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| author | Jovina Dsouza | 2014-06-18 12:43:07 +0530 |
|---|---|---|
| committer | Jovina Dsouza | 2014-06-18 12:43:07 +0530 |
| commit | 206d0358703aa05d5d7315900fe1d054c2817ddc (patch) | |
| tree | f2403e29f3aded0caf7a2434ea50dd507f6545e2 /Elements_of_Electromagnetics | |
| parent | c6f0d6aeb95beaf41e4b679e78bb42c4ffe45a40 (diff) | |
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diff --git a/Elements_of_Electromagnetics/README.txt b/Elements_of_Electromagnetics/README.txt new file mode 100644 index 00000000..d2a7204f --- /dev/null +++ b/Elements_of_Electromagnetics/README.txt @@ -0,0 +1,10 @@ +Contributed By: Vishal MV +Course: others +College/Institute/Organization: Indian Institute of Engineering Bombay +Department/Designation: Physics +Book Title: Elements of Electromagnetics +Author: M. N. O. Sadiku +Publisher: Oxford University Press +Year of publication: 2001 +Isbn: 195686233 +Edition: 3rd
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_1.ipynb b/Elements_of_Electromagnetics/chapter_1.ipynb new file mode 100644 index 00000000..ae4e75b3 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_1.ipynb @@ -0,0 +1,420 @@ +{
+ "metadata": {
+ "name": "chapter_1.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 1: Vector Analysis<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1.1, Page number: 8<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "If A= 1Oa_x - 4a_y + 6a_z and B == 2a_x + a_y, find: \n",
+ "(a) the component of A along a_y ,\n",
+ "(b) the magnitude of 3A - B,\n",
+ "(c) a unit vector along A + 2B. '''\n",
+ "\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "A=array([10,-4,6]) \n",
+ "B=array([2,1,0])\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Ay=dot(A,ay) #Component of A along y direction\n",
+ "l=scipy.sqrt(dot(3*A-B,3*A-B)) #Magnitude of the vector 3A-B\n",
+ "\n",
+ " #Defining the x,y and z components of the unit vector along A+2B\n",
+ " \n",
+ "ux=round(dot(A+2*B,ax)/scipy.sqrt(dot(A+2*B,A+2*B)),4)\n",
+ "uy=round(dot(A+2*B,ay)/scipy.sqrt(dot(A+2*B,A+2*B)),4)\n",
+ "uz=round(dot(A+2*B,az)/scipy.sqrt(dot(A+2*B,A+2*B)),4)\n",
+ "\n",
+ "u=array([ux,uy,uz])\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The component of A along y direction is',Ay\n",
+ "print 'Magnitude of 3A-B =',round(l,2)\n",
+ "print 'Unit vector along A+2B is',u"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The component of A along y direction is -4\n",
+ "Magnitude of 3A-B = 35.74\n",
+ "Unit vector along A+2B is [ 0.9113 -0.1302 0.3906]\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1.2, Page number: 9<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Points P and Q are located at (0, 2, 4) and (- 3, 1, 5). Calculate \n",
+ "(a) The position vector P \n",
+ "(b) The distance vector from P to Q \n",
+ "(c) The distance between P and Q \n",
+ "(d) A vector parallel to PQ with magntude of 10 '''\n",
+ "\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "P=array([0,2,4]) \n",
+ "Q=array([-3,1,5])\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "origin=array([0,0,0]) #Defining the origin\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "PosP=P-origin #The position vector P\n",
+ "Dpq=Q-P #The distance vector from P to Q\n",
+ "l=scipy.sqrt(dot(Dpq,Dpq)) #Magnitude of the distance vector from P to Q\n",
+ "\n",
+ " #Defining the x,y and z components of the unit vector along Dpq\n",
+ " \n",
+ "ux=round(dot(Dpq,ax)/l,4)\n",
+ "uy=round(dot(Dpq,ay)/l,4)\n",
+ "uz=round(dot(Dpq,az)/l,4)\n",
+ "\n",
+ "vect=array([ux*10,uy*10,uz*10]) #Vector parallel to PQ with magntude of 10\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The position vector P is',PosP\n",
+ "print 'The distance vector from P to Q is',Dpq\n",
+ "print 'The distance between P and Q is',round(l,3)\n",
+ "print 'Vector parallel to PQ with magntude of 10 is',vect"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The position vector P is [0 2 4]\n",
+ "The distance vector from P to Q is [-3 -1 1]\n",
+ "The distance between P and Q is 3.317\n",
+ "Vector parallel to PQ with magntude of 10 is [-9.045 -3.015 3.015]\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1.3, Page number: 10<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A river flows southeast at 10 km/hr and a boat flows upon it \n",
