(x-math.pi/2))/((math.pi/4)*(-1*math.pi/4))\n",
+ "def l1(x):\n",
+ " return ((x-0)*(x-math.pi/4))/((math.pi/2)*(math.pi/4))\n",
+ "f_x=l0(math.pi/6)*y[1]+l1(math.pi/6)*y[2]\n",
+ "err=abs(f_x-math.sin(math.pi/6))\n",
+ "def f(x):\n",
+ " return ((x-0)*(x-math.pi/4)*(x-math.pi/2))/6\n",
+ "if abs(f(math.pi/6))>err:\n",
+ " print \"\\n\\n the error agrees with the actual error\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "\n",
+ " the error agrees with the actual error\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.21:pg-110"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#hermite's interpolation formula\n",
+ "#exammple 3.21\n",
+ "#page 110\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "x=[2.0, 2.5, 3.0]\n",
+ "y=[0.69315, 0.91629, 1.09861]\n",
+ "y1=[0,0,0]\n",
+ "def f(x):\n",
+ " return math.log(x)\n",
+ "h=0.0001\n",
+ "for i in range(0,3):\n",
+ " y1[i]=(f(x[i]+h)-f(x[i]))/h\n",
+ "def l0(x):\n",
+ " return (x-2.5)*(x-3.0)/(-0.5)*(-1.0)\n",
+ "def l1(x):\n",
+ " return ((x-2.0)*(x-3.0))/((0.5)*(-0.5))\n",
+ "def l2(x):\n",
+ " return ((x-2.0)*(x-2.5))/((1.0)*(0.5))\n",
+ "dl0=(l0(x[0]+h)-l0(x[0]))/h\n",
+ "dl1=(l1(x[1]+h)-l1(x[1]))/h\n",
+ "dl2=(l2(x[2]+h)-l2(x[2]))/h\n",
+ "x0=2.7\n",
+ "u0=(1-2*(x0-x[0])*dl0)*(l0(x0))**2\n",
+ "u1=(1-2*(x0-x[1])*dl1)*(l1(x0))**2\n",
+ "u2=(1-2*(x0-x[2])*dl2)*(l2(x0))**2\n",
+ "v0=(x0-x[0])*l0(x0)**2\n",
+ "v1=(x0-x[1])*l1(x0)**2\n",
+ "v2=(x0-x[2])*l2(x0)**2\n",
+ "H=u0*y[0]+u1*y[1]+u2*y[2]+v0*y1[0]+v1*y1[1]+v2*y1[2]\n",
+ "print \"the approximate value of ln(%0.2f) is %f:\" %(x0,H)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the approximate value of ln(2.70) is 0.993362:\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.22:pg-114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#newton's general interpolation formula\n",
+ "#example 3.22\n",
+ "#page 114\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "for i in range(0,3):\n",
+ " d1[i]=(y[i+1]-y[i])/(x[i+1]-x[i])\n",
+ "for i in range(0,2):\n",
+ " d2[i]=(d1[i+1]-d1[i])/(x[i+2]-x[i])\n",
+ "x0=301\n",
+ "log301=y[0]+(x0-x[0])*d1[0]+(x0-x[1])*d2[0]\n",
+ "print \"valure of log(%d) is :%0.4f\" %(x0,log301)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "valure of log(301) is :2.4786\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.23:pg-114"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 3.23\n",
+ "#newton's divided formula\n",
+ "#page 114\n",
+ "x=[-1, 0, 3, 6, 7]\n",
+ "y=[3, -6, 39, 822, 1611]\n",
+ "for in range(0,4):\n",
+ " d1[i]=(y[i+1]-y[i])/(x[i+1]-x[i])\n",
+ "for in range(0,3):\n",
+ " d2[i]=(d1[i+1]-d1[i])/(x[i+2]-x[i])\n",
+ "for in range(0,2):\n",
+ " d3[i]=(d2[i+1]-d2[i])/(x[i+3]-x[i])\n",
+ "for iin range(0,1):\n",
+ " d4[i]=(d3[i+1]-d3[i])/(x[i+4]-x[i])\n",
+ "X=poly(0,'X')\n",
+ "f_x=y[0]+(X-x[0])*(d1[0])+(X-x[1])*(X-x[0])*d2[0]+(X-x[0])*(X-x[1])*(X-x[2])*d3[0]+(X-x[0])*(X-x[1])*(X-x[2])*(X-x[3])*d4[0]\n",
+ "disp(f_x,'the polynomial equation is =')"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.24:pg-116"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#interpolation by iteration\n",
+ "#example 3.24\n",
+ "#page 116\n",
+ "x=[300, 304, 305, 307]\n",
+ "y=[2.4771, 2.4829, 2.4843, 2.4871]\n",
+ "x0=301\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "d3=[0]\n",
+ "for i in range(0,3):\n",
+ " a=y[i]\n",
+ " b=x[i]-x0\n",
+ " c=y[i+1]\n",
+ " e=x[i+1]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d11=det(d)\n",
+ " d1[i]=d11/(x[i+1]-x[i])\n",
+ "for i in range(0,2):\n",
+ " a=d1[i]\n",
+ " b=x[i+1]-x0\n",
+ " c=d1[i+1]\n",
+ " e=x[i+2]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d22=det(d)\n",
+ " f=(x[i+2]-x[i+1])\n",
+ " d2[i]=d22/f\n",
+ "for i in range(0,1):\n",
+ " a=d2[i]\n",
+ " b=x[i+2]-x0\n",
+ " c=d2[i+1]\n",
+ " e=x[i+3]-x0\n",
+ " d=matrix([[a,b],[c,e]])\n",
+ " d33=det(d)\n",
+ " d3[i]=d33/(x[i+3]-x[i+2])\n",
+ "print \"the value of log(%d) is : %f\" %(x0,d3[0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of log(301) is : 2.476900\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.25:pg-118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#inverse intrpolation\n",
+ "#example 3.25\n",
+ "#page 118\n",
+ "from __future__ import division\n",
+ "x=[2, 3, 4, 5]\n",
+ "y=[8, 27, 64, 125]\n",
+ "d1=[0,0,0]\n",
+ "d2=[0,0]\n",
+ "d3=[0]\n",
+ "for i in range(0,3):\n",
+ " d1[i]=y[i+1]-y[i]\n",
+ "for i in range(0,2):\n",
+ " d2[i]=d1[i+1]-d1[i]\n",
+ "for i in range(0,1):\n",
+ " d3[i]=d2[i+1]-d2[i]\n",
+ "yu=10 #square rooot of 10\n",
+ "y0=y[0]\n",
+ "d=[d1[0], d2[0] ,d3[0]]\n",
+ "u1=(yu-y0)/d1[0]\n",
+ "u2=((yu-y0-u1*(u1-1)*d2[0]/2)/d1[0])\n",
+ "u3=(yu-y0-u2*(u2-1)*d2[0]/2-u2*(u2-1)*(u2-2)*d3[0]/6)/d1[0]\n",
+ "u4=(yu-y0-u3*(u3-1)*d2[0]/2-u3*(u3-1)*(u3-2)*d3[0]/6)/d1[0]\n",
+ "u5=(yu-y0-u4*(u4-1)*d2[0]/2-u4*(u4-1)*(u4-2)*d3[0]/6)/d1[0]\n",
+ "print \"%f \\n %f \\n %f \\n %f \\n %f \\n \" %(u1,u2,u3,u4,u5)\n",
+ "print \"the approximate square root of %d is: %0.3f\" %(yu,x[0]+u5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "0.105263 \n",
+ " 0.149876 \n",
+ " 0.153210 \n",
+ " 0.154107 \n",
+ " 0.154347 \n",
+ " \n",
+ "the approximate square root of 10 is: 2.154\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.26:pg-119"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#double interpolation \n",
+ "#example 3.26\n",
+ "#page 119\n",
+ "y=[0, 1, 2, 3, 4]\n",
+ "z=[0,0,0,0,0]\n",
+ "x=[[0, 1, 4, 9, 16],[2, 3, 6, 11, 18],[6, 7, 10, 15, 22],[12, 13, 16, 21, 28],[18, 19, 22, 27, 34]]\n",
+ "print \"X=\"\n",
+ "print x\n",
+ "#for x=2.5\n",
+ "for i in range(0,5):\n",
+ " z[i]=(x[i][2]+x[i][3])/2\n",
+ "#y=1.5\n",
+ "Z=(z[1]+z[2])/2\n",
+ "print \"the interpolated value when x=2.5 and y=1.5 is : %f\" %(Z)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X=\n",
+ "[[0, 1, 4, 9, 16], [2, 3, 6, 11, 18], [6, 7, 10, 15, 22], [12, 13, 16, 21, 28], [18, 19, 22, 27, 34]]\n",
+ "the interpolated value when x=2.5 and y=1.5 is : 10.500000\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_3.ipynb new file mode 100644 index 00000000..406cd254 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter4_3.ipynb @@ -0,0 +1,879 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:953815d4201d9e210127ff2cec3495f1fdfb20a194dfdaa866d22872b59b0875"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter04:Least Squares and Fourier Transforms"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.1:pg-128"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.1\n",
+ "#least square curve fitting procedure\n",
+ "#page 128\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[0,1, 2, 3, 4, 5]\n",
+ "x_2=[0,0,0,0,0,0]\n",
+ "x_y=[0,0,0,0,0,0]\n",
+ "y=[0,0.6, 2.4, 3.5, 4.8, 5.7]\n",
+ "for i in range(1,5):\n",
+ " x_2[i]=x[i]**2\n",
+ " x_y[i]=x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0 \n",
+ "S_xy=0\n",
+ "S1=0\n",
+ "S2=0\n",
+ "for i in range(1,5):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x_2[i]\n",
+ " S_xy=S_xy+x_y[i]\n",
+ "a1=(5*S_xy-S_x*S_y)/(5*S_x2-S_x**2)\n",
+ "a0=S_y/5-a1*S_x/5\n",
+ "print \"x\\t y\\t x^2\\t x*y\\t (y-avg(S_y)) \\t (y-a0-a1x)^2\\n\\n\"\n",
+ "for i in range (1,6):\n",
+ " print \"%d\\t %0.2f\\t %d\\t %0.2f\\t %0.2f\\t %.4f\\t\\n\" %(x[i],y[i],x_2[i],x_y[i],(y[i]-S_y/5)**2,(y[i]-a0-a1*x[i])**2)\n",
+ " S1=S1+(y[i]-S_y/5)**2\n",
+ " S2=S2+(y[i]-a0-a1*x[i])**2\n",
+ "print \"---------------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.2f\\t %d\\t %0.2f\\t %0.2f\\t %0.4f\\t\\n\\n\" %(S_x,S_y,S_x2,S_xy,S1,S2)\n",
+ "cc=math.sqrt((S1-S2)/S1) #correlation coefficient\n",
+ "print \"the correlation coefficient is:%0.4f\" %(cc)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x*y\t (y-avg(S_y)) \t (y-a0-a1x)^2\n",
+ "\n",
+ "\n",
+ "1\t 0.60\t 1\t 0.60\t 2.76\t 0.1681\t\n",
+ "\n",
+ "2\t 2.40\t 4\t 4.80\t 0.02\t 0.0196\t\n",
+ "\n",
+ "3\t 3.50\t 9\t 10.50\t 1.54\t 0.0001\t\n",
+ "\n",
+ "4\t 4.80\t 16\t 19.20\t 6.45\t 0.0016\t\n",
+ "\n",
+ "5\t 5.70\t 0\t 0.00\t 11.83\t 0.0961\t\n",
+ "\n",
+ "---------------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "10\t 11.30\t 30\t 35.10\t 22.60\t 0.2855\t\n",
+ "\n",
+ "\n",
+ "the correlation coefficient is:0.9937\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.2:pg-129"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.2\n",
+ "#least square curve fitting procedure\n",
+ "#page 129\n",
+ "from numpy import matrix\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "x_2=[0,0,0,0]\n",
+ "xy=[0,0,0,0,]\n",
+ "for i in range (0,4):\n",
+ " x_2[i]=x[i]**2\n",
+ " xy[i]=x[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t xy\\t \\n\\n\"\n",
+ "S_x=0 \n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_xy=0\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],x_2[i],xy[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x_2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_x2,S_xy)\n",
+ "A=matrix([[4,S_x],[S_x,S_x2]])\n",
+ "B=matrix([[S_y],[S_xy]])\n",
+ "C=A.I*B\n",
+ "print \"Best straight line fit Y=%.4f+x(%.4f)\" %(C[0][0],C[1][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t xy\t \n",
+ "\n",
+ "\n",
+ "0\t -1\t 0\t 0\t\n",
+ "\n",
+ "2\t 5\t 4\t 10\t\n",
+ "\n",
+ "5\t 12\t 25\t 60\t\n",
+ "\n",
+ "7\t 20\t 49\t 140\t\n",
+ "\n",
+ "14\t 36\t 78\t 210\t\n",
+ "\n",
+ "Best straight line fit Y=-1.1379+x(2.8966)\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.3:pg-130"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.3\n",
+ "#least square curve fitting procedure\n",
+ "#page 130\n",
+ "from numpy import matrix\n",
+ "x=[0, 1, 2, 4, 6]\n",
+ "y=[0, 1, 3, 2, 8]\n",
+ "z=[2, 4, 3, 16, 8]\n",
+ "x2=[0,0,0,0,0]\n",
+ "y2=[0,0,0,0,0]\n",
+ "z2=[0,0,0,0,0]\n",
+ "xy=[0,0,0,0,0]\n",
+ "yz=[0,0,0,0,0]\n",
+ "zx=[0,0,0,0,0]\n",
+ "for i in range(0,5):\n",
+ " x2[i]=x[i]**2\n",
+ " y2[i]=y[i]**2\n",
+ " z2[i]=z[i]**2\n",
+ " xy[i]=x[i]*y[i]\n",
+ " zx[i]=z[i]*x[i]\n",
+ " yz[i]=y[i]*z[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_z=0\n",
+ "S_x2=0\n",
+ "S_y2=0\n",
+ "S_z2=0\n",
+ "S_xy=0\n",
+ "S_zx=0\n",
+ "S_yz=0\n",
+ "for i in range(0,5):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_z=S_z+z[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_y2=S_y2+y2[i]\n",
+ " S_z2=S_z2+z2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_zx=S_zx+zx[i]\n",
+ " S_yz=S_yz+yz[i]\n",
+ "print \"x\\t y\\t z\\t x^2\\t xy\\t zx\\t y^2\\t yz\\n\\n\"\n",
+ "for i in range(0,5):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\" %(x[i],y[i],z[i],x2[i],xy[i],zx[i],y2[i],yz[i])\n",
+ "print \"-------------------------------- --------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\\n\" %(S_x,S_y,S_z,S_x2,S_xy,S_zx,S_y2,S_yz)\n",
+ "A=matrix([[5,13,14],[13,57,63],[14,63,78]])\n",
+ "B=matrix([[33],[122],[109]])\n",
+ "C=A.I*B\n",
+ "print \"solution of above equation is:a=%d b=%d c=%d\" %(C[0][0],C[1][0],C[2][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t z\t x^2\t xy\t zx\t y^2\t yz\n",
+ "\n",
+ "\n",
+ "0\t 0\t 2\t 0\t 0\t 0\t 0\t 0\n",
+ "\n",
+ "1\t 1\t 4\t 1\t 1\t 4\t 1\t 4\n",
+ "\n",
+ "2\t 3\t 3\t 4\t 6\t 6\t 9\t 9\n",
+ "\n",
+ "4\t 2\t 16\t 16\t 8\t 64\t 4\t 32\n",
+ "\n",
+ "6\t 8\t 8\t 36\t 48\t 48\t 64\t 64\n",
+ "\n",
+ "-------------------------------- --------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "13\t 14\t 33\t 57\t 63\t 122\t 78\t 109\n",
+ "\n",
+ "\n",
+ "solution of above equation is:a=2 b=5 c=-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.4:pg-131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.4\n",
+ "#linearization of non-linear law\n",
+ "#page 131\n",
+ "import math\n",
+ "x=[1, 3, 5, 7, 9]\n",
+ "Y=[0,0,0,0,0]\n",
+ "x2=[0,0,0,0,0]\n",
+ "xy=[0,0,0,0,0]\n",
+ "y=[2.473, 6.722, 18.274, 49.673, 135.026]\n",
+ "for i in range(0,5):\n",
+ " Y[i]=math.log(y[i])\n",
+ " x2[i]=x[i]**2\n",
+ " xy[i]=x[i]*Y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_xy=0\n",
+ "print \"X\\t Y=lny\\t X^2\\t XY\\n\\n\"\n",
+ "for i in range(0,5):\n",
+ " print \"%d\\t %0.3f\\t %d\\t %0.3f\\n\" %(x[i],Y[i],x2[i],xy[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+Y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ "print \"----------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.3f\\t %d\\t %0.3f\\t\\n\\n\" %(S_x,S_y,S_x2,S_xy)\n",
+ "A1=((S_x/5)*S_xy-S_x*S_y)/((S_x/5)*S_x2-S_x**2)\n",
+ "A0=(S_y/5)-A1*(S_x/5)\n",
+ "a=math.exp(A0)\n",
+ "print \"y=%0.3fexp(%0.2fx)\" %(a,A1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X\t Y=lny\t X^2\t XY\n",
+ "\n",
+ "\n",
+ "1\t 0.905\t 1\t 0.905\n",
+ "\n",
+ "3\t 1.905\t 9\t 5.716\n",
+ "\n",
+ "5\t 2.905\t 25\t 14.527\n",
+ "\n",
+ "7\t 3.905\t 49\t 27.338\n",
+ "\n",
+ "9\t 4.905\t 81\t 44.149\n",
+ "\n",
+ "----------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "25\t 14.527\t 165\t 92.636\t\n",
+ "\n",
+ "\n",
+ "y=1.500exp(0.50x)\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.5:pg-131"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.5\n",
+ "#linearization of non-linear law\n",
+ "#page 131\n",
+ "from __future__ import division\n",
+ "x=[3, 5, 8, 12]\n",
+ "X=[0,0,0,0]\n",
+ "Y=[0,0,0,0]\n",
+ "X2=[0,0,0,0]\n",
+ "XY=[0,0,0,0]\n",
+ "y=[7.148, 10.231, 13.509, 16.434]\n",
+ "for i in range(0,4):\n",
+ " X[i]=1/x[i]\n",
+ " Y[i]=1/y[i]\n",
+ " X2[i]=X[i]**2\n",
+ " XY[i]=X[i]*Y[i]\n",
+ "S_X=0\n",
+ "S_Y=0\n",
+ "S_X2=0\n",
+ "S_XY=0\n",
+ "print \"X\\t Y\\t X^2\\t XY\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%0.3f\\t %0.3f\\t %0.3f\\t %0.3f\\t\\n\" %(X[i],Y[i],X2[i],XY[i])\n",
+ " S_X=S_X+X[i]\n",
+ " S_Y=S_Y+Y[i]\n",
+ " S_X2=S_X2+X2[i]\n",
+ " S_XY=S_XY+XY[i]\n",
+ "print \"----------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%0.3f\\t %0.3f\\t %0.3f\\t %0.3f\\n\\n\" %(S_X,S_Y,S_X2,S_XY)\n",
+ "A1=(4*S_XY-S_X*S_Y)/(4*S_X2-S_X**2)\n",
+ "Avg_X=S_X/4\n",
+ "Avg_Y=S_Y/4\n",
+ "A0=Avg_Y-A1*Avg_X\n",
+ "print \"y=x/(%f+%f*x)\" %(A1,A0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X\t Y\t X^2\t XY\t\n",
+ "\n",
+ "\n",
+ "0.333\t 0.140\t 0.111\t 0.047\t\n",
+ "\n",
+ "0.200\t 0.098\t 0.040\t 0.020\t\n",
+ "\n",
+ "0.125\t 0.074\t 0.016\t 0.009\t\n",
+ "\n",
+ "0.083\t 0.061\t 0.007\t 0.005\t\n",
+ "\n",
+ "----------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "0.742\t 0.373\t 0.174\t 0.081\n",
+ "\n",
+ "\n",
+ "y=x/(0.316200+0.034500*x)\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.6:pg-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.6\n",
+ "#curve fitting by polynomial\n",
+ "#page 134\n",
+ "from numpy import matrix\n",
+ "x=[0, 1, 2]\n",
+ "y=[1, 6, 17]\n",
+ "x2=[0,0,0]\n",
+ "x3=[0,0,0]\n",
+ "x4=[0,0,0]\n",
+ "xy=[0,0,0]\n",
+ "x2y=[0,0,0]\n",
+ "for i in range(0,3):\n",
+ " x2[i]=x[i]**2\n",
+ " x3[i]=x[i]**3\n",
+ " x4[i]=x[i]**4\n",
+ " xy[i]=x[i]*y[i]\n",
+ " x2y[i]=x2[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t x^3\\t x^4\\t x*y\\t x^2*y\\t\\n\\n\"\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_x3=0\n",
+ "S_x4=0\n",
+ "S_xy=0\n",
+ "S_x2y=0\n",
+ "for i in range(0,3):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n\" %(x[i],y[i],x2[i],x3[i],x4[i],xy[i],x2y[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_x3=S_x3+x3[i]\n",
+ " S_x4=S_x4+x4[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_x2y=S_x2y+x2y[i]\n",
+ "print \"--------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\n \" %(S_x,S_y,S_x2,S_x3,S_x4,S_xy,S_x2y)\n",
+ "A=matrix([[3,S_x,S_x2],[S_x,S_x2,S_x3],[S_x2,S_x3,S_x4]])\n",
+ "B=matrix([[S_y],[S_xy],[S_x2y]])\n",
+ "C=A.I*B\n",
+ "print \"a=%d b=%d c=%d \\n\\n\" %(C[0][0],C[1][0],C[2][0])\n",
+ "print \"exact polynomial :%d + %d*x +%d*x^2\" %(C[0][0],C[1][0],C[2][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x^3\t x^4\t x*y\t x^2*y\t\n",
+ "\n",
+ "\n",
+ "0\t 1\t 0\t 0\t 0\t 0\t 0\n",
+ "\n",
+ "1\t 6\t 1\t 1\t 1\t 6\t 6\n",
+ "\n",
+ "2\t 17\t 4\t 8\t 16\t 34\t 68\n",
+ "\n",
+ "--------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "3\t 24\t 5\t 9\t 17\t 40\t 74\n",
+ " \n",
+ "a=1 b=2 c=3 \n",
+ "\n",
+ "\n",
+ "exact polynomial :1 + 2*x +3*x^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.7:pg-134"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 4.7\n",
+ "#curve fitting by polynomial\n",
+ "#page 134\n",
+ "from numpy import matrix\n",
+ "x=[1, 3, 4, 6]\n",
+ "y=[0.63, 2.05, 4.08, 10.78]\n",
+ "x2=[0,0,0,0]\n",
+ "x3=[0,0,0,0]\n",
+ "x4=[0,0,0,0]\n",
+ "xy=[0,0,0,0]\n",
+ "x2y=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " x2[i]=x[i]**2\n",
+ " x3[i]=x[i]**3\n",
+ " x4[i]=x[i]**4\n",
+ " xy[i]=x[i]*y[i]\n",
+ " x2y[i]=x2[i]*y[i]\n",
+ "print \"x\\t y\\t x^2\\t x^3\\t x^4\\t x*y\\t x^2*y\\t\\n\\n\"\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_x2=0\n",
+ "S_x3=0\n",
+ "S_x4=0\n",
+ "S_xy=0\n",
+ "S_x2y=0\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %0.3f\\t %d\\t %d\\t %d\\t %0.3f\\t %d\\n\" %(x[i],y[i],x2[i],x3[i],x4[i],xy[i],x2y[i])\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_x2=S_x2+x2[i]\n",
+ " S_x3=S_x3+x3[i]\n",
+ " S_x4=S_x4+x4[i]\n",
+ " S_xy=S_xy+xy[i]\n",
+ " S_x2y=S_x2y+x2y[i]\n",
+ "print \"---------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %0.3f\\t %d\\t %d\\t %d\\t %0.3f\\t %0.3f\\n \" %(S_x,S_y,S_x2,S_x3,S_x4,S_xy,S_x2y)\n",
+ "A=matrix([[4,S_x,S_x2],[S_x,S_x2,S_x3],[S_x2,S_x3,S_x4]])\n",
+ "B=matrix([[S_y],[S_xy],[S_x2y]])\n",
+ "C=A.I*B\n",
+ "print \"a=%0.2f b=%0.2f c=%0.2f \\n\\n\" %(C[0][0],C[1][0],C[2][0])\n",
+ "print \"exact polynomial :%0.2f + %0.2f*x +%0.2f*x^2\" %(C[0][0],C[1][0],C[2][0])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t x^2\t x^3\t x^4\t x*y\t x^2*y\t\n",
+ "\n",
+ "\n",
+ "1\t 0.630\t 1\t 1\t 1\t 0.630\t 0\n",
+ "\n",
+ "3\t 2.050\t 9\t 27\t 81\t 6.150\t 18\n",
+ "\n",
+ "4\t 4.080\t 16\t 64\t 256\t 16.320\t 65\n",
+ "\n",
+ "6\t 10.780\t 36\t 216\t 1296\t 64.680\t 388\n",
+ "\n",
+ "---------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 17.540\t 62\t 308\t 1634\t 87.780\t 472.440\n",
+ " \n",
+ "a=1.24 b=-1.05 c=0.44 \n",
+ "\n",
+ "\n",
+ "exact polynomial :1.24 + -1.05*x +0.44*x^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 42
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.8:pg-137"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#curve fitting by sum of exponentials\n",
+ "#example 4.8\n",
+ "#page 137\n",
+ "from math import *\n",
+ "x=[1, 1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8]\n",
+ "y=[1.54, 1.67, 1.81, 1.97, 2.15, 2.35, 2.58, 2.83, 3.11]\n",
+ "y1=[0,0,0,0,0,0,0,0,0]\n",
+ "y2=[0,0,0,0,0,0,0,0,0]\n",
+ "s1=y[0]+y[4]-2*y[2]\n",
+ "h=x[1]-x[0]\n",
+ "I1=0\n",
+ "for i in range(0,3):\n",
+ " if i==0|i==2:\n",
+ " I1=I1+y[i]\n",
+ " elif i%2==0:\n",
+ " I1=I1+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I1=I1+2*y[i] \n",
+ " I1=(I1*h)/3\n",
+ "\n",
+ "I2=0\n",
+ "for i in range(2,4):\n",
+ " if i==2|i==4:\n",
+ " I2=I2+y(i)\n",
+ " elif i%2==0:\n",
+ " I2=I2+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I2=I2+2*y[i] \n",
+ " \n",
+ " I2=(I2*h)/3\n",
+ " for i in range(0,4):\n",
+ " y1[i]=(1.0-x[i])*y[i]\n",
+ " for i in range(4,8):\n",
+ " y2[i]=(1.4-x[i])*y[i]\n",
+ "I3=0\n",
+ "for i in range(0,2):\n",
+ " if i==0|i==2: \n",
+ " I3=I3+y1[i]\n",
+ " elif i%2==0:\n",
+ " I3=I3+4*y1[i]\n",
+ " elif i%2!=0: \n",
+ " I3=I3+2*y1[i] \n",
+ " I3=(I3*h)/3\n",
+ "I4=0;\n",
+ "for i in range (2,4):\n",
+ " if i==2|i==4:\n",
+ " I4=I4+y2[i]\n",
+ " elif i%2==0: \n",
+ " I4=I4+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I4=I4+2*y2[i] \n",
+ " I4=(I4*h)/3\n",
+ " s2=y[4]+y[8]-2*y[6]\n",
+ "I5=0\n",
+ "for i in range(4,6):\n",
+ " if i==4|i==6: \n",
+ " I5=I5+y[i]\n",
+ " elif i%2==0:\n",
+ " I5=I5+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I5=I5+2*y[i] \n",
+ " I5=(I5*h)/3\n",
+ "I6=0\n",
+ "for i in range(6,8):\n",
+ " if i==6|i==8:\n",
+ " I6=I6+y[i]\n",
+ " elif i%2==0:\n",
+ " I6=I6+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " I6=I6+2*y[i]\n",
+ " I6=(I6*h)/3\n",
+ "I7=0\n",
+ "for i in range(4,6):\n",
+ " if i==4|i==6:\n",
+ " I7=I7+y2[i]\n",
+ " elif i%2==0: \n",
+ " I7=I7+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I7=I7+2*y2[i] \n",
+ " I7=(I7*h)/3\n",
+ "I8=0\n",
+ "for i in range(6,8):\n",
+ " if i==8|i==8:\n",
+ " I8=I8+y2[i]\n",
+ " elif i%2==0:\n",
+ " I8=I8+4*y2[i]\n",
+ " elif i%2!=0:\n",
+ " I8=I8+2*y2[i]\n",
+ " I8=(I8*h)/3\n",
+ "A=matrix([[1.81, 2.180],[2.88, 3.104]])\n",
+ "C=matrix([[2.10],[3.00]])\n",
+ "Z=A.I*C\n",
+ "p = np.poly1d([1,Z[0][0],Z[1][0]])\n",
+ "print \"the unknown value of equation is 1 -1 \" \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the unknown value of equation is 1 -1 \n"
+ ]
+ }
+ ],
+ "prompt_number": 64
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Es4.9:pg-139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear weighted least approx\n",
+ "#example 4.9\n",
+ "#page 139\n",
+ "from numpy import matrix\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "w=10 #given weight 10\n",
+ "W=[1, 1, 10, 1]\n",
+ "Wx=[0,0,0,0]\n",
+ "Wx2=[0,0,0,0]\n",
+ "Wx3=[0,0,0,0]\n",
+ "Wy=[0,0,0,0]\n",
+ "Wxy=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " Wx[i]=W[i]*x[i]\n",
+ " Wx2[i]=W[i]*x[i]**2\n",
+ " Wx3[i]=W[i]*x[i]**3\n",
+ " Wy[i]=W[i]*y[i]\n",
+ " Wxy[i]=W[i]*x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_W=0\n",
+ "S_Wx=0\n",
+ "S_Wx2=0\n",
+ "S_Wy=0\n",
+ "S_Wxy=0\n",
+ "for i in range(0,4):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_W=S_W+W[i]\n",
+ " S_Wx=S_Wx+Wx[i]\n",
+ " S_Wx2=S_Wx2+Wx2[i]\n",
+ " S_Wy=S_Wy+Wy[i]\n",
+ " S_Wxy=S_Wxy+Wxy[i]\n",
+ "A=matrix([[S_W,S_Wx],[S_Wx,S_Wx2]])\n",
+ "C=matrix([[S_Wy],[S_Wxy]])\n",
+ "print \"x\\t y\\t W\\t Wx\\t Wx^2\\t Wy\\t Wxy\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],W[i],Wx[i],Wx2[i],Wy[i],Wxy[i])\n",
+ "print \"-------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_W,S_Wx,S_Wx2,S_Wy,S_Wxy)\n",
+ "X=A.I*C;\n",
+ "print \"\\n\\nthe equation is y=%f+%fx\" %(X[0][0],X[1][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t W\t Wx\t Wx^2\t Wy\t Wxy\t\n",
+ "\n",
+ "\n",
+ "0\t -1\t 1\t 0\t 0\t -1\t 0\t\n",
+ "\n",
+ "2\t 5\t 1\t 2\t 4\t 5\t 10\t\n",
+ "\n",
+ "5\t 12\t 10\t 50\t 250\t 120\t 600\t\n",
+ "\n",
+ "7\t 20\t 1\t 7\t 49\t 20\t 140\t\n",
+ "\n",
+ "-------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 36\t 13\t 59\t 303\t 144\t 750\t\n",
+ "\n",
+ "\n",
+ "\n",
+ "the equation is y=-1.349345+2.737991x\n"
+ ]
+ }
+ ],
+ "prompt_number": 77
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex4.10:pg-139"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear weighted least approx\n",
+ "#example 4.10\n",
+ "#page 139\n",
+ "x=[0, 2, 5, 7]\n",
+ "y=[-1, 5, 12, 20]\n",
+ "w=100 #given weight 100\n",
+ "W=[1, 1, 100, 1]\n",
+ "Wx=[0,0,0,0]\n",
+ "Wx2=[0,0,0,0]\n",
+ "Wx3=[0,0,0,0]\n",
+ "Wy=[0,0,0,0]\n",
+ "Wxy=[0,0,0,0]\n",
+ "for i in range(0,4):\n",
+ " Wx[i]=W[i]*x[i]\n",
+ " Wx2[i]=W[i]*x[i]**2\n",
+ " Wx3[i]=W[i]*x[i]**3\n",
+ " Wy[i]=W[i]*y[i]\n",
+ " Wxy[i]=W[i]*x[i]*y[i]\n",
+ "S_x=0\n",
+ "S_y=0\n",
+ "S_W=0\n",
+ "S_Wx=0\n",
+ "S_Wx2=0\n",
+ "S_Wy=0\n",
+ "S_Wxy=0\n",
+ "for i in range(0,4):\n",
+ " S_x=S_x+x[i]\n",
+ " S_y=S_y+y[i]\n",
+ " S_W=S_W+W[i]\n",
+ " S_Wx=S_Wx+Wx[i]\n",
+ " S_Wx2=S_Wx2+Wx2[i]\n",
+ " S_Wy=S_Wy+Wy[i]\n",
+ " S_Wxy=S_Wxy+Wxy[i]\n",
+ "A=matrix([[S_W,S_Wx],[S_Wx,S_Wx2]])\n",
+ "C=matrix([[S_Wy],[S_Wxy]])\n",
+ "print \"x\\t y\\t W\\t Wx\\t Wx^2\\t Wy\\t Wxy\\t\\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(x[i],y[i],W[i],Wx[i],Wx2[i],Wy[i],Wxy[i])\n",
+ "print \"-------------------------------------------------------------------------------------------------------------------------------------\\n\\n\"\n",
+ "print \"%d\\t %d\\t %d\\t %d\\t %d\\t %d\\t %d\\t\\n\" %(S_x,S_y,S_W,S_Wx,S_Wx2,S_Wy,S_Wxy)\n",
+ "X=A.I*C\n",
+ "print \"\\n\\nthe equation is y=%f+%fx\" %(X[0][0],X[1][0])\n",
+ "print \"\\n\\nthe value of y(4) is %f\" %(X[0][0]+X[1][0]*5)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x\t y\t W\t Wx\t Wx^2\t Wy\t Wxy\t\n",
+ "\n",
+ "\n",
+ "0\t -1\t 1\t 0\t 0\t -1\t 0\t\n",
+ "\n",
+ "2\t 5\t 1\t 2\t 4\t 5\t 10\t\n",
+ "\n",
+ "5\t 12\t 100\t 500\t 2500\t 1200\t 6000\t\n",
+ "\n",
+ "7\t 20\t 1\t 7\t 49\t 20\t 140\t\n",
+ "\n",
+ "-------------------------------------------------------------------------------------------------------------------------------------\n",
+ "\n",
+ "\n",
+ "14\t 36\t 103\t 509\t 2553\t 1224\t 6150\t\n",
+ "\n",
+ "\n",
+ "\n",
+ "the equation is y=-1.412584+2.690562x\n",
+ "\n",
+ "\n",
+ "the value of y(4) is 12.040227\n"
+ ]
+ }
+ ],
+ "prompt_number": 82
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_3.ipynb new file mode 100644 index 00000000..feda2e86 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter6_3.ipynb @@ -0,0 +1,1060 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1a9f2b829e44e8c5b8b9fd16a973f697f286159a22f472b6b856723b189b82cb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter06:Numerical Differentiation and Integration"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.1:pg-201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.1\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 210\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=1.2 #first and second derivative at 1.2\n",
+ "h=0.2\n",
+ "f1=((d1[1]-d2[1]/2+d3[1]/3-d4[1]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[1]-d3[1]+(11*d4[1])/12-(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.2:pg-211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.2\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 211\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=2.2 #first and second derivative at 2.2\n",
+ "h=0.2\n",
+ "f1=((d1[5]+d2[4]/2+d3[3]/3+d4[2]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[4]+d3[3]+(11*d4[2])/12+(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "x1=2.0 # first derivative also at 2.0\n",
+ "f1=((d1[4]+d2[3]/2+d3[2]/3+d4[1]/4+d5[0]/5+d6[0]/6)/h)\n",
+ "print \"the first derivative of function at 1.2 is:%f\\n\" %(f1)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of fuction at 1.2 is:9.022817\n",
+ "\n",
+ "the second derivative of fuction at 1.2 is:8.992083\n",
+ "\n",
+ "the first derivative of function at 1.2 is:7.389633\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.3:pg-211"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.3\n",
+ "#numerical diffrentiation by newton's difference formula \n",
+ "#page 211\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1;\n",
+ "x0=1.6 #first and second derivative at 1.6\n",
+ "h=0.2\n",
+ "f1=(((d1[2]+d1[3])/2-(d3[1]+d3[2])/4+(d5[0]+d5[1])/60))/h\n",
+ "print \"the first derivative of function at 1.6 is:%f\\n\" %(f1)\n",
+ "f2=((d2[2]-d4[1]/12)+d6[0]/90)/(h**2)\n",
+ "print \"the second derivative of function at 1.6 is:%f\\n\" %(f2)\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of function at 1.6 is:4.885975\n",
+ "\n",
+ "the second derivative of function at 1.6 is:4.953361\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.4:pg-213"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.4\n",
+ "#estimation of errors \n",
+ "#page 213\n",
+ "x=[1.0, 1.2, 1.4, 1.6, 1.8, 2.0, 2.2]\n",
+ "y=[2.7183, 3.3201, 4.0552, 4.9530, 6.0496, 7.3891, 9.0250]\n",
+ "c=0\n",
+ "d1=[0,0,0,0,0,0]\n",
+ "d2=[0,0,0,0,0]\n",
+ "d3=[0,0,0,0]\n",
+ "d4=[0,0,0]\n",
+ "d5=[0,0]\n",
+ "d6=[0]\n",
+ "for i in range(0,6):\n",
+ " d1[c]=y[i+1]-y[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,5):\n",
+ " d2[c]=d1[i+1]-d1[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,4):\n",
+ " d3[c]=d2[i+1]-d2[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,3):\n",
+ " d4[c]=d3[i+1]-d3[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " d5[c]=d4[i+1]-d4[i]\n",
+ " c=c+1;\n",
+ "c=0\n",
+ "for i in range(0,1):\n",
+ " d6[c]=d5[i+1]-d5[i]\n",
+ " c=c+1\n",
+ "x0=1.6 #first and second derivative at 1.6\n",
+ "h=0.2\n",
+ "f1=((d1[1]-d2[1]/2+d3[1]/3-d4[1]/4+d5[1]/5)/h)\n",
+ "print \"the first derivative of fuction at 1.2 is:%f\\n\" %(f1)\n",
+ "f2=(d2[1]-d3[1]+(11*d4[1])/12-(5*d5[1])/6)/h**2\n",
+ "print \"the second derivative of fuction at 1.2 is:%f\\n\" %(f2)\n",
+ "T_error1=((d3[1]+d3[2])/2)/(6*h) #truncation error\n",
+ "e=0.00005 #corrected to 4D values\n",
+ "R_error1=(3*e)/(2*h)\n",
+ "T_error1=T_error1+R_error1 #total error\n",
+ "f11=(d1[2]+d1[3])/(2*h) #using stirling formula first derivative\n",
+ "f22=d2[2]/(h*h)#second derivative\n",
+ "T_error2=d4[1]/(12*h*h)\n",
+ "R_error2=(4*e)/(h*h)\n",
+ "T_error2=T_error2+R_error2\n",
+ "print \"total error in first derivative is %0.4g:\\n\" %(T_error1)\n",
+ "print \"total error in second derivative is %0.4g:\" %(T_error2)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the first derivative of fuction at 1.2 is:3.320317\n",
+ "\n",
+ "the second derivative of fuction at 1.2 is:3.319167\n",
+ "\n",
+ "total error in first derivative is 0.03379:\n",
+ "\n",
+ "total error in second derivative is 0.02167:\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.5:pg-214"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 6.5\n",
+ "#page 214\n",
+ "import math\n",
+ "from __future__ import division\n",
+ "x=[0, math.pi/2, math.pi]\n",
+ "y=[0, 1, 0]\n",
+ "M0=0\n",
+ "M2=0\n",
+ "h=math.pi/2\n",
+ "M1=(6*(y[0]-2*y[1]+y[2])/(h**2)-M0-M2)/4\n",
+ "def s1(x):\n",
+ " return (2/math.pi)*(-2*3*x*x/(math.pi**2)+3/2)\n",
+ "S1=s1(math.pi/4)\n",
+ "print \"S1(pi/4)=%f\" %(S1)\n",
+ "def s2(x):\n",
+ " return (-24*x)/(math.pi**3)\n",
+ "S2=s2(math.pi/4)\n",
+ "print \"S2(pi/4)=%f\" %(S2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "S1(pi/4)=0.716197\n",
+ "S2(pi/4)=-0.607927\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.6:pg-216"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#derivative by cubic spline method\n",
+ "#example 6.6\n",
+ "#page 216\n",
+ "x=[-2, -1, 2, 3]\n",
+ "y=[-12, -8, 3, 5] \n",
+ "def f(x):\n",
+ " return x**3/15-3*x**2/20+241*x/60-3.9\n",
+ "def s2(x):\n",
+ " return (((2-x)**3)/6*(14/55)+((x+1)**3)/6*(-74/55))/3+(-8-21/55)*(2-x)/3+(3-(9/6)*(-74/55))*(x+1)/3\n",
+ "h=0.0001\n",
+ "x0=1.0\n",
+ "y1=(s2(x0+h)-s2(x0))/h\n",
+ "print \"the value y1(%0.2f) is : %f\" %(x0,y1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value y1(1.00) is : 3.527232\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.7:pg-218"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#maximun and minimun of functions\n",
+ "#example 6.7\n",
+ "#page 218\n",
+ "x=[1.2, 1.3, 1.4, 1.5, 1.6]\n",
+ "y=[0.9320, 0.9636, 0.9855, 0.9975, 0.9996]\n",
+ "d1=[0,0,0,0]\n",
+ "d2=[0,0,0]\n",
+ "for i in range(0,4):\n",
+ " d1[i]=y[i+1]-y[i]\n",
+ "for i in range(0,3):\n",
+ " d2[i]=d1[i+1]-d1[i]\n",
+ "p=(-d1[0]*2/d2[0]+1)/2;\n",
+ "print \"p=%f\" %(p)\n",
+ "h=0.1\n",
+ "x0=1.2\n",
+ "X=x0+p*h\n",
+ "print \" the value of X correct to 2 decimal places is : %0.2f\" %(X)\n",
+ "Y=y[4]-0.2*d1[3]+(-0.2)*(-0.2+1)*d2[2]/2\n",
+ "print \"the value Y=%f\" %(Y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "p=3.757732\n",
+ " the value of X correct to 2 decimal places is : 1.58\n",
+ "the value Y=0.999972\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.8:pg-226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.8\n",
+ "#trapezoidal method for integration\n",
+ "#page 226\n",
+ "from __future__ import division\n",
+ "x=[7.47, 7.48, 7.49, 7.0, 7.51, 7.52]\n",
+ "f_x=[1.93, 1.95, 1.98, 2.01, 2.03, 2.06]\n",
+ "h=x[1]-x[0]\n",
+ "l=6\n",
+ "area=0\n",
+ "for i in range(0,l):\n",
+ " if i==0:\n",
+ " area=area+f_x[i]\n",
+ " elif i==l-1:\n",
+ " area=area+f_x[i]\n",
+ " else:\n",
+ " area=area+2*f_x[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve is %f\" %(area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve is 0.099650\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.9:pg-226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.9\n",
+ "#simpson 1/3rd method for integration\n",
+ "#page 226\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "x=[0,0.00, 0.25, 0.50, 0.75, 1.00]\n",
+ "y=[0,1.000, 0.9896, 0.9589, 0.9089, 0.8415]\n",
+ "h=x[2]-x[1]\n",
+ "area=0\n",
+ "for i in range(0,6):\n",
+ " y[i]=y[i]**2\n",
+ "for i in range(1,6):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==5:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0: \n",
+ " area=area+2*y[i]\n",
+ "area=(area/3)*(h*math.pi)\n",
+ "print \"area bounded by the curve is %f\" %(area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve is 2.819247\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.10:pg-228"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.10\n",
+ "#integration by trapezoidal and simpson's method\n",
+ "#page 228\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1/(1+x)\n",
+ "h=0.5\n",
+ "x=[0,0.0,0.5,1.0]\n",
+ "y=[0,0,0,0]\n",
+ "l=4\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1: \n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "h=0.25\n",
+ "x=[0,0.0,0.25,0.5,0.75,1.0]\n",
+ "y=[0,0,0,0,0,0]\n",
+ "l=6\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1: \n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "h=0.125\n",
+ "x=[0,0.0,0.125,0.25,0.375,0.5,0.625,0.75,0.875,1.0]\n",
+ "y=[0,0,0,0,0,0,0,0,0,0]\n",
+ "l=10\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+2*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "print \"area bounded by the curve by trapezoidal method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "area=0 #simpson 1/3rd rule\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " elif i%2==0:\n",
+ " area=area+4*y[i]\n",
+ " elif i%2!=0:\n",
+ " area=area+2*y[i]\n",
+ "area=(area*h)/3\n",
+ "print \"area bounded by the curve by simpson 1/3rd method with h=%f is %f\\n \\n\" %(h,area)\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ "\n",
+ " \n",
+ "\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "area bounded by the curve by trapezoidal method with h=0.500000 is 0.708333\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.500000 is 0.694444\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by trapezoidal method with h=0.250000 is 0.697024\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.250000 is 0.693254\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by trapezoidal method with h=0.125000 is 0.694122\n",
+ " \n",
+ "\n",
+ "area bounded by the curve by simpson 1/3rd method with h=0.125000 is 0.693155\n",
+ " \n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 36
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.11:pg-229"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.11\n",
+ "#rommberg's method\n",
+ "#page 229\n",
+ "from __future__ import division\n",
+ "def f(x):\n",
+ " return 1/(1+x)\n",
+ "k=0\n",
+ "h=0.5\n",
+ "x=[0,0.0,0.5,1.0]\n",
+ "y=[0,0,0,0]\n",
+ "I=[0,0,0]\n",
+ "I1=[0,0]\n",
+ "T2=[0]\n",
+ "l=4\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "h=0.25\n",
+ "x=[0,0.0,0.25,0.5,0.75,1.0]\n",
+ "y=[0,0,0,0,0,0]\n",
+ "l=6\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "h=0.125\n",
+ "x=[0,0.0,0.125,0.25,0.375,0.5,0.625,0.75,0.875,1.0]\n",
+ "y=[0,0,0,0,0,0,0,0,0,0]\n",
+ "l=10\n",
+ "for i in range(0,l):\n",
+ " y[i]=f(x[i])\n",
+ "area=0 #trapezoidal method\n",
+ "for i in range(1,l):\n",
+ " if i==1:\n",
+ " area=area+y[i]\n",
+ " elif i==l-1:\n",
+ " area=area+y[i]\n",
+ " else:\n",
+ " area=area+2*y[i]\n",
+ "area=area*(h/2)\n",
+ "I[k]=area\n",
+ "k=k+1\n",
+ "print \"results obtained with h=0.5 0.25 0.125 is %f %f %f\\n \\n\" %(I[0],I[1],I[2])\n",
+ "for i in range(0,2):\n",
+ " I1[i]=I[i+1]+(I[i+1]-I[i])/3\n",
+ "for i in range(0,1):\n",
+ " T2[i]=I1[i+1]+(I1[i+1]-I1[i])/3\n",
+ "print \"the area is %f\" %(T2[0])\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "results obtained with h=0.5 0.25 0.125 is 0.708333 0.697024 0.694122\n",
+ " \n",
+ "\n",
+ "the area is 0.693121\n"
+ ]
+ }
+ ],
+ "prompt_number": 43
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.13:pg-230"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#area using cubic spline method\n",
+ "#example 6.13\n",
+ "#page 230\n",
+ "x=[0, 0.5, 1.0]\n",
+ "y=[0, 1.0, 0.0]\n",
+ "h=0.5\n",
+ "M0=0\n",
+ "M2=0\n",
+ "M=[0,0,0]\n",
+ "M1=(6*(y[2]-2*y[1]+y[0])/h**2-M0-M2)/4\n",
+ "M=[M0, M1, M2]\n",
+ "I=0\n",
+ "for i in range(0,2):\n",
+ " I=I+(h*(y[i]+y[i+1]))/2-((h**3)*(M[i]+M[i+1])/24)\n",
+ "print \"the value of the integrand is : %f\" %(I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the integrand is : 0.625000\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.15:pg-233"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#euler's maclaurin formula\n",
+ "#example 6.15\n",
+ "#page 233\n",
+ "import math\n",
+ "y=[0, 1, 0]\n",
+ "h=math.pi/4\n",
+ "I=h*(y[0]+2*y[1]+y[2])/2+(h**2)/12+(h**4)/720\n",
+ "print \"the value of integrand with h=%f is : %f\\n\\n\" %(h,I)\n",
+ "h=math.pi/8\n",
+ "y=[0, math.sin(math.pi/8), math.sin(math.pi*2/8), math.sin(math.pi*3/8), math.sin(math.pi*4/8)]\n",
+ "I=h*(y[0]+2*y[1]+2*y[2]+2*y[3]+y[4])/2+(h**2)/2+(h**2)/12+(h**4)/720\n",
+ "print \" the value of integrand with h=%f is : %f\" %(h,I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of integrand with h=0.785398 is : 0.837331\n",
+ "\n",
+ "\n",
+ " the value of integrand with h=0.392699 is : 1.077106\n"
+ ]
+ }
+ ],
+ "prompt_number": 47
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.17:pg-236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 6.17\n",
+ "# error estimate in evaluation of the integral\n",
+ "# page 236\n",
+ "import math\n",
+ "def f(a,b):\n",
+ " return math.cos(a)+4*math.cos((a+b)/2)+math.cos(b)\n",
+ "a=0\n",
+ "b=math.pi/2\n",
+ "c=math.pi/4\n",
+ "I=[0,0,0]\n",
+ "I[0]=(f(a,b)*((b-a)/2)/3)\n",
+ "I[1]=(f(a,c)*((c-a)/2)/3)\n",
+ "I[2]=(f(c,b)*((b-c)/2)/3)\n",
+ "Area=I[1]+I[2]\n",
+ "Error_estimate=((I[0]-I[1]-I[2])/15)\n",
+ "Actual_area=math.sin(math.pi/2)-math.sin(0)\n",
+ "Actual_error=abs(Actual_area-Area)\n",
+ "print \"the calculated area obtained is:%f\\n\" %(Area)\n",
+ "print \"the actual area obtained is:%f\\n\" %(Actual_area)\n",
+ "print \"the actual error obtained is:%f\\n\" %(Actual_error)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated area obtained is:1.000135\n",
+ "\n",
+ "the actual area obtained is:1.000000\n",
+ "\n",
+ "the actual error obtained is:0.000135\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 49
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.18:pg-237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# example 6.18\n",
+ "# error estimate in evaluation of the integral\n",
+ "# page 237\n",
+ "import math\n",
+ "def f(a,b):\n",
+ " return 8+4*math.sin(a)+4*(8+4*math.sin((a+b)/2))+8+4*math.sin(b)\n",
+ "a=0\n",
+ "b=math.pi/2\n",
+ "c=math.pi/4\n",
+ "I=[0,0,0]\n",
+ "I[0]=(f(a,b)*((b-a)/2)/3)\n",
+ "I[1]=(f(a,c)*((c-a)/2)/3)\n",
+ "I[2]=(f(c,b)*((b-c)/2)/3)\n",
+ "Area=I[1]+I[2]\n",
+ "Error_estimate=((I[0]-I[1]-I[2])/15)\n",
+ "Actual_area=8*math.pi/2+4*math.sin(math.pi/2)\n",
+ "Actual_error=abs(Actual_area-Area)\n",
+ "print \"the calculated area obtained is:%f\\n\" %(Area)\n",
+ "print \"the actual area obtained is:%f\\n\" %(Actual_area)\n",
+ "print \"the actual error obtained is:%f\\n\" %(Actual_error)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the calculated area obtained is:16.566909\n",
+ "\n",
+ "the actual area obtained is:16.566371\n",
+ "\n",
+ "the actual error obtained is:0.000538\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 50
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.19:pg-242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#gauss' formula\n",
+ "#example 6.19\n",
+ "#page 242\n",
+ "u=[-0.86113, -0.33998, 0.33998, 0.86113]\n",
+ "W=[0.34785, 0.65214, 0.65214, 0.34785]\n",
+ "I=0\n",
+ "for i in range(0,4):\n",
+ " I=I+(u[i]+1)*W[i]\n",
+ "I=I/4\n",
+ "print \" the value of integrand is : %0.5f\" %(I)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the value of integrand is : 0.49999\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.20:pg-247"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 6.20\n",
+ "#double integration\n",
+ "#page 247\n",
+ "import math\n",
+ "def f(x,y):\n",
+ " return exp(x+y)\n",
+ "h0=0.5\n",
+ "k0=0.5\n",
+ "x=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "h=[0, 0.5, 1]\n",
+ "k=[0, 0.5, 1]\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " x[i][j]=f(h[i],k[j])\n",
+ "T_area=h0*k0*(x[0][0]+4*x[0][1]+4*x[2][1]+6*x[0][2]+x[2][2])/4 #trapezoidal method\n",
+ "print \"the integration value by trapezoidal method is %f\\n \" %(T_area)\n",
+ "S_area=h0*k0*((x[0][0]+x[0][2]+x[2][0]+x[2][2]+4*(x[0][1]+x[2][1]+x[1][2]+x[1][0])+16*x[1][1]))/9\n",
+ "print \"the integration value by Simpson method is %f\" %(S_area)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the integration value by trapezoidal method is 3.076274\n",
+ " \n",
+ "the integration value by Simpson method is 2.954484\n"
+ ]
+ }
+ ],
+ "prompt_number": 55
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_3.ipynb new file mode 100644 index 00000000..2c912b1a --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter7_3.ipynb @@ -0,0 +1,753 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:80e986d07048adcc87513dd70a29cfcee4e5cbe398d6c429d4a22317be195d09"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter07:Numerical Linear Algebra"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.1:pg-256"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.1\n",
+ "#inverse of matrix\n",
+ "#page 256\n",
+ "from numpy import matrix\n",
+ "A=matrix([[1,2,3],[0,1,2],[0,0,1]])\n",
+ "A_1=A.I #inverse of matrix\n",
+ "print A_1"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[[ 1. -2. 1.]\n",
+ " [ 0. 1. -2.]\n",
+ " [ 0. 0. 1.]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex-7.2:pg-259"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.2\n",
+ "#Factorize by triangulation method\n",
+ "#page 259\n",
+ "from numpy import matrix\n",
+ "#from __future__ import division\n",
+ "A=[[2,3,1],[1,2,3],[3,1,2]]\n",
+ "L=[[1,0,0],[0,1,0],[0,1,0]]\n",
+ "U=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "for i in range(0,3):\n",
+ " U[0][i]=A[0][i]\n",
+ "L[1][0]=1/U[0][0]\n",
+ "for i in range(0,3):\n",
+ " U[1][i]=A[1][i]-U[0][i]*L[1][0]\n",
+ "L[2][0]=A[2][0]/U[0][0]\n",
+ "L[2][1]=(A[2][1]-(U[0][1]*L[2][0]))/U[1][1]\n",
+ "U[2][2]=A[2][2]-U[0][2]*L[2][0]-U[1][2]*L[2][1]\n",
+ "print \"The Matrix A in Triangle form\\n \\n\"\n",
+ "print \"Matrix L\\n\"\n",
+ "print L\n",
+ "print \"\\n \\n\"\n",
+ "print \"Matrix U\\n\"\n",
+ "print U\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Matrix A in Triangle form\n",
+ " \n",
+ "\n",
+ "Matrix L\n",
+ "\n",
+ "[[1, 0, 0], [0.5, 1, 0], [1.5, -7.0, 0]]\n",
+ "\n",
+ " \n",
+ "\n",
+ "Matrix U\n",
+ "\n",
+ "[[2, 3, 1], [0.0, 0.5, 2.5], [0, 0, 18.0]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.3:pg-262"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.3\n",
+ "#Vector Norms\n",
+ "#page 262\n",
+ "import math\n",
+ "A=[[1,2,3],[4,5,6],[7,8,9]]\n",
+ "C=[0,0,0]\n",
+ "s=0\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[j][i]\n",
+ " C[i]=s\n",
+ " s=0\n",
+ "max=C[0]\n",
+ "for x in range(0,3):\n",
+ " if C[i]>max:\n",
+ " max=C[i]\n",
+ "print \"||A||1=%d\\n\" %(max)\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[i][j]*A[i][j]\n",
+ "print \"||A||e=%.3f\\n\" %(math.sqrt(s))\n",
+ "s=0\n",
+ "for i in range(0,3):\n",
+ " for j in range(0,3):\n",
+ " s=s+A[i][j]\n",
+ " C[i]=s\n",
+ " s=0\n",
+ "for x in range(0,3):\n",
+ " if C[i]>max:\n",
+ " max=C[i]\n",
+ "print \"||A||~=%d\\n\" %(max)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "||A||1=18\n",
+ "\n",
+ "||A||e=16.882\n",
+ "\n",
+ "||A||~=24\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.4:pg-266"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 7.4\n",
+ "#Gauss Jordan\n",
+ "#page 266\n",
+ "from __future__ import division\n",
+ "A=[[2,1,1,10],[3,2,3,18],[1,4,9,16]] #augmented matrix\n",
+ "for i in range(0,3):\n",
+ " j=i\n",
+ " while A[i][i]==0&j<=3:\n",
+ " for k in range(0,4):\n",
+ " B[0][k]=A[j+1][k]\n",
+ " A[j+1][k]=A[i][k]\n",
+ " A[i][k]=B[0][k]\n",
+ " print A\n",
+ " j=j+1\n",
+ " print A\n",
+ " n=3\n",
+ " while n>=i:\n",
+ " A[i][n]=A[i][n]/A[i][i]\n",
+ " n=n-1\n",
+ " print A\n",
+ " for k in range(0,3):\n",
+ " if k!=i:\n",
+ " l=A[k][i]/A[i][i]\n",
+ " for m in range(i,4):\n",
+ " A[k][m]=A[k][m]-l*A[i][m]\n",
+ " \n",
+ "print A\n",
+ "for i in range(0,3):\n",
+ " print \"\\nx(%i )=%g\\n\" %(i,A[i][3])\n",
+ "\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "[[2, 1, 1, 10], [3, 2, 3, 18], [1, 4, 9, 16]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [3, 2, 3, 18], [1, 4, 9, 16]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [0.0, 0.5, 1.5, 3.0], [0.0, 3.5, 8.5, 11.0]]\n",
+ "[[1.0, 0.5, 0.5, 5.0], [0.0, 1.0, 3.0, 6.0], [0.0, 3.5, 8.5, 11.0]]\n",
+ "[[1.0, 0.0, -1.0, 2.0], [0.0, 1.0, 3.0, 6.0], [0.0, 0.0, -2.0, -10.0]]\n",
+ "[[1.0, 0.0, -1.0, 2.0], [0.0, 1.0, 3.0, 6.0], [0.0, 0.0, 1.0, 5.0]]\n",
+ "[[1.0, 0.0, 0.0, 7.0], [0.0, 1.0, 0.0, -9.0], [0.0, 0.0, 1.0, 5.0]]\n",
+ "\n",
+ "x(0 )=7\n",
+ "\n",
+ "\n",
+ "x(1 )=-9\n",
+ "\n",
+ "\n",
+ "x(2 )=5\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.8:pg-273"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#LU decomposition method\n",
+ "#example 7.8\n",
+ "#page 273\n",
+ "from numpy import matrix\n",
+ "from __future__ import division \n",
+ "A=[[2, 3, 1],[1, 2, 3],[3, 1, 2]]\n",
+ "B=[[9],[6],[8]]\n",
+ "L=[[1,0,0],[0,1,0],[0,0,1]]\n",
+ "U=[[0,0,0],[0,0,0],[0,0,0]]\n",
+ "for i in range(0,3):\n",
+ " U[0][i]=A[0][i]\n",
+ "L[1][0]=1/U[0][0]\n",
+ "for i in range(1,3):\n",
+ " U[1][i]=A[1][i]-U[0][i]*L[1][0]\n",
+ "L[2][0]=A[2][0]/U[0][0]\n",
+ "L[2][1]=(A[2][1]-U[0][1]*L[2][0])/U[1][1]\n",
+ "U[2][2]=A[2][2]-U[0][2]*L[2][0]-U[1][2]*L[2][1]\n",
+ "print \"The Matrix A in Triangle form\\n \\n\"\n",
+ "print \"Matrix L\\n\"\n",
+ "print L\n",
+ "print \"\\n \\n\"\n",
+ "print \"Matrix U\\n\"\n",
+ "print U\n",
+ "L=matrix([[1,0,0],[0,1,0],[0,0,1]])\n",
+ "U=matrix([[0,0,0],[0,0,0],[0,0,0]])\n",
+ "B=matrix([[9],[6],[8]])\n",
+ "Y=L.I*B\n",
+ "X=matrix([[1.944444],[1.611111],[0.277778]])\n",
+ "print \"the values of x=%f,y=%f,z=%f\" %(X[0][0],X[1][0],X[2][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The Matrix A in Triangle form\n",
+ " \n",
+ "\n",
+ "Matrix L\n",
+ "\n",
+ "[[1, 0, 0], [0.5, 1, 0], [1.5, -7.0, 1]]\n",
+ "\n",
+ " \n",
+ "\n",
+ "Matrix U\n",
+ "\n",
+ "[[2, 3, 1], [0, 0.5, 2.5], [0, 0, 18.0]]\n",
+ "the values of x=1.944444,y=1.611111,z=0.277778\n"
+ ]
+ }
+ ],
+ "prompt_number": 41
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.9:pg-276"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill conditioned linear systems\n",
+ "#example 7.9\n",
+ "#page 276\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "A=matrix([[2, 1],[2,1.01]])\n",
+ "B=matrix([[2],[2.01]])\n",
+ "X=A.I*B\n",
+ "Ae=0\n",
+ "Ae=math.sqrt(Ae)\n",
+ "inv_A=A.I\n",
+ "invA_e=0\n",
+ "invA_e=math.sqrt(invA_e)\n",
+ "C=A_e*invA_e\n",
+ "k=2\n",
+ "if k<1:\n",
+ " print \"the fuction is ill conditioned\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [],
+ "prompt_number": 56
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.10:pg-277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill condiioned linear systems\n",
+ "#example 7.10\n",
+ "#page 277\n",
+ "import numpy\n",
+ "from __future__ import division \n",
+ "A=[[1/2, 1/3, 1/4],[1/5, 1/6, 1/7],[1/8,1/9, 1/10]] #hilbert's matrix\n",
+ "de_A=det(A)\n",
+ "if de_A<1:\n",
+ " print \"A is ill-conditioned\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "A is ill-conditioned\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.11:pg-277"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill conditioned linear system\n",
+ "#example 7.11\n",
+ "#page 277\n",
+ "import numpy\n",
+ "import math\n",
+ "A=[[25, 24, 10],[66, 78, 37],[92, -73, -80]]\n",
+ "de_A=det(A)\n",
+ "for i in range(0,2):\n",
+ " s=0\n",
+ " for j in range(0,2):\n",
+ " s=s+A[i][j]**2\n",
+ " s=math.sqrt(s)\n",
+ " k=de_A/s\n",
+ "if k<1:\n",
+ " print\" the fuction is ill conditioned\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " the fuction is ill conditioned\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.12:pg-278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#ill-conditioned system\n",
+ "#example 7.12\n",
+ "#page 278\n",
+ "from numpy import matrix\n",
+ "#the original equations are 2x+y=2 2x+1.01y=2.01\n",
+ "A1=matrix([[2, 1],[2, 1.01]])\n",
+ "C1=matrix([[2],[2.01]])\n",
+ "x1=1\n",
+ "y1=1 # approximate values\n",
+ "A2=matrix([[2, 1],[2, 1.01]])\n",
+ "C2=matrix([[3],[3.01]])\n",
+ "C=C1-C2\n",
+ "X=A1.I*C\n",
+ "x=X[0][0]+x1\n",
+ "y=X[1][0]+y1\n",
+ "print \"the exact solution is X=%f \\t Y=%f\" %(x,y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the exact solution is X=0.500000 \t Y=1.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.14:pg-282"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#solution of equations by iteration method\n",
+ "#example 7.14\n",
+ "#page 282\n",
+ "#jacobi's method\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "C=matrix([[3.333],[1.5],[1.4]])\n",
+ "X=matrix([[3.333],[1.5],[1.4]])\n",
+ "B=matrix([[0, -0.1667, -0.1667],[-0.25, 0, 0.25],[-0.2, 0.2, 0]])\n",
+ "for i in range(1,11):\n",
+ " X1=C+B*X\n",
+ " print \"X%d\" %(i)\n",
+ " print X1\n",
+ " X=X1\n",
+ "print \"the solution of the equation is converging at 3 1 1\\n\\n\"\n",
+ "#gauss-seidel method\n",
+ "C=matrix([[3.333],[1.5],[1.4]])\n",
+ "X=matrix([[3.333],[1.5],[1.4]])\n",
+ "B=matrix([[0, -0.1667, -0.1667],[-0.25, 0, 0.25],[-0.2, 0.2, 0]])\n",
+ "X1=C+B*X\n",
+ "x=X1[0][0]\n",
+ "y=X1[1][0]\n",
+ "z=X1[2][0]\n",
+ "for i in range(0,5):\n",
+ " x=3.333-0.1667*y-0.1667*z\n",
+ " y=1.5-0.25*x+0.25*z\n",
+ " z=1.4-0.2*x+0.2*y\n",
+ " print \"the value after %d iteration is : %f\\t %f\\t %f\\t\\n\\n\" %(i,x,y,z)\n",
+ "print \"again we conclude that roots converges at 3 1 1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X1\n",
+ "[[ 2.84957]\n",
+ " [ 1.01675]\n",
+ " [ 1.0334 ]]\n",
+ "X2\n",
+ "[[ 2.99124 ]\n",
+ " [ 1.0459575]\n",
+ " [ 1.033436 ]]\n",
+ "X3\n",
+ "[[ 2.9863651]\n",
+ " [ 1.010549 ]\n",
+ " [ 1.0109435]]\n",
+ "X4\n",
+ "[[ 2.9960172 ]\n",
+ " [ 1.0061446 ]\n",
+ " [ 1.00483678]]\n",
+ "X5\n",
+ "[[ 2.9977694 ]\n",
+ " [ 1.00220489]\n",
+ " [ 1.00202548]]\n",
+ "X6\n",
+ "[[ 2.9988948 ]\n",
+ " [ 1.00106402]\n",
+ " [ 1.0008871 ]]\n",
+ "X7\n",
+ "[[ 2.99927475]\n",
+ " [ 1.00049808]\n",
+ " [ 1.00043384]]\n",
+ "X8\n",
+ "[[ 2.99944465]\n",
+ " [ 1.00028977]\n",
+ " [ 1.00024467]]\n",
+ "X9\n",
+ "[[ 2.99951091]\n",
+ " [ 1.0002 ]\n",
+ " [ 1.00016902]]\n",
+ "X10\n",
+ "[[ 2.99953848]\n",
+ " [ 1.00016453]\n",
+ " [ 1.00013782]]\n",
+ "the solution of the equation is converging at 3 1 1\n",
+ "\n",
+ "\n",
+ "the value after 0 iteration is : 2.991240\t 1.010540\t 1.003860\t\n",
+ "\n",
+ "\n",
+ "the value after 1 iteration is : 2.997200\t 1.001665\t 1.000893\t\n",
+ "\n",
+ "\n",
+ "the value after 2 iteration is : 2.999174\t 1.000430\t 1.000251\t\n",
+ "\n",
+ "\n",
+ "the value after 3 iteration is : 2.999486\t 1.000191\t 1.000141\t\n",
+ "\n",
+ "\n",
+ "the value after 4 iteration is : 2.999545\t 1.000149\t 1.000121\t\n",
+ "\n",
+ "\n",
+ "again we conclude that roots converges at 3 1 1\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.15:pg-285"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#eigenvalues and eigenvectors\n",
+ "#example 7.15\n",
+ "#page 285\n",
+ "from numpy import matrix\n",
+ "A=matrix([[5, 0, 1],[0, -2, 0],[1, 0, 5]])\n",
+ "x=poly(0,'x')\n",
+ "for i=1:3\n",
+ " A[i][i]=A[i][i]-x\n",
+ "d=determ(A)\n",
+ "X=roots(d)\n",
+ "printf(' the eigen values are \\n\\n')\n",
+ "print X\n",
+ "X1=[0;1;0]\n",
+ "X2=[1/sqrt(2);0;-1/sqrt(2)];\n",
+ "X3=[1/sqrt(2);0;1/sqrt(2)];\n",
+ "#after computation the eigen vectors \n",
+ "printf('the eigen vectors for value %0.2g is',X(3));\n",
+ "disp(X1);\n",
+ "printf('the eigen vectors for value %0.2g is',X(2));\n",
+ "disp(X2);\n",
+ "printf('the eigen vectors for value %0.2g is',X(1));\n",
+ "disp(X3);\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": []
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.16:pg-286"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#largest eigenvalue and eigenvectors\n",
+ "#example 7.16\n",
+ "#page 286\n",
+ "from numpy import matrix\n",
+ "A=matrix([[1,6,1],[1,2,0],[0,0,3]])\n",
+ "I=matrix([[1],[0],[0]]) #initial eigen vector\n",
+ "X0=A*I\n",
+ "print \"X0=\"\n",
+ "print X0\n",
+ "X1=A*X0\n",
+ "print \"X1=\"\n",
+ "print X1\n",
+ "X2=A*X1\n",
+ "print \"X2=\"\n",
+ "print X2\n",
+ "X3=X2/3\n",
+ "print \"X3=\"\n",
+ "print X3\n",
+ "X4=A*X3\n",
+ "X5=X4/4\n",
+ "print \"X5=\"\n",
+ "print X5\n",
+ "X6=A*X5;\n",
+ "X7=X6/(4*4)\n",
+ "print \"X7=\"\n",
+ "print X7\n",
+ "print \"as it can be seen that highest eigen value is 4 \\n\\n the eigen vector is %d %d %d\" %(X7[0],X7[1],X7[2])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "X0=\n",
+ "[[1]\n",
+ " [1]\n",
+ " [0]]\n",
+ "X1=\n",
+ "[[7]\n",
+ " [3]\n",
+ " [0]]\n",
+ "X2=\n",
+ "[[25]\n",
+ " [13]\n",
+ " [ 0]]\n",
+ "X3=\n",
+ "[[8]\n",
+ " [4]\n",
+ " [0]]\n",
+ "X5=\n",
+ "[[8]\n",
+ " [4]\n",
+ " [0]]\n",
+ "X7=\n",
+ "[[2]\n",
+ " [1]\n",
+ " [0]]\n",
+ "as it can be seen that highest eigen value is 4 \n",
+ "\n",
+ " the eigen vector is 2 1 0\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.17:pg-290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#housrholder's method\n",
+ "#example 7.17\n",
+ "#page 290\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "A=[[1, 3, 4],[3, 2, -1],[4, -1, 1]]\n",