+ "with its bow pointed in the direction of travel. A man walks upon\n",
+ "the deck at 2 km/hr in a direction to the right and perpendicular\n",
+ "to the direction of the boat's movement. \n",
+ "Find the velocity of the man with respect to the earth. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "vriver=10 #Speed of river in km/hr\n",
+ "vman=2 #Speed of man relative to boat in km/hr\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ " #Velocity of boat\n",
+ "\n",
+ "Vboat=vriver*(scipy.cos(scipy.pi/4)*ax- \n",
+ " scipy.sin(scipy.pi/4)*ay) \n",
+ " \n",
+ " #Relative velocity of man with respect to boat\n",
+ "\n",
+ "Vrelman=vman*(-scipy.cos(scipy.pi/4)*ax- \n",
+ " scipy.sin(scipy.pi/4)*ay) \n",
+ " \n",
+ " #Absolute velocity of man\n",
+ " \n",
+ "Vabs=Vboat+Vrelman\n",
+ " \n",
+ "mag=scipy.sqrt(dot(Vabs,Vabs)) #Magnitude of the velocity of man\n",
+ "Vabsx=dot(Vabs,ax) #X component of absolute velocity\n",
+ "Vabsy=dot(Vabs,ay) #Y component of absolute velocity\n",
+ "angle=scipy.arctan(Vabsy/Vabsx)*180/scipy.pi #Angle made with east in degrees\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The velocity of the man is',round(mag,1),'km/hr at an angle of'\n",
+ "print -round(angle,1),'degrees south of east'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The velocity of the man is 10.2 km/hr at an angle of\n",
+ "56.3 degrees south of east\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1.4, Page number: 17<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Given vectors A = 3a_x + 4a_y + a_z and B = 2a_y - 5a_z , \n",
+ "find the angle between A and B. '''\n",
+ " \n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "A=array([3,4,1])\n",
+ "B=array([0,2,-5])\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "magA=scipy.sqrt(dot(A,A)) #Magnitude of A\n",
+ "magB=scipy.sqrt(dot(B,B)) #Magnitude of B\n",
+ "angle=scipy.arccos(dot(A,B)/(magA*magB)) #Angle between A and B in radians\n",
+ "angled=angle*180/scipy.pi #Angle between A and B in degrees\n",
+ "\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The angle between A and B =',round(angled,2),'degrees'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The angle between A and B = 83.73 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1.5, Page number: 17<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Three field quantities are given by \n",
+ "P = 2a_x - a_z \n",
+ "Q = 2a_x - a_y + 2a_z \n",
+ "R = 2a_x - 3a_y + a_z\n",
+ "\n",
+ "Determine \n",
+ "\n",
+ "(a) (P + Q) X (P - Q) \n",
+ "(b) Q.(R X P)\n",
+ "(c) P.(Q X R)\n",
+ "(d) Sin(theta_QR) \n",
+ "(e) P X (Q X R) \n",
+ "(f) A unit vector perpendicular to both Q and R \n",
+ "(g) The component of P along Q '''\n",
+ "\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "P=array([2,0,-1])\n",
+ "Q=array([2,-1,2])\n",
+ "R=array([2,-3,1])\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "ansa=cross((P+Q),(P-Q))\n",
+ "ansb=dot(Q,cross(R,P))\n",
+ "ansc=dot(P,cross(Q,R))\n",
+ "lqxr=scipy.sqrt(dot(cross(Q,R),cross(Q,R))) #Magnitude of QXR\n",
+ "lq=scipy.sqrt(dot(Q,Q)) #Magnitude of Q\n",
+ "lr=scipy.sqrt(dot(R,R)) #Magnitude of R\n",
+ "ansd=lqxr/(lq*lr)\n",
+ "anse=cross(P,cross(Q,R))\n",
+ "\n",
+ "#Finding unit vector perpendicular to Q and R\n",
+ "\n",
+ "ux=dot(cross(Q,R),ax)/lqxr #X component of the unit vector\n",
+ "uy=dot(cross(Q,R),ay)/lqxr #Y component of the unit vector\n",
+ "uz=dot(cross(Q,R),az)/lqxr #Z component of the unit vector\n",
+ "\n",
+ "ansf=array([round(ux,3),round(uy,3),round(uz,3)])\n",
+ "\n",
+ "ansg=round((float(dot(P,Q))/dot(Q,Q)),4)*Q\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print '(P+Q)X(P-Q) =',ansa\n",
+ "print 'Q.(R X P) =',ansb\n",
+ "print 'P.(Q X R) =',ansc\n",
+ "print 'Sin(theta_QR) =',round(ansd,4)\n",
+ "print 'P X (Q X R) =',anse\n",
+ "print 'A unit vector perpendicular to both Q and R =',ansf\n",
+ "print 'The component of P along Q =',ansg\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "(P+Q)X(P-Q) = [ 2 12 4]\n",
+ "Q.(R X P) = 14\n",
+ "P.(Q X R) = 14\n",
+ "Sin(theta_QR) = 0.5976\n",
+ "P X (Q X R) = [2 3 4]\n",
+ "A unit vector perpendicular to both Q and R = [ 0.745 0.298 -0.596]\n",
+ "The component of P along Q = [ 0.4444 -0.2222 0.4444]\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 1.7, Page number: 21<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Show that points P1(5,2,-4), P2(1,1,2), and P3( -3,0,8) all \n",
+ "lie on a straight line. Determine the shortest distance between \n",
+ "the line and point P4(3,- 1,0). '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "P1=array([5,2,-4])\n",
+ "P2=array([1,1,2])\n",
+ "P3=array([-3,0,8])\n",
+ "P4=array([3,-1,0])\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R12=P2-P1; #Distance vector from P1 to P2\n",
+ "R13=P3-P1; #Distance vector from P1 to P3\n",
+ "R14=P4-P1; #Distance vector from P1 to P4\n",
+ "s=cross(R12,R13)\n",
+ "x=cross(R14,R12)\n",
+ "d=scipy.sqrt(dot(x,x))/scipy.sqrt(dot(R12,R12)) #Distance from line to P4 \n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The cross product of the distance vectors R12 and R14 =',s\n",
+ "print 'So they are along same direction and hence P1, P2 and P3 are collinear.'\n",
+ "print 'The shortest distance from the line to point P4 =',round(d,3)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The cross product of the distance vectors R12 and R14 = [0 0 0]\n",
+ "So they are along same direction and hence P1, P2 and P3 are collinear.\n",
+ "The shortest distance from the line to point P4 = 2.426\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_10.ipynb b/Elements_of_Electromagnetics/chapter_10.ipynb new file mode 100644 index 00000000..ed7744f3 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_10.ipynb @@ -0,0 +1,566 @@ +{
+ "metadata": {
+ "name": "chapter_10.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 10: Electromagnetic Wave Propagation<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10.1, Page number: 416<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The elcctric field in free space is given by \n",
+ "E=50 cos(10^8t + Bx) a_y V/m \n",
+ "(a) Find the direction of wave propagation. \n",
+ "(b) Calculate B and the time it takes to travel a distance of lambda/2. \n",
+ "(c) Sketch the wave at t = 0, T/4, and T/2. '''\n",
+ "\n",
+ "import scipy\n",
+ "from pylab import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "w=10**8 \n",
+ "c=3.0*10**8\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=2*scipy.pi/w #timeperiod of the wave in sec\n",
+ "B=(w/c) #in rad/m\n",
+ "lam=2*scipy.pi/B #wavelength in m\n",
+ "t1=lam*10**9/(2*c) #time taken to travel half the wavelength in ns \n",
+ "\n",
+ "x=arange(-6*scipy.pi,6*scipy.pi,0.1)\n",
+ "\n",
+ "t=0\n",
+ "E=50*scipy.cos(10**8*t+x*w/c)\n",
+ "\n",
+ "subplot(3,1,1)\n",
+ "xlabel(\"x\")\n",
+ "ylabel(\"E for t=0\")\n",
+ "plot(x,E,'r')\n",
+ "\n",
+ "subplot(3,1,2)\n",
+ "t=T/4\n",
+ "E=50*scipy.cos(10**8*t+x*w/c)\n",
+ "xlabel(\"x\")\n",
+ "ylabel(\"E for t=T/4\")\n",
+ "plot(x,E)\n",
+ "\n",
+ "subplot(3,1,3)\n",
+ "t=T/2\n",
+ "E=50*scipy.cos(10**8*t+x*w/c)\n",
+ "xlabel(\"x\")\n",
+ "ylabel(\"E for t=T/2\")\n",
+ "plot(x,E,'g')\n",
+ "show()\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Since the argument of cosine function is positive, '\n",
+ "print 'the wave is propagating in the negative x direction.'\n",
+ "print' B =',round(B,4),'rad/m'\n",
+ "print 'Time taken to travel a distance of lambda/2 =',round(t1,2),'n sec'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "display_data",
+ "png": 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06tRJaFmVIqan8uoQ67WZk5ODMWPGYNWqVWjRooVi51NFy2A1kpubSxMnTiQz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\n",
+ "text": [
+ "<matplotlib.figure.Figure at 0x60587d0>"
+ ]
+ },
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Since the argument of cosine function is positive, \n",
+ "the wave is propagating in the negative x direction.\n",
+ " B = 0.3333 rad/m\n",
+ "Time taken to travel a distance of lambda/2 = 31.42 n sec\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10.2, Page number: 428<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A lossy dielectric has an intrinsic impedance of 200 /30degree ohms at\n",
+ "a particular frequency. If, at that frequency, the plane wave propagating \n",
+ "through the dielectric has the magnetic field component \n",
+ "\n",
+ "H = 10 e^-ax cos(wt-0.5x)a_y A/m \n",
+ "find E and a. Determine the skin depth and wave polarization. '''\n",
+ "\n",
+ "import cmath\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Ho=10 \n",