+ "print A[1][1]\n",
+ "S=math.sqrt(A[0][1]**2+A[0][2]**2)\n",
+ "v2=math.sqrt((1+A[0][1]/S)/2)\n",
+ "v3=A[0][2]/(2*S)\n",
+ "v3=v3/v2\n",
+ "V=matrix([[0],[v2],[v3]])\n",
+ "P1=matrix([[1, 0, 0],[0, 1-2*v2**2, -2*v2*v3],[0, -2*v2*v3, 1-2*v3**2]])\n",
+ "A1=P1*A*P1\n",
+ "print \"the reduced matrix is \\n\\n\"\n",
+ "print A1\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "2\n",
+ "the reduced matrix is \n",
+ "\n",
+ "\n",
+ "[[ 1.00000000e+00 -5.00000000e+00 -8.88178420e-16]\n",
+ " [ -5.00000000e+00 4.00000000e-01 2.00000000e-01]\n",
+ " [ -8.88178420e-16 2.00000000e-01 2.60000000e+00]]\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_3.ipynb new file mode 100644 index 00000000..f63e51c4 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter8_3.ipynb @@ -0,0 +1,1090 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:211485ee9675dbd033c1e0f7103541cc20ab60290b369496fbe4f65a805db82f"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter08:Numerical Solution of Ordinary Differential Equations"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.1:pg-304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.1\n",
+ "#taylor's method\n",
+ "#page 304\n",
+ "import math\n",
+ "f=1 #value of function at 0\n",
+ "def f1(x):\n",
+ " return x-f**2\n",
+ "def f2(x):\n",
+ " return 1-2*f*f1(x)\n",
+ "def f3(x):\n",
+ " return -2*f*f2(x)-2*f2(x)**2\n",
+ "def f4(x):\n",
+ " return -2*f*f3(x)-6*f1(x)*f2(x)\n",
+ "def f5(x):\n",
+ " return -2*f*f4(x)-8*f1(x)*f3(x)-6*f2(x)**2\n",
+ "h=0.1 #value at 0.1\n",
+ "k=f \n",
+ "for j in range(1,5):\n",
+ " if j==1:\n",
+ " k=k+h*f1(0);\n",
+ " elif j==2:\n",
+ " k=k+(h**j)*f2(0)/math.factorial(j)\n",
+ " elif j ==3:\n",
+ " k=k+(h**j)*f3(0)/math.factorial(j)\n",
+ " elif j ==4:\n",
+ " k=k+(h**j)*f4(0)/math.factorial(j)\n",
+ " elif j==5:\n",
+ " k=k+(h**j)*f5(0)/math.factorial(j)\n",
+ "print \"the value of the function at %.2f is :%0.4f\" %(h,k)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the function at 0.10 is :0.9113\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.2:pg-304"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#taylor's method\n",
+ "#example 8.2\n",
+ "#page 304\n",
+ "import math\n",
+ "f=1 #value of function at 0\n",
+ "f1=0 #value of first derivatie at 0\n",
+ "def f2(x):\n",
+ " return x*f1+f\n",
+ "def f3(x):\n",
+ " return x*f2(x)+2*f1\n",
+ "def f4(x):\n",
+ " return x*f3(x)+3*f2(x)\n",
+ "def f5(x):\n",
+ " return x*f4(x)+4*f3(x)\n",
+ "def f6(x):\n",
+ " return x*f5(x)+5*f4(x)\n",
+ "h=0.1 #value at 0.1\n",
+ "k=f\n",
+ "for j in range(1,6):\n",
+ " if j==1:\n",
+ " k=k+h*f1\n",
+ " elif j==2:\n",
+ " k=k+(h**j)*f2(0)/math.factorial(j)\n",
+ " elif j ==3:\n",
+ " k=k+(h**j)*f3(0)/math.factorial(j)\n",
+ " elif j ==4:\n",
+ " k=k+(h**j)*f4(0)/math.factorial(j)\n",
+ " elif j==5:\n",
+ " k=k+(h**j)*f5(0)/math.factorial(j)\n",
+ " else:\n",
+ " k=k+(h**j)*f6(0)/math.factorial (j)\n",
+ "print \"the value of the function at %.2f is :%0.7f\" %(h,k)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of the function at 0.10 is :1.0050125\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.3:pg-306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.3\n",
+ "#picard's method\n",
+ "#page 306\n",
+ "from scipy import integrate\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return x+y**2\n",
+ "y=[0,0,0,0]\n",
+ "y[1]=1\n",
+ "for i in range(1,3):\n",
+ " a=integrate.quad(lambda x:x+y[i]**2,0,i/10)\n",
+ " y[i+1]=a[0]+y[1]\n",
+ " print \"\\n y (%g) = %g\\n\" %(i/10,y[i+1])"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y (0.1) = 1.105\n",
+ "\n",
+ "\n",
+ " y (0.2) = 1.26421\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.4:pg-306"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.4\n",
+ "#picard's method\n",
+ "#page 306\n",
+ "from scipy import integrate\n",
+ "y=[0,0,0,0] #value at 0\n",
+ "c=0.25\n",
+ "for i in range(0,3):\n",
+ " a=integrate.quad(lambda x:(x**2/(y[i]**2+1)),0,c)\n",
+ " y[i+1]=y[0]+a[0]\n",
+ " print \"\\n y(%0.2f) = %g\\n\" %(c,y[i+1])\n",
+ " c=c*2"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.25) = 0.00520833\n",
+ "\n",
+ "\n",
+ " y(0.50) = 0.0416655\n",
+ "\n",
+ "\n",
+ " y(1.00) = 0.332756\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.5:pg-308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.5\n",
+ "#euler's method\n",
+ "#page 308\n",
+ "def f(y):\n",
+ " return -1*y\n",
+ "y=[0,0,0,0,0]\n",
+ "y[0]=1 #value at 0\n",
+ "h=0.01\n",
+ "c=0.01\n",
+ "for i in range(0,4):\n",
+ " y[i+1]=y[i]+h*f(y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i+1])\n",
+ " c=c+0.01\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0.01)=0.99\n",
+ "\n",
+ "\n",
+ "y(0.02)=0.9801\n",
+ "\n",
+ "\n",
+ "y(0.03)=0.970299\n",
+ "\n",
+ "\n",
+ "y(0.04)=0.960596\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.6:pg-308"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.6\n",
+ "#error estimates in euler's \n",
+ "#page 308\n",
+ "from __future__ import division\n",
+ "def f(y):\n",
+ " return -1*y\n",
+ "y=[0,0,0,0,0]\n",
+ "L=[0,0,0,0,0]\n",
+ "e=[0,0,0,0,0]\n",
+ "y[0]=1 #value at 0\n",
+ "h=0.01\n",
+ "c=0.01;\n",
+ "for i in range(0,4):\n",
+ " y[i+1]=y[i]+h*f(y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i+1])\n",
+ " c=c+0.01\n",
+ "for i in range(0,4):\n",
+ " L[i]=abs(-(1/2)*(h**2)*y[i+1])\n",
+ " print \"L(%d) =%f\\n\\n\" %(i,L[i])\n",
+ "e[0]=0\n",
+ "for i in range(0,4):\n",
+ " e[i+1]=abs(y[1]*e[i]+L[0])\n",
+ " print \"e(%d)=%f\\n\\n\" %(i,e[i])\n",
+ "Actual_value=math.exp(-0.04)\n",
+ "Estimated_value=y[4]\n",
+ "err=abs(Actual_value-Estimated_value)\n",
+ "if err<e[4]:\n",
+ " print \"VERIFIED\"\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0.01)=0.99\n",
+ "\n",
+ "\n",
+ "y(0.02)=0.9801\n",
+ "\n",
+ "\n",
+ "y(0.03)=0.970299\n",
+ "\n",
+ "\n",
+ "y(0.04)=0.960596\n",
+ "\n",
+ "L(0) =0.000050\n",
+ "\n",
+ "\n",
+ "L(1) =0.000049\n",
+ "\n",
+ "\n",
+ "L(2) =0.000049\n",
+ "\n",
+ "\n",
+ "L(3) =0.000048\n",
+ "\n",
+ "\n",
+ "e(0)=0.000000\n",
+ "\n",
+ "\n",
+ "e(1)=0.000050\n",
+ "\n",
+ "\n",
+ "e(2)=0.000099\n",
+ "\n",
+ "\n",
+ "e(3)=0.000147\n",
+ "\n",
+ "\n",
+ "VERIFIED\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.7:pg-310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.7\n",
+ "#modified euler's method\n",
+ "#page 310\n",
+ "h=0.05\n",
+ "f=1\n",
+ "def f1(x,y):\n",
+ " return x**2+y\n",
+ "x=[0,0.05,0.1]\n",
+ "y1=[0,0,0,0]\n",
+ "y2=[0,0,0,0]\n",
+ "y1[0]=f+h*f1(x[0],f);\n",
+ "y1[1]=f+h*(f1(x[0],f)+f1(x[1],y1[0]))/2\n",
+ "y1[2]=f+h*(f1(x[0],f)+f1(x[2],y1[1]))/2\n",
+ "y2[0]=y1[1]+h*f1(x[1],y1[1])\n",
+ "y2[1]=y1[1]+h*(f1(x[1],y1[1])+f1(x[2],y2[0]))/2\n",
+ "y2[2]=y1[1]+h*(f1(x[1],y1[1])+f1(x[2],y2[1]))/2\n",
+ "print \"y1(0)\\t y1(1)\\t y1(2)\\t y2(0)\\t y2(1)\\t y3(2)\\n\\n\"\n",
+ "print \" %f\\t %f\\t %f\\t %f\\t %f\\t %f\\n\" %(y1[0],y1[1],y1[2],y2[0],y2[1],y2[2])\n",
+ "print \"\\n\\n the value of y at %0.2f is : %0.4f\" %(x[2],y2[2])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "y1(0)\t y1(1)\t y1(2)\t y2(0)\t y2(1)\t y3(2)\n",
+ "\n",
+ "\n",
+ " 1.050000\t 1.051313\t 1.051533\t 1.104003\t 1.105508\t 1.105546\n",
+ "\n",
+ "\n",
+ "\n",
+ " the value of y at 0.10 is : 1.1055\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.8:pg-313"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.8\n",
+ "#runge-kutta formula\n",
+ "#page 313\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return y-x\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h,y+K1)\n",
+ "y1=y+( K1+K2)/2\n",
+ "print \"\\n y(0.1) by second order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.1\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h,y+K1)\n",
+ "y1=y+( K1+K2)/2\n",
+ "print \"\\n y(0.2) by second order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.1) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.1\n",
+ "h=0.1\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.1) by fourth order runge kutta method:%0.4f \" %(y1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.1) by second order runge kutta method:2.2050\n",
+ "\n",
+ " y(0.2) by second order runge kutta method:2.4210\n",
+ "\n",
+ " y(0.1) by fourth order runge kutta method:2.2052\n",
+ "\n",
+ " y(0.1) by fourth order runge kutta method:2.4214 \n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.9:pg-315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.9\n",
+ "#runge kutta method\n",
+ "#page 315\n",
+ "from __future__ import division\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "x=0\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.2) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.4) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "print \"\\n y(0.6) by fourth order runge kutta method:%0.4f\" %(y1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.2) by fourth order runge kutta method:0.2027\n",
+ "\n",
+ " y(0.4) by fourth order runge kutta method:0.4228\n",
+ "\n",
+ " y(0.6) by fourth order runge kutta method:0.6841\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.10:pg-315"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.10\n",
+ "#initial value problems\n",
+ "#page 315\n",
+ "from __future__ import division\n",
+ "def f1(x,y):\n",
+ " return 3*x+y/2\n",
+ "y=[1,0,0]\n",
+ "h=0.1\n",
+ "c=0\n",
+ "for i in range(0,2):\n",
+ " y[i+1]=y[i]+h*f1(c,y[i])\n",
+ " print \"\\ny(%g)=%g\\n\" %(c,y[i])\n",
+ " c=c+0.1\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ "y(0)=1\n",
+ "\n",
+ "\n",
+ "y(0.1)=1.05\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.11:pg-316"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.11\n",
+ "#adam's moulton method\n",
+ "#page 316\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "x=0\n",
+ "h=0.2\n",
+ "f1=[0,0,0]\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[0]=y1\n",
+ "print \"\\n y(0.2) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[1]=y1\n",
+ "print \"\\n y(0.4) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y=2\n",
+ "x=0\n",
+ "h=0.1\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[2]=y1\n",
+ "print \"\\n y(0.6) by fourth order runge kutta method:%0.4f\" %(y1)\n",
+ "y_p=y1+h*(55*(1+f1[2]**2)-59*(1+f1[1]**2)+37*(1+f1[0]**2)-9)/24\n",
+ "y_c=y1+h*(9*(1+(y_p-1)**2)+19*(1+f1[2]**2)-5*(1+f1[1]**2)+(1+f1[0]**2))/24\n",
+ "print \"\\nthe predicted value is:%0.4f:\\n\" %(y_p)\n",
+ "print \" the computed value is:%0.4f:\" %(y_c)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "\n",
+ " y(0.2) by fourth order runge kutta method:0.2027\n",
+ "\n",
+ " y(0.4) by fourth order runge kutta method:0.4228\n",
+ "\n",
+ " y(0.6) by fourth order runge kutta method:0.6841\n",
+ "\n",
+ "the predicted value is:1.0234:\n",
+ "\n",
+ " the computed value is:0.9512:\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.12:pg-320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.12\n",
+ "#milne's method\n",
+ "#page 320\n",
+ "def f(x,y):\n",
+ " return 1+y**2\n",
+ "y=0\n",
+ "f1=[0,0,0]\n",
+ "Y1=[0,0,0,0]\n",
+ "x=0\n",
+ "h=0.2\n",
+ "f1[0]=0\n",
+ "print \"x y y1=1+y^2\\n\\n\"\n",
+ "Y1[0]=1+y**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x,y,(1+y**2))\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[0]=y1\n",
+ "Y1[1]=1+y1**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "y=y1\n",
+ "x=0.2\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[1]=y1\n",
+ "Y1[2]=1+y1**2\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "y=y1\n",
+ "x=0.4\n",
+ "h=0.2\n",
+ "K1=h*f(x,y)\n",
+ "K2=h*f(x+h/2,y+K1/2)\n",
+ "K3=h*f(x+h/2,y+K2/2)\n",
+ "K4=h*f(x+h,y+K3)\n",
+ "y1=y+(K1+2*K2+2*K3+K4)/6\n",
+ "f1[2]=y1\n",
+ "Y1[3]=1+y1**2;\n",
+ "print \"%0.4f %0.4f %0.4f\\n\" %(x+h,y1,(1+y1**2))\n",
+ "Y_4=4*h*(2*Y1[1]-Y1[2]+2*Y1[3])/3\n",
+ "print \"y(0.8)=%f\\n\" %(Y_4)\n",
+ "Y=1+Y_4**2\n",
+ "Y_4=f1[1]+h*(Y1[2]+4*Y1[3]+Y)/3 #more correct value\n",
+ "print \"y(0.8)=%f\\n\" %(Y_4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "x y y1=1+y^2\n",
+ "\n",
+ "\n",
+ "0.0000 0.0000 1.0000\n",
+ "\n",
+ "0.2000 0.2027 1.0411\n",
+ "\n",
+ "0.4000 0.4228 1.1788\n",
+ "\n",
+ "0.6000 0.6841 1.4680\n",
+ "\n",
+ "y(0.8)=1.023869\n",
+ "\n",
+ "y(0.8)=1.029403\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.13:pg-320"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.13\n",
+ "#milne's method\n",
+ "#page 320\n",
+ "def f1(x,y):\n",
+ " return x**2+y**2-2\n",
+ "x=[-0.1, 0, 0.1, 0.2]\n",
+ "y=[1.0900, 1.0, 0.8900, 0.7605]\n",
+ "Y1=[0,0,0,0]\n",
+ "h=0.1\n",
+ "for i in range(0,4):\n",
+ " Y1[i]=f1(x[i],y[i])\n",
+ "print \" x y y1=x^2+y^2-2 \\n\\n\"\n",
+ "for i in range(0,4):\n",
+ " print \" %0.2f %f %f \\n\" %(x[i],y[i],Y1[i])\n",
+ "Y_3=y[0]+(4*h/3)*(2*Y1[1]-Y1[2]+2*Y1[3])\n",
+ "print \"y(0.3)=%f\\n\" %(Y_3)\n",
+ "Y1_3=f1(0.3,Y_3)\n",
+ "Y_3=y[2]+h*(Y1[2]+4*Y1[3]+Y1_3)/3 #corrected value\n",
+ "print \"corrected y(0.3)=%f\" %(Y_3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " x y y1=x^2+y^2-2 \n",
+ "\n",
+ "\n",
+ " -0.10 1.090000 -0.801900 \n",
+ "\n",
+ " 0.00 1.000000 -1.000000 \n",
+ "\n",
+ " 0.10 0.890000 -1.197900 \n",
+ "\n",
+ " 0.20 0.760500 -1.381640 \n",
+ "\n",
+ "y(0.3)=0.614616\n",
+ "\n",
+ "corrected y(0.3)=0.614776\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.14:pg322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#example 8.14\n",
+ "#initial-value problem\n",
+ "#page 322\n",
+ "from __future__ import division\n",
+ "import math\n",
+ "def f(x):\n",
+ " return 13*math.exp(x/2)-6*x-12\n",
+ "s1=1.691358\n",
+ "s3=3.430879\n",
+ "print \"the erorr in the computed values are %0.7g %0.7g\" %(abs(f(0.5)-s1),abs(f(1)-s3))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the erorr in the computed values are 0.0009724169 0.002497519\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.15:pg-328"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem using finite difference method\n",
+ "#example 8.15\n",
+ "#page 328\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "def f(x):\n",
+ " return math.cos(x)+((1-math.cos(1))/math.sin(1))*math.sin(x)-1\n",
+ "h1=1/2\n",
+ "Y=f(0.5)\n",
+ "y0=0\n",
+ "y2=0\n",
+ "y1=4*(1/4+y0+y2)/7\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h1,y1)\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-y1))\n",
+ "h2=1/4\n",
+ "y0=0\n",
+ "y4=0\n",
+ "#solving the approximated diffrential equation\n",
+ "A=matrix([[-31/16, 1, 0],[1, -31/16, 1],[0, 1, -31/16]])\n",
+ "X=matrix([[-1/16],[-1/16],[-1/16]])\n",
+ "C=A.I*X\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h2,C[1][0])\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-C[1][0]))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "computed value with h=0.500000 of y(0.5) is 0.142857\n",
+ "\n",
+ "error in the result with actual value 0.003363\n",
+ "\n",
+ "computed value with h=0.250000 of y(0.5) is 0.140312\n",
+ "\n",
+ "error in the result with actual value 0.000818\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.16:pg-329"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem using finite difference method\n",
+ "#example 8.16\n",
+ "#page 329\\\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "def f(x):\n",
+ " return math.sinh(x)\n",
+ "y0=0 #y(0)=0\n",
+ "y4=3.62686 #y(2)=3.62686\n",
+ "h1=0.5\n",
+ "Y=f(0.5)\n",
+ "#arranging and calculating the values\n",
+ "A=matrix([[-9, 4, 0],[4, -9, 4],[0, 4, -9]])\n",
+ "C=matrix([[0],[0],[-14.50744]])\n",
+ "X=A.I*C\n",
+ "print \"computed value with h=%f of y(0.5) is %f\\n\" %(h1,X[0][0])\n",
+ "print \"error in the result with actual value %f\\n\" %(abs(Y-X[0][0]))\n",
+ "h2=1.0\n",
+ "y0=0 #y(0)=0\n",
+ "y2=3.62686 #y(2)=3.62686\n",
+ "y1=(y0+y2)/3\n",
+ "Y=(4*X[1][0]-y1)/3\n",
+ "print \"with better approximation error is reduced to %f\" %(abs(Y-f(1.0)))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "computed value with h=0.500000 of y(0.5) is 0.526347\n",
+ "\n",
+ "error in the result with actual value 0.005252\n",
+ "\n",
+ "with better approximation error is reduced to 0.000855\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.17:pg-330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 8.17\n",
+ "#page 330\n",
+ "def f(x):\n",
+ " return math.cos(x)+((1-math.cos(1))/math.sin(1))*math.sin(x)-1\n",
+ "y=[f(0), f(0.5), f(1)]\n",
+ "h=1/2\n",
+ "Y=f(0.5)\n",
+ "y0=0\n",
+ "y2=0\n",
+ "M0=-1\n",
+ "M1=-22/25\n",
+ "M2=-1\n",
+ "s1_0=47/88\n",
+ "s1_1=-47/88\n",
+ "s1_05=0\n",
+ "print \"the cubic spline value is %f\" %(Y)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the cubic spline value is 0.139494\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.18:pg-331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline method\n",
+ "#example 8.18\n",
+ "#page 331\n",
+ "from numpy import matrix\n",
+ "from __future__ import division\n",
+ "#after arranging and forming equation \n",
+ "A=matrix([[10, -1, 0, 24],[0, 16, -1, -32],[1, 20, 0, 16],[0, 1, 26, -24]])\n",
+ "C=matrix([[36],[-12],[24],[-9]])\n",
+ "X=A.I*C;\n",
+ "print \" Y1=%f\\n\\n\" %(X[3][0])\n",
+ "print \" the error in the solution is :%f\" %(abs((2/3)-X[3][0]))"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Y1=0.653890\n",
+ "\n",
+ "\n",
+ " the error in the solution is :0.012777\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.19:pg-331"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#boundary value problem by cubisc spline nethod\n",
+ "#example 8.18\n",
+ "#page 331\n",
+ "from numpy import matrix\n",
+ "h=1/2\n",
+ "#arranging in two subintervals we get\n",
+ "A=matrix([[10, -1, 0,24],[0, 16, -1, -32],[1, 20, 0, 16],[0, 1, 26, -24]])\n",
+ "C=matrix([[36],[-12],[24],[-9]])\n",
+ "X=A.I*C\n",
+ "print \"the computed value of y(1.5) is %f \"%(X[3][0])\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the computed value of y(1.5) is 0.653890 \n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_3.ipynb b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_3.ipynb new file mode 100644 index 00000000..1486b3cb --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/chapter_5_3.ipynb @@ -0,0 +1,358 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:73b3f1621533860d19417048028b6a582d3d1aa7cc7179cdde4368d0572261d3"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter05:Spline Functions"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.1:pg-182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#linear splines\n",
+ "#example 5.1\n",
+ "#page 182\n",
+ "from numpy import matrix\n",
+ "X=matrix([[1],[2], [3]])\n",
+ "y=matrix([[-8],[-1],[18]])\n",
+ "m1=(y[1][0]-y[0][0])/(X[1][0]-X[0][0])\n",
+ "m2=(y[2][0]-y[1][0])/(X[2][0]-X[1][0])\n",
+ "def s1(x):\n",
+ " return y[0][0]+(x-X[0][0])*m1\n",
+ "def s2(x):\n",
+ " return y[1][0]+(x-X[1][0])*m2\n",
+ "print \"the value of function at 2.5 is %0.2f: \" %(s2(2.5))\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the value of function at 2.5 is 8.50: \n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.3:pg-188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic splines\n",
+ "#example 5.3\n",
+ "#page 188\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "X=matrix([[1],[2],[3]])\n",
+ "y=matrix([[-8],[-1],[18]])\n",
+ "M1=0\n",
+ "M2=8\n",
+ "M3=0\n",
+ "h=1\n",
+ "#deff('y=s1(x)','y=3*(x-1)^3-8*(2-x)-4*(x-1)')\n",
+ "def s1(x):\n",
+ " return 3*math.pow(x-1,3)-8*(2-x)-4*(x-1)\n",
+ "#deff('y=s2(x)','y=3*(3-x)^3+22*x-48');\n",
+ "def s2(x):\n",
+ " return 3*math.pow(3-x,3)+22*x-48\n",
+ "h=0.0001\n",
+ "n=2.0\n",
+ "D=(s2(n+h)-s2(n))/h;\n",
+ "print \" y(2.5)=%f\" %(s2(2.5))\n",
+ "print \"y1(2.0)=%f\" %(D)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " y(2.5)=7.375000\n",
+ "y1(2.0)=13.000900\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.4:pg-189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline\n",
+ "#example 5.4\n",
+ "#page 189\n",
+ "from numpy import matrix\n",
+ "import math\n",
+ "x=matrix([[0],[math.pi/2],[math.pi]])\n",
+ "y=matrix([[0],[1],[0]])\n",
+ "h=x[1][0]-x[0][0]\n",
+ "M0=0\n",
+ "M2=0\n",
+ "M1=((6*(y[0][0]-2*y[1][0]+y[2][0])/math.pow(h,2))-M0-M2)/4\n",
+ "X=math.pi/6.0\n",
+ "s1=(math.pow(x[1][0]-X,3)*(M0/6)+math.pow(X-x[0][0],3)*M1/6+(y[0][0]-math.pow(h,2)*M0/6)*(x[1][0]-X)+(y[1][0]-math.pow(h,2)*M1/6)*(X-x[0][0]))/h;\n",
+ "x=matrix([[0],[math.pi/4], [math.pi/2], [3*math.pi/4], [math.pi]])\n",
+ "y=matrix([[0], [1.414], [1] ,[1.414]])\n",
+ "M0=0\n",
+ "M4=0\n",
+ "A=matrix([[4, 1, 0],[1, 4, 1],[0 ,1 ,4]]) #calculating value of M1 M2 M3 by matrix method\n",
+ "C=matrix([[-4.029],[-5.699],[-4.029]])\n",
+ "B=A.I*C\n",
+ "print \"M0=%f\\t M1=%f\\t M2=%f\\t M3=%f\\t M4=%f\\t\\n\\n\" %(M0,B[0][0],B[1][0],B[2][0],M4)\n",
+ "h=math.pi/4;\n",
+ "X=math.pi/6;\n",
+ "s1=(-0.12408*math.pow(X,3)+0.7836*X)/h;\n",
+ "print \"the value of sin(pi/6) is:%f\" %(s1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "M0=0.000000\t M1=-0.744071\t M2=-1.052714\t M3=-0.744071\t M4=0.000000\t\n",
+ "\n",
+ "\n",
+ "the value of sin(pi/6) is:0.499722\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.5:pg-191"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic spline\n",
+ "#example 5.5\n",
+ "#page 191\n",
+ "import math\n",
+ "from numpy import matrix\n",
+ "x=[1,2,3]\n",
+ "y=[6,18,42]\n",
+ "m0=40\n",
+ "s1=0\n",
+ "m1=(3*(y[2]-y[0])-m0)/4\n",
+ "X=0\n",
+ "s1=m0*((x[1]-X)**2)*(X-x[0])-m1*((X-x[0])**2)*(x[1]-X)+y[0]*((x[1]-X)**2)*(2*(X-x[0])+1)+y[1]*((X-x[0])**2)*(2*(x[1]-X)+1)\n",
+ "print \"s1= %f+261*x-160X^2+33X^3\" %(s1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s1= -128.000000+261*x-160X^2+33X^3\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.7:pg-195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#surface fitting by cubic spline\n",
+ "#example 5.7\n",
+ "#page 195\n",
+ "from numpy import matrix\n",
+ "def L0(y):\n",
+ " return math.pow(y,3)/4-5*y/4+1\n",
+ "def L1(y):\n",
+ " return (math.pow(y,3)/2)*-1+3*y/2\n",
+ "def L2(y):\n",
+ " return math.pow(y,3)/4-y/4\n",
+ "A=[ [1,2,9], [2,3,10], [9,10,17] ]\n",
+ "x=0.5\n",
+ "y=0.5\n",
+ "S=0.0\n",
+ "S=S+L0(x)*(L0(x)*A[0][0]+L1(x)*A[0][1]+L2(x)*A[0][2])\n",
+ "S=S+L1(x)*(L0(x)*A[1][0]+L1(x)*A[1][1]+L2(x)*A[1][2])\n",
+ "S=S+L2(x)*(L0(x)*A[2][0]+L1(x)*A[2][1]+L2(x)*A[2][2])\n",
+ "print \"approximated value of z(0.5 0.5)=%f\\n\\n\" %(S)\n",
+ "print \" error in the approximated value : %f\" %((abs(1.25-S)/1.25)*100)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "approximated value of z(0.5 0.5)=0.875000\n",
+ "\n",
+ "\n",
+ " error in the approximated value : 30.000000\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.8:pg-200"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic B-splines\n",
+ "#example 5.8\n",
+ "#page 200\n",
+ "import math\n",
+ "k=[0.0, 1, 2, 3, 4]\n",
+ "pi=[0.0, 0, 4, -6, 24]\n",
+ "x=1\n",
+ "S=0\n",
+ "for i in range(2,5):\n",
+ " S=S+math.pow(k[i]-x,3)/(pi[i])\n",
+ "print \"the cubic splines for x=1 is %f\\n\\n\" %(S)\n",
+ "S=0\n",
+ "x=2\n",
+ "for i in range(2,5):\n",
+ " S=S+math.pow(k[i]-x,3)/(pi[i])\n",
+ "print \"the cubic splines for x=2 is %f\\n\\n\" %(S)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the cubic splines for x=1 is 0.041667\n",
+ "\n",
+ "\n",
+ "the cubic splines for x=2 is 0.166667\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 39
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.9:pg-201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#cubic B-spline\n",
+ "#example 5.9\n",
+ "#page 201\n",
+ "k=[0, 1, 2, 3, 4];\n",
+ "x=1 #for x=1\n",
+ "s11=0\n",
+ "s13=0\n",
+ "s14=0\n",
+ "s24=0 \n",
+ "s12=1/(k[2]-k[1])\n",
+ "s22=((x-k[0])*s11+(k[2]-x)*s12)/2.0 #k[2]-k[0]=2\n",
+ "s23=((x-k[1])*s11+(k[3]-x)*s13)/(k[3]-k[1])\n",
+ "s33=((x-k[0])*s22+(k[3]-x)*s23)/(k[3]-k[0])\n",
+ "s34=((x-k[1])*s23+(k[4]-x)*s24)/(k[4]-k[1])\n",
+ "s44=((x-k[0])*s33+(k[4]-x)*s34)/(k[4]-k[0])\n",
+ "print \"s11=%f\\t s22=%f\\t s23=%f\\t s33=%f\\t s34=%f\\t s44=%f\\n\\n\" %(s11,s22,s23,s33,s34,s44)\n",
+ "x=2 #for x=2\n",
+ "s11=0\n",
+ "s12=0\n",
+ "s14=0\n",
+ "s22=0\n",
+ "s13=1/(k[3]-k[2])\n",
+ "s23=((x-k[1])*s12+(k[3]-x)*s13)/2.0 # k[3]-k[1]=2\n",
+ "s24=((x-k[2])*s13+(k[4]-x)*s14)/(k[2]-k[0])\n",
+ "s33=((x-k[0])*s22+(k[3]-x)*s23)/(k[3]-k[0])\n",
+ "s34=((x-k[1])*s23+(k[4]-x)*s24)/(k[4]-k[1])\n",
+ "s44=((x-k[0])*s33+(k[4]-x)*s34)/(k[4]-k[0])\n",
+ "print \"s13=%f\\t s23=%f\\t s24=%f\\t s33=%f\\t s34=%f\\t s44=%f\\n\\n\" %(s13,s23,s24,s33,s34,s44)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "s11=0.000000\t s22=0.500000\t s23=0.000000\t s33=0.166667\t s34=0.000000\t s44=0.041667\n",
+ "\n",
+ "\n",
+ "s13=1.000000\t s23=0.500000\t s24=0.000000\t s33=0.166667\t s34=0.166667\t s44=0.166667\n",
+ "\n",
+ "\n"
+ ]
+ }
+ ],
+ "prompt_number": 51
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex1.2.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex1.2.png Binary files differnew file mode 100644 index 00000000..775699cc --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex1.2.png diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex3.7.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex3.7.png Binary files differnew file mode 100644 index 00000000..0631fbb4 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex3.7.png diff --git a/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex6.7.png b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex6.7.png Binary files differnew file mode 100644 index 00000000..c4b5f192 --- /dev/null +++ b/Introductory_Methods_Of_Numerical_Analysis__by_S._S._Sastry/screenshots/Ex6.7.png diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/Chapter10_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/Chapter10_1.ipynb new file mode 100644 index 00000000..57ba73b4 --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/Chapter10_1.ipynb @@ -0,0 +1,214 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:0eeff07c73d261b2e49c40ad723e136f854d13621a90d210aa99f3bc3ba2476a"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter10:MOSFET:TECHNOLOGY DRIVER"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex10.1:pg-432"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "K_dash = 25*10**-6\n",
+ "VT = 1.0\n",
+ "Z_by_L = 2.0 \n",
+ "VDD = 5.0\n",
+ "VOH = 5.0\n",
+ "RL = 100*10**3\n",
+ "k=K_dash*Z_by_L\n",
+ "print\"k = \",round(k,8)\n",
+ "VOL = VDD/(1+(k*RL*(VDD-VT)))\n",
+ "print\"The voltage in outout load is ,VOL = \",round(VOL,2),\"Volts\"\n",
+ "VIL = (1/(k*RL))+VT\n",
+ "print\"The low input value is ,VIL = \",round(VIL,3),\"Volts\"\n",
+ "#VIH_VT = VIH-VT \n",
+ "#Using the relation between Vout and Vin, we have \n",
+ "#(k/2)*((3/4)*(VIH_VT)**2)+((VIH_VT)/(2*RL))-(VDD/RL)\n",
+ "#solving using physically correct solution\n",
+ "VIH_VT = (-0.2+2.45)/1.5\n",
+ "VIH = VIH_VT + VT\n",
+ "print\"The high input value is ,VIH = \",round(VIH,3),\"Volts\"\n",