+ "n=200*scipy.exp(1)**(1j*scipy.pi/6) \n",
+ "b=0.5\n",
+ "\n",
+ "#Calclations\n",
+ "\n",
+ "Eo=n*Ho #amplitude of electric field in kV/m\n",
+ "P=scipy.arctan(scipy.sqrt(3)) \n",
+ "a=b*((scipy.sqrt(((1+(scipy.tan(P))**2)**0.5)-1))/(scipy.sqrt(((1+(scipy.tan(P)\n",
+ ")**2)**0.5)+1)))\n",
+ "delta=1/a\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'E has the same form as H except for amplitude and phase.'\n",
+ "print 'The amplitude and phase of E =',Eo,'kV/m'\n",
+ "print '= magnitude of 2000 and angle of pi/6'\n",
+ "print 'a =',round(a,4),'Np/m'\n",
+ "print 'Skin depth =',round(delta,3),'m'\n",
+ "print 'The polarization of wave is in z direction since it has an z component.'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "E has the same form as H except for amplitude and phase.\n",
+ "The amplitude and phase of E = (1732.05080757+1000j) kV/m\n",
+ "= magnitude of 2000 and angle of pi/6\n",
+ "a = 0.2887 Np/m\n",
+ "Skin depth = 3.464 m\n",
+ "The polarization of wave is in z direction since it has an z component.\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10.3, Page number: 430<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "In a loss]ess medium for which eta = 60 pi, mu_r = 1 and \n",
+ "H= -0.1 Cos(Wt - z)a_x + 0.5 sin(wt - z)a_y A/m, \n",
+ "calculate epison_r, w and E. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "B=1\n",
+ "n=60*scipy.pi \n",
+ "Ur=1 #relative permeability\n",
+ "Eo=10**-9/(36*scipy.pi) #permittivity of free space\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Er=Uo*Ur/(n**2*Eo) #relative permittivity\n",
+ "w=B/scipy.sqrt(Eo*Er*Uo*Ur) #in rad/sec\n",
+ "eps=Eo*Er #permittivity of the medium in Farad/m\n",
+ "H1o=-0.1\n",
+ "H2o=0.5\n",
+ "Ex=H2o/(eps*w) #amplitude of x component of E in V/m\n",
+ "Ey=H1o/(eps*w) #amplitude of y component of E in V/m\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'er =',Er\n",
+ "print 'w =',w,'rad/sec'\n",
+ "print 'E =',round(Ex,2),'sin(wt-z)ax +',round(-Ey,2),'cos(wt-z)ay V/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "er = 4.0\n",
+ "w = 150000000.0 rad/sec\n",
+ "E = 94.25 sin(wt-z)ax + 18.85 cos(wt-z)ay V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10.4, Page number: 432<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A uniform plane wave propagating in a medium has \n",
+ "E = 2e^(-az) sin(10^8t-Bz)a_y V/m. \n",
+ "If the medium is characterized by epsilon_r=1, mu_r=20 and sigma=3 mhos/m, \n",
+ "find a,B and H. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "E=2 #amplitude of E in V/m\n",
+ "sigma=3 #in mhos/m\n",
+ "w=10**8 #in rad/sec\n",
+ "Ur=20 #relative permeability\n",
+ "Eo=10**-9/(36*scipy.pi) #permittivity of free space in Farad/m\n",
+ "Er=1 #relative permittivity\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "a=round(scipy.sqrt(Uo*Ur*w*sigma/2),1) #in Np/m\n",
+ "B=a #rad/m\n",
+ "theta=scipy.arctan(sigma/(w*Eo*Er))*0.5 #in radians\n",
+ "thetad=round(theta*180/scipy.pi,0) #in degrees\n",
+ "H=E/(scipy.sqrt(Uo*Ur*w/sigma))*10**3 #amplitude of H in mA/m\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'alpha =',a,'Np/m'\n",
+ "print 'beta =',B,'rad/m'\n",
+ "print 'H =',round(H,1),'e^ (',a,'z ) sin(wt - Bz -',thetad,') mA/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "alpha = 61.4 Np/m\n",
+ "beta = 61.4 rad/m\n",
+ "H = 69.1 e^ ( 61.4 z ) sin(wt - Bz - 45.0 ) mA/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10.6, Page number: 434<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "For the copper coaxial cable of Figure 7.12, let a = 2 mm, b = 6 mm\n",
+ "and t = 1 mm. Calculate the resistance of 2 m length of the cablc at dc\n",
+ "and at 100 MHz. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ " \n",
+ "a=2*10**-3 #in m\n",
+ "b=6*10**-3 #in m \n",
+ "t=10**-3 #in m\n",
+ "l=2 #in m\n",
+ "c=5.8*10**7 #conductivity in seimens\n",
+ "f=100*10**6 #frequency in Hz\n",
+ "mu=4*scipy.pi*10**-7 #permeability of free space\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Ri=l/(c*scipy.pi*a*a) #dc resistance of inner cable in ohms\n",
+ "Ro=l/(c*scipy.pi*((b+t)**2-b**2)) #dc resistance of outer cable in ohms\n",
+ "Rdc=Ro+Ri #total dc resistance in ohms\n",
+ "\n",
+ "Ria=round(l/(2*scipy.pi*a)*scipy.sqrt(scipy.pi*f*mu/c),1)\n",
+ "Roa=round(l/(2*scipy.pi*b)*scipy.sqrt(scipy.pi*f*mu/c),4)\n",
+ "Rac=Ria+Roa #ac resistance in ohms\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Rdc =',round(Rdc*10**3,3),'m ohms'\n",
+ "print 'Rac =',round(Rac,4),'ohms'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rdc = 3.588 m ohms\n",
+ "Rac = 0.5384 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10.7, Page number: 439<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "In a nonmagnetic medium \n",
+ "E= 4 sin (2pi X 10^7t - 0.8x) a_z V/m \n",
+ "Find\n",
+ "(a) epsilon_r,eta\n",
+ "(b) The time-average power carried by the wave \n",
+ "(c) The total power crossing 100 cm^2 of plane 2x + y = 5 '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "a=0 #alpha in m^-1\n",
+ "b=0.8 #beta in m^-1\n",
+ "Eo=10**-9/(36*scipy.pi) #permittivity of free space in farad/m\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "Ur=1 #relative permeability of medium\n",
+ "w=2*scipy.pi*10**7 #omega in rad/s\n",
+ "Eamp=4 #amplitude of the field in V/m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Er=b**2/(Uo*Eo*w*w) #relative permittivity of the medium\n",
+ "n=scipy.sqrt(Uo/(Eo*Er)) #eta in ohms\n",
+ "Pav=Eamp**2/(2*n)*ax #average power in W/m^2\n",
+ "an=(2*ax+ay)/scipy.sqrt(5) #normal to the plane\n",
+ "S=100*10**-4*an #area in m^2\n",
+ "P=dot(Pav,S)*10**6 #power through the plane in micro W\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Er=',round(Er,2)\n",
+ "print 'eta= ',round(n,1),'ohms'\n",
+ "print 'The time-average power =',round(dot(Pav,ax)*10**3,0),'ax mW/m^2'\n",
+ "print 'The total power crossing 100 cm^2 of the plane =',round(P,2),'micro W'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Er= 14.59\n",
+ "eta= 98.7 ohms\n",
+ "The time-average power = 81.0 ax mW/m^2\n",
+ "The total power crossing 100 cm^2 of the plane = 725.0 micro W\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 10.10, Page number: 458<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "An EM wave travels in free space with the electric field component \n",
+ "E =:100 e^j(0.866x-0.5z) ax V/m \n",
+ "Determine \n",
+ "(a) w and lambda\n",
+ "(b) The magnetic field component \n",
+ "(c) The time average power in the wave '''\n",
+ "\n",
+ "import scipy \n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "kx=0 #in m^-1\n",
+ "ky=0.866 #in m^-1\n",
+ "kz=0.5 #in m^-1\n",
+ "Eo=10**-9/(36*scipy.pi) #permittivity of free space in farad/m\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "c=1/(scipy.sqrt(Uo*Eo)) #speed of light in m/s\n",
+ "kvect=kx*ax+ky*ay+kz*az #propogation vector in m^-1\n",
+ "Eo=100 #amplitude of electric field\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "k=round(scipy.sqrt(kx*kx+ky*ky+kz*kz),0) #magnitude of k in m^-1\n",
+ "w=k*c #omega in rad/sec\n",
+ "lam=2*scipy.pi/k #wavelength in m\n",
+ "Ho=cross(kvect,Eo*ax*10)/(Uo*w) #amplitude of magnetic field in mA/m\n",
+ "Hoy=round(dot(Ho,ay),2) #y component of Ho\n",
+ "Hoz=round(dot(Ho,az),1) #z component of Ho\n",
+ "Hr=array([0,Hoy,Hoz]) #Ho with components rounded off\n",
+ "P=Eo**2/(2*120*scipy.pi)*kvect #average power in W/m^2\n",
+ "Py=round(dot(P,ay),2) #y component of P\n",
+ "Pz=round(dot(P,az),3) #z component of P\n",
+ "Pr=array([0,Py,Pz]) #P with components rounded off\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'w =',w,'rad/sec'\n",
+ "print 'lambda =',round(lam,3),'m'\n",
+ "print 'The magnetic field component =',Hr,'e^j(0.866x-0.5z) mA/m'\n",
+ "print 'The time average power in the wave =',Pr,'W/m^2'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "w = 300000000.0 rad/sec\n",
+ "lambda = 6.283 m\n",
+ "The magnetic field component = [ 0. 1.33 -2.3 ] e^j(0.866x-0.5z) mA/m\n",
+ "The time average power in the wave = [ 0. 11.49 6.631] W/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 10.11, Page number: 459"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A uniform plane wave in air with \n",
+ "\u001f\u001d",
+ "E = 8 cos(wt - 4x - 3z) ay V/m \n",
+ "\u001f\u001d",
+ "is incident on a dielectric slah (z>= 0) with mur= 1.0\u001f epsilonr = 2.5\u001f ,sigma=0. Find \n",
+ "(a) The polarization of the wave \n",
+ "(b) The angle of incidence \n",
+ "(c) The reflected E field \n",
+ "(d) The transmitted H field '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "Ei=8 #incident wave amplitude\n",
+ "k=5 #propogation constant\n",