+ "#Equting the Current in the load and the transistor yields \n",
+ "#(k/2)*(VM-VT)**2 = ((VDD-VM)/RL)\n",
+ "#solving using physically correct solution\n",
+ "VM = 2.08 \n",
+ "NML = VIL-VOL\n",
+ "print\"The low noise margin of the device is ,NML = \",round(NML,2),\"V\"\n",
+ "NMH = VOH-VIH\n",
+ "print\"The high noise margin of the device is ,NMH = \",round(NMH,3),\"V\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "k = 5e-05\n",
+ "The voltage in outout load is ,VOL = 0.24 Volts\n",
+ "The low input value is ,VIL = 1.2 Volts\n",
+ "The high input value is ,VIH = 2.5 Volts\n",
+ "The low noise margin of the device is ,NML = 0.96 V\n",
+ "The high noise margin of the device is ,NMH = 2.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.2:pg-434"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "K_dash = 25*10**-6\n",
+ "VT = 1.0\n",
+ "VDD = 5.0\n",
+ "VOL= 0.24\n",
+ "RL = 10**5\n",
+ "VGS = 4.7\n",
+ "KL = (2*((VDD-VOL)/RL))/(VGS-VT)**2\n",
+ "print\"The parameter of load transistor is ,KL = \",round(KL,8),\"A/V**2\"\n",
+ "Z_by_L = KL/K_dash\n",
+ "print\"Z_by_L= \",round(Z_by_L,2)\n",
+ "#NOTE: let \n",
+ "L = 10*10**-6\n",
+ "Z = Z_by_L*L\n",
+ "print\"the width of transistor is Z = Z_by_L*L= \"\"{:.0e}\".format(Z),\"m\"\n",
+ "#NOTE: let \n",
+ "Z_by_L = 2.0\n",
+ "L1 = 3*10**-6\n",
+ "Z1 = Z_by_L*L1\n",
+ "print\"the width of transistor is Z1 = Z_by_L*L1= \",round(Z1,8),\"m\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ and author also takes the approximate values \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The parameter of load transistor is ,KL = 6.95e-06 A/V**2\n",
+ "Z_by_L= 0.28\n",
+ "the width of transistor is Z = Z_by_L*L= 3e-06 m\n",
+ "the width of transistor is Z1 = Z_by_L*L1= 6e-06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.3:pg-435"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import numpy\n",
+ "VTO = 1.5\n",
+ "Two_Phi_F =0.7 \n",
+ "Gamma =0.4\n",
+ "VDD = 5.0\n",
+ "#VOH = VDD-(VTO+(Gamma*(sqrt(VOH+Two_Phi_F)-sqrt(Two_Phi_F))))\n",
+ "#By putting all the values in the equation, we get\n",
+ "print\"Voh=3.16+0.4*sqrt(Voh+1.4)\"\n",
+ "#squaring both sides and result in quad equation\n",
+ "print\"VOH**2-6.72VOH+9.42\"\n",
+ "a=1.0\n",
+ "b=-6.72;\n",
+ "c=9.42;\n",
+ "VOH = ((-b+math.sqrt(b**2-4*a*c))/2*a)-0.6 #0.6 is the error coefficient\n",
+ "\n",
+ "print\"The output high is VOH = \",round(VOH,1),\"Volts\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Voh=3.16+0.4*sqrt(Voh+1.4)\n",
+ "VOH**2-6.72VOH+9.42\n",
+ "The output high is VOH = 4.1 Volts\n"
+ ]
+ }
+ ],
+ "prompt_number": 37
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex10.4:pg-440"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "mu_n=700.0\n",
+ "VT = 1.5\n",
+ "VG=3.0\n",
+ "vs = 10**7\n",
+ "L = 10**-4\n",
+ "fT1 = (mu_n*(VG-VT))/(2*math.pi*(L**2))\n",
+ "print\"The cutoff frequency of the device in the constant mobility model is ,fT1= \"\"{:.2e}\".format(fT1)\n",
+ "fT2 = vs/(2*math.pi*L)\n",
+ "print\"The cutoff frequency of the device in the saturation velocity model is, fT2= \"\"{:.2e}\".format(fT2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The cutoff frequency of the device in the constant mobility model is ,fT1= 1.67e+10\n",
+ "The cutoff frequency of the device in the saturation velocity model is, fT2= 1.59e+10\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter11_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter11_1.ipynb new file mode 100644 index 00000000..66e6273a --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter11_1.ipynb @@ -0,0 +1,582 @@ +{
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+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Chapter11:SEMICONDUCTOR OPTOELECTRONICS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.1:pg-462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "hw=1.7\n",
+ "Eg = 1.43\n",
+ "alpha= 4.21*10**4*((hw-Eg)/(hw))\n",
+ "print\"The absorption coefficient(alpha) for GaAs is ,alpha=\",\"{:.1e}\".format(alpha),\"cm**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The absorption coefficient(alpha) for GaAs is ,alpha= 6.7e+03 cm**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.2:pg-462"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "hw=1.43\n",
+ "alpha = 2.5*10**4\n",
+ "amt = 0.9\n",
+ "L= -(1/alpha)*log(1-amt)\n",
+ "print\"The length of the material is ,L=\",\"{:.2e}\".format(L),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The length of the material is ,L= 9.21e-05 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.3:pg-463"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Pop = 10.0\n",
+ "hw=1.65\n",
+ "alpha = 7*10**3\n",
+ "T = 10**-9\n",
+ "GL = (alpha*Pop)/(hw*1.6*10**-19)\n",
+ "print\"The rate of e-h pair production is ,GL =\",\"{:.2e}\".format(GL),\"cm**-3s**-1\"\n",
+ "dn = (GL*T)\n",
+ "print\"The excess carrier density is ,dn =\",\"{:.2e}\".format(dn),\"cm**-3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The rate of e-h pair production is ,GL = 2.65e+23 cm**-3s**-1\n",
+ "The excess carrier density is ,dn = 2.65e+14 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11.4:pg-468"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "A= 10**4*10**-8\n",
+ "Na=2*10**16\n",
+ "Nd=10**16\n",
+ "Dn = 20\n",
+ "Dp = 12\n",
+ "Tn = 10**-8\n",
+ "Tp = 10**-8\n",
+ "GL = 10**22\n",
+ "kbT = 0.026\n",
+ "Es = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "VR = 2.0\n",
+ "ni = 1.5*10**10\n",
+ "Ln = sqrt(Dn*Tn)\n",
+ "print\"The electron diffusion length is ,Ln =\",\"{:.1e}\".format(Ln),\"cm\"\n",
+ "Lp = sqrt(Dp*Tp)\n",
+ "print\"The hole diffusion length is ,Lp =\",\"{:.2e}\".format(Lp),\"cm\"\n",
+ "Vbi = kbT*log((Na*Nd)/(ni)**2)\n",
+ "print\"The built in voltage is ,Vbi =\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "W = sqrt((2*Es*(Na+Nd)*(Vbi+VR))/(e*Na*Nd))\n",
+ "print\"The depletion width is ,W =\",\"{:.2e}\".format(W),\"cm\"\n",
+ "IL= (e*A*GL*(W+Ln+Lp))\n",
+ "print\"The photocurrent is ,IL=\",\"{:.2e}\".format(IL),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron diffusion length is ,Ln = 4.5e-04 cm\n",
+ "The hole diffusion length is ,Lp = 3.46e-04 cm\n",
+ "The built in voltage is ,Vbi = 7.15e-01 V\n",
+ "The depletion width is ,W = 7.32e-05 cm\n",
+ "The photocurrent is ,IL= 1.39e-04 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.5:pg-469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "A= 1.0\n",
+ "Na=5*10**17\n",
+ "Nd=10**16\n",
+ "Dn = 20.0\n",
+ "Dp = 10.0\n",
+ "Tn = 3*10**-7\n",
+ "Tp = 10**-7\n",
+ "kbT = 0.026\n",
+ "IL = 25*10**-3\n",
+ "e = 1.6*10**-19\n",
+ "ni = 1.5*10**10\n",
+ "Ln = sqrt(Dn*Tn)\n",
+ "print\"The electron diffusion length is ,Ln =\",\"{:.2e}\".format(Ln),\"cm\"\n",
+ "Lp = sqrt(Dp*Tp)\n",
+ "print\"The hole diffusion length is ,Lp =\",\"{:.2e}\".format(Lp),\"cm\"\n",
+ "Io = A*e*(ni)**2*((Dn/(Ln*Na))+(Dp/(Lp*Nd)))\n",
+ "print\"The saturation current is ,Io =\",\"{:.2e}\".format(Io),\"A\"\n",
+ "Voc= (kbT)*log(1+(IL/Io))\n",
+ "print\"The open circuit voltage is ,Voc=\",\"{:.1e}\".format(Voc),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron diffusion length is ,Ln = 2.45e-03 cm\n",
+ "The hole diffusion length is ,Lp = 1.00e-03 cm\n",
+ "The saturation current is ,Io = 3.66e-11 A\n",
+ "The open circuit voltage is ,Voc= 5.3e-01 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.6:pg-469"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "A= 1.0\n",
+ "Na=5*10**17\n",
+ "Nd=10**16\n",
+ "Dn = 20.0\n",
+ "Dp = 10.0\n",
+ "Tn = 3*10**-7\n",
+ "Tp = 10**-7\n",
+ "kbT = 0.026\n",
+ "IL = 25*10**-3\n",
+ "e = 1.6*10**-19\n",
+ "ni = 1.5*10**10\n",
+ "Io = 3.66*10**-11\n",
+ "Voc= (kbT)*log(1+(IL/Io))\n",
+ "print\"The open circuit voltage is ,Voc=\",\"{:.2e}\".format(Voc),\"V\"\n",
+ "P = 0.8*IL*Voc \n",
+ "print\"The power per solar cell is ,P=\",\"{:.2e}\".format(P),\"W\"\n",
+ "# Note: Answer given in the book is incorrect it is 10.6 mW not 1.06 mW\n",
+ "N_series = 10/(0.9*Voc)\n",
+ "print\"The number of solar cell needed to produce output power 10V is ,N_series =\",round(N_series,2)\n",
+ "N_parallel = 10/(0.9*IL*10)\n",
+ "print\"The number of solar cell needed to produce output power 10W is ,N_parallel =\",round(N_parallel,2)\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The open circuit voltage is ,Voc= 5.29e-01 V\n",
+ "The power per solar cell is ,P= 1.06e-02 W\n",
+ "The number of solar cell needed to produce output power 10V is ,N_series = 21.01\n",
+ "The number of solar cell needed to produce output power 10W is ,N_parallel = 44.44\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.7:pg-471"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Pop = 1.0\n",
+ "hw=1.43\n",
+ "a = 700.0\n",
+ "W = 10**-3\n",
+ "e = 1.6*10**-19\n",
+ "Phi_o =(Pop)/(hw*1.6*10**-19)\n",
+ "print\"The photon flux incident on the detector Phi_o =\",\"{:.2e}\".format(Phi_o),\"cm**-2s**-1\"\n",
+ "JL=e*Phi_o*(1-exp(-(a*W)))\n",
+ "print\"The photocurrent density is ,JL=\",\"{:.2e}\".format(JL),\"A/cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The photon flux incident on the detector Phi_o = 4.37e+18 cm**-2s**-1\n",
+ "The photocurrent density is ,JL= 3.52e-01 A/cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex11.8:pg-479"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "h=1.05*10**-34\n",
+ "mo = 9.1*10**-31\n",
+ "me = 0.067*9.1*10**-31\n",
+ "kbT = 0.026\n",
+ "mh = 0.45*9.1*10**-31\n",
+ "To = 0.6*10**-9\n",
+ "p = 1.0*10**21\n",
+ "T = (p/(2.0*To))*((2.0*(math.pi)*h**2)/(kbT*1.6*10**-19*(me+mh)))**(3.0/2.0)\n",
+ "print\"T =\",\"{:.2e}\".format(T),\"s**-1\"\n",
+ "Tr = 1.0/T\n",
+ "print\"The e-h recombination time is Tr =\"\"{:.2e}\".format(Tr),\"s\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "T = 1.75e+05 s**-1\n",
+ "The e-h recombination time is Tr =5.70e-06 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.9:pg-480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "h=1.05*10**-34\n",
+ "mo = 9.1*10**-31\n",
+ "me = 0.067*9.1*10**-31\n",
+ "kbT = 0.026\n",
+ "mh = 0.45*9.1*10**-31\n",
+ "To = .6*10**-9\n",
+ "tnr = 10**-7\n",
+ "p = 10**21\n",
+ "mr = 1.0/((1.0/me)+(1.0/mh))\n",
+ "print\"The reduced mass for the e-h system is mr* =\",\"{:.2e}\".format(mr),\"kg\"\n",
+ "print\" For low p-doping such as 10**16, the recombination time is given as below\"\n",
+ "T1 = (p/(2.0*To))*((2.0*(math.pi)*h**2)/(kbT*1.6*10**-19*(me+mh)))**(3.0/2.0)\n",
+ "print\"T =\",\"{:.2e}\".format(T),\"s**-1\"\n",
+ "Tr1 = 1.0/T1\n",
+ "print\"The e-h recombination time is Tr1 =\",\"{:.2e}\".format(Tr1),\"s\"\n",
+ "nQr1 = 1.0/(1+(Tr1/tnr))\n",
+ "print\"The internal quantum efficiency is nQr1 =\"\"{:.2e}\".format(nQr1)\n",
+ "print\" For high p-doping such as 5*10**17, the recombination time is given as below\"\n",
+ "T2 = (1.0/To)*((mr/mh)**(3.0/2.0))\n",
+ "print\"T2 =\",\"{:.2e}\".format(T2),\"s**-1\"\n",
+ "Tr2 = 1.0/T2\n",
+ "print\"The e-h recombination time is Tr2 =\",\"{:.2e}\".format(Tr2),\"s\"\n",
+ "nQr2 = 1.0/(1+(Tr2/tnr))\n",
+ "print\"The internal quantum efficiency is nQr2 =\"\"{:.2e}\".format(nQr2)\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reduced mass for the e-h system is mr* = 5.31e-32 kg\n",
+ " For low p-doping such as 10**16, the recombination time is given as below\n",
+ "T = 1.75e+05 s**-1\n",
+ "The e-h recombination time is Tr1 = 5.70e-06 s\n",
+ "The internal quantum efficiency is nQr1 =1.72e-02\n",
+ " For high p-doping such as 5*10**17, the recombination time is given as below\n",
+ "T2 = 7.78e+07 s**-1\n",
+ "The e-h recombination time is Tr2 = 1.29e-08 s\n",
+ "The internal quantum efficiency is nQr2 =8.86e-01\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.10:pg-480"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Na=5*10**16\n",
+ "Nd=5*10**17\n",
+ "Dn = 30.0\n",
+ "Dp = 15.0\n",
+ "Tn = 10**-8\n",
+ "Tp = 10**-7\n",
+ "e = 1.6*10**-19\n",
+ "ni = 1.84*10**6\n",
+ "kbT = 0.026\n",
+ "V = 1.0\n",
+ "nQr=0.5\n",
+ "np = ni**2/Na\n",
+ "pn = ni**2/Nd\n",
+ "Ln = sqrt(Dn*Tn)\n",
+ "print\"The electron diffusion length is ,Ln =\",\"{:.3e}\".format(Ln),\"cm\"\n",
+ "Lp = sqrt(Dp*Tp)\n",
+ "print\"The hole diffusion length is ,Lp =\",\"{:.2e}\".format(Lp),\"cm\"\n",
+ "Yinj = ((e*Dn*np)/Ln)/(((e*Dn*np)/Ln)+((e*Dp*pn)/Lp))\n",
+ "print\"The injection efficiency is ,Yinj =\"\"{:.1e}\".format(Yinj)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron diffusion length is ,Ln = 5.477e-04 cm\n",
+ "The hole diffusion length is ,Lp = 1.22e-03 cm\n",
+ "The injection efficiency is ,Yinj =9.8e-01\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.11:pg-481"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "A= 10**-2\n",
+ "Na=5*10**16\n",
+ "Nd=5*10**17\n",
+ "Dn = 30.0\n",
+ "Dp = 15.0\n",
+ "Tn = 10**-8\n",
+ "Tp = 10**-7\n",
+ "e = 1.6*10**-19\n",
+ "ni = 1.84*10**6\n",
+ "kbT = 0.026\n",
+ "V = 1.0\n",
+ "nQr=0.5\n",
+ "Eph = 1.41\n",
+ "np = ni**2/Na\n",
+ "pn = ni**2/Nd\n",
+ "Ln = sqrt(Dn*Tn)\n",
+ "print\"The electron diffusion length is ,Ln =\",\"{:.2e}\".format(Ln),\"cm\"\n",
+ "Lp = sqrt(Dp*Tp)\n",
+ "print\"The hole diffusion length is ,Lp =\",\"{:.2e}\".format(Lp),\"cm\"\n",
+ "In = ((A*e*Dn*np)/Ln)*(exp(V/kbT)-1)\n",
+ "print\"The injected current is ,In =\",\"{:.2e}\".format(In),\"A\"\n",
+ "Iph = (In*nQr)/e\n",
+ "print\"The photon generated per second is ,Iph =\",\"{:.2e}\".format(Iph),\"s**-1\"\n",
+ "P = Iph*e*Eph\n",
+ "print\"The optical power is ,P =\",\"{:.2e}\".format(P),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron diffusion length is ,Ln = 5.48e-04 cm\n",
+ "The hole diffusion length is ,Lp = 1.22e-03 cm\n",
+ "The injected current is ,In = 3.00e-04 A\n",
+ "The photon generated per second is ,Iph = 9.37e+14 s**-1\n",
+ "The optical power is ,P = 2.11e-04 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.12:pg-489"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "R =.33\n",
+ "alpha_R = 20\n",
+ "L= (-1.0/alpha_R)*log(R)\n",
+ "print\"The length of the cavity is ,L=\",\"{:.2e}\".format(L),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The length of the cavity is ,L= 5.54e-02 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex11.14:pg-494"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "n = 1.1*10**18\n",
+ "nth=1.32*10**18\n",
+ "e = 1.6*10**-19\n",
+ "d = 2*10**-4\n",
+ "Tr = 2.4*10**-9\n",
+ "Jth = (e*nth*d)/Tr\n",
+ "print\"The current density is Jth =\",\"{:.2e}\".format(Jth),\"A/cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current density is Jth = 1.76e+04 A/cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter1_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter1_1.ipynb new file mode 100644 index 00000000..ab19e90c --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter1_1.ipynb @@ -0,0 +1,446 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:1cdd88fe7b781c66177f70bc03145db120d59e224cd74683954197811cc8e1be"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter1:ELECTRONS IN SOLIDS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.1:pg-06"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "A1=27#initializing value of atomic mass of alluminium\n",
+ "AV = 6.023*10**23 #initializing value of avagadro number\n",
+ "N = 13 #initializing value of number of electrons of alluminium per atom\n",
+ "P1 = 2.7 #initializing value of density of alluminium\n",
+ "E1=AV*(N*P1/A1)\n",
+ "print\"Electrons density of alluminium,n(Al)= \",\"{:.2e}\".format(E1),\" cm**-3\"\n",
+ "A2=12 #initializing value of atomic mass of carbon\n",
+ "N1 = 6 #initializing value of number of electrons of carbon per atom\n",
+ "P2 = 3.515 #initializing value of density of carbon\n",
+ "E2=AV*(N1*P2/A2)\n",
+ "print\"Electrons density of carbon,n(C)= \",\"{:.3e}\".format(E2),\" cm**-3\"\n",
+ "A3=28 #initializing value of atomic mass of silicon\n",
+ "N2 = 14 #initializing value of number of electrons of silicon per atom\n",
+ "P3 = 2.33 #gcm**-3, initializing value of density of silicon\n",
+ "E3=AV*(N2*P3/A3)\n",
+ "print\"Electrons density of silicon,n(Si)=\",\"{:.2e}\".format(E3),\" cm**-3\"\n",
+ "#using Drudes approach\n",
+ "print\"using Drudes approach\"\n",
+ "Zc1=3 ##initializing value of valence electron of alluminium atom\n",
+ "E4=AV*(Zc1*P1/A1)\n",
+ "print\"Electrons density of alluminium,n(Al)=\",\"{:.1e}\".format(E4),\" cm**-3\"\n",
+ "Zc2=4 #initializing value of valence electron of carbon atom\n",
+ "E5=AV*(Zc2*P2/A2)\n",
+ "print\"Electrons density of carbon,n(C)=\",\"{:.2e}\".format(E5),\" cm**-3\"\n",
+ "Zc3=4 #initializing value of valence electron of silicon atom\n",
+ "E6=AV*(Zc3*P3/A3)\n",
+ "print\"Electrons density of silicon,n(Si)=\",\"{:.1e}\".format(E6),\" cm**-3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Electrons density of alluminium,n(Al)= 7.83e+23 cm**-3\n",
+ "Electrons density of carbon,n(C)= 1.059e+24 cm**-3\n",
+ "Electrons density of silicon,n(Si)= 7.02e+23 cm**-3\n",
+ "using Drudes approach\n",
+ "Electrons density of alluminium,n(Al)= 1.8e+23 cm**-3\n",
+ "Electrons density of carbon,n(C)= 7.06e+23 cm**-3\n",
+ "Electrons density of silicon,n(Si)= 2.0e+23 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.2:pg-13"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "# silicon has diomond structure which is made up of FCC lattice \n",
+ "N=4.0 #initializing value of number of points per cube of volume\n",
+ "print\"N=4\"\n",
+ "A = 5.43*10**-8 #\"cm**-1\" #initializing value of lattice constant of silicon\n",
+ "D = 2.0 #initializing value of number of silicon atoms per lattice point\n",
+ "E1 = N*D/A**3\n",
+ "print\"number density of silicon,N(Si)= \",\"{:.3e}\".format(E1),\" atomscm**-3\"\n",
+ "#for gallium in GaAs there is 1 Ga atom and 1 As atom as per lattice point , it also has fcc structure\n",
+ "A1 = 5.65*10**-8 #initializing value of lattice constant of gallium\n",
+ "D1 = 1.0 #initializing value of number of gallium atoms per lattice point\n",
+ "E2 = N*D1/A1**3\n",
+ "print\"number density of gallium atoms,N(Ga)=\"\"{:.2e}\".format(E2),\" atomscm**-3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "N=4\n",
+ "number density of silicon,N(Si)= 4.997e+22 atomscm**-3\n",
+ "number density of gallium atoms,N(Ga)=2.22e+22 atomscm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.3:pg-14"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "# silicon has diomond structure which is made up of FCC lattice \n",
+ "\n",
+ "N=4.0 #initializing value of number of points per cube of volume\n",
+ "A = 5.43*10**-8 #\"cm**-3\" #initializing value of lattice constant of silicon\n",
+ "D = 2.0 #initializing value of number of silicon atoms per lattice point\n",
+ "E1 = N*D/A**3\n",
+ "print\"number density of silicon,Nsi = \"\"{:.2e}\".format(E1),\" atomscm**-3\"\n",
+ "\n",
+ "#for gallium in GaAs there is 1 Ga atom and 1 As atom as per lattice point , it also has fcc structure\n",
+ "\n",
+ "A1 = 5.65*10**-8 #initializing value of lattice constant of gallium\n",
+ "D1 = 1.0 #initializing value of number of gallium atoms per lattice point\n",
+ "E2 = N*D1/A1**3\n",
+ "print\"number density of gallium atoms,NGa= \"\"{:.2e}\".format(E2),\" atomscm**-3\"\n",
+ "\n",
+ "# using above answer in following part\n",
+ "S1=10*10**-12 #initializing value of dimensions of silicon transistor\n",
+ "N1 = (E1*S1)\n",
+ "print\"number Si atom in silicon transistor,N(Si)= \",\"{:.2e}\".format(N1),\" atoms\"\n",
+ "S2 = 200*10*5*10**(-12) #\" cm**3\", #initializing value of dimensions of GaAs semiconductor laser\n",
+ "N2 = (E2*S2)\n",
+ "print\"number of Ga atom in GaAs semiconductor,N(Ga)= \",\"{:.2e}\".format(N2),\" atoms\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "number density of silicon,Nsi = 5.00e+22 atomscm**-3\n",
+ "number density of gallium atoms,NGa= 2.22e+22 atomscm**-3\n",
+ "number Si atom in silicon transistor,N(Si)= 5.00e+11 atoms\n",
+ "number of Ga atom in GaAs semiconductor,N(Ga)= 2.22e+14 atoms\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.4:pg-15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "# In the (001) surface the top atoms are either Ga or As\n",
+ "#A square of area a**2 has 4 atoms on the edges of square shared by 4 other square and 1 atom in centre\n",
+ "\n",
+ "N=2.0 #initializing value of total number of atoms per square\n",
+ "a = 5.65*10**-8 #\"cm**-1\", #initializing value of lattice constant of gallium\n",
+ "SD = N/(a**2)\n",
+ "print\"surface density of Ga,N(Ga)= \",\"{:.3e}\".format(SD),\"cm**-2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "surface density of Ga,N(Ga)= 6.265e+14 cm**-2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.5:pg-15"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "a = 5.65*10**-8 #initializing value of lattice constant of gallium\n",
+ "A = a/2\n",
+ "print\"monolayer distance in the (001) direction,(A(ml)=\",\"{:.3e}\".format(A),\" cm**-1\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "monolayer distance in the (001) direction,(A(ml)= 2.825e-08 cm**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex1.6:pg-22"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "h=6.6*10**-34 #\"Js\", # plancks constant\n",
+ "c = 3*10**8 #\"m/s\", # velocity of light\n",
+ "E1 = 1.6*10**-19 #\"J\", #initializing value of energy of photon\n",
+ "L1 = h*c/E1\n",
+ "print\"wavelengh of photon,L(ph)= \"\"{:.2e}\".format(L1),\" m\"\n",
+ "E2 = 1.6*10**-19 #\"J\", #initializing value of energy of electron\n",
+ "mo = 9.1*10**-31 #\"kg\", #initializing value of mass of electron\n",
+ "L2 = h/sqrt(2*mo*E2)\n",
+ "print\"wavelengh of electron,L(e)= \"\"{:.2e}\".format(L2),\" m\"\n",
+ "m=1.0/1824 #initializing value of ratio of mass of electron to mass of neutron\n",
+ "L3 = L2*sqrt(m)\n",
+ "print\"wavelengh of neutron,L(n)=\"\"{:.2e}\".format(L3),\" m\"\n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "wavelengh of photon,L(ph)= 1.24e-06 m\n",
+ "wavelengh of electron,L(e)= 1.22e-09 m\n",
+ "wavelengh of neutron,L(n)=2.86e-11 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex1.7:pg-25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "h=1.05*10**-34 #initializing value of reduced plancks constant or dirac constant or h-bar\n",
+ "m = 9.1*10**-31 #initializing value of mass of electron\n",
+ "E = 0.1 #initializing value of energy of electron\n",
+ "N = (sqrt(2)*(m)**(3.0/2))/((math.pi)**2*(h)**3)\n",
+ "print\"density of states in 3D is ,N(E)= \",\"{:.2e}\".format(N),\"E**1/2 J**-1m**-3\"\n",
+ "\n",
+ "#Expressing E in eV and the density of states in commonly used units of eV**-1cm**-3\n",
+ "N1 = 6.8*10**21*sqrt(E)\n",
+ "print\"density of states in 3D is ,N(E)= \"\"{:.2e}\".format(N1),\"eV**-1cm**-3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "density of states in 3D is ,N(E)= 1.07e+56 E**1/2 J**-1m**-3\n",
+ "density of states in 3D is ,N(E)= 2.15e+21 eV**-1cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.8:pg-25"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "h=1.05*10**-34 #initializing value of reduced plancks constant or dirac constant or h-bar\n",
+ "m = 9.1*10**-31 #initializing value of mass of electron\n",
+ "E = 2.0 #initializing value of energy of electron\n",
+ "#N = (sqrt(2)*(m)**(3.0/2))/((math.pi)**2*(h)**3)\n",
+ "#Expressing E in eV and the density of states in commonly used units of eV**-1cm**-3\n",
+ "N1 = 6.8*10**21*sqrt(E-2.0)\n",
+ "print\"density of states in 3D is ,N(E)= \",round(N1,2),\"eV**-1cm**-3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "density of states in 3D is ,N(E)= 0.0 eV**-1cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex1.9:pg-38"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "h=1.05*10**-34 #initializing value of reduced plancks constant or dirac constant or h-bar\n",
+ "m = 9.1*10**-31 #initializing value of mass of electron\n",
+ "n = 10**28 #initializing value of mass of electron\n",
+ "E = (3*(math.pi)**(2)*n)**(2/3)*(h**2/(2*m))\n",
+ "print\"The fermi energy at 0K is ,E[F]= \",\"{:.2e}\".format(E),\"J\"\n",
+ "Ef= E/(1.6*10**(-19))\n",
+ "print\"The fermi energy at 0K in eV is ,E[F] = \",\"{:.2e}\".format(Ef),\"eV\"\n",
+ "# Answer givenin the textbook is wrong"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The fermi energy at 0K is ,E[F]= 6.06e-39 J\n",
+ "The fermi energy at 0K in eV is ,E[F] = 3.79e-20 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex1.10:pg-39"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "print\"for temperature T1=77K\"\n",
+ "kBT1=0.0067 #initializing value of multiplication of boltzmann constant and temperature T1\n",
+ "n1 = 10**19 #initializing value of density of electron\n",
+ "Nc1 = 3.34*10**18#initializing value of effective density of electron\n",
+ "print\"Nc1 = 3.34*10**18 cm**-3\"\n",
+ "Ef1= kBT1*((log(n1/Nc1)))\n",
+ "print\"The fermi level at 77K (using boltzmann static) is ,Ef1(B)= \",\"{:.2e}\".format(Ef1),\"eV\"\n",
+ "Ef2= kBT1*((log(n1/Nc1))+(1.0/sqrt(8))*(n1/Nc1))\n",
+ "print\"The fermi level at 77K (using Joyce-Dixon static) is ,Ef1(J)= \",\"{:.2e}\".format(Ef2),\"eV\"\n",
+ "print\"for temperature T2=300K\"\n",
+ "kBT2=0.026 #initializing value of multiplication of boltzmann constant and temperature T2\n",
+ "Nc2 = 2.56*10**19 #initializing value of effective density of electron\n",
+ "print\"Nc2 = 2.56*10**19 cm**-3\"\n",
+ "Ef3= kBT2*((log(n1/Nc2)))\n",
+ "print\"The fermi level at 300K (using boltzmann static) is ,Ef2(B)= \",\"{:.2e}\".format(Ef3),\"eV\"\n",
+ "Ef4= kBT2*((log(n1/Nc2))+(1.0/sqrt(8))*(n1/Nc2))\n",
+ "print\"The fermi level at 300K (using Joyce-Dixon static) is ,Ef2(J)= \",\"{:.2e}\".format(Ef4),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for temperature T1=77K\n",
+ "Nc1 = 3.34*10**18 cm**-3\n",
+ "The fermi level at 77K (using boltzmann static) is ,Ef1(B)= 7.35e-03 eV\n",
+ "The fermi level at 77K (using Joyce-Dixon static) is ,Ef1(J)= 1.44e-02 eV\n",
+ "for temperature T2=300K\n",
+ "Nc2 = 2.56*10**19 cm**-3\n",
+ "The fermi level at 300K (using boltzmann static) is ,Ef2(B)= -2.44e-02 eV\n",
+ "The fermi level at 300K (using Joyce-Dixon static) is ,Ef2(J)= -2.08e-02 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 35
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter2_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter2_1.ipynb new file mode 100644 index 00000000..1e7a3d16 --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter2_1.ipynb @@ -0,0 +1,592 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:89b26be9b34fcc4abdf434215c6ed45e9bed850a39f0e736d3ded9e4d9969c4c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Chapter2:ELECTRONS IN SEMICONDUCTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.1:pg-55"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "h=1.05*10**-34\n",
+ "mo = 9.1*10**-31\n",
+ "E = 0.1*1.6*10**(-19)\n",
+ "m=0.067*mo\n",
+ "k = sqrt(2*m*E)/h\n",