+ "Eo=10**-9/36*scipy.pi #permittivity of free space\n",
+ "Erel=2.5 #relative permittivity\n",
+ "muo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "mur=1 #relative permeability\n",
+ "c=3*10**8 #speed of light\n",
+ "etao=377\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "w=k*c #frequency in rad\n",
+ "theta=scipy.arctan(4/3.0) #angle of incidence in rad\n",
+ "eta1=etao\n",
+ "eta2=377/scipy.sqrt(2.5)\n",
+ "thetai=scipy.arcsin(sin(theta)/scipy.sqrt(2.5))\n",
+ "gamm=(eta2*cos(theta)-eta1*cos(thetai))/(eta2*cos(theta)+eta1*cos(thetai))\n",
+ "Er=Ei*gamm #reflected E field amplitude in V/m\n",
+ "kt=w*scipy.sqrt(mur*Erel)/c\n",
+ "tao=2*eta2*cos(theta)/((eta2*cos(theta)+eta1*cos(thetai)))\n",
+ "Et=tao*Ei*ay\n",
+ "Ht=cross((4*ax+6.819*az)/(eta2*kt),Et)*10**3\n",
+ "Htx=round(dot(Ht,ax),2)\n",
+ "Hty=round(dot(Ht,ay),2)\n",
+ "Htz=round(dot(Ht,az),2)\n",
+ "Htc=array([Htx,Hty,Htz]) #transmitted H field amplitude\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Polarisation is perpendicular polarization'\n",
+ "print 'Angle of incidence is ',round(180*theta/scipy.pi,2),'degrees'\n",
+ "print 'Er =',round(Er,3),'cos(',w,'t - 4x + 3z) V/m'\n",
+ "print 'Ht =',Htc,'cos(',w,'t - 4x - 6.819z) mA/m'\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Polarisation is perpendicular polarization\n",
+ "Angle of incidence is 53.13 degrees\n",
+ "Er = -3.112 cos( 1500000000 t - 4x + 3z) V/m\n",
+ "Ht = [-17.68 0. 10.37] cos( 1500000000 t - 4x - 6.819z) mA/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 25
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_11.ipynb b/Elements_of_Electromagnetics/chapter_11.ipynb new file mode 100644 index 00000000..28188144 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_11.ipynb @@ -0,0 +1,673 @@ +{
+ "metadata": {
+ "name": "chapter_11.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 11: Transmission Lines<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.1, Page number: 482<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "An air line has characteristic impedance of 70 ohms and phase constant\n",
+ "of 3 rad/m at 100 MHz. Calculate the inductance per meter and the \n",
+ "capacitance per meter of the line. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "R=0\n",
+ "G=0\n",
+ "a=0\n",
+ "Ro=70 #characteristic impedence in ohms\n",
+ "B=3 #phase constant in rad/sec\n",
+ "f=100*10**6 #frequency in Hz\n",
+ "w=2*scipy.pi*f #omega in rad/sec\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "C=B/(w*Ro) #capacitance in F/m\n",
+ "L=Ro*Ro*C #inductance in H/m\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'inductance per meter =',round(L*10**9,1),'nH/m'\n",
+ "print 'capacitance per meter =',round(C*10**12,1),'pF/m'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "inductance per meter = 334.2 nH/m\n",
+ "capacitance per meter = 68.2 pF/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.2, Page number: 483<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A distortion less line has Z_o = 60 ohms, a = 20 mNp/m, u = 0.6c, where c\n",
+ "is the speed of light in a vacuum. Find R, L, G, C and lambda at 100 MHz. '''\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Zo=60 #in ohms\n",
+ "a=20*10**-3 #in Np/m\n",
+ "u=0.6*3*10**8 #in m/sec\n",
+ "f=100*10**6 #in Hz\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R=a*Zo #resistance in ohms/m\n",
+ "L=Zo/u #inductance in H/m\n",
+ "G=a*a/R #conductivity in S/m\n",
+ "C=1/(u*Zo) #capacitance in F/m\n",
+ "lam=u/f #wavelentgh in m\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'R =',R,'ohm/m'\n",
+ "print 'L =',round(L*10**9,0),'nH/m'\n",
+ "print 'G =',round(G*10**6,0),'micro S/m'\n",
+ "print 'C =',round(C*10**12,2),'pF/m'\n",
+ "print 'lambda =',lam,'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "R = 1.2 ohm/m\n",
+ "L = 333.0 nH/m\n",
+ "G = 333.0 micro S/m\n",
+ "C = 92.59 pF/m\n",
+ "lambda = 1.8 m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.3, Page number: 490<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A certain transmission line operating at w = 10^6 rad/sec has a=8 dB/m,\n",
+ "B = 1 rad/m, and Zo = 60 + j40 ohms, and is 2 m long. If the line is \n",
+ "connected to a source of 10/0 V, Zg = 40 ohms and terminated by a load of\n",
+ "20 + j50 ohms, deternline \n",
+ "\n",
+ "(a) The input impedance \n",