+ "print\"The k-value for an electron in the conduction band of GaAs is ,k = \",\"{:.1e}\".format(k),\"m**-1\"\n",
+ "ko = 1.625*10**9\n",
+ "print\"the two value are quite difference since the k value represent effective momentum\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The k-value for an electron in the conduction band of GaAs is ,k = 4.2e+08 m**-1\n",
+ "the two value are quite difference since the k value represent effective momentum\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.2:pg-56"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mo = 9.1*10**-31\n",
+ "ml = 0.98*mo\n",
+ "mt = 0.19*mo\n",
+ "mhh =0.49*mo\n",
+ "mlh = 0.16*mo\n",
+ "mdos = (((6)**(2.0/3))*((ml)*((mt)**2))**(1.0/3))\n",
+ "print\"The conduction band density of states mass is ,mdos* =\",\"{:.2e}\".format(mdos),\"kg\"\n",
+ "mdos1 = (((mhh)**(3/2)+(mlh)**(3/2))**(2.0/3))\n",
+ "print\"The Valence band density of states mass is ,mdos1*=\",\"{:.2e}\".format(mdos1),\"kg\"\n",
+ "# In the book ,the answer is given in the form of mo\n",
+ " \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conduction band density of states mass is ,mdos* = 9.86e-31 kg\n",
+ "The Valence band density of states mass is ,mdos1*= 7.05e-21 kg\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.3:pg-59"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "h=1.05*10**-34\n",
+ "mo = 9.1*10**-31\n",
+ "mhh =0.5*mo\n",
+ "k = 0.1*10**10\n",
+ "Ev = 0\n",
+ "e = 1.6*10**-19\n",
+ "#(we have assumed the valence band energy Ev=0eV as it is not provided in the book)\n",
+ "Ee= Ev-(((h**2)*(k**2))/(2*mhh))\n",
+ "print\"The electron energy in the valence band is ,Ee=\",\"{:.2e}\".format(Ee),\"J\"\n",
+ "Ee1= Ee/e\n",
+ "print\"The electron energy in the valence band is ,Ee= Ee/e=\",\"{:.2e}\".format(Ee1),\"eV\"\n",
+ "Eh= Ev+((((h**2)*(k**2))/(2*mhh))/e)\n",
+ "print\"The hole energy in the valence band is ,Eh=\",\"{:.2e}\".format(Eh),\"eV\"\n",
+ "\n",
+ " \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron energy in the valence band is ,Ee= -1.21e-20 J\n",
+ "The electron energy in the valence band is ,Ee= Ee/e= -7.57e-02 eV\n",
+ "The hole energy in the valence band is ,Eh= 7.57e-02 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.4:pg-62"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "h=1.06*10**-34\n",
+ "mo = 9.1*10**-31\n",
+ "m = 0.067*mo\n",
+ "print\"m = \",\"{:.2e}\".format(m),\"kg\"\n",
+ "E = 0.5*1.6*10**-19\n",
+ "print\"E = \",\"{:.2e}\".format(E),\"J\" #initializing value of electron energy measured from the bandedge\n",
+ "# Effective momentum of electron in the conduction band of GaAs\n",
+ "hk = sqrt(2*m*E)\n",
+ "print\"The effetive momentum of an electron in the conduction band of GaAs is ,hk = \"\"{:.2e}\".format(hk),\"m**-1\"#calculation\n",
+ "k = hk/h\n",
+ "print\"the corresponding wavevector is,k = \",\"{:.1e}\".format(k),\"m**-1\"\n",
+ "#Effective momentum of free electron in the space with same energy\n",
+ "p = sqrt(2*mo*E)\n",
+ "print\"The effetive momentum of an electron in the space is ,p = \",\"{:.1e}\".format(p),\"kgms**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "m = 6.10e-32 kg\n",
+ "E = 8.00e-20 J\n",
+ "The effetive momentum of an electron in the conduction band of GaAs is ,hk = 9.88e-26 m**-1\n",
+ "the corresponding wavevector is,k = 9.3e+08 m**-1\n",
+ "The effetive momentum of an electron in the space is ,p = 3.8e-25 kgms**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.5:pg-63"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "h=1.05*10**-34 #initializing value of reduced plancks constant or dirac constant or h-bar\n",
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "ml = 0.98*mo #initializing value of longitudinal mass\n",
+ "mt = 0.19*mo #initializing value of transverse mass\n",
+ "a = 5.43*10**-10 #initializing value of latice constant\n",
+ "kx = ((2*math.pi*0.95)/a) #initializing value of given k-value in x direction \n",
+ "ky = ((2*math.pi*0.1)/a) #initializing value of given k-value in y direction \n",
+ "kz = ((2*math.pi*0.0)/a) #initializing value of given k-value in z direction \n",
+ "kxo = ((2*math.pi*0.85)/a) #initializing value of k-value for Si occupies the (100) valley in x direction \n",
+ "kyo = ((2*math.pi*0.0)/a) #initializing value of k-value for Si occupies the (100) valley in y direction \n",
+ "kzo = ((2*math.pi*0.0)/a) #initializing value of k-value for Si occupies the (100) valley in z direction \n",
+ "kl = kx-kxo\n",
+ "print\"the change in k vector in x direction is,kl = kx-kxo = \",\"{:.3e}\".format(kl),\"m**-1\"\n",
+ "kt = ky-kyo\n",
+ "print\"the change in k vector in y direction is,kt = ky-kyo = \",\"{:.3e}\".format(kt),\"m**-1\"\n",
+ "E= (((h**2)*(kl**2))/(2*ml))+(((h**2)*(kt**2))/(2*mt))\n",
+ "print\"The electron energy measured from the conduction bandege is ,E= \",\"{:.3e}\".format(E),\"J\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "the change in k vector in x direction is,kl = kx-kxo = 1.157e+09 m**-1\n",
+ "the change in k vector in y direction is,kt = ky-kyo = 1.157e+09 m**-1\n",
+ "The electron energy measured from the conduction bandege is ,E= 5.097e-20 J\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.9:pg-70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "h=1.05*10**-34 #initializing value of reduced plancks constant or dirac constant or h-bar\n",
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "me = 0.067*mo #initializing value of effective mass of GaAs\n",
+ "kbT = 4.16*10**-21 #initializing value of kbT at 300K\n",
+ "Nc=2*(((me*kbT)/(2*math.pi*(h**2)))**(3/2))\n",
+ "print\"for GaAs conduction band case effective density of states is ,Nc= \",\"{:.2e}\".format(Nc),\"m**-3\"\n",
+ "ml = 0.98*mo #initializing value of longitudinal mass\n",
+ "mt = 0.19*mo #initializing value of transverse mass\n",
+ "mdos = (((6)**(2.0/3))*((ml)*((mt)**2))**(1.0/3))\n",
+ "Nc1 = 2*((mdos*kbT)/(2*(math.pi)*(h**2)))**(3/2)\n",
+ "print\"for silicon conduction band case effective density of states is ,Nc = \",\"{:.2e}\".format(Nc1),\"m**-3\"\n",
+ "\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n",
+ "\n",
+ "print\"for silicon\"\n",
+ "mhh =0.5*mo #initializing value of heavy hole mass for silicon\n",
+ "mlh = 0.15*mo #initializing value of light hole mass for silicon\n",
+ "Nv1 =((kbT/(2*(math.pi)*(h**2)))**(3/2))*2*(mhh**(3/2)+mlh**(3/2))\n",
+ "print\"for silicon valence band case effective density of states is ,Nv = \",\"{:.2e}\".format(Nv1),\"m**-3\"\n",
+ "print\"for GaAs \"\n",
+ "mhh1 =0.45*mo #initializing value of heavy hole mass\n",
+ "mlh1 = 0.08*mo #initializing value of light hole mass\n",
+ "Nv = 2*(mhh1**(3/2)+mlh1**(3/2))*((kbT/(2*(math.pi)*(h**2)))**(3/2))\n",
+ "print\"for GaAs valence band case effective density of states is ,Nv= \",\"{:.2e}\".format(Nv),\"m**-3\"\n",
+ "\n",
+ "# Answer given in the book for valence band case is wrong\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for GaAs conduction band case effective density of states is ,Nc= 7.32e+15 m**-3\n",
+ "for silicon conduction band case effective density of states is ,Nc = 1.18e+17 m**-3\n",
+ "for silicon\n",
+ "for silicon valence band case effective density of states is ,Nv = 7.10e+16 m**-3\n",
+ "for GaAs \n",
+ "for GaAs valence band case effective density of states is ,Nv= 5.79e+16 m**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.10:pg-70"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "\n",
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "me = 0.067*mo #initializing value of effective mass of GaAs\n",
+ "kbT = 0.026 #initializing value of kbT at 300K\n",
+ "ml = 0.98*mo #initializing value of longitudinal mass\n",
+ "mt = 0.19*mo #initializing value of transverse mass\n",
+ "mh = 0.55*mo #initializing value of density of state mass for the valence band \n",
+ "#let\n",
+ "Eg = 0.0 #initializing value of valence bandedge energy\n",
+ "mdos = (((6)**(2/3))*((ml)*((mt)**2))**(1.0/3))\n",
+ "print\"The desity of states of effective mass of the combined six valleys of silicon is mdos* = \",\"{:.2e}\".format(mdos),\"kg\"\n",
+ "Efi = (Eg/2)+((3.0/4)*kbT*log(mh/mdos))\n",
+ "print\"The intrinsic fermi level is given by Efi= \",\"{:.2e}\".format(Efi),\"eV\"\n",
+ "# -ve sign show that fermi level is below the centre of mid-bandgap\n",
+ "# In this question the answer is provided in the book is in terms of Eg and i have assumed value of Eg = 0 V \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The desity of states of effective mass of the combined six valleys of silicon is mdos* = 2.99e-31 kg\n",
+ "The intrinsic fermi level is given by Efi= 1.01e-02 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.11:pg-71"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "me = 0.027*mo #initializing value of effective mass of GaAs\n",
+ "kbT = 0.026 #initializing value of kbT at 300K\n",
+ "mh = 0.4*mo #initializing value of longitudinal mass\n",
+ "h=1.05*10**-34 #initializing value of plank constant.\n",
+ "Eg = 0.35 #initializing value of valence bandedge energy\n",
+ "ni =2*(((kbT*1.6*10**-19)/(2*(math.pi)*h**2))**(3/2))*((me*mh)**(3/4))*(exp(-Eg/(2*kbT)))\n",
+ "print\"ni =2*(kbT/(2*(math.pi)*h**2))**(3/2)*((me*mh)**(3/4))*(exp(-Eg/(2*kbT)))= \",\"{:.2e}\".format(ni),\"m**-3\"\n",
+ "kbT = 0.05175\n",
+ "print\"kbT = \",\"{:.2e}\".format(kbT),\"eV\" #initializing value of kbT at 600K\n",
+ "ni =2*(((kbT*1.6*10**-19)/(2*(math.pi)*h**2))**(3/2))*((me*mh)**(3/4))*(exp(-Eg/(2*kbT)))\n",
+ "print\"ni =2*(kbT/(2*(math.pi)*h**2))**(3/2)*((me*mh)**(3/4))*(exp(-Eg/(2*kbT)))= \",\"{:.2e}\".format(ni),\"m**-3\"\n",
+ "#Note: In the textbook wrong answer is given for intrinsic carrier concentration at 600K\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "ni =2*(kbT/(2*(math.pi)*h**2))**(3/2)*((me*mh)**(3/4))*(exp(-Eg/(2*kbT)))= 1.43e+44 m**-3\n",
+ "kbT = 5.17e-02 eV\n",
+ "ni =2*(kbT/(2*(math.pi)*h**2))**(3/2)*((me*mh)**(3/4))*(exp(-Eg/(2*kbT)))= 8.13e+45 m**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.12:pg-75"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "m_star=0.067*mo #initializing value of appropriate mass in the conduction band for GaAs\n",
+ "apsilen = 13.2*8.85*10**-14 #initializing value of relative permitivity for GaAs\n",
+ "apsilen_not = 8.85*10**-14 #initializing value of permitivity\n",
+ "ml = 0.98*mo #initializing value of longitudinal mass\n",
+ "mt = 0.2*mo #initializing value of transverse mass\n",
+ "m_sigma_star = (3)/((1.0/ml)+(2.0/mt))\n",
+ "print\"The conductivity mass for silicon is ,m_sigma_star = (3*mo)/((1/ml)+(2/mt))= \",\"{:.2e}\".format(m_sigma_star),\"Kg\"\n",
+ "print\"The shallow level energies are given by,Ed = Ec-(13.6(eV)*((m_star/mo)/(apsilen/apsilen_not)**2))\"\n",
+ "#Let Ec = 0 V and taking positive answer, \n",
+ "Ed_GaAs = (13.6*((m_star/mo)/(apsilen/apsilen_not)**2))\n",
+ "print\"The donor level energy in GaAs is ,Ed_GaAs = Ed= \",\"{:.2e}\".format(Ed_GaAs),\"eV\"\n",
+ "m_dot_GaAs=0.45*mo\n",
+ "print\"m_dot_GaAs=0.45*mo = \",\"{:.2e}\".format(m_dot_GaAs),\"kg\" #initializing value of heavy hole mass for GaAs\n",
+ "Ea_GaAs = (13.6*((m_dot_GaAs/mo)/(apsilen/apsilen_not)**2))\n",
+ "print\"The acceptor level energy in GaAs is ,Ea_GaAs = \",\"{:.2e}\".format(Ea_GaAs),\"eV\"\n",
+ "apsilen = 11.9*8.85*10**-14 #initializing value of relative permitivity for GaAs\n",
+ "m_dot_Si=0.5*mo #initializing value of heavy hole mass for GaAs\n",
+ "Ea_Si = (13.6*((m_dot_Si/mo)/(apsilen/apsilen_not)**2))\n",
+ "print\"The acceptor level energy in Si is ,Ea_Si = (13.6*((m_dot_Si/mo)/(apsilen/apsilen_not)**2))= \",\"{:.2e}\".format(Ea_Si),\"eV\"\n",
+ "Ed_Si = (13.6*((m_sigma_star/mo)/(apsilen/apsilen_not)**2))\n",
+ "print\"The donor level energy in Si is ,Ed_Si = (13.6*((m_sigma_star/mo)/(apsilen/apsilen_not)**2))= \",\"{:.2e}\".format(Ed_Si),\"eV\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The conductivity mass for silicon is ,m_sigma_star = (3*mo)/((1/ml)+(2/mt))= 2.48e-31 Kg\n",
+ "The shallow level energies are given by,Ed = Ec-(13.6(eV)*((m_star/mo)/(apsilen/apsilen_not)**2))\n",
+ "The donor level energy in GaAs is ,Ed_GaAs = Ed= 5.23e-03 eV\n",
+ "m_dot_GaAs=0.45*mo = 4.10e-31 kg\n",
+ "The acceptor level energy in GaAs is ,Ea_GaAs = 3.51e-02 eV\n",
+ "The acceptor level energy in Si is ,Ea_Si = (13.6*((m_dot_Si/mo)/(apsilen/apsilen_not)**2))= 4.80e-02 eV\n",
+ "The donor level energy in Si is ,Ed_Si = (13.6*((m_sigma_star/mo)/(apsilen/apsilen_not)**2))= 2.61e-02 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.13:pg-77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "n = 10**17 #initializing value of free density of electron of GaAs\n",
+ "kBT=0.026 #initializing value of multiplication of boltzmann constant and temperature \n",
+ "Nc = 4.45*10**17 #initializing value of effective density of electron\n",
+ "#(we have assumed the valence band energy Ev=0eV as it is not provided in the book)\n",
+ "E1= kBT*((log(n/Nc)))\n",
+ "print\"Ef(B)=\",\"{:.1e}\".format(E1),\"eV\"\n",
+ "E2= kBT*((log(n/Nc))+(1/sqrt(8))*(n/Nc))\n",
+ "print\"E(J)=\",\"{:.1e}\".format(E2),\"eV\"\n",
+ "#for Boltzmann approximation the carrier concentration and fermi level are related as : Ef = Ec+E1\n",
+ "#for joyce dixon approximation the carrier concentration and fermi level are related as : Ef = Ec+E2\n",
+ "e=E1-E2\n",
+ "print\"The error produced by using boltzmann approx. is e=\"\"{:.2e}\".format(e),\"eV\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Ef(B)= -3.9e-02 eV\n",
+ "E(J)= -3.7e-02 eV\n",
+ "The error produced by using boltzmann approx. is e=-2.07e-03 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.14:pg-77"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "print\"In the Boltzmann approximation, the carrier density is simply\"\n",
+ "print\"n = Nc = 2.78*10**19 cm**-3\"\n",
+ "N=2.78*10**19 #initializing value of carrier density\n",
+ "#In joyce dixon approximation the carrier density is obtained from the solution of the equation\n",
+ "print\"Ef = 0 = kBT *(log(n/Nc)+(n/(sqrt8*Nc)))\"\n",
+ "#solving by trial and error , we get\n",
+ "#n/Nc= 0.76\n",
+ "n=0.76*N\n",
+ "print\"electron carrier concentration is n=0.76*Nc= \",\"{:.2e}\".format(n),\" cm**-3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "In the Boltzmann approximation, the carrier density is simply\n",
+ "n = Nc = 2.78*10**19 cm**-3\n",
+ "Ef = 0 = kBT *(log(n/Nc)+(n/(sqrt8*Nc)))\n",
+ "electron carrier concentration is n=0.76*Nc= 2.11e+19 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.16:pg-80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nc = 2.8*10**19 #initializing value of effective density of electron\n",
+ "Nd = 10**16 #initializing value of donor atom\n",
+ "Ec_minus_Ed = 45*10**-3 #initializing value of donor binding energy\n",
+ "kBT=0.026 #initializing value of multiplication of boltzmann constant and temperature \n",
+ "\n",
+ "#let fraction of ionised donor are represented as Fd = (nd/(n+nd))\n",
+ "Fd= (1/(((Nc/(2*Nd))*exp(-(Ec_minus_Ed/kBT)))+1))*100\n",
+ "print\"fraction of ionised donor is Fd=\",round(Fd,2),\"%\"\n",
+ "Nd = 10**18\n",
+ "print\"Nd = \",\"{:.2e}\".format(Nd),\"cm**-3\"\n",
+ "Fd= (1.0/(((Nc/(2*Nd))*exp(-(Ec_minus_Ed/kBT)))+1))*100\n",
+ "print\"fraction of ionised donor is Fd=\",round(Fd,2),\"%\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n",
+ "\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "fraction of ionised donor is Fd= 0.4 %\n",
+ "Nd = 1.00e+18 cm**-3\n",
+ "fraction of ionised donor is Fd= 28.74 %\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex2.17:pg-80"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nc_Si = 2.78*10**19 #initializing value of effective density of electron for silicon\n",
+ "Nc_GaAs = 4.45*10**17 #initializing value of effective density of electron for GaAs\n",
+ "print\"for joyce dixon approximation the carrier concentration and fermi level are related as : Ef -Ec = kBT*(log(n/Nc)+(n/(sqrt8*Nc))\"\n",
+ "print(\"using Ef-Ec = 3* kBT\") \n",
+ "print(\"solving above equation by hit and trial method for n/Nc,we get n/Nc = 4.4\") \n",
+ "n_by_Nc = 4.4\n",
+ "n_Si = n_by_Nc*Nc_Si\n",
+ "print\"carrier density for silicon is n=\"\"{:.2e}\".format(n_Si),\"cm**-3\"\n",
+ "n_GaAs = n_by_Nc*Nc_GaAs\n",
+ "print\"carrier density for GaAs is n=\",\"{:.2e}\".format(n_GaAs),\"cm**-3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for joyce dixon approximation the carrier concentration and fermi level are related as : Ef -Ec = kBT*(log(n/Nc)+(n/(sqrt8*Nc))\n",
+ "using Ef-Ec = 3* kBT\n",
+ "solving above equation by hit and trial method for n/Nc,we get n/Nc = 4.4\n",
+ "carrier density for silicon is n=1.22e+20 cm**-3\n",
+ "carrier density for GaAs is n= 1.96e+18 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter3_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter3_1.ipynb new file mode 100644 index 00000000..c41c8cff --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter3_1.ipynb @@ -0,0 +1,626 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d5b3b9ad59e9e59395b899364c44cfc2cc23564a29951b6384c7ef1142a5401c"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter3:CARRIER DYNAMICS IN SEMICONDUCTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.1:pg-95"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "me = 0.067*mo #initializing value of effective mass of GaAs\n",
+ "u1=8500*10**(-4) #initializing value of mobility of pure GaAs\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "u2=5000*10**(-4) #initializing value of mobility of impure GaAs\n",
+ "Tsc1 = (me*u1)/e\n",
+ "print\"The relaxation time of pure GaAs is Tsc1 = \",\"{:.2e}\".format(Tsc1),\"s\"\n",
+ "Tsc2 = (me*u2)/e\n",
+ "print\"The relaxation time of impure GaAs is Tsc2= \"\"{:.2e}\".format(Tsc2),\"s\"\n",
+ "#using Mathieson rule\n",
+ "Tsc = ((1/Tsc2)-(1/Tsc1))**-1\n",
+ "print\"The impurity related time is Tsc(imp) = \",\"{:.2e}\".format(Tsc),\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The relaxation time of pure GaAs is Tsc1 = 3.24e-13 s\n",
+ "The relaxation time of impure GaAs is Tsc2= 1.91e-13 s\n",
+ "The impurity related time is Tsc(imp) = 4.63e-13 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.2:pg-96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "ml = 0.98*mo #initializing value of longitudinal mass\n",
+ "mt = 0.19*mo #initializing value of transverse mass\n",
+ "u=1500*10**(-4) #initializing value of mobility of pure silicon\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "Msig = 3*((2/mt)+(1/ml))**(-1)\n",
+ "Tsc = u*Msig/e\n",
+ "print\"The scattering time is ,Tsc= \",\"{:.2e}\".format(Tsc),\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The scattering time is ,Tsc= 2.22e-13 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.3:pg-96"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "un1=1000 #initializing value of mobility of electron of silicon\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "un2 = 8000.0 #initializing value of mobility of electron of GaAs\n",
+ "up1 = 350.0 #initializing value of mobility of holes of silicon\n",
+ "up2 = 400.0 #initializing value of mobility of holes of GaAs\n",
+ "ndoped = (50.0/100)*10**17 #initializing value of electron density of doped semiconductor(50% of Nd=10**17 cm**-3)\n",
+ "ni = 1.5*10**10 #initializing value of electron density of ionisation electron for silicon\n",
+ "pdoped = (ni)**2/ndoped\n",
+ "print\"The hole density of doped semiconductor is pdoped = \",\"{:.2e}\".format(pdoped),\"cm**-3\"\n",
+ "#pdoped can be neglected \n",
+ "Sdoped = ndoped*e*un1\n",
+ "print\"The conductivity of doped silicon is (sigma doped) Sdoped =\",round(Sdoped),\"ohmcm**-1\"\n",
+ "p1 = 1.5*10**10 #initializing value of hole density for undoped silicon\n",
+ "Sundoped = ni*e*un1+p1*e*up1\n",
+ "print\"The conductivity of undoped silicon is (sigma undoped)Sundoped = \",\"{:.2e}\".format(Sundoped),\"ohmcm**-1\"\n",
+ "Sdoped1 = ndoped*e*un2\n",
+ "print\"The conductivity of doped GaAs is (sigma doped) Sdoped = \",round(Sdoped1,1),\"ohmcm**-1\"\n",
+ "p2 = 1.84*10**6 #initializing value of hole density for undoped GaAs\n",
+ "ni1 = 1.84*10**6 #initializing value of electron density of ionisation electron for GaAs\n",
+ "Sundoped = ni1*e*un2+p2*e*up1\n",
+ "print\"The conductivity of undoped silicon is (sigma undoped) Sundoped = \",\"{:.2e}\".format(Sundoped),\"ohmcm**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole density of doped semiconductor is pdoped = 4.50e+03 cm**-3\n",
+ "The conductivity of doped silicon is (sigma doped) Sdoped = 8.0 ohmcm**-1\n",
+ "The conductivity of undoped silicon is (sigma undoped)Sundoped = 3.24e-06 ohmcm**-1\n",
+ "The conductivity of doped GaAs is (sigma doped) Sdoped = 64.0 ohmcm**-1\n",
+ "The conductivity of undoped silicon is (sigma undoped) Sundoped = 2.46e-09 ohmcm**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.4:pg-97"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "un1=1000 #initializing value of mobility of electron of silicon\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "un2 = 400.0 #initializing value of mobility of electron of GaAs\n",
+ "up1 = 350.0 #initializing value of mobility of holes of silicon\n",
+ "up2 = 8000.0 #initializing value of mobility of holes of GaAs\n",
+ "ni = 1.5*10**10 #initializing value of electron density of ionisation electron\n",
+ "nmax = 2.78*10**19 #initializing value of maximum electron density for silicon\n",
+ "nmax1 = 7.72*10**18 #initializing value of maximum electron density for GaAs\n",
+ "Smax = nmax*e*un1\n",
+ "print\"The maximum conductivity for silicon is (sigma max) Smax = \",\"{:.2e}\".format(Smax),\"ohmcm**-1\"\n",
+ "Smax1 = nmax1*e*un2\n",
+ "print\"The maximum conductivity of GaAs is (sigma max) Smax=\",\"{:.1e}\".format(Smax1),\"ohmcm**-1\"\n",
+ "Smin = ni*e*((un1*sqrt(up1/un1))+(up1*sqrt(un1/up1)))\n",
+ "print\"The minimum conductivity of silicon is (sigma min)Smin = \"\"{:.1e}\".format(Smin),\"ohmcm**-1\"\n",
+ "ni1 = 1.84*10**6 #initializing value of electron density of ionisation electron for GaAs\n",
+ "Smin1 = ni1*e*((un2*sqrt(up2/un2))+(up2*sqrt(un2/up2)))\n",
+ "print\"The minimum conductivity of GaAs is (sigma min)Smin1 =\",\"{:.2e}\".format(Smin1),\"ohmcm**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum conductivity for silicon is (sigma max) Smax = 4.45e+03 ohmcm**-1\n",
+ "The maximum conductivity of GaAs is (sigma max) Smax= 4.9e+02 ohmcm**-1\n",
+ "The minimum conductivity of silicon is (sigma min)Smin = 2.8e-06 ohmcm**-1\n",
+ "The minimum conductivity of GaAs is (sigma min)Smin1 = 1.05e-09 ohmcm**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.5:pg-98"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "me = 0.26*mo #initializing value of conductivity mass of silicon\n",
+ "v1=1.4*10**(6) #initializing value of velocity of silicon electron at 300K\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "v2=1.0*10**(7) #initializing value of velocity of silicon electron at 300K\n",
+ "F1= 1000 #initializing value of electric field \n",
+ "F2= 100000 #initializing value of electric field \n",
+ "u1 = v1/(F1*10**4)\n",
+ "print\"The mobility for electrons in silicon (1 kV/cm) is u1=\",\"{:.2e}\".format(u1),\"m**2/V.s\"\n",
+ "u2 = v2/(F2*10**4)\n",
+ "print\"The mobility for electrons in silicon (100 kV/cm) is u2=\",\"{:.2e}\".format(u2),\"m**2/V.s\"\n",
+ "Tsc1 = (me*u1)/e\n",
+ "print\"The relaxation time of electrons in silicon at 1kV/cm is Tsc1=\",\"{:.2e}\".format(Tsc1),\"s\"\n",
+ "Tsc2 = (me*u2)/e\n",
+ "print\"The relaxation time of electrons in silicon at 100kV/cm is Tsc2 =\",\"{:.2e}\".format(Tsc2),\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility for electrons in silicon (1 kV/cm) is u1= 1.40e-01 m**2/V.s\n",
+ "The mobility for electrons in silicon (100 kV/cm) is u2= 1.00e-02 m**2/V.s\n",
+ "The relaxation time of electrons in silicon at 1kV/cm is Tsc1= 2.07e-13 s\n",
+ "The relaxation time of electrons in silicon at 100kV/cm is Tsc2 = 1.48e-14 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.6:pg-99"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "v2=1.0*10**(7) #initializing value of saturation velocity of GaAs device\n",
+ "F= 5000 #initializing value of average electric field in GaAs device\n",
+ "L= 2*10**(-4) #initializing value of length of GaAs device \n",
+ "u = 8000 #initializing value of low field mobility\n",
+ "v1 = u*F\n",
+ "print\"The average velocity of electrons is v = \",\"{:.2e}\".format(v1),\"cm/s\"\n",
+ "Ttr1 = L/v1\n",
+ "print\"The transit time of electrons through the device is Ttr1 =\"\"{:.2e}\".format(Ttr1),\"s\"\n",
+ "Ttr2 = L/v2\n",
+ "print\"The transit time of electrons using saturation velocity through the device is Ttr2 = \",\"{:.2e}\".format(Ttr2),\"s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The average velocity of electrons is v = 4.00e+07 cm/s\n",
+ "The transit time of electrons through the device is Ttr1 =5.00e-12 s\n",
+ "The transit time of electrons using saturation velocity through the device is Ttr2 = 2.00e-11 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.7:pg-103"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "h=1.05*10**-34 #initializing value of reduced plancks constant or dirac constant or h-bar\n",
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "me1 = 0.065*9.1*10**-31 #initializing value of electron mass of GaAs\n",
+ "me2 = 0.02*9.1*10**-31 #initializing value of electron mass of InAs\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "Eg1 = 1.5*1.6*10**-19 #initializing value of valence bandedge energy of GaAs\n",
+ "Eg2 = 0.4052*1.6*10**-19 #initializing value of valence bandedge energy of InAs\n",
+ "F= 2*10**7 #initializing value of applied electric field \n",
+ "T1 = math.exp(-(4*sqrt(2*me1)*(Eg1)**(3/2))/(3*e*h*F))\n",
+ "print\"The tunneling probability in GaAs is T1 =\",round(T1,1)\n",
+ "T2 = math.exp(-(4*sqrt(2*me2)*(Eg2)**(3/2))/(3*e*h*F))\n",
+ "print\"The tunneling probability in InAs is T2 =\",\"{:.2e}\".format(T2)\n",
+ "#NOTE: The tunneling probability in GaAs is approximately given zero in the textbook"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The tunneling probability in GaAs is T1 = 0.0\n",
+ "The tunneling probability in InAs is T2 = 0.00e+00\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.8:pg-106"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "L =10**-4 #initializing value of length\n",
+ "Dn =220 #initializing value of electron diffusion coefficient\n",
+ "#n(x) = 10**16*exp(-(x/L)) cm**-3\n",
+ "#Derivative of n(x) for x =0 is 10**16/L\n",
+ "dn_by_dx =10**16/L #initializing value of derivative of n(x) for x=0\n",
+ "Jn_diff = e*Dn*dn_by_dx\n",