+ "(b) The sending-end current \n",
+ "(c) The current at the middle of the line '''\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "w=10**6 #omega in rad/sec\n",
+ "B=1 #phase factor in rad/m\n",
+ "a=8.0/8.686 #alpha in Np/m\n",
+ "Y=a+1j #in m^-1\n",
+ "l=2 #length in m\n",
+ "Vg=10 #source voltage in volts\n",
+ "Zo=60+40j #in ohms\n",
+ "Zg=40 #in ohms\n",
+ "Zl=20+50j #load impedance in ohms\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "s=scipy.tanh(Y*l)\n",
+ "Zin=Zo*(Zl+Zo*s)/(Zo+Zl*s) #input impedance in ohms\n",
+ "Zinr=round(Zin.real,2) #real part of Zin rounded to 2 decimal places\n",
+ "Zini=round(Zin.imag,2) #imaginary part of Zin rounded to 2 decimal places\n",
+ "Io=Vg/(Zin+Zg) #in A\n",
+ "absIo=round(abs(Io),6) #absolute value of Io rounded to 6 decimal place\n",
+ "Ior=Io.real #real part of Io\n",
+ "Ioi=Io.imag #imaginary part of Io\n",
+ "angIo=scipy.arctan(Ioi/Ior)*180/scipy.pi \n",
+ " #in degrees\n",
+ "Vo=Zin*Io\n",
+ "Vop=(Vo+Zo*Io)/2\n",
+ "Vom =(Vo-Zo*Io)/2\n",
+ "Im=((Vop*scipy.e**(-Y)/Zo))-((Vom*scipy.e**Y)/Zo)\n",
+ " #current at the middle in A\n",
+ "absIm=round(abs(Im),5) #absolute value of Im rounded to 6 decimal place\n",
+ "Imr=Im.real #real part of Im \n",
+ "Imi=Im.imag #imaginary part of Im\n",
+ "angIm=360+scipy.arctan(Imi/Imr)*180/scipy.pi \n",
+ " #in degrees\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The input impedance =',Zinr,'+',Zini,'j ohms'\n",
+ "print 'The sending-end current is'\n",
+ "print 'mod =',absIo*10**3,'mA, angle =',round(angIo,2),'degrees'\n",
+ "print 'The current at the middle is'\n",
+ "print 'mod =',absIm*10**3,'mA, angle =',round(angIm,0),'degrees'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The input impedance = 60.25 + 38.79 j ohms\n",
+ "The sending-end current is\n",
+ "mod = 93.03 mA, angle = -21.15 degrees\n",
+ "The current at the middle is\n",
+ "mod = 34.92 mA, angle = 281.0 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.4, Page number: 499<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A 30-m long loss less transmission line with Zo = 50 ohms operating at 2 MHz\n",
+ "is terminated with a load Zl = 60 + j40 ohms. If u = 0.6c on the line, find \n",
+ "\n",
+ "(a)The reflection coefficient r \n",
+ "(b)The standing wave ratio s \n",
+ "(c)The input impedance '''\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "l=30 #length in m\n",
+ "Zo=50 #in ohms\n",
+ "f=2*10**6 #frequency in Hz\n",
+ "Zl=60+40j #load impedence in ohms\n",
+ "u=0.6*3*10**8 #in m/s\n",
+ "w=2*scipy.pi*f #omega in rad/sec\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=(Zl-Zo)/(Zl+Zo) #reflection coefficient\n",
+ "ang=scipy.arctan(T.imag/T.real)*180/scipy.pi #argument of T is degrees\n",
+ "s=(1+abs(T))/(1-abs(T)) #standing wave ratio \n",
+ "B=w/u #propogation vector in m^-1\n",
+ "Zin=Zo*(Zl+Zo*scipy.tan(B*l)*1j)/(Zo+Zl*scipy.tan(B*l)*1j)\n",
+ "Zinr=round(Zin.real,2) #real part of Zin rounded to 2 decimal places\n",
+ "Zini=round(Zin.imag,2) #imaginary part of Zin rounded to 2 decimal places\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The reflection coefficient is'\n",
+ "print 'mod =',round(abs(T),4),'angle =',round(ang,0),'degrees'\n",
+ "print 'The standing wave ratio s =',round(s,3)\n",
+ "print 'The input impedance =',Zinr,'+',Zini,'j ohms'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The reflection coefficient is\n",
+ "mod = 0.3523 angle = 56.0 degrees\n",
+ "The standing wave ratio s = 2.088\n",
+ "The input impedance = 23.97 + 1.35 j ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.5, Page number: 501<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A 100 + j150-ohms load is connected to a 75-ohm lossless line. Find: \n",
+ "\n",
+ "(a) r \n",
+ "(b) s \n",
+ "(c) The load admittance Yl\n",
+ "(d) Zin at O.4 lambda from the load \n",
+ "(e) The locations of V max and V min with respect to the load if the line is\n",
+ "O.6 lambda long \n",
+ "(f) Zin at the generator '''\n",
+ "\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Zl=100+150j #load impedance in ohms\n",
+ "Zo=75 #impedance of line in ohms\n",
+ "B=2*scipy.pi \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "T=(Zl-Zo)/(Zl+Zo)\n",
+ "angT=scipy.arctan(T.imag/T.real)*180/scipy.pi \n",
+ "s=(1+abs(T))/(1-abs(T))\n",
+ "Yl=(1/Zl)*10**3 #admittance in mS\n",