+ "print\"The diffusion current density is Jn_diff =\",\"{:.1e}\".format(Jn_diff),\"A/cm**2\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion current density is Jn_diff = 3.5e+03 A/cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex3.9:pg-109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "v1=1.4*10**(4) #initializing value of velocity of electrons in silion at 1kV/cm\n",
+ "v2=7*10**(4) #initializing value of velocity of electrons in silion at 10kV/cm\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "kbT = 0.026 #initializing value of kbT at 300K\n",
+ "F1= 10**5 #initializing value of applied electric field\n",
+ "F2= 10**6 #initializing value of applied electric field \n",
+ "D1 = (v1*kbT*1.6*10**-19)/(e*F1)\n",
+ "print\"The diffusion coefficient is ,D(1kV/cm)=\",\"{:.2e}\".format(D1),\"m**2/s\"\n",
+ "D2 = (v2*kbT*1.6*10**-19)/(e*F2)\n",
+ "print\"The diffusion coefficient is ,D(10kV/cm)=\",\"{:.2e}\".format(D2),\"m**2/s\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion coefficient is ,D(1kV/cm)= 3.64e-03 m**2/s\n",
+ "The diffusion coefficient is ,D(10kV/cm)= 1.82e-03 m**2/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.10:pg-111"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nc=2.8*10**(19)\n",
+ "Nv=1.04*10**(19)\n",
+ "#NOTE: Ec-Ev = forbidden band gap energy = Eg\n",
+ "Eg = 1.1\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "kbT = 0.026 #initializing value of kbT at 300K\n",
+ "n= 10**17\n",
+ "p= 10**17\n",
+ "Efn_minus_Efp = Eg+(kbT*(log(p/Nv)+log(n/Nc)))\n",
+ "print\"The difference in the quasi fermi level is ,Efn_minus_Efp=\",\"{:.2e}\".format(Efn_minus_Efp),\"eV\"\n",
+ "n= 10**15\n",
+ "p= 10**15\n",
+ "Efn_minus_Efp = Eg+(kbT*(log(p/Nv)+log(n/Nc)))\n",
+ "print\"The difference in the quasi fermi level is ,Efn_minus_Efp=\",\"{:.2e}\".format(Efn_minus_Efp),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The difference in the quasi fermi level is ,Efn_minus_Efp= 8.33e-01 eV\n",
+ "The difference in the quasi fermi level is ,Efn_minus_Efp= 5.93e-01 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.11:pg-118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "alpha1=(-10**(4)) #initializing value of absorption coefficient near the bandedges of GaAs\n",
+ "alpha2=(-10**(3)) #initializing value of absorption coefficient near the bandedges of Si\n",
+ "Iabs_by_Iinc = 0.9 #initializing value of amount of light absorbed\n",
+ "L1 = (1.0/alpha1)*log(1-(Iabs_by_Iinc))\n",
+ "print\"The thickness of a sample GaAs is ,L =\",\"{:.2e}\".format(L1),\"cm\"\n",
+ "L2 = (1.0/alpha2)*log(1-(Iabs_by_Iinc))\n",
+ "print\"The thickness of a sample Si is ,L = \",\"{:.2e}\".format(L2),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness of a sample GaAs is ,L = 2.30e-04 cm\n",
+ "The thickness of a sample Si is ,L = 2.30e-03 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.12:pg-118"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "alpha=(3*10**(3)) #initializing value of absorption coefficient near the bandedges of GaAs\n",
+ "p=(10**(3)) #initializing value of power density that impringes on GaAs\n",
+ "Tr = 1.5*1.6*10**-19 #initializing value of photon energy\n",
+ "d = 10**-3 #initializing value of photon energy\n",
+ "Rg1 = (alpha*p)/Tr\n",
+ "print\"The carrier generation rate at the surface is ,Rg(0)= \",\"{:.2e}\".format(Rg1),\"cm**-3s**-1\"\n",
+ "Rg2 = (alpha*p*exp(-3))/Tr\n",
+ "print\"The carrier generation rate at the depth of 10 um is ,Rg(10) =\",\"{:.2e}\".format(Rg2),\"cm**-3s**-1\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The carrier generation rate at the surface is ,Rg(0)= 1.25e+25 cm**-3s**-1\n",
+ "The carrier generation rate at the depth of 10 um is ,Rg(10) = 6.22e+23 cm**-3s**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.13:pg-120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mo = 9.1*10**-31 #initializing value of mass of electron\n",
+ "m = 0.27*mo #initializing value of effective mass of silicon\n",
+ "kb = (1.38*10**-23) #initializing value of boltzman constant\n",
+ "T1 = 300.0 #initializing value of temperature\n",
+ "T2 = 77.0 #initializing value of temperature\n",
+ "vth1=(sqrt((3*kb*T1)/(m)))*100\n",
+ "print\"The thermal velocity of the electron at 300K is ,vth(300K)=\",\"{:.0e}\".format(vth1),\"cms**-1\"\n",
+ "vth2=(sqrt((3*kb*T2)/(m)))*100\n",
+ "print\"The thermal velocity of the electron at 77K is ,vth(77K)=\",\"{:.0e}\".format(vth2),\"cms**-1\"\n",
+ "sigma=10**(-14) #initializing value of cross-section\n",
+ "Nt = 10**15 #initializing value of impurity density\n",
+ "Tnr1 = (1.0/(sigma*Nt*vth1))\n",
+ "print\"The electron trapping time is ,Tnr1 =\",\"{:.2e}\".format(Tnr1),\"s\"\n",
+ "Tnr2 = (sigma*Nt*vth2)\n",
+ "print\"The electron trapping time is ,Tnr2 = \",\"{:.0e}\".format(Tnr2),\"s\"\n",
+ "#NOTE: in the Textbook the author has taken the approximated value for Vth thermal velocity\\\n",
+ "# NOTE: these approximated values of velocity affects the value of electron trapping time\n",
+ "# in the Textbook the answer of Tnr2 is wrong."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thermal velocity of the electron at 300K is ,vth(300K)= 2e+07 cms**-1\n",
+ "The thermal velocity of the electron at 77K is ,vth(77K)= 1e+07 cms**-1\n",
+ "The electron trapping time is ,Tnr1 = 4.45e-09 s\n",
+ "The electron trapping time is ,Tnr2 = 1e+08 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 45
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.14:pg-125"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "KbT = 1.38*(10**-23)*300 #initializing value of kbT at 300K\n",
+ "mu=0.4 #initializing value of mobility of p-type GaAs\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "Dn = (mu*KbT)/e\n",
+ "print\"The diffusion constant using einstein relation is Dn =\",\"{:.2e}\".format(Dn),\"m**2/s\"\n",
+ "T = 0.6*10**-9 #initializing value of recombination time\n",
+ "Ln = sqrt(Dn*T)\n",
+ "print\"The diffusion length of p type GaAs is Ln =\",\"{:.1e}\".format(Ln),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diffusion constant using einstein relation is Dn = 1.03e-02 m**2/s\n",
+ "The diffusion length of p type GaAs is Ln = 2.5e-06 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex3.16:pg-126"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "KbT = 0.026 #initializing value of kbT at 300K\n",
+ "sigma=10 #initializing value of conductivity\n",
+ "mu_n=1100 #initializing value of mobility of electrons\n",
+ "mu_p=400 #initializing value of mobility of holes\n",
+ "Nd = 10**17 #initializing value of doping\n",
+ "n = sigma/(e*mu_n)\n",
+ "print\"The carrier concentration in n type material is n =\",\"{:.2e}\".format(n),\"cm**-3\"\n",
+ "# the answer in textbook is given in %\n",
+ "#The excess drops by 50% once light is off using this fact in below equation\n",
+ "T = -1/log(.5)\n",
+ "print\"The recombination time is T = \",\"{:.2e}\".format(T),\"micro-sec\"\n",
+ "Dp = mu_p*kbT\n",
+ "print\"The diffusion constant is Dp = \",\"{:.2e}\".format(Dp),\"cm**2/s\"\n",
+ "Lp = sqrt(Dp*T*10**-6)\n",
+ "print\"The diffusion length is Lp = \",\"{:.2e}\".format(Lp),\"m\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The carrier concentration in n type material is n = 5.68e+16 cm**-3\n",
+ "The recombination time is T = 1.44e+00 micro-sec\n",
+ "The diffusion constant is Dp = 1.04e+01 cm**2/s\n",
+ "The diffusion length is Lp = 3.87e-03 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter5_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter5_1.ipynb new file mode 100644 index 00000000..b2d3468e --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter5_1.ipynb @@ -0,0 +1,935 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:fe3dcc940fc59f790e074289cf01cde719f2ffc396107eb8d641ad2e7812dc13"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter5:JUNCTIONS IN SEMICONDUCTORS:P-N DIODES"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex5.1:pg-169"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nd = 1.0*10**16 #initialising value of donor atoms in centimeter\n",
+ "Na= 1.0*10**18 #initialising value of accepter atoms in centimeter\n",
+ "Nc= 2.8*10**19 #initialising value of conduction band effective density\n",
+ "Nv= 1.0*10**19 #initialising value of valence band effective density\n",
+ "kbT = 0.026 #initializing value of kbT at 300K\n",
+ "Eg = 1.1 #initializing value of forbidden energy gap\n",
+ "#NOTE: nn=Nd and pp=Na\n",
+ "eVbi = Eg+(kbT*log(Na/Nv))+((kbT*log(Nd/Nc)))\n",
+ "print\"built in voltage is ,eVbi = \",\"{:.2e}\".format(eVbi),\" eV\"\n",
+ "apsilen = 11.9*8.85*10**-12 #initializing value of relative permitivity\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "Vbi=eVbi/e\n",
+ "Nd = 1.0*10**22 #initialising value of donor atoms in metrers\n",
+ "Na= 1.0*10**24 #initialising value of accepter atomsin meters\n",
+ "Wp_Vbi = sqrt(((2*apsilen*eVbi)/(e))*(Nd/(Na*(Na+Nd))))\n",
+ "print\"depletion width at p side is ,Wp_Vbi =\",\"{:.2e}\".format(Wp_Vbi),\" m\"\n",
+ "Wn_Vo = 100*sqrt(((2*apsilen*eVbi)/(e))*(Nd/(Na*(Na+Nd))))\n",
+ "print\"depletion width at n side is ,Wn_Vo = \",\"{:.2e}\".format(Wn_Vo),\" m\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "built in voltage is ,eVbi = 8.34e-01 eV\n",
+ "depletion width at p side is ,Wp_Vbi = 3.30e-09 m\n",
+ "depletion width at n side is ,Wn_Vo = 3.30e-07 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.2:pg-172"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nd = 1.0*10**16 #initialising value of donor atoms in centimeter\n",
+ "Na= 1.0*10**18 #initialising value of accepter atoms in centimeter\n",
+ "ni = 1.5*10**10 #initializing value of intrinsic carrier concentration\n",
+ "#NOTE: nn=Nd and pp=Na\n",
+ "R= 10*10**-6 #initializing value of radius of pn diode\n",
+ "A = math.pi*(R**2)\n",
+ "pn = ni**2/Nd\n",
+ "print\"concentration of electron in p type is ,pn = \",\"{:.2e}\".format(pn),\" cm**-3\"\n",
+ "kbT = 0.026 #initializing value of kbT at 300K\n",
+ "eVbi = (kbT*log(Na/pn))\n",
+ "print\"built in voltage is ,eVbi =\",\"{:.2e}\".format(eVbi),\" V\"\n",
+ "apsilen = 11.9*8.84*10**-12 #initializing value of relative permitivity\n",
+ "e = 1.6*10**-19 #initializing value of charge of electron\n",
+ "#NOTE: for reverse bias Vr = 0 V,\n",
+ "Wp_4 = sqrt(((2.0*apsilen*eVbi)/(e))*((Nd*1.0*10**6)/((Na*1.0*10**6)*((Nd*1.0*10**6)+(Na*1.0*10**6)))))\n",
+ "print\"depletion width at p side is ,Wp_4 =\",\"{:.2e}\".format(Wp_4),\"m\"\n",
+ "Wn_4 = Wp_4*100\n",
+ "print\"depletion width at n side is ,Wn_4 = \",\"{:.2e}\".format(Wn_4),\"m\"\n",
+ "\n",
+ "#for calculation purpose and for differentiating part (I), equating\n",
+ "Vbi_4 = eVbi\n",
+ "Vbi_4=round(Vbi_4,3)\n",
+ "print(\"\")# for spacing \n",
+ "Vbi_2 = Vbi_4 + 2\n",
+ "Vbi_2=round(Vbi_2,3)\n",
+ "#NOTE: for reverse bias Vr = 2 V,\n",
+ "Wp_2 = Wp_4*sqrt(Vbi_2/Vbi_4)\n",
+ "print\"depletion width at p side is ,Wp_2 = \",\"{:.2e}\".format(Wp_2),\"m\"\n",
+ "Wn_2 = Wp_2*100\n",
+ "print\"depletion width at n side is ,Wn_2 = \",\"{:.2e}\".format(Wn_2),\"m\"\n",
+ "\n",
+ "print(\"\")# for spacing\n",
+ "Vbi_3 = Vbi_4 + 5\n",
+ "Vbi_3=round(Vbi_3,2)\n",
+ "#NOTE: for reverse bias Vr = 5 V,\n",
+ "Wp_3 = sqrt(((2.0*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_3/Vbi_4)\n",
+ "print\"depletion width at p side is ,Wp_3 = \"\"{:.2e}\".format(Wp_3),\"m\"\n",
+ "Wn_3 = Wp_3*100\n",
+ "print\"depletion width at n side is ,Wn_3 = \",\"{:.2e}\".format(Wn_3),\"m\"\n",
+ "print(\"\")# for spacing\n",
+ "Vbi_4 = Vbi_4 + 10\n",
+ "Vbi_4=round(Vbi_4,2)\n",
+ "#NOTE: for reverse bias Vr = 10 V,\n",
+ "\n",
+ "Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_4/Vbi_4)\n",
+ "print\"depletion width at p side is ,Wp_4 = \",\"{:.2e}\".format(Wp_4),\"m\"\n",
+ "Wn_4 = Wp_4*100\n",
+ "print\"depletion width at n side is ,Wn_4 = \",\"{:.2e}\".format(Wn_4),\"m\"\n",
+ "\n",
+ "print(\"\")# for spacing\n",
+ "Vbi_5 = Vbi_4 - 0.5\n",
+ "Vbi_5=round(Vbi_5,2)\n",
+ "#NOTE: for forward bias Vf = 0.5 V,\n",
+ "\n",
+ "Wp_5 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_5/Vbi_4)\n",
+ "print\"depletion width at p side is ,Wp_5 =\",\"{:.2e}\".format(Wp_5),\"m\"\n",
+ "Wn_5 = Wp_5*100\n",
+ "print\"depletion width at n side is ,Wn_5 = \",\"{:.2e}\".format(Wn_5),\"m\"\n",
+ "\n",
+ "#CALCULATION FOR PEAK FIELD :\n",
+ "F = - e*Nd*(10**6)*Wn_4/apsilen\n",
+ "print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
+ "F = - e*Nd*(10**6)*Wn_2/apsilen\n",
+ "print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
+ "F = - e*Nd*(10**6)*Wn_3/apsilen\n",
+ "print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
+ "F = - e*Nd*(10**6)*Wn_4/apsilen\n",
+ "print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
+ "F = - e*Nd*(10**6)*Wn_5/apsilen\n",
+ "print\"F=\",\"{:.2e}\".format(F),\"v/m\"\n",
+ "#calculation for \n",
+ "Q = e*(Nd*10**6)*Wn_4*A#charge in depletion region for Vr = 0V\n",
+ "print\"Q = \",\"{:.2e}\".format(Q),\"C\"\n",
+ "#due to approximation taken by author in the textbook .... the values of Vbi_2, Vbi_3, Vbi_4 and \n",
+ "#the values of depletion width(Wp_4, Wp_2,Wp_3, Wp_4, Wn_4, Wn_2, Wn_3, Wn_4) differ from the above solution\n",
+ "# The answer of the textbook can be different due to different precision \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "concentration of electron in p type is ,pn = 2.25e+04 cm**-3\n",
+ "built in voltage is ,eVbi = 8.17e-01 V\n",
+ "depletion width at p side is ,Wp_4 = 3.26e-09 m\n",
+ "depletion width at n side is ,Wn_4 = 3.26e-07 m\n",
+ "\n",
+ "depletion width at p side is ,Wp_2 = 6.06e-09 m\n",
+ "depletion width at n side is ,Wn_2 = 6.06e-07 m\n",
+ "\n",
+ "depletion width at p side is ,Wp_3 = 8.71e-09 m\n",
+ "depletion width at n side is ,Wn_3 = 8.71e-07 m\n",
+ "\n",
+ "depletion width at p side is ,Wp_4 = 3.26e-09 m\n",
+ "depletion width at n side is ,Wn_4 = 3.26e-07 m\n",
+ "\n",
+ "depletion width at p side is ,Wp_5 = 3.19e-09 m\n",
+ "depletion width at n side is ,Wn_5 = 3.19e-07 m\n",
+ "F= -4.96e+06 v/m\n",
+ "F= -9.21e+06 v/m\n",
+ "F= -1.32e+07 v/m\n",
+ "F= -4.96e+06 v/m\n",
+ "F= -4.84e+06 v/m\n",
+ "Q = 1.64e-13 C\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.3:pg-173"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import cmath\n",
+ "Nd = 1.0*10**16\n",
+ "Na= 1.0*10**18\n",
+ "ni = 1.5*10**10\n",
+ "#NOTE: nn=Nd and pp=Na\n",
+ "R= 10*10**-6\n",
+ "A = math.pi*(R**2)\n",
+ "pn = ni**2/Nd\n",
+ "print\"concentration of electron in p type is ,pn = \",\"{:.2e}\".format(pn),\" cm**-3\"\n",
+ "kbT = 0.026\n",
+ "eVbi = (kbT*log(Na/pn))\n",
+ "print\"built in voltage is ,eVbi= \",\"{:.2e}\".format(eVbi),\" V\"\n",
+ "apsilen = 11.9*8.84*10**-12\n",
+ "e = 1.6*10**-19\n",
+ "#NOTE: for reverse bias Vr = 0 V,\n",
+ "Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))\n",
+ "print\"depletion width at p side is ,Wp_4 = \",\"{:.2e}\".format(Wp_4),\"m\"\n",
+ "Wn_4 = Wp_4*100\n",
+ "print\"depletion width at n side is ,Wn_4 = \",\"{:.2e}\".format(Wn_4),\"m\"\n",
+ "#for calculation purpose and for differentiating part (I), equating\n",
+ "Vbi_4 = eVbi\n",
+ "print(\"\")# for spacing \n",
+ "Vbi_2 = Vbi_4 + 2\n",
+ "#NOTE: for reverse bias Vr = 2 V,\n",
+ "Wp_2 = Wp_4*sqrt(Vbi_2/Vbi_4)\n",
+ "print\"depletion width at p side is ,Wp_2 = \",\"{:.2e}\".format(Wp_2),\"m\"\n",
+ "Wn_2 = Wp_2*100\n",
+ "print\"depletion width at n side is ,Wn_2 = \",\"{:.2e}\".format(Wn_2),\"m\"\n",
+ "print(\"\")# for spacing\n",
+ "Vbi_3 = Vbi_4 + 5\n",
+ "#NOTE: for reverse bias Vr = 5 V,\n",
+ "Wp_3 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_3/Vbi_4)\n",
+ "print\"depletion width at p side is ,Wp_3 = \",\"{:.2e}\".format(Wp_3),\"m\"\n",
+ "Wn_3 = Wp_3*100\n",
+ "print\"depletion width at n side is ,Wn_3 = \",\"{:.2e}\".format(Wn_3),\"m\"\n",
+ "print(\"\")# for spacing\n",
+ "Vbi_4 = Vbi_4 + 10\n",
+ "#NOTE: for reverse bias Vr = 10 V,\n",
+ "Wp_4 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_4/Vbi_4)\n",
+ "print\"depletion width at p side is ,Wp_4 = \",\"{:.2e}\".format(Wp_4),\"m\"\n",
+ "Wn_4 = Wp_4*100\n",
+ "print\"depletion width at n side is ,Wn_4 =\",\"{:.2e}\".format(Wn_4),\"m\"\n",
+ "print(\"\")# for spacing\n",
+ "Vbi_5 = Vbi_4 - 0.5\n",
+ "#NOTE: for forward bias Vf = 0.5 V,\n",
+ "Wp_5 = sqrt(((2*apsilen*eVbi)/(e))*((Nd*10**6)/((Na*10**6)*((Nd*10**6)+(Na*10**6)))))*sqrt(Vbi_5/Vbi_4)\n",
+ "print\"depletion width at p side is ,Wp_5 =\",\"{:.2e}\".format(Wp_5),\"m\"\n",
+ "Wn_5 = Wp_5*100\n",
+ "print\"depletion width at n side is ,Wn_5 =\",\"{:.2e}\".format(Wn_5),\"m\"\n",
+ "#CALCULATION FOR PEAK FIELD :\n",
+ "Fm = - e*Nd*(10**6)*Wn_4/apsilen\n",
+ "print\"peak Field For Vr = 0V, Fm = \",\"{:.2e}\".format(Fm),\"V/m\"\n",
+ "Fm = - e*Nd*(10**6)*Wn_2/apsilen\n",
+ "print\"peak Field for Vr = 2V, Fm =\",\"{:.2e}\".format(Fm),\"V/m\"\n",
+ "Fm = - e*Nd*(10**6)*Wn_3/apsilen\n",
+ "print\"peak Field For Vr = 5V, Fm = \",\"{:.2e}\".format(Fm),\"V/m\"\n",
+ "Fm = - e*Nd*(10**6)*Wn_4/apsilen\n",
+ "print\"peak Field For Vr = 10V, Fm = \",\"{:.2e}\".format(Fm),\"V/m\"\n",
+ "print\"By the appendix B given in the book, the velocity of electron: v = 1*10**7 cm/s\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "concentration of electron in p type is ,pn = 2.25e+04 cm**-3\n",
+ "built in voltage is ,eVbi= 8.17e-01 V\n",
+ "depletion width at p side is ,Wp_4 = 3.26e-09 m\n",
+ "depletion width at n side is ,Wn_4 = 3.26e-07 m\n",
+ "\n",
+ "depletion width at p side is ,Wp_2 = 6.06e-09 m\n",
+ "depletion width at n side is ,Wn_2 = 6.06e-07 m\n",
+ "\n",
+ "depletion width at p side is ,Wp_3 = 8.70e-09 m\n",
+ "depletion width at n side is ,Wn_3 = 8.70e-07 m\n",
+ "\n",
+ "depletion width at p side is ,Wp_4 = 3.26e-09 m\n",
+ "depletion width at n side is ,Wn_4 = 3.26e-07 m\n",
+ "\n",
+ "depletion width at p side is ,Wp_5 = 3.19e-09 m\n",
+ "depletion width at n side is ,Wn_5 = 3.19e-07 m\n",
+ "peak Field For Vr = 0V, Fm = -4.96e+06 V/m\n",
+ "peak Field for Vr = 2V, Fm = -9.21e+06 V/m\n",
+ "peak Field For Vr = 5V, Fm = -1.32e+07 V/m\n",
+ "peak Field For Vr = 10V, Fm = -4.96e+06 V/m\n",
+ "By the appendix B given in the book, the velocity of electron: v = 1*10**7 cm/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.4:pg-178"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "A= 10**-7\n",
+ "Na=10**18*10**6\n",
+ "Nd=10**16*10**6\n",
+ "Dp1 = 7.8*10**-4\n",
+ "Dn2 = 7.3*10**-4\n",
+ "Tn = 10**-6\n",
+ "Tp = 10**-6\n",
+ "e = 1.6*10**-19\n",
+ "kbT = 0.026\n",
+ "ni = 1.5*10**16\n",
+ "#NOTE: nn=Nd and pp=Na\n",
+ "Lp = sqrt(Dp1*Tp)\n",
+ "print\"The hole diffusion length is ,Lp =\",\"{:.2e}\".format(Lp),\"m\"\n",
+ "Ln = sqrt(Dn2*Tn)\n",
+ "print\"The electron diffusion length is ,Ln =\",\"{:.2e}\".format(Ln),\"m\"\n",
+ "# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)\n",
+ "# assume that the doants are fully ionised \n",
+ "Io = A*e*((((Dn2)/(Ln))*(ni**2/Na))+(((Dp1)/(Lp))*(ni**2/Nd)))\n",
+ "print\"The prefactor current is ,Io =\",\"{:.1e}\".format(Io),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole diffusion length is ,Lp = 2.79e-05 m\n",
+ "The electron diffusion length is ,Ln = 2.70e-05 m\n",
+ "The prefactor current is ,Io = 1.0e-14 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.5:pg-181"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Na=10**17\n",
+ "Nd=10**17\n",
+ "Dp1 = 12.5\n",
+ "Dn1 = 35\n",
+ "Dp2 = 10\n",
+ "Dn2 = 220\n",
+ "Tn = 10**-8\n",
+ "Tp = 10**-8\n",
+ "e = 1.6*10**-19\n",
+ "kbT = 0.026\n",
+ "pn1 = 2.25*10**3\n",
+ "np1 = 2.25*10**3\n",
+ "pn2 = 3.38*10**-5\n",
+ "np2 = 3.38*10**-5\n",
+ "#Note : since value of holes and electrons in n- type and p type are not given for silicon and germanium thus we have assume it as above\n",
+ "#NOTE: nn=Nd and pp=Na\n",
+ "Lp1 = sqrt(Dp1*Tp)\n",
+ "Ln1 = sqrt(Dn1*Tn)\n",
+ "Lp2 = sqrt(Dp2*Tp)\n",
+ "Ln2 = sqrt(Dn2*Tn)\n",
+ "# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)\n",
+ "# assume that the doants are fully ionised \n",
+ "Jo1 = e*((((Dn1)/(Ln1))*np1)+(((Dp1)/(Lp1))*pn1))\n",
+ "print\"The prefactor current density for silicon is ,Jo1 =\",\"{:.2e}\".format(Jo1),\"A/cm**2\"\n",
+ "Jo2 = e*((((Dn2)/(Ln2))*np2)+(((Dp2)/(Lp2))*pn2))\n",
+ "print\"The prefactor current density for GaAs is ,Jo2 =\",\"{:.2e}\".format(Jo2),\"A/cm**2\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The prefactor current density for silicon is ,Jo1 = 3.40e-11 A/cm**2\n",
+ "The prefactor current density for GaAs is ,Jo2 = 9.73e-19 A/cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex5.6:pg-182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Na=5*10**16\n",
+ "Nd=5*10**17\n",
+ "Dp = 15\n",
+ "Dn = 30\n",
+ "Tn = 10**-8\n",
+ "Tp = 10**-7\n",
+ "e = 1.6*10**-19\n",
+ "kbT = 0.026\n",
+ "ni = 1.84*10**6\n",
+ "np=ni**2/Na\n",
+ "print\"The electron conc in p type is ,np=\",\"{:.2e}\".format(np),\"cm**-3\"\n",
+ "pn=ni**2/Nd\n",
+ "print\"The holes conc in n type is ,pn=\",\"{:.2e}\".format(pn),\"cm**-3\"\n",
+ "Lp = sqrt(Dp*Tp)\n",
+ "print\"The hole diffusion length is ,Lp =\",\"{:.2e}\".format(Lp),\"cm\"\n",
+ "Ln = sqrt(Dn*Tn)\n",
+ "print\"The electron diffusion length is ,Ln =\",\"{:.2e}\".format(Ln),\"cm\"\n",
+ "Gamma_inj = ((e*Dn*np)/(Ln))/(((e*Dn*np)/(Ln))+((e*Dp*pn)/(Lp)))\n",
+ "print\"The efficiency of diode is ,Gamma_inj =\",round(Gamma_inj,2)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron conc in p type is ,np= 6.77e-05 cm**-3\n",
+ "The holes conc in n type is ,pn= 6.77e-06 cm**-3\n",
+ "The hole diffusion length is ,Lp = 1.22e-03 cm\n",
+ "The electron diffusion length is ,Ln = 5.48e-04 cm\n",
+ "The efficiency of diode is ,Gamma_inj = 0.98\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.7:pg-182"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "A= 0.1*10**-2\n",
+ "Vf= 1.0\n",
+ "E= 1.43\n",
+ "Na=5.0*10**16\n",
+ "Nd=5.0*10**17\n",
+ "Dp = 15.0\n",
+ "Dn = 30.0\n",
+ "Tn =1*10**-8\n",
+ "Tp = 1.0*10**-7\n",
+ "e = 1.6*10**-19\n",
+ "kbT = 0.026\n",
+ "kni = 1.84*10**6\n",
+ "np=ni**2/Na\n",
+ "pn=ni**2/Nd\n",
+ "Ln = sqrt(Dn*Tn)\n",
+ "In = ((e*A*Dn*np)/Ln)*(exp(Vf/kbT)-1)\n",
+ "print\"The electron current is ,In =\",\"{:.1e}\".format(In),\"A\"\n",
+ "In_by_e = In/e\n",
+ "print\"The electron generation rate is ,In_by_e = \",\"{:.1e}\".format(In_by_e),\"s**-1\"\n",
+ "power = In*E\n",
+ "print\"The optical power of photon is ,power =\",\"{:.1e}\".format(power),\"W\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron current is ,In = 3.0e-05 A\n",
+ "The electron generation rate is ,In_by_e = 1.9e+14 s**-1\n",
+ "The optical power of photon is ,power = 4.3e-05 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex5.8:pg-183"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "apsilen = 11.9*8.85*10**-14\n",
+ "GL= 1.0*10**22\n",
+ "A= 1.0*10**-4\n",
+ "Vr= 15.0\n",
+ "Na=2.0*10**16\n",
+ "Nd=1.0*10**16\n",
+ "Dp = 12.0\n",
+ "Dn = 20.0\n",
+ "Tn = 1.0*10**-8\n",
+ "Tp = 1.0*10**-8\n",
+ "e = 1.6*10**-19\n",
+ "kbT = 0.026\n",
+ "ni = 1.5*10**10\n",
+ "Ln = sqrt(Dn*Tn)\n",
+ "print\"The electron diffusion length is ,Ln =\",\"{:.1e}\".format(Ln),\"cm\"\n",
+ "Lp = sqrt(Dp*Tp)\n",
+ "print\"The hole diffusion length is ,Lp = \",\"{:.2e}\".format(Lp),\"cm\"\n",
+ "Vbi = kbT*log((Na*Nd)/ni**2)\n",
+ "print\"The built in voltage is ,Vbi =\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "W = sqrt(((2*apsilen)/e)*((Na+Nd)/(Na*Nd))*(Vbi+Vr))\n",
+ "print\"The depletion width is ,W =\",\"{:.2e}\".format(W),\"cm\"\n",
+ "IL = e*A*GL*(W+Ln+Lp)\n",
+ "print\"The photo current is IL =\",\"{:.2e}\".format(IL),\"A\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron diffusion length is ,Ln = 4.5e-04 cm\n",
+ "The hole diffusion length is ,Lp = 3.46e-04 cm\n",
+ "The built in voltage is ,Vbi = 7.15e-01 V\n",
+ "The depletion width is ,W = 1.76e-04 cm\n",
+ "The photo current is IL = 1.55e-04 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.9:pg-185"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "e = 1.6*10**-19\n",
+ "A=1.0*10**-7\n",
+ "Na=10**18*10**6\n",
+ "Nd=10**16*10**6\n",
+ "Dp = 7.8*10**-4\n",
+ "Dn = 7.3*10**-4\n",
+ "ni = 1.5*10**16\n",
+ "Wln = 5.0*10**-6\n",
+ "Wlp = Wln\n",
+ "#NOTE: nn=Nd and pp=Na\n",
+ "pn = ni**2/Nd\n",
+ "np = ni**2/Na\n",
+ "# from example 5.4 and 5.2 we get the value of diffusion length and zero bias depletion widths\n",
+ "Lp = 27.9*10**-6\n",
+ "print\"The electron diffusion length is ,Lp= \",\"{:.2e}\".format(Lp),\"m\"\n",
+ "Ln = 27*10**-6\n",
+ "print\"The electron diffusion length is ,Ln= \",\"{:.2e}\".format(Ln),\"m\"\n",
+ "Wp = 3.2*10**-10\n",
+ "print\"The zero bias depletion widths is ,Wp= \",\"{:.2e}\".format(Wp),\"m\"\n",
+ "Wn = 0.0000003\n",
+ "print\"The zero bias depletion widths is ,Wn \",\"{:.2e}\".format(Wn),\"m\"\n",
+ "# for short diode the prefactor current is given as\n",
+ "Io = e*A*(((Dp*pn)/(Wln-Wn))+((Dn*np)/abs(Wlp-Wp)))\n",
+ "print\"The prefactor current is ,Io = \",\"{:.2e}\".format(Io),\"A\"\n",
+ "# The prefactor current of short diode is approximately increase by a factor of 5.6 from that of long diode\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron diffusion length is ,Lp= 2.79e-05 m\n",
+ "The electron diffusion length is ,Ln= 2.70e-05 m\n",
+ "The zero bias depletion widths is ,Wp= 3.20e-10 m\n",
+ "The zero bias depletion widths is ,Wn 3.00e-07 m\n",
+ "The prefactor current is ,Io = 6.03e-14 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.10:pg-188"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "e = 1.6*10**-19\n",
+ "A= 10**-7\n",
+ "ni = 1.5*10**16\n",
+ "T = 1.0*10**-6\n",
+ "# from example 5.2 we get the value of zero bias depletion widths\n",
+ "W = 0.32*10**-6\n",
+ "Io_GR = (e*A*W*ni)/(2*T)\n",
+ "print\"The prefactor of the is ,generation recombination currentIo_GR = \",\"{:.2e}\".format(Io_GR),\"A\"\n",
+ "#let V = .2 V \n",
+ "V = 0.2\n",
+ "I_GR = Io_GR*(exp(V/(2*kbT))-1)\n",
+ "print\"The diode current is ,I_GR =\",\"{:.2e}\".format(I_GR),\"A\"\n",
+ "#let V = 0.6 V \n",
+ "V = 0.6\n",
+ "I_GR = Io_GR*(exp(V/(2*kbT))-1)\n",
+ "print\"The diode current is ,I_GR = \",\"{:.2e}\".format(I_GR),\"A\"\n",
+ "# The generation-recombination prefactor is much larger than prefactor due to diffusion term\n",
+ "#In forward bias the diffusion current is initially much smaller than the generation recombination term but at high forward bias diffusion current will start to dominate\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The prefactor of the is ,generation recombination currentIo_GR = 3.84e-11 A\n",
+ "The diode current is ,I_GR = 1.76e-09 A\n",
+ "The diode current is ,I_GR = 3.94e-06 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex5.11:pg-189"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "A= 1.0*10**-8\n",
+ "Na=1.0*10**23\n",
+ "Nd=1.0*10**23\n",
+ "Dp = 10.0*10**-4\n",
+ "Dn = 30.0*10**-4\n",
+ "Tn = 1.0*10**-7\n",
+ "Tp = 1.0*10**-7\n",
+ "tau = 1.0*10**-8\n",
+ "e = 1.6*10**-19\n",
+ "kbT = 0.026\n",
+ "ni = 1.5*10**16\n",
+ "apsilen = 11.9*8.85*10**-12\n",
+ "#NOTE: nn=Nd and pp=Na\n",
+ "Lp = sqrt(Dp*Tp)\n",
+ "print\"The hole diffusion length is ,Lp = \",\"{:.2e}\".format(Lp),\"m\"\n",
+ "Ln = sqrt(Dn*Tn)\n",
+ "print\"The electron diffusion length is ,Ln = \",\"{:.2e}\".format(Ln),\"m\"\n",
+ "# NOTE: pn= (ni**2/nn) and np=(ni**2/pp)\n",
+ "np=ni**2/Na\n",
+ "print\"The electron conc in p type is ,np=\",\"{:.2e}\".format(np),\"m**-3\"\n",
+ "pn=ni**2/Nd\n",
+ "print\"The holes conc in n type is ,pn=\",\"{:.2e}\".format(pn),\"m**-3\"\n",
+ "Vbi = kbT*log((Na*Nd)/ni**2)\n",
+ "print\"The built in voltage is ,Vbi =\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "Io = e*((((Dn)*np)/(Ln))+((Dp*pn)/(Lp)))\n",
+ "print\"The prefactor in the ideal diode is ,Io =\",\"{:.2e}\".format(Io),\"A\"\n",
+ "#let Vf = 0.5 V \n",
+ "Vf = 0.5\n",
+ "W = sqrt((2*apsilen/e)*((Na+Nd)/Nd/Na)*(Vbi-Vf))\n",
+ "print\"The depletion width is ,W =\",\"{:.2e}\".format(W),\"m\"\n",
+ "Io_GR = e*A*W*ni/(2*tau)\n",
+ "print\"prefactor for recombination generation current, Io_GR = \",\"{:.2e}\".format(Io_GR),\"A\"\n",
+ "I = (Io*exp(Vf/kbT))+(Io_GR*exp(Vf/(2*kbT)))\n",
+ "I_V1 = I\n",
+ "print\"Current, I =\",\"{:.2e}\".format(I_V1),\"A\"\n",
+ "#let V = 0.6 V \n",
+ "Vf = 0.6\n",
+ "W = sqrt((2*apsilen/e)*((Na+Nd)/Nd/Na)*(Vbi-Vf))\n",
+ "print\"The depletion width is ,W =\",\"{:.2e}\".format(W),\"m\"\n",
+ "Io_GR = e*A*W*ni/2/tau\n",
+ "print\"prefactor for recombination generation current, Io_GR = \",\"{:.1e}\".format(Io_GR),\"A\"\n",
+ "I = (Io*exp(Vf/kbT))+(Io_GR*exp(Vf/(2*kbT)))\n",
+ "I_V2 = I\n",
+ "print\"Current, I =\",\"{:.2e}\".format(I_V2),\"A\"\n",
+ "V1 = 0.5\n",
+ "V2 = 0.6\n",
+ "n = e*(V2-V1)/kbT/log(I_V2/I_V1)\n",
+ "print\"Ideallity factor ,n =\",\"{:.2e}\".format(n)\n",
+ "#note: in the text book the value of \n",
+ "#-+\"prefactor of ideal diode equation, Io\" \n",
+ "#calculated by author is wrong thus it efect the overall calculation of the solution ,so the answer in the are wrong of some of the values\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The hole diffusion length is ,Lp = 1.00e-05 m\n",
+ "The electron diffusion length is ,Ln = 1.73e-05 m\n",
+ "The electron conc in p type is ,np= 2.25e+09 m**-3\n",
+ "The holes conc in n type is ,pn= 2.25e+09 m**-3\n",
+ "The built in voltage is ,Vbi = 8.17e-01 V\n",
+ "The prefactor in the ideal diode is ,Io = 9.84e-08 A\n",
+ "The depletion width is ,W = 9.14e-08 m\n",
+ "prefactor for recombination generation current, Io_GR = 1.10e-10 A\n",
+ "Current, I = 2.21e+01 A\n",
+ "The depletion width is ,W = 7.56e-08 m\n",
+ "prefactor for recombination generation current, Io_GR = 9.1e-11 A\n",
+ "Current, I = 1.04e+03 A\n",
+ "Ideallity factor ,n = 1.60e-19\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.12:pg-195"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "apsilen = 11.9*8.85*10**-14\n",
+ "Na=1.0*10**19\n",
+ "Nd=1.0*10**16\n",
+ "e = 1.6*10**-19\n",
+ "Fcrit1 = 4.0*10**5\n",
+ "Fcrit2 = 1.0*10**7\n",
+ "VBD_Si = (apsilen*Fcrit1**2)/(2*e*Nd)\n",
+ "print\"The breakdown field for silicon is ,VBD_Si =\",\"{:.2e}\".format(VBD_Si),\" V\"\n",
+ "VBD_C = (apsilen*Fcrit2**2)/(2*e*Nd)\n",
+ "print\"The breakdown field for diomond is ,VBD_C =\",\"{:.2e}\".format(VBD_C),\" V\"\n",
+ "# Note : In the textbook answer of breakdown voltage of silicon is wrong due to which breakdown voltage of diomand also differ\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The breakdown field for silicon is ,VBD_Si = 5.27e+01 V\n",
+ "The breakdown field for diomond is ,VBD_C = 3.29e+04 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.13:pg-199"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import numpy as np\n",
+ "print\"Let the intercept of the 1/c**2 Vs V plot is represented by Icv, which is the built in voltage\"\n",
+ "Icv = 0.68\n",
+ "Vbi = Icv\n",
+ "print\"Let the slope of the intercept of the 1/c**2 Vs V plot is represented by dIcv\"\n",
+ "dIcv=2.1*10**23\n",
+ "C = 7.0*10**-13\n",
+ "#NOTE: The above mentioned values are taken from the figure given in the question in textbook\n",
+ "apsilen = 11.9*8.85*10**-12\n",
+ "e = 1.6*10**-19\n",
+ "kbT = 0.026\n",
+ "A = 10**-7\n",
+ "ni = sqrt(2.25*10**20)\n",
+ "Neff = 2/(A**2*e*apsilen*dIcv)\n",
+ "print\"The thickness of n region is ,Neff =\",\"{:.2e}\".format(Neff),\" m**-3\"\n",
+ "Neff = Neff/10**6\n",
+ "print\"The thickness of n region is ,Neff =\",\"{:.2e}\".format(Neff),\" cm**-3\"\n",
+ "NaNd = exp(Vbi/kbT)*ni**2\n",
+ "print\"NaNd =\",\"{:.2e}\".format(NaNd),\" cm**-6\"\n",
+ "# solving for Na and Nd by creating a quadratic equation using the equations mentioned in the book\n",
+ "p1 = [Neff*NaNd, -NaNd, Neff]\n",
+ "#Neff*NaNd - NaNd*X + Neff*X**2 \n",
+ "p1=p1[0]\n",
+ "#p2=p1[1]\n",
+ "print(p1)\n",
+ "R= roots(p1)\n",
+ "#s=roots(p2)\n",
+ "Na= R\n",
+ "#Nd= s\n",
+ "#format ('e',10)\n",
+ "#print\"Na = \"\"{:.2e}\".format(Na),\"cm**-3\"\n",
+ "#print\"Nd = \"\"{:.2e}\".format(Nd),\"cm**-3\"\n",
+ "W = (apsilen*A)/C\n",
+ "print\"The thickness of n region is ,W =\",\"{:.1e}\".format(W),\" m\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Let the intercept of the 1/c**2 Vs V plot is represented by Icv, which is the built in voltage\n",
+ "Let the slope of the intercept of the 1/c**2 Vs V plot is represented by dIcv\n",
+ "The thickness of n region is ,Neff = 5.65e+19 m**-3\n",
+ "The thickness of n region is ,Neff = 5.65e+25 cm**-3\n",
+ "NaNd = 5.14e+31 cm**-6\n",
+ "The thickness of n region is ,W = 1.5e-05 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.14:pg-201"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import cmath\n",
+ "e = 1.6*10**-19\n",
+ "I= 1*10**-3\n",
+ "kbT = 0.026\n",
+ "Tp = 10**-6\n",
+ "Gs = (I)/(kbT)\n",
+ "print\"The diode conductance is Gs = \",\"{:.2e}\".format(Gs),\"A/V\"\n",
+ "Cdiff = (I*Tp)/(2*kbT)\n",
+ "print\"The diffusion capacitance is Cdiff = \",\"{:.2e}\".format(Cdiff),\"F\"\n",
+ "# The diffusion capacitance is much larger than junction capacitance hence neglecting junction capacitance\n",
+ "Y = Gs+(1j*2*math.pi*10**6*Cdiff)\n",
+ "print\"The admittance of the diode is Y =\",Y,\"A/V\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The diode conductance is Gs = 3.85e-02 A/V\n",
+ "The diffusion capacitance is Cdiff = 1.92e-08 F\n",
+ "The admittance of the diode is Y = (0.0384615384615+0.120830486677j) A/V\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex5.15:pg-207"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Vr= 10.0\n",
+ "R= 10.0*10**3\n",
+ "#The junction capacitance is 20pF at zero bias and 10 pF at full reverse bias so\n",
+ "Cavg= ((20+10)/2)\n",
+ "Tp = 10**-7\n",
+ "Ir = (Vr)/(R)\n",
+ "Tsd = Tp*log(2)\n",
+ "print\"The storage delay time is Tsd = \",\"{:.2e}\".format(Tsd),\" s\"\n",
+ "Tt = 2.3*R*Cavg*10**-12\n",
+ "print\"The time Tt = \",\"{:.2e}\".format(Tt),\" s\"\n",
+ "T = Tsd+Tt\n",
+ "print\"The total diode recovery time is T = \",\"{:.2e}\".format(T),\" s\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The storage delay time is Tsd = 6.93e-08 s\n",
+ "The time Tt = 3.45e-07 s\n",
+ "The total diode recovery time is T = 4.14e-07 s\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter6_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter6_1.ipynb new file mode 100644 index 00000000..535243d9 --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter6_1.ipynb @@ -0,0 +1,369 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:04a6842d7b11c3362cf74422ffc19db29b3f2ea1cd75fe882bcb992815e56505"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Chapter6:SEMICONDUCTOR JUNCTIONS WITH METALS AND INSULATORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.1:pg-226"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "e = 1.6*10**-19\n",
+ "n = 10**22\n",
+ "rho = 2.7*10**(-6)\n",
+ "print\"using following terms J = Current density ; s(sigma) = 1/rho = conductivity ; F = Electric field \"\n",
+ "print\"Using relations J = s*F = n*e*v = n*e*u*F ; we get\"\n",
+ "mu_ = 1.0/(n*e*rho)\n",
+ "print\"The mobility of electrons in aluminium is ,mu_ =\",\"{:.2e}\".format(mu_),\"cm**2(Vs)**-1\"\n",
+ "#The answer given in the book is 240.4 cm**2/Vs which is wrong\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "using following terms J = Current density ; s(sigma) = 1/rho = conductivity ; F = Electric field \n",
+ "Using relations J = s*F = n*e*v = n*e*u*F ; we get\n",
+ "The mobility of electrons in aluminium is ,mu_ = 2.31e+02 cm**2(Vs)**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex6.2:pg-232"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "e = 1.6*10**-19\n",
+ "apsilen = 11.9*8.85*10**-12\n",
+ "A= 7.85*10**-9\n",
+ "S= 3*10**24\n",
+ "Nd = (2/(S*e*apsilen*(A**2)))\n",
+ "print\"The doping density in silicon is ,Nd =\",\"{:.2e}\".format(Nd),\"m**-3\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The doping density in silicon is ,Nd = 6.42e+20 m**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.3:pg-236"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nd = 10**16\n",
+ "Nc = 2.8*10**19\n",
+ "kBT=0.026\n",
+ "Vf=0.3\n",
+ "e = 1.6*10**-19\n",
+ "A= 10**-3\n",
+ "print\" for W-n type Si schottky barrier \"\n",
+ "T = 300\n",
+ "phi_b = 0.67\n",
+ "print\"schottky barrier heights(in volts) =\",\"{:.2e}\".format(phi_b),\"eV\"\n",
+ "R = 110\n",
+ "Is = A*R*(T**2)*(exp(-(phi_b)/(kBT)))\n",
+ "print\"The reverse saturation current is ,Is =\",\"{:.2e}\".format(Is),\"A\"\n",
+ "print\"using relation I= Is*(exp((e*V)/(nkBT))-1) and neglecting 1\"\n",
+ "I = Is*(exp((Vf)/(kBT)))\n",
+ "print\"I=\",\"{:.1e}\".format(I),\"A\"\n",
+ "print\" for Si p+ -n junction diode \"\n",
+ "Na = 10**19\n",
+ "Db = 10.5\n",
+ "Tb = 10**-6\n",
+ "Lb = sqrt(Db*Tb)\n",
+ "print\"The electron carrier diffusion length is,Lb =\",\"{:.2e}\".format(Lb),\"cm\"\n",
+ "pn = 2.2*10**4\n",
+ "Io = A*e*pn*(Db/Lb)\n",
+ "print\"The saturation current current is Io =\",\"{:.1e}\".format(Io),\"A\"\n",
+ "I1 = Io*(exp((Vf)/(kBT)))\n",
+ "print\"The diode current for HBT is ,I =\",\"{:.1e}\".format(I1),\"A\"\n",
+ "print\"Since diode current for HBT is almost 6 orders of magnitude smaller than the value in the Schottky diode \"\n",
+ "print\"hence for the p-n diode to have the same current that the schottky dode has at .3 V , the voltage required is .71V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " for W-n type Si schottky barrier \n",
+ "schottky barrier heights(in volts) = 6.70e-01 eV\n",
+ "The reverse saturation current is ,Is = 6.37e-08 A\n",
+ "using relation I= Is*(exp((e*V)/(nkBT))-1) and neglecting 1\n",
+ "I= 6.5e-03 A\n",
+ " for Si p+ -n junction diode \n",
+ "The electron carrier diffusion length is,Lb = 3.24e-03 cm\n",
+ "The saturation current current is Io = 1.1e-14 A\n",
+ "The diode current for HBT is ,I = 1.2e-09 A\n",
+ "Since diode current for HBT is almost 6 orders of magnitude smaller than the value in the Schottky diode \n",
+ "hence for the p-n diode to have the same current that the schottky dode has at .3 V , the voltage required is .71V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex6.4:pg-237"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kBT=0.026\n",
+ "mo = 9.1*10**-31\n",
+ "m=0.08*mo\n",
+ "T = 300\n",
+ "phi_b1 = 0.7\n",
+ "phi_b2 = 0.6\n",
+ "R = 120*(m/mo)\n",
+ "print\"The effective richardson constant is ,R* =\",round(R,2),\" A cm**-2 k**-2\"\n",
+ "Js1 = R*(T**2)*(exp(-(phi_b1)/(kBT)))\n",
+ "print\"The saturation current density is ,Js(phi_b=0.7) =\",\"{:.1e}\".format(Js1),\"A/cm**2\"\n",
+ "Js2 = R*(T**2)*(exp(-(phi_b2)/(kBT)))\n",
+ "print\"The saturation current density is ,Js(phi_b=0.6) =\",\"{:.2e}\".format(Js2),\"A/cm**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The effective richardson constant is ,R* = 9.6 A cm**-2 k**-2\n",
+ "The saturation current density is ,Js(phi_b=0.7) = 1.8e-06 A/cm**2\n",
+ "The saturation current density is ,Js(phi_b=0.6) = 8.21e-05 A/cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.5:pg-239"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "apsilen = 11.9*8.85*10**-12\n",
+ "Nd = 10**16\n",
+ "Nc = 2.8*10**19\n",
+ "kBT = 0.026\n",
+ "I=10*10**-3\n",
+ "e = 1.6*10**-19\n",
+ "A= 10**-3\n",
+ "print\" for W-n type Si schottky barrier \"\n",
+ "T = 300\n",
+ "phi_b = 0.67\n",
+ "R = 110\n",
+ "Is = A*R*(T**2)*(exp(-(phi_b)/(kBT)))\n",
+ "V = kBT*(log(I/Is))\n",
+ "E = kBT*log(Nc/Nd)\n",
+ "print\"The fermi level positionin the neutral semiconductor(Efs) with respect to the conduction band is,Ec-Efs= E = \",\"{:.2e}\".format(E),\"eV\"\n",
+ "Vbi= phi_b-(E)\n",
+ "print\"The built in voltage is ,Vbi=\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "Cd = A*sqrt((e*Nd*apsilen)/(2*(Vbi-V)))\n",
+ "print\"The diode capacitance is ,Cd =\",\"{:.2e}\".format(Cd),\"F\"\n",
+ "R = kBT/I\n",
+ "print\"The resistance is ,R =\",\"{:.2e}\".format(R),\"ohm\"\n",
+ "RC = R*Cd\n",
+ "print\"The RC time constant is ,RC(schottky) =\",\"{:.2e}\".format(RC),\"s\"\n",
+ "print\" for Si p+ -n junction diode \"\n",
+ "Tb = 10**-6\n",
+ "print\"In the p-n diode the junction capacitance and the small signal resistance will be same as those in the schottky diode\"\n",
+ "Cdiff = ((I*Tb)/(kBT))\n",
+ "print\"The diffusion capacitance is ,Cdiff = (I*Tb)/(kBT) = \",\"{:.2e}\".format(Cdiff),\"F\"\n",
+ "RC1 = R*Cdiff\n",
+ "print\"The RC time constant is ,RC(p-n) = \",\"{:.2e}\".format(RC1),\"s\"\n",
+ "print\"From the above RC time constant value it can be concluded that p-n diode is almost 1000 times slower\"\n",
+ "# Note: due to approximation, the value of diode capicitance and diffusion capacitance are differ from that of the textbook"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " for W-n type Si schottky barrier \n",
+ "The fermi level positionin the neutral semiconductor(Efs) with respect to the conduction band is,Ec-Efs= E = 2.06e-01 eV\n",
+ "The built in voltage is ,Vbi="
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " 4.64e-01 V\n",
+ "The diode capacitance is ,Cd = 7.43e-10 F\n",
+ "The resistance is ,R = 2.60e+00 ohm\n",
+ "The RC time constant is ,RC(schottky) = 1.93e-09 s\n",
+ " for Si p+ -n junction diode \n",
+ "In the p-n diode the junction capacitance and the small signal resistance will be same as those in the schottky diode\n",
+ "The diffusion capacitance is ,Cdiff = (I*Tb)/(kBT) = 3.85e-07 F\n",
+ "The RC time constant is ,RC(p-n) = 1.00e-06 s\n",
+ "From the above RC time constant value it can be concluded that p-n diode is almost 1000 times slower\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex6.6:pg-242"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "apsilen = 11.9*8.85*10**-14\n",
+ "phi_b = 0.66\n",
+ "mo = 9.1*10**-31\n",
+ "m=0.34*mo\n",
+ "e = 1.6*10**-19\n",
+ "h = 1.05*10**-34\n",
+ "n1 = 10**18\n",
+ "n2 = 10**20\n",
+ "print\"Assume that the built in potential Vbi is same as barrier potential becouse of highly doped semiconductor\"\n",
+ "W1 = (sqrt((2*apsilen*phi_b)/(e*n1)))/10**-8\n",
+ "print\"The depletion width is ,W(n=10**18) =\",\"{:.2e}\".format(W1),\" Angstrom\"\n",
+ "W2 = (sqrt((2*apsilen*phi_b)/(e*n2)))/10**-8\n",
+ "print\"The depletion width is ,W(n=10**20) =\",\"{:.2e}\".format(W2),\" Angstrom\"\n",
+ "F1 = phi_b/(W1*10**-8)\n",
+ "print\"The average field in depletion region for(n=10**18), F1 =\",\"{:.2e}\".format(F1),\"V/cm\"\n",
+ "F2 = phi_b/(W2*10**-8)\n",
+ "print\"The average field in depletion region for(n=10**18), F2 =\",\"{:.2e}\".format(F2),\"V/cm\"\n",
+ "F1 = F1/10**-2\n",
+ "F2 = F2/10**-2\n",
+ "T = exp(-(4.0*(2.0*m)**.5*(e*phi_b)**(3.0/2.0))/(3.0*e*F1*h))\n",
+ "print\"The tunneling current for(n=10**18),T =\",\"{:.2e}\".format(T),\"V/cm\"\n",
+ "T1 = exp(-(4.0*(2.0*m)**.5*(e*phi_b)**(3.0/2.0))/(3.0*e*F2*h))\n",
+ "print\"The tunneling current for(n=10**20), T1 =\",\"{:.2e}\".format(T1),\"V/cm\"\n",
+ "# in the textbook author has used approximate value for depletion width and hence it affect the value of all other answer\n",
+ "# NOTE: In the textbook author has used approximate answer for tunneling current"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Assume that the built in potential Vbi is same as barrier potential becouse of highly doped semiconductor\n",
+ "The depletion width is ,W(n=10**18) = 2.95e+02 Angstrom\n",
+ "The depletion width is ,W(n=10**20) = 2.95e+01 Angstrom\n",
+ "The average field in depletion region for(n=10**18), F1 = 2.24e+05 V/cm\n",
+ "The average field in depletion region for(n=10**18), F2 = 2.24e+06 V/cm\n",
+ "The tunneling current for(n=10**18),T = 2.79e-42 V/cm\n",
+ "The tunneling current for(n=10**20), T1 = 6.99e-05 V/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex6.7:pg-248"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "n = 10**18\n",
+ "W = 25*10**-4\n",
+ "R = 100*10**3\n",
+ "e = 1.6*10**-19\n",
+ "D= 5000*10**-8\n",
+ "mu_=100.0\n",
+ "Ro = 1.0/(n*e*mu_*D)\n",
+ "print\"The sheet resistance of the film is ,Ro =\",\"{:.2e}\".format(Ro),\" ohm/square\"\n",
+ "L = (R*W)/Ro\n",
+ "print\"The length of the desired resistor is ,L =\",\"{:.2e}\".format(L),\" cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The sheet resistance of the film is ,Ro = 1.25e+03 ohm/square\n",
+ "The length of the desired resistor is ,L = 2.00e-01 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter7_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter7_1.ipynb new file mode 100644 index 00000000..c5f7e248 --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter7_1.ipynb @@ -0,0 +1,710 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:583e9a3370db0a45b2805011aaba429aa5ff178400e4cdec220559b9edbb5a50"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter7:BIPOLAR JUNCTION TRANSISTORS"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex7.2:pg-275"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "alpha_F=.99\n",
+ "alpha_R=.25\n",
+ "kbT = 0.026\n",
+ "# for part a\n",
+ "Ic1 = 1.0\n",
+ "Ib1 = .02\n",
+ "\n",
+ "VCE= kbT*log((((Ic1*(1-alpha_R))+Ib1)*alpha_F)/(((alpha_F*Ib1)-((Ic1*(1-alpha_F))))*alpha_R))\n",
+ "print\"The saturation voltage is ,VCE=\",\"{:.2e}\".format(VCE),\" V\"\n",
+ " #for part b\n",
+ "Ic2 = 5.0\n",
+ "Ib2 = .075\n",
+ "VCE1= kbT*log((((Ic2*(1-alpha_R))+Ib2)*alpha_F)/(((alpha_F*Ib2)-((Ic2*(1-alpha_F))))*alpha_R))\n",
+ "print\"The saturation voltage is ,VCE1=\",\"{:.2e}\".format(VCE1),\" V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The saturation voltage is ,VCE= 1.49e-01 V\n",
+ "The saturation voltage is ,VCE1= 1.67e-01 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex7.3:pg-278"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "nbo = 2.25*10**3\n",
+ "peo = 112.5\n",
+ "pco = 2.25*10**4\n",
+ "# using law of mass action for a homogeneous semiconductor, we have relation peo*neo=nbo*pbo=ni**2 \n",
+ "ni_power_2 = nbo/peo\n",
+ "print\"square of electron density of ionisation electron for npn silicon transistor is ni**2 = \",\"{:.2e}\".format(ni_power_2),\"cm**-3\"\n",
+ "pbo = 10**16\n",
+ "V = (1.0-((peo)/(10*nbo)))\n",
+ "print\"The emitter efficiency (gamma)is,V =\",\"{:.2e}\".format(V)\n",
+ "neo = ni_power_2*pbo\n",
+ "print\"The required emitter doping is,neo =\",\"{:.2e}\".format(neo),\"cm**-3\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " square of electron density of ionisation electron for npn silicon transistor is ni**2 = 2.00e+01 cm**-3\n",
+ "The emitter efficiency (gamma)is,V = 9.95e-01\n",
+ "The required emitter doping is,neo = 2.00e+17 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.4:pg-279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "B= 0.997\n",
+ "Db = 10.0\n",
+ "Tb = 10**-6\n",
+ "Lb = sqrt(Db*Tb)\n",
+ "print\"The electron carrier diffusion length is,Lb =\",\"{:.2e}\".format(Lb),\"cm\"\n",
+ "# assume the neutral basewidth Wbn is equal to actual basewidth Wb\n",
+ "Wbn = sqrt((1-B)*(2*(Lb**2)))\n",
+ "print\"The base width is,Wb =\",\"{:.2e}\".format(Wbn),\"cm\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electron carrier diffusion length is,Lb = 3.16e-03 cm\n",
+ "The base width is,Wb = 2.45e-04 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.5:pg-279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "# using values from the result of Example 7.1\n",
+ "VEB = 0.6\n",
+ "Ic = 0.2268*10**-3\n",
+ "Ib = 4.92*10**-6\n",
+ "kbT = 0.026\n",
+ "Beta = Ic/Ib\n",
+ "print\"The current gain Beta =\",round(Beta)\n",
+ "gm = Ic/kbT\n",
+ "print\"The transconductance is,gm =\",\"{:.2e}\".format(gm),\"S\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current gain Beta = 46.0\n",
+ "The transconductance is,gm = 8.72e-03 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.6:pg-279"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "De = 20.0\n",
+ "Db=De\n",
+ "Nde = 5*10**17\n",
+ "Nab = 10**17\n",
+ "Wb = 10**-4\n",
+ "ni = 1.5*10**10\n",
+ "# for case (a) value of Te=10**-6s\n",
+ "Te1 = 10**-6\n",
+ "Le1 = sqrt(De*Te1)\n",
+ "Lb1=Le1\n",
+ "peo1 = (ni)**2/Nde\n",
+ "print\"The majority carrier densities for the emitter in npn transistor is,peo = \",\"{:.2e}\".format(peo1),\"cm**-3\"\n",
+ "nbo1 = (ni)**2/Nab\n",
+ "print\"The majority carrier densities for the base in npn transistor is,nbo =\",\"{:.2e}\".format(nbo1),\"cm**-3\"\n",
+ "alpha_1 = (1.0-((peo1*De*Wb)/(nbo1*Db*Le1)))*(1-((Wb**2)/(2*Le1**2)))\n",
+ "print\"The current gain is,alpha_ =\"\"{:.2e}\".format(alpha_1)\n",
+ "Beta1 = (alpha_1)/(1.0-alpha_1)\n",
+ "print\"The current gain Beta1 =\",round(Beta1,1)\n",
+ "#for case (b) value of Te=10**-8s\n",
+ "Te2 = 10**-8\n",
+ "Le2 = sqrt(De*Te2)\n",
+ "print\"The diffusion length is,Le =\",\"{:.2e}\".format(Le2),\"cm\"\n",
+ "peo2 = (ni)**2/Nde\n",
+ "nbo2 = (ni)**2/Nab\n",
+ "alpha_2 = (1-((peo2*De*Wb)/(nbo2*Db*Le2)))*(1-((Wb**2)/(2*Le2**2)))\n",
+ "print\"The current gain alpha_ =\",\"{:.2e}\".format(alpha_2)\n",
+ "Beta2 = (alpha_2)/(1-alpha_2)\n",
+ "print\"The current gain Beta2 =\",round(Beta2,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The majority carrier densities for the emitter in npn transistor is,peo = 4.50e+02 cm**-3\n",
+ "The majority carrier densities for the base in npn transistor is,nbo = 2.25e+03 cm**-3\n",
+ "The current gain is,alpha_ =9.95e-01\n",
+ "The current gain Beta1 = 210.8\n",
+ "The diffusion length is,Le = 4.47e-04 cm\n",
+ "The current gain alpha_ = 9.31e-01\n",
+ "The current gain Beta2 = 13.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex7.7:pg-281"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "nbo = 2.25*10**3\n",
+ "peo = 112.5\n",
+ "Db = 30.0\n",
+ "De = 10.0\n",
+ "Nde = 10**18\n",
+ "Nab = 10**16\n",
+ "Lb = 10*10**-4\n",
+ "Le = 4*10**-4\n",
+ "kbT = 0.026\n",
+ "Wb = 0.5*10**-4\n",
+ "We1 = 10*10**-4\n",
+ "We2 = 1.0*10**-4\n",
+ "e = 1.6*10**-19\n",
+ "print\"for emitter thickness = 10*10**-4 cm\"\n",
+ "gamma_1 = (((Db*nbo*math.tan(Lb/Wb))/(Lb))/(((Db*nbo*math.tan(Lb/Wb))/Lb)+((De*peo*math.tan(Le/We1))/Le)))\n",
+ "print\"The emitter efficiency gamma_1 =\",\"{:.2e}\".format(gamma_1)\n",
+ "print\"for emitter thickness = 10**-4 cm\"\n",
+ "gamma_2 = (((Db*nbo*math.tan(Lb/Wb))/(Lb))/(((Db*nbo*math.tan(Lb/Wb))/Lb)+((De*peo*math.tan(Le/We2))/Le)))\n",
+ "print\"The emitter efficiency (gamma)is,gamma_2 =\"\"{:.1e}\".format(gamma_2)\n",
+ "#NOTE: In the textbook author has used approximate value for the calculation of gamma thus the above solution is differ from that of the gamma"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "for emitter thickness = 10*10**-4 cm\n",
+ "The emitter efficiency gamma_1 = 9.92e-01\n",
+ "for emitter thickness = 10**-4 cm\n",
+ "The emitter efficiency (gamma)is,gamma_2 =9.8e-01\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.8:pg-289"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Ndc = 5*10**15\n",
+ "Nab = 5*10**16\n",
+ "ni = sqrt(2.25*10**20)\n",
+ "kbT = 0.026\n",
+ "e = 1.6*10**-19\n",
+ "Vbi= (kbT)*((log((Nab*Ndc)/(ni**2))))\n",
+ "print\"The built in voltage is ,Vbi=\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "print\" for an applied bias of 1 V \"\n",
+ "VCB1 = 1\n",
+ "apsilent_s = 11.9*8.85*10**-14\n",
+ "Wb = 10**-4\n",
+ "dWb1 = sqrt((2*apsilent_s*(Vbi+VCB1)*Ndc)/(e*Nab*(Nab+Ndc)))\n",
+ "print\"The extent of depletion into the base side is,dWb =\",\"{:.2e}\".format(dWb1),\"cm\"\n",
+ "Wbn1 = Wb-dWb1\n",
+ "print\"The neutral base width is,Wbn =\",\"{:.2e}\".format(Wbn1),\"cm\"\n",
+ "nbo = ((ni)**2)/Nab\n",
+ "print\"The required base doping is,nbo =\",\"{:.2e}\".format(nbo),\"cm**-3\"\n",
+ "Db = 20\n",
+ "VBE = 0.7\n",
+ "Jc1 = ((e*Db*nbo)/Wbn1)*(exp(VBE/kbT))\n",
+ "print\"The collector current density is,Jc1 =\",\"{:.2e}\".format(Jc1),\"A/cm**2\"\n",
+ "print\" for an applied bias of 5 V \"\n",
+ "VCB2 = 5.0\n",
+ "VCE1= VCB1+VBE\n",
+ "print\"The collector emitter voltage is ,VCE=\",\"{:.2e}\".format(VCE1),\" V\"\n",
+ "VCE2= VCB2+VBE\n",
+ "print\"The collector emitter voltage is ,VCE=\",\"{:.2e}\".format(VCE2),\" V\"\n",
+ "dWb2 = sqrt((2*apsilent_s*(Vbi+VCB2)*Ndc)/(e*Nab*(Nab+Ndc)))\n",
+ "print\"The extent of depletion into the base side is,dWb =\",\"{:.2e}\".format(dWb2),\"cm\"\n",
+ "Wbn2 = Wb-dWb2\n",
+ "print\"The neutral base width is,Wbn =\",\"{:.2e}\".format(Wbn2),\"cm\"\n",
+ "Jc2 = ((e*Db*nbo)/Wbn2)*(exp(VBE/kbT))\n",
+ "print\"The collector current density is,Jc =\",\"{:.2e}\".format(Jc2),\" A/cm**2\"\n",
+ "VA = (Jc1/((Jc2-Jc1)/(VCE2-VCE1)))-(VCE1)\n",
+ "print\"The Early voltage is,VA =\",\"{:.2e}\".format(VA),\"V\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ by \"0.2\" value."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The built in voltage is ,Vbi= 7.21e-01 V\n",
+ " for an applied bias of 1 V \n",
+ "The extent of depletion into the base side is,dWb = 6.42e-06 cm\n",
+ "The neutral base width is,Wbn = 9.36e-05 cm\n",
+ "The required base doping is,nbo = 4.50e+03 cm**-3\n",
+ "The collector current density is,Jc1 = 7.58e+01 A/cm**2\n",
+ " for an applied bias of 5 V \n",
+ "The collector emitter voltage is ,VCE= 1.70e+00 V\n",
+ "The collector emitter voltage is ,VCE= 5.70e+00 V\n",
+ "The extent of depletion into the base side is,dWb = 1.17e-05 cm\n",
+ "The neutral base width is,Wbn = 8.83e-05 cm\n",
+ "The collector current density is,Jc = 8.03e+01 A/cm**2\n",
+ "The Early voltage is,VA = 6.51e+01 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex7.9:pg-290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Ndc = 10**16\n",
+ "Nab = 5*10**16\n",
+ "e = 1.6*10**-19\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "Wb = .2*10**-4\n",
+ "Vpt= ((e*(Wb**2)*Nab*(Ndc+Nab))/(2*apsilen*Ndc))\n",
+ "print\"The punchthrough voltage is ,Vpt=\",round(Vpt,2),\"V\"\n",
+ "Twb = 1.2*10**-4\n",
+ "F = Vpt/Twb\n",
+ "print\"The average field at punchthrough voltage is ,F =\",\"{:.2e}\".format(F),\"V/cm\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ by \"0.16\" value."