+ "Ylr=round(Yl.real,2) #real part of Yl rounded to 2 decimal places\n",
+ "Yli=round(Yl.imag,2) #imaginary part of Yl rounded to 2 decimal places\n",
+ "l1=0.4\n",
+ "Zin=Zo*(Zl+Zo*scipy.tan(B*l1)*1j)/(Zo+Zl*scipy.tan(B*l1)*1j)\n",
+ "Zinr=round(Zin.real,2) #real part of Zin rounded to 2 decimal places\n",
+ "Zini=round(Zin.imag,2) #imaginary part of Zin rounded to 2 decimal places\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'r is mod =',round(abs(T),3),',angle =',round(angT,0),'degrees'\n",
+ "print 's =',round(s,2)\n",
+ "print 'The load admittance Yl =',Ylr,'+',Yli,'j mS'\n",
+ "print 'Zin at O.4 lambda from the load =',Zinr,'+',Zini,'j ohms'\n",
+ "#part (e) and (f) don't require computations"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "r is mod = 0.66 ,angle = 40.0 degrees\n",
+ "s = 4.88\n",
+ "The load admittance Yl = 3.08 + -4.62 j mS\n",
+ "Zin at O.4 lambda from the load = 21.96 + 47.61 j ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.6, Page number: 509<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "With an unknown load connected to a slotted air line, s = 2 is recorded by \n",
+ "a standing wave indicator and minima are found at 11 cm, 19 cm, . . . on \n",
+ "the scale. When the load is replaced by a short circuit, the minima are at\n",
+ "16 cm, 24 cm,. . . . If Zo = 50 ohms, calculate lambda,f, and ZL '''\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "s=2\n",
+ "l1=11 \n",
+ "l2=19\n",
+ "ma=24 \n",
+ "mi=16\n",
+ "u=3*10**8 #speed of wave in m/s\n",
+ "Zo=50 #in ohms\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "l=(l2-l1)*2 #lambda in cm\n",
+ "f=(u/l)*10**-7 #frequency in GHz\n",
+ "L=(24-19)/l #Let us assume load is at 24cm\n",
+ "zl=1.4+0.75j # by smith chart\n",
+ "Zl=Zo*zl #ZL in ohms\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'lambda =',l,'cm'\n",
+ "print 'f =',f,'GHz'\n",
+ "print 'ZL =',Zl,'ohms'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "lambda = 16 cm\n",
+ "f = 1.875 GHz\n",
+ "ZL = (70+37.5j) ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.7, Page number: 510<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Antenna with impedance 40 + j30 ohms is to be matched to a 100-ohm lossless \n",
+ "line with a shorted stub. Determine \n",
+ "(a) The required stub admittance \n",
+ "(b) The distance between the stub and the antenna \n",
+ "(c) The stub length \n",
+ "(d) The standing wave ratio on each ratio of the system '''\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Zo=100 #in ohms\n",
+ "Zl=40+30j #in ohms\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Yo=1.0/Zo #in S\n",
+ "yl=Zo/Zl\n",
+ "ys1=1.04j #By smith chart\n",
+ "ys2=-1.04j #By smith chart\n",
+ "Ys1=Yo*ys1 #in S\n",
+ "Ys2=Yo*ys2 #in S\n",
+ "la=round(0.5-(62-(-39))/720.0,2) #in units of lambda\n",
+ "lb=round((62-39)/720.0,3) #in units of lambda\n",
+ "da=round(88/720.0,4) #in units of lambda\n",
+ "db=round(272/720.0,4) #in units of lambda\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The required stub admittance values in mS are',Ys1*1000,'and',Ys2*1000\n",
+ "print 'The distance between stub and antenna at A =',la,'in units of lambda'\n",
+ "print 'The distance between stub and antenna at B =',lb,'in units of lambda'\n",
+ "print 'The stub lengths =',da,'and',db,'in units of lambda'\n",
+ "print 'Part (d) is done using smith chart'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The required stub admittance values in mS are 10.4j and -10.4j\n",
+ "The distance between stub and antenna at A = 0.36 in units of lambda\n",
+ "The distance between stub and antenna at B = 0.032 in units of lambda\n",
+ "The stub lengths = 0.1222 and 0.3778 in units of lambda\n",
+ "Part (d) is done using smith chart\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 11.9, Page number: 521"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A 75 ohm transmission line of length 60 m is terminated by a 100 ohm load. If a rectangular \n",
+ "pulse of width 5 micro sec and magnitude 4V is sent out by the generator connected to the line, \n",
+ "sketch I(0, t) and I( l, t) for 0 < t < 15 microS. Take Zg = 25 ohm and u = 0.1 c. '''\n",
+ "\n",
+ "import scipy\n",
+ "import matplotlib.pyplot as plt\n",
+ "\n",
+ "#Variable Declarataion\n",
+ "\n",
+ "zo=75 #in ohms\n",
+ "zg=25 #in ohms\n",
+ "