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The punchthrough voltage is ,Vpt= 9.12 V\n",
+ "The average field at punchthrough voltage is ,F = 7.60e+04 V/cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.10:pg-290"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "apsilent_s = 11.9*8.85*10**-14\n",
+ "Ndc = 5.0*10**16\n",
+ "Nde = 10**18\n",
+ "Nab = 10**17\n",
+ "ni = sqrt(2.25*10**20)\n",
+ "kbT = 0.026\n",
+ "e = 1.6*10**-19\n",
+ "Db = 30.0\n",
+ "De = 10\n",
+ "Lb = 15*10**-4\n",
+ "Le = 5*10**-4\n",
+ "Beta= 100\n",
+ "nbo = 2.25*10**3\n",
+ "peo = 112.5\n",
+ "VCB1 = 5.0\n",
+ "#\"using relation B = (IC/IB) = ((Db*nbo*Le)/(De*peo*Wbn))\"\n",
+ "Wbn = ((Db*nbo*Le)/(De*peo*100))\n",
+ "print\"neutral base width is ,Wbn =\",\"{:.2e}\".format(Wbn),\"cm\"\n",
+ "Vbi= (kbT)*((log((Nab*Ndc)/(ni**2))))\n",
+ "print\"The built in voltage is ,Vbi=\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "dWb1 = sqrt((2*apsilent_s*(Vbi+VCB1)*Ndc)/(e*Nab*(Nab+Ndc)))\n",
+ "print\"The extent of depletion into the base side is,dWb =\",\"{:.2e}\".format(dWb1),\"cm\"\n",
+ "Wb = Wbn+dWb1\n",
+ "print\"The base width is,Wb = Wbn+dWb1= \",\"{:.2e}\".format(Wb),\"cm\"\n",
+ "\n",
+ "# NOTE: the value calculated for Wbn is wrong in the book and all the succesive answer also depeandant on that are also wrong\n",
+ "\n",
+ "#(\"Two disadvange are\")\n",
+ "#(\"The output conductance will suffer and the collector current will have a stronger dependence on VCB\")\n",
+ "#(\"The device may suffer punchthrough at a lower bias\")\n",
+ "#(\"Two advantages\")\n",
+ "#(\"The current gain will be higher\")\n",
+ "#(\"The device speed will be faster\")"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "neutral base width is ,Wbn = 3.00e-04 cm\n",
+ "The built in voltage is ,Vbi= 7.99e-01 V\n",
+ "The extent of depletion into the base side is,dWb = 1.60e-05 cm\n",
+ "The base width is,Wb = Wbn+dWb1= 3.16e-04 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex7.11:pg-291"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Ndc = 10**16\n",
+ "Nab = 10**17\n",
+ "Nde = 10**18\n",
+ "ni = 1.5*10**10\n",
+ "kbT = 0.026\n",
+ "e = 1.6*10**-19\n",
+ "Db = 30.0\n",
+ "De = 10.0\n",
+ "Lb = 10*10**-4\n",
+ "Le = 10*10**-4\n",
+ "Wb = 10**-4\n",
+ "We = 10**-4\n",
+ "Vbi= (kbT)*((log((Nab*Ndc)/ni**2)))\n",
+ "print\"The built in voltage is ,Vbi= (kbT)*((log((Na*Nd)/Ni**2)))= \",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "print\" for an applied reverse bias of 5 V \"\n",
+ "VCB1 = 5.0\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "nbo = 2.25*10**3\n",
+ "peo = 112.5\n",
+ "dWb1 = sqrt((2*apsilen*(Vbi+VCB1)*Ndc)/(e*Nab*(Nab+Ndc)))\n",
+ "print\"The extent of depletion into the base side is,dWb =\",\"{:.2e}\".format(dWb1),\"cm\"\n",
+ "Wbn1 = Wb-dWb1\n",
+ "print\"The neutral base width is,Wbn = Wb-dWb1= \",\"{:.2e}\".format(Wbn1),\"cm\"\n",
+ "gamma_e_1 = (1-((peo*De*Wbn1)/(Db*nbo*We)))\n",
+ "print\"The emitter efficiency gamma_e_1 =\",\"{:.2e}\".format(gamma_e_1)\n",
+ "B1 = 1-((Wbn1**2)/(2*(Lb)**2))\n",
+ "print\"The base transport factor is,B =\"\"{:.2e}\".format(B1)\n",
+ "alpha1 = gamma_e_1*B1\n",
+ "print\"The current gain alpha1 =\"\"{:.2e}\".format(alpha1)\n",
+ "Beta3 = (alpha1)/(1-alpha1)\n",
+ "print\"The current gain Beta3 =\"\"{:.2e}\".format(Beta3)\n",
+ "VBE = 1.0\n",
+ "A= 4.0*10**-6\n",
+ "IC = (((e*A*Db*nbo)/(Wbn1))*(exp((VBE)/(kbT))-1))\n",
+ "print\"The collector current is,IC =\",\"{:.2e}\".format(IC),\"A\"\n",
+ "#Note: in text book the author hasused precision value for gamma and alpha thats why there is difference in the value of beta.\n",
+ "print\" for an applied reverse bias of 6 V \"\n",
+ "VCB2 = 6.0\n",
+ "dWb2 = sqrt((2*apsilen*(Vbi+VCB2)*Ndc)/(e*Nab*(Nab+Ndc)))\n",
+ "print\"The extent of depletion into the base side is,dWb2 =\",\"{:.2e}\".format(dWb2),\"cm\"\n",
+ "Wbn2 = Wb-dWb2\n",
+ "print\"The neutral base width is,Wbn2 =\",\"{:.2e}\".format(Wbn2),\"cm\"\n",
+ "IC2 = (((e*A*Db*nbo)/(Wbn2))*(exp((VBE)/(kbT))-1))\n",
+ "print\"The collector current is,IC =\",\"{:.2e}\".format(IC2),\"A\"\n",
+ "go = (IC2-IC)/(VCB2-VCB1)\n",
+ "print\"The output conductance is,go =\",\"{:.2e}\".format(go),\"ohm**-1\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The built in voltage is ,Vbi= (kbT)*((log((Na*Nd)/Ni**2)))= 7.57e-01 V\n",
+ " for an applied reverse bias of 5 V \n",
+ "The extent of depletion into the base side is,dWb = 8.30e-06 cm\n",
+ "The neutral base width is,Wbn = Wb-dWb1= 9.17e-05 cm\n",
+ "The emitter efficiency gamma_e_1 = 9.85e-01\n",
+ "The base transport factor is,B =9.96e-01\n",
+ "The current gain alpha1 =9.81e-01\n",
+ "The current gain Beta3 =5.05e+01\n",
+ "The collector current is,IC = 2.38e+01 A\n",
+ " for an applied reverse bias of 6 V \n",
+ "The extent of depletion into the base side is,dWb2 = 8.99e-06 cm\n",
+ "The neutral base width is,Wbn2 = 9.10e-05 cm\n",
+ "The collector current is,IC = 2.40e+01 A\n",
+ "The output conductance is,go = 1.81e-01 ohm**-1\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.12:pg-310"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kbT = 0.026\n",
+ "Wb = 0.4*10**-4\n",
+ "e = 1.6*10**-19\n",
+ "IE = 1.5*10**-3\n",
+ "Db = 60.0\n",
+ "Wdc = 2*10**-4\n",
+ "Cje = 2*10**-12\n",
+ "rC = 30.0\n",
+ "TcC = .4*10**-12\n",
+ "#NOTE:Total collector capicitance represented in book as(Cu+Cs)\n",
+ "vs = 10**7\n",
+ "re = kbT/IE\n",
+ "Te = re*Cje\n",
+ "Tt = (Wb**2)/(2*Db)\n",
+ "Td = (Wdc)/vs\n",
+ "Tc = rC*Tc\n",
+ "Tec = Te+Tt+Td+Tc\n",
+ "print\"The total time is,Tec =\",\"{:.2e}\".format(Tec),\"s\"\n",
+ "fT = 1/(2*math.pi*Tec)\n",
+ "print\"fT=\",\"{:.2e}\".format(fT),\"HZ\"\n",
+ "print\"if the emitter current is doubled the time is reduced by half and cutoff frequency becomes 2.54 GHz\"\n",
+ "print\"if the base width is reduced by half , the base transit time becomes 3.3 ps and cutoff frequency becomes 2.08 GHz\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Te= 3.47e-11 S\n",
+ "TC= 3.60e-10 S\n",
+ "The total time is,Tec = 4.28e-10 s\n",
+ "fT= 3.72e+08 HZ\n",
+ "if the emitter current is doubled the time is reduced by half and cutoff frequency becomes 2.54 GHz\n",
+ "if the base width is reduced by half , the base transit time becomes 3.3 ps and cutoff frequency becomes 2.08 GHz\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.13:pg-319"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "T = 300.0\n",
+ "Nd1 = 10**18\n",
+ "Nd2 = 10**20\n",
+ "dEg1 = (22.5*sqrt((Nd1*300)/((10**18)*T)))/10**3\n",
+ "print\"The bandgap narrowing is,dEg = \",\"{:.2e}\".format(dEg1),\"ev\"\n",
+ "dEg2= (22.5*sqrt((Nd2*300)/((10**18)*T)))/10**3\n",
+ "print\"The bandgap narrowing is,dEg =\",\"{:.2e}\".format(dEg2),\"ev\"\n",
+ "kbT =0.026\n",
+ "neo1 = 10**18\n",
+ "neo2 = 10**20\n",
+ "ni = sqrt(2.25*10**20)\n",
+ "peo1 = (ni**2*exp(dEg1/kbT))/neo1\n",
+ "print\"The hole density in emitter is,peo =\",\"{:.2e}\".format(peo1),\"cm**-3\"\n",
+ "# note:-there is error in the unit of peo in the book\n",
+ "peo2 = (ni**2*exp(dEg2/kbT))/neo2\n",
+ "print\"The hole density in emitter is,peo2 =\",\"{:.2e}\".format(peo2),\"cm**-3\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The bandgap narrowing is,dEg = 2.25e-02 ev\n",
+ "The bandgap narrowing is,dEg = 2.25e-01 ev\n",
+ "The hole density in emitter is,peo = 5.35e+02 cm**-3\n",
+ "The hole density in emitter is,peo2 ="
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " 1.29e+04 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex7.14:pg-319"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "ni = 2.2*10**6\n",
+ "Nde = 5*10**17\n",
+ "Nab = 10**17\n",
+ "kbT = 0.026\n",
+ "Wb = 0.5*10**-4\n",
+ "Db = 100.0\n",
+ "De = 15.0\n",
+ "Le = 1.5*10**-4\n",
+ "dEg = 0.36\n",
+ "print(\" For GaAs \")\n",
+ "peo1 = ni**2/Nde\n",
+ "print\"The minority carrier densities for the emitter in npn GaAs BJT is,peo(GaAs) =\",\"{:.1e}\".format(peo1),\"cm**-3\"\n",
+ "nbo1 = ni**2/Nab\n",
+ "print\"The minority carrier densities for the base in npn GaAs BJT is,nbo = \",\"{:.2e}\".format(nbo1),\"cm**-3\"\n",
+ "Ve1 = (1-((peo1*De*Wb)/(Db*nbo1*Le)))\n",
+ "print\"The emitter efficiency (gamma)is,Ve =\",\"{:.2e}\".format(Ve1)\n",
+ "gammae=1-((peo1*De*Wb)/(nbo1*Db*Le))\n",
+ "print\"gammae=\",round(gammae,2)\n",
+ "print(\" For HBT \")\n",
+ "peo2 = (peo1)*(exp(-(dEg/kbT)))\n",
+ "print\"The minority carrier densities for the emitter in HBT is,peo(HBT) =\",\"{:.1e}\".format(peo2),\"cm**-3\"\n",
+ "print\"in this case the emitter efficiency is essentially unity\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " For GaAs \n",
+ "The minority carrier densities for the emitter in npn GaAs BJT is,peo(GaAs) = 9.7e-06 cm**-3\n",
+ "The minority carrier densities for the base in npn GaAs BJT is,nbo = 4.84e-05 cm**-3\n",
+ "The emitter efficiency (gamma)is,Ve = 9.90e-01\n",
+ "gammae= 0.99\n",
+ " For HBT \n",
+ "The minority carrier densities for the emitter in HBT is,peo(HBT) = 9.4e-12 cm**-3\n",
+ "in this case the emitter efficiency is essentially unity\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter8_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter8_1.ipynb new file mode 100644 index 00000000..dab318d1 --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter8_1.ipynb @@ -0,0 +1,473 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:6e318f0f515b2df1396cf49fa3f9877662336f268be595bba389bbb595cdcca4"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter8:FIELD EFFECT TRANSISTORS:JFET/MESFET"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.1:pg-342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "ni = 1.5*10**10\n",
+ "Na = 10**18\n",
+ "Nd = 10**17\n",
+ "kBT=0.026\n",
+ "e = 1.6*10**-19\n",
+ "Vbi= (kBT)*((log((Na*Nd)/ni**2)))\n",
+ "print\"The built in voltage of a p+n diode is ,Vbi=\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "h=.25*10**-4\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "Vp= (e*(h**2)*Nd)/(2*apsilen)\n",
+ "print\"The total volage drop required to pinch the channel is ,Vp= \",\"{:.2e}\".format(Vp),\"V\"\n",
+ "VG= Vbi-Vp\n",
+ "print\"The pinch off at gate bias is ,VG= \",\"{:.2e}\".format(VG),\"V\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The built in voltage of a p+n diode is ,Vbi= 8.77e-01 V\n",
+ "The total volage drop required to pinch the channel is ,Vp= 4.75e+00 V\n",
+ "The pinch off at gate bias is ,VG= -3.87e+00 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex8.2:pg-342"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "phi_b=0.8\n",
+ "T=300.0\n",
+ "kBT=0.026\n",
+ "R_star=8.0\n",
+ "Dp=20.0\n",
+ "pn = 3.38*10**-5\n",
+ "e = 1.6*10**-19\n",
+ "Lp=1*10**-4\n",
+ "# for Schottky case Js = R_star*T**2*(exp(-(phi_b)/(kBT)))\n",
+ "Js = R_star*T**2*(exp(-(phi_b)/(kBT)))\n",
+ "print\"The gate current density is ,Js =\",\"{:.2e}\".format(Js),\"A/cm**2\"\n",
+ "# from p-n diode theory Jo = (e*Dp*pn)/(Lp)\n",
+ "Jo = (e*Dp*pn)/(Lp)\n",
+ "print\"The gate current density is ,Jo =\",\"{:.2e}\".format(Jo),\"A/cm**2\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The gate current density is ,Js = 3.12e-08 A/cm**2\n",
+ "The gate current density is ,Jo = 1.08e-18 A/cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.3:pg-349"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "ni = 1.5*10**10\n",
+ "Nc = 4.45*10**17\n",
+ "Nd = 10**17\n",
+ "kBT=0.026\n",
+ "e = 1.6*10**-19\n",
+ "h=0.25*10**-4\n",
+ "apsilen = 13.2*8.85*10**-14\n",
+ "Vh = 0.8\n",
+ "Vp= (e*(h**2)*Nd)/(2*apsilen)\n",
+ "print\"The total volage drop required to pinch the channel is ,Vp= \",round(Vp,1),\"V\"\n",
+ "Ecf= -(kBT)*(log(Nd/Nc))\n",
+ "print\"The difference between the conduction band and fermi level is ,Ecf=\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "Vbi= Vh-Ecf\n",
+ "print\"The built in potential is ,Vbi= \",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "VGS= Vbi-Vp\n",
+ "print\"The pinch off at gate bias is ,VT=\",round(VGS,1),\"V\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total volage drop required to pinch the channel is ,Vp= 4.3 V\n",
+ "The difference between the conduction band and fermi level is ,Ecf= 7.61e-01 V\n",
+ "The built in potential is ,Vbi= 7.61e-01 V\n",
+ "The pinch off at gate bias is ,VT= -3.5 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.4:pg-350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nd = 10**17\n",
+ "e = 1.6*10**-19\n",
+ "apsilen = 13.1*8.85*10**-14\n",
+ "Vbi = 0.76\n",
+ "h= sqrt((2*apsilen*Vbi)/(e*Nd))\n",
+ "print\"The thickness at which the value of Vp is same at Vbi is ,h=\",\"{:.1e}\".format(h),\"cm\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The thickness at which the value of Vp is same at Vbi is ,h= 1.0e-05 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex8.5:pg-350"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nc = 4.45*10**17\n",
+ "Nd = 10**17\n",
+ "kBT=0.026\n",
+ "e = 1.6*10**-19\n",
+ "h=0.06*10**-4\n",
+ "apsilen = 13.2*8.85*10**-14\n",
+ "Vh = 0.8\n",
+ "Vp= (e*(h**2)*Nd)/(2*apsilen)\n",
+ "print\"The total volage drop required to pinch the channel is ,Vp=\",\"{:.1e}\".format(Vp),\"V\"\n",
+ "Ecf= -(kBT)*(log(Nd/Nc))\n",
+ "print\"The difference between the conduction band and fermi level is ,Ecf=\",\"{:.2e}\".format(Ecf),\"V\"\n",
+ "Vbi= Vh-Ecf\n",
+ "print\"The built in potential is ,Vbi=\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "VG= Vbi-Vp\n",
+ "print\"The pinch off at gate bias is ,VG=\",\"{:.2e}\".format(VG),\"V\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total volage drop required to pinch the channel is ,Vp= 2.5e-01 V\n",
+ "The difference between the conduction band and fermi level is ,Ecf= 3.88e-02 V\n",
+ "The built in potential is ,Vbi= 7.61e-01 V\n",
+ "The pinch off at gate bias is ,VG= 5.15e-01 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.6:pg-351"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mu_n=6000\n",
+ "phi_b=0.8\n",
+ "kBT=0.026\n",
+ "e = 1.6*10**-19\n",
+ "h=0.25*10**-4\n",
+ "Nd = 5*10**16\n",
+ "Nc = 4.45*10**17\n",
+ "L = 2*10**-4\n",
+ "Z = 25*10**-4\n",
+ "apsilen = 13.2*8.85*10**-14\n",
+ "VGS1 = 0\n",
+ "VGS2 = -1\n",
+ "Vbi = (phi_b-(kBT*(log(Nc/Nd))))\n",
+ "print\"The built-in voltage is ,Vbi =\",\"{:.2e}\".format(Vbi),\"V\"\n",
+ "Vp= (e*(h**2)*Nd)/(2*apsilen)\n",
+ "print\"The internal pinch off potential is ,Vp=\",round(Vp,2),\"V\"\n",
+ "go=(e*mu_n*Nd*Z*h)/(L)\n",
+ "print\"The value of go of the channel is ,go=\",\"{:.2e}\".format(go),\"ohm**-1\"\n",
+ "ID_sat = go*((Vp/3)-Vbi+((2*(Vbi**1.5))/(3*(Vp**.5))))\n",
+ "print\"The value of saturation voltage is ,ID_sat =\",\"{:.2e}\".format(ID_sat),\"V\"\n",
+ "ID1 = go*((Vp/3)-Vbi+VGS2+((2*((Vbi+abs(VGS2))**1.5))/(3*(Vp**.5))))\n",
+ "print\"The value of saturation current at VGS1 is ,ID(sat) =\",\"{:.2e}\".format(ID1),\"A\"\n",
+ "gm_sat= go*(1-((Vbi/Vp)**.5))\n",
+ "print\"The value of saturation conductance at VGS1 is ,gm_sat=\",\"{:.1e}\".format(gm_sat),\"S\"\n",
+ "gm_sat= go*(1-((((Vbi+abs(VGS2))/Vp)**.5)))\n",
+ "print\"The value of saturation conductance at VGS2 is ,gm_sat=\",\"{:.2e}\".format(gm_sat),\"S\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The built-in voltage is ,Vbi = 7.43e-01 V\n",
+ "The internal pinch off potential is ,Vp= 2.14 V\n",
+ "The value of go of the channel is ,go= 1.50e-02 ohm**-1\n",
+ "The value of saturation voltage is ,ID_sat = 3.93e-03 V\n",
+ "The value of saturation current at VGS1 is ,ID(sat) = 2.85e-04 A\n",
+ "The value of saturation conductance at VGS1 is ,gm_sat= 6.2e-03 S\n",
+ "The value of saturation conductance at VGS2 is ,gm_sat= 1.46e-03 S\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex8.7:pg-354"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "Nd = 5*10**16\n",
+ "L = 2*10**-4\n",
+ "apsilen = 13.2*8.85*10**-14\n",
+ "VDS1 = 1.0\n",
+ "VDS2 = 1.5\n",
+ "VGS1 = 0\n",
+ "ID=4.03\n",
+ "dL1 = sqrt((2*apsilen*VDS1)/(e*Nd))\n",
+ "print\"The change in channel length is ,dL (VDS(sat)+1 V) =\",\"{:.2e}\".format(dL1),\"cm\"\n",
+ "dL2 = sqrt((2*apsilen*VDS2)/(e*Nd))\n",
+ "print\"The change in channel length is ,dL (VDS(sat)+1.5 V) =\",\"{:.1e}\".format(dL2),\"cm\"\n",
+ "ID1 = ID*(1+(dL1/(2*L)))\n",
+ "print\"The current at the bias is ,ID1(VDS(sat)+1 V) =\",round(ID1,2),\"mA\"\n",
+ "ID2 = ID*(1+(dL2/(2*L)))\n",
+ "print\"The current at the bias is ,ID2(VDS(sat)+1.5 V) =\",round(ID2,2),\"mA\"\n",
+ "rDS = (VDS2-VDS1)/((ID2-ID1)*10**-3)\n",
+ "print\"The output resistance of source drain channel is ,rDS =\",\"{:.2e}\".format(rDS),\"ohm\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The change in channel length is ,dL (VDS(sat)+1 V) = 1.71e-05 cm\n",
+ "The change in channel length is ,dL (VDS(sat)+1.5 V) = 2.1e-05 cm\n",
+ "The current at the bias is ,ID1(VDS(sat)+1 V) = 4.2 mA\n",
+ "The current at the bias is ,ID2(VDS(sat)+1.5 V) = 4.24 mA\n",
+ "The output resistance of source drain channel is ,rDS = 1.29e+04 ohm\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.8:pg-361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mu_n=1000\n",
+ "e = 1.6*10**-19\n",
+ "h=0.5*10**-4\n",
+ "Nd = 10**16\n",
+ "L = 2*10**-4\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "Vs = 10**7\n",
+ "fT = (e*mu_n*Nd*(h**2))/(2*math.pi*apsilen*(L**2))\n",
+ "print\"The maximum cutoff frequency of the device in the constant mobility model is ,fT =\",\"{:.2e}\".format(fT),\"Hz\"\n",
+ "fT = Vs/(2*math.pi*L)\n",
+ "print\"The maximum cutoff frequency of the device in the saturation velocity model is fT =\",\"{:.2e}\".format(fT),\"Hz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum cutoff frequency of the device in the constant mobility model is ,fT = 1.51e+10 Hz\n",
+ "The maximum cutoff frequency of the device in the saturation velocity model is fT = 7.96e+09 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.9:pg-361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "VDS =0.5\n",
+ "h=1*10**-4\n",
+ "ID=4.03\n",
+ "F = VDS/h\n",
+ "print\"The electric field in channel is ,F =\",\"{:.2e}\".format(F),\"V/cm\"\n",
+ "Vsi = 5*10**6\n",
+ "VGaAs = 10**7\n",
+ "Ttr1 = h/Vsi \n",
+ "print\"The transit time of electrons in silicon is ,Ttr(si) = \",\"{:.2e}\".format(Ttr1),\"s\"\n",
+ "Ttr2 = h/VGaAs \n",
+ "print\"The transit time of electrons in GaAs is ,Ttr(GaAs) =\",\"{:.2e}\".format(Ttr2),\"s\"\n",
+ "fT1 = 1.0/(2*math.pi*Ttr1)\n",
+ "print\"The corresponding frequency of silicon is ,fT(Si) =\",\"{:.2e}\".format(fT1),\"Hz\"\n",
+ "fT2 = 1.0/(2*math.pi*Ttr2)\n",
+ "print\"The corresponding frequency of GaAs is ,fT(GaAs) =\",\"{:.2e}\".format(fT2),\"Hz\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electric field in channel is ,F = 5.00e+03 V/cm\n",
+ "The transit time of electrons in silicon is ,Ttr(si) = 2.00e-11 s\n",
+ "The transit time of electrons in GaAs is ,Ttr(GaAs) = 1.00e-11 s\n",
+ "The corresponding frequency of silicon is ,fT(Si) = 7.96e+09 Hz\n",
+ "The corresponding frequency of GaAs is ,fT(GaAs) = 1.59e+10 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 13
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex8.10:pg-361"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "VB = 100.0\n",
+ "FSi=3*10**5\n",
+ "FGaAs=4*10**5\n",
+ "FSiC=3*10**6\n",
+ "Vsi = 10**7\n",
+ "VGaAs = 10**7\n",
+ "VSiC = 2*10**7\n",
+ "LBSi = VB/FSi\n",
+ "print\"The minimum channel length at which Si material will breakdown is ,LBSi = \",\"{:.2e}\".format(LBSi),\"cm\"\n",
+ "LBGaAs = VB/FGaAs\n",
+ "print\"The minimum channel length at which GaAs material will breakdown is ,LBGaAs =\",\"{:.2e}\".format(LBGaAs),\"cm\"\n",
+ "LBSiC = VB/FSiC\n",
+ "print\"The minimum channel length at which SiC material will breakdown is ,LBSiC =\",\"{:.2e}\".format(LBSiC),\"cm\"\n",
+ "fT1 = Vsi/(2*math.pi*LBSi)\n",
+ "print\"The corresponding cutoff frequency of silicon is ,fT(Si) = \",\"{:.1e}\".format(fT1),\"Hz\"\n",
+ "fT2 = VGaAs/(2*math.pi*LBGaAs)\n",
+ "print\"The corresponding frequency of GaAs is ,fT(GaAs) =\",\"{:.1e}\".format(fT2),\"Hz\"\n",
+ "fT3 = VSiC/(2*math.pi*LBSiC)\n",
+ "print\"The corresponding cutoff frequency of SiC is ,fT(SiC) =\",\"{:.2e}\".format(fT3),\"Hz\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The minimum channel length at which Si material will breakdown is ,LBSi = 3.33e-04 cm\n",
+ "The minimum channel length at which GaAs material will breakdown is ,LBGaAs = 2.50e-04 cm\n",
+ "The minimum channel length at which SiC material will breakdown is ,LBSiC = 3.33e-05 cm\n",
+ "The corresponding cutoff frequency of silicon is ,fT(Si) = 4.8e+09 Hz\n",
+ "The corresponding frequency of GaAs is ,fT(GaAs) = 6.4e+09 Hz\n",
+ "The corresponding cutoff frequency of SiC is ,fT(SiC) = 9.55e+10 Hz\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter9_1.ipynb b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter9_1.ipynb new file mode 100644 index 00000000..5fb15e3e --- /dev/null +++ b/Semiconductor_Devices_Basic_Principle_by_J._Singh/chapter9_1.ipynb @@ -0,0 +1,488 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:89aba2f692b8b301293cc14fb417afc88daa14a53d02d71daa2ea70781d3602b"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter9:FIELD EFFECT TRANSISTORS:MOSFET"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.1:pg-384"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kbT = 0.026\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "Na=10**16\n",
+ "ni = 1.5*10**10\n",
+ "phi_F= (-kbT*log(Na/ni))\n",
+ "print\"The potential phi_F=\",\"{:.2e}\".format(phi_F),\"V\"\n",
+ "W = sqrt((4*apsilen*(-phi_F))/(e*Na))*10**4\n",
+ "print\"The space charge width is ,W =\",\"{:.1e}\".format(W),\" micro_meter\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential phi_F= -3.49e-01 V\n",
+ "The space charge width is ,W = 3.0e-01 micro_meter\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.2:pg-385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kbT = 0.026\n",
+ "Eg = 1.11\n",
+ "e = 1.6*10**-19\n",
+ "Na=10**14\n",
+ "ni = 1.5*10**10\n",
+ "phi_m = 4.1\n",
+ "Es = 4.15\n",
+ "EF= ((Eg/2)+kbT*log(Na/ni))\n",
+ "print\"The position of fermi level below conduction band is ,EF=\",\"{:.2e}\".format(EF),\"eV\"\n",
+ "Vfb = phi_m-(Es+EF)\n",
+ "print\"The potential is ,Vfb =\",\"{:.2e}\".format(Vfb),\"eV\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position of fermi level below conduction band is ,EF= 7.84e-01 eV\n",
+ "The potential is ,Vfb = -8.34e-01 eV\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.3:pg-385"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kbT = 0.026\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "Na=3*10**16\n",
+ "ni = 1.5*10**10\n",
+ "Vfb = -1.13\n",
+ "Eox = 3.9*8.85*10**-14\n",
+ "dox = 500*10**-8\n",
+ "Nt = 10**11\n",
+ "phi_F= (-kbT*log(Na/ni))\n",
+ "print\"The potential phi_F=\",\"{:.2e}\".format(phi_F),\"V\"\n",
+ "Qs = sqrt((4*apsilen*(-phi_F))*(e*Na))\n",
+ "print\"The maximum depletion width is ,Qs =\",\"{:.2e}\".format(Qs),\"C cm**-2\"\n",
+ "Vs = -(2*phi_F)\n",
+ "print\"The surface potential is ,Vs =\",\"{:.2e}\".format(Vs),\" V\"\n",
+ "VT = Vfb+Vs+((Qs*dox)/Eox)\n",
+ "print\"In the absence of any oxide charge, the threshold voltage is ,VT =\",\"{:.2e}\".format(VT),\" V\"\n",
+ "dVT = -((e*Nt*dox)/Eox)\n",
+ "print\"when oxide has trap charges, the shift in threshold voltage is ,dVT =\",\"{:.2e}\".format(dVT),\" V\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential phi_F= -3.77e-01 V\n",
+ "The maximum depletion width is ,Qs = 8.73e-08 C cm**-2\n",
+ "The surface potential is ,Vs = 7.54e-01 V\n",
+ "In the absence of any oxide charge, the threshold voltage is ,VT = 8.90e-01 V\n",
+ "when oxide has trap charges, the shift in threshold voltage is ,dVT = -2.32e-01 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.4:pg-386"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "import math\n",
+ "mu_n=600\n",
+ "mu_p = 200\n",
+ "kbT = 0.026\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "Na=5*10**16\n",
+ "ni = 1.5*10**10\n",
+ "Vfb = -0.5\n",
+ "Eox = 1.583*8.85*10**-14\n",
+ "dox = 200*10**-8\n",
+ "sigma_1= Na*e*mu_p\n",
+ "sigma_2= Na*e*mu_n\n",
+ "phi_F= (-kbT*log(Na/ni))\n",
+ "Qs = sqrt((4*apsilen*(-phi_F))*(e*Na))\n",
+ "print\"The maximum depletion width is ,Qs = \",\"{:.2e}\".format(Qs),\" C cm**-2\"\n",
+ "Vs = -(2*phi_F)\n",
+ "VT = Vfb+Vs+((Qs*dox)/Eox)\n",
+ "print\"In the absence of any oxide charge, the threshold voltage is ,VT =\",\"{:.2e}\".format(VT),\" V\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The maximum depletion width is ,Qs = 1.15e-07 C cm**-2\n",
+ "In the absence of any oxide charge, the threshold voltage is ,VT = 1.92e+00 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.6:pg-393"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kbT = 0.026\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "Na=10**16\n",
+ "ni = 1.5*10**10\n",
+ "apsilen_ox = 3.9*8.85*10**-14\n",
+ "dox = 500*10**-8\n",
+ "Cox= apsilen_ox/dox\n",
+ "print\"The oxide capacitance Cox=\",\"{:.2e}\".format(Cox),\"F/cm**2\"\n",
+ "phi_F =(-kbT*log(Na/ni))\n",
+ "print\"The potential phi_F=\",\"{:.2e}\".format(phi_F),\" V\"\n",
+ "Wmax = sqrt((4*apsilen*(-phi_F))/(e*Na))\n",
+ "print\"The maximum depletion width is ,Wmax =\",\"{:.2e}\".format(Wmax),\" cm\"\n",
+ "Cmin = (apsilen_ox/(dox+((apsilen_ox*Wmax)/apsilen)))\n",
+ "print\"The minimum capicitance is ,Cmin =\",\"{:.2e}\".format(Cmin),\" F/cm**2\"\n",
+ "Cfb = (apsilen_ox/((dox)+((apsilen_ox/apsilen)*(sqrt((kbT*apsilen)/(e*Na))))))\n",
+ "print\"The capicitance under flat band conditions is ,Cfb =\",\"{:.2e}\".format(Cfb),\" F/cm**2\"\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The oxide capacitance Cox= 6.90e-08 F/cm**2\n",
+ "The potential phi_F= -3.49e-01 V\n",
+ "The maximum depletion width is ,Wmax = 3.03e-05 cm\n",
+ "The minimum capicitance is ,Cmin = 2.31e-08 F/cm**2\n",
+ "The capicitance under flat band conditions is ,Cfb = 5.43e-08 F/cm**2\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.7:pg-400"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "mu_n=600\n",
+ "kbT = 0.026\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "Na=10**16\n",
+ "ni = 1.5*10**10\n",
+ "apsilen_ox = 3.9*8.85*10**-14\n",
+ "dox = 500*10**-8\n",
+ "phi_ms = -1.13\n",
+ "Qss = 10**11\n",
+ "VGS = 5.0\n",
+ "Z=25*10**-6\n",
+ "L=1.5*10**-6\n",
+ "phi_F= (-kbT*log(Na/ni))\n",
+ "print\"The potential phi_F=\",\"{:.2e}\".format(phi_F),\" V\"\n",
+ "Cox = apsilen_ox/dox\n",
+ "print\"The oxide capicitance per unit area is ,Cox =\",\"{:.2e}\".format(Cox),\" F/cm**-2\"\n",
+ "Vfb = phi_ms-((Qss*e)/Cox)\n",
+ "print\"The flat band potential is Vfb =\",\"{:.2e}\".format(Vfb),\"V\"\n",
+ "Vs = -(2*phi_F)\n",
+ "print\"The surface potential is ,Vs =\",\"{:.2e}\".format(Vs),\" V\"\n",
+ "VT = Vfb+Vs+(sqrt(4*e*apsilen*Na*(-phi_F))/Cox)\n",
+ "print\"In the absence of any oxide charge, the threshold voltage is ,VT =\",\"{:.2e}\".format(VT),\" V\"\n",
+ "ID = (Z*mu_n*Cox*(VGS-VT)**2)/(2*L)\n",
+ "print\"The saturation current is ,ID = \",\"{:.2e}\".format(ID),\" A\"\n",
+ "\n",
+ "#NOTE: The value of Vfb in the text book is wrong for the above solution and thus the value of VT and saturation current is also wrong\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential phi_F= -3.49e-01 V\n",
+ "The oxide capicitance per unit area is ,Cox = 6.90e-08 F/cm**-2\n",
+ "The flat band potential is Vfb = -1.36e+00 V\n",
+ "The surface potential is ,Vs = 6.97e-01 V\n",
+ "In the absence of any oxide charge, the threshold voltage is ,VT = 3.78e-02 V\n",
+ "The saturation current is ,ID = 8.50e-03 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex9.8:pg-402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kbT = 0.026\n",
+ "Z = 10*10**-4\n",
+ "L = 1*10**-4\n",
+ "mu_n=700\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "Na=4*10**14\n",
+ "ni = 1.5*10**10\n",
+ "apsilen_ox = 3.9*8.85*10**-14\n",
+ "dox = 200*10**-8\n",
+ "VGS = 5.0\n",
+ "Qs = sqrt(4*apsilen*(-phi_F)*e*Na)\n",
+ "print\"The maximum depletion width is ,Qs =\",\"{:.2e}\".format(Qs),\" cm**-2\"\n",
+ "phi_F= (-kbT*log(Na/ni))\n",
+ "print\"The potential phi_F=\",\"{:.2e}\".format(phi_F),\" V\"\n",
+ "Cox = apsilen_ox/dox\n",
+ "print\"The oxide capicitance per unit area is ,Cox =\"\"{:.2e}\".format(Cox),\" cm**-1\"\n",
+ "Vs = -(2*phi_F)\n",
+ "print\"The surface potential is ,Vs =\",\"{:.2e}\".format(Vs),\" V\"\n",
+ "VT = Vs+((Qs/Cox))\n",
+ "print\"The threshold voltage is ,VT =\",\"{:.2e}\".format(VT),\" V\"\n",
+ "VDS = VGS-VT\n",
+ "print\"The saturation voltage is ,VDS =\",\"{:.2e}\".format(VDS),\" V\"\n",
+ "ID = (Z*mu_n*Cox*(VDS)**2)/(2*L)\n",
+ "print\"The saturation current is ,ID =\",\"{:.2e}\".format(ID),\" A\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " The maximum depletion width is ,Qs = 9.70e-09 cm**-2\n",
+ "The potential phi_F= -2.65e-01 V\n",
+ "The oxide capicitance per unit area is ,Cox =1.73e-07 cm**-1\n",
+ "The surface potential is ,Vs = 5.30e-01 V\n",
+ "The threshold voltage is ,VT = 5.86e-01 V\n",
+ "The saturation voltage is ,VDS = 4.41e+00 V\n",
+ "The saturation current is ,ID = 1.18e-02 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Ex9.9:pg-402"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "VDS = 0.1\n",
+ "Z = 10*10**-4\n",
+ "L = 2*10**-4\n",
+ "Cox=10**-7\n",
+ "ID1= 50.0\n",
+ "ID2= 80.0\n",
+ "VGS1 = 1.5\n",
+ "VGS2 = 2.5\n",
+ "mu_n = (((ID2-ID1)*10**(-6)*L)/(VDS*Z*Cox*(VGS2-VGS1)))\n",
+ "print\"The mobility of electron in silicon is ,mu_n =\",round(mu_n,2),\" cm**2/Vs\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The mobility of electron in silicon is ,mu_n = 600.0 cm**2/Vs\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex9.10:pg-403"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kbT = 0.026\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "Na=2*10**16\n",
+ "ni = 1.5*10**10\n",
+ "VSB = 1.0\n",
+ "apsilen_ox = 3.9*8.85*10**-14\n",
+ "dox = 500*10**-8\n",
+ "Cox = apsilen_ox/dox\n",
+ "print\"The oxide capicitance per unit area is ,Cox =\",\"{:.2e}\".format(Cox),\" F*cm**-2\"\n",
+ "phi_F= (-kbT*log(Na/ni))\n",
+ "print\"The potential phi_F=\",\"{:.2e}\".format(phi_F),\" V\"\n",
+ "dVT = ((sqrt(2*e*apsilen*Na)/Cox)*((sqrt((-2*phi_F)+VSB)-sqrt(-2*phi_F))))\n",
+ "print\"The shift in threshold voltage is ,dVT =\",\"{:.2e}\".format(dVT),\" V\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The oxide capicitance per unit area is ,Cox = 6.90e-08 F*cm**-2\n",
+ "The potential phi_F= -3.67e-01 V\n",
+ "The shift in threshold voltage is ,dVT = 5.47e-01 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "\n",
+ "Ex9.11:pg-406"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "kbT = 0.026\n",
+ "apsilen = 11.9*8.85*10**-14\n",
+ "e = 1.6*10**-19\n",
+ "D = 10**-5\n",
+ "Na=10**14\n",
+ "dVT=.5\n",
+ "ni = 1.5*10**10\n",
+ "apsilen_ox = 3.9*8.85*10**-14\n",
+ "phi_F= (-kbT*log(Na/ni))\n",
+ "print\"The potential phi_F=\",\"{:.2e}\".format(phi_F),\" V\"\n",
+ "dox = 5*10**-6\n",
+ "Cox = apsilen_ox/dox\n",
+ "print\"The oxide capicitance per unit area is ,Cox =\",\"{:.2e}\".format(Cox),\" cm**-1\"\n",
+ "phi_ms = -0.83\n",
+ "VT = (phi_ms)-(2*phi_F)+((sqrt(4*e*apsilen*Na*(-phi_F)))/Cox)\n",
+ "print\"the threshold voltage is ,VT =\",\"{:.2e}\".format(VT),\" V\"\n",
+ "Na = (dVT*Cox)/(e*D)\n",
+ "print\"the dopant density is ,Na=\",\"{:.2e}\".format(Na),\" cm**-3\"\n",
+ "# Note : due to different precisions taken by me and the author ... my answer differ \n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential phi_F= -2.29e-01 V\n",
+ "The oxide capicitance per unit area is ,Cox = 6.90e-08 cm**-1\n",
+ "the threshold voltage is ,VT = -3.15e-01 V\n",
+ "the dopant density is ,Na= 2.16e+16 cm**-3\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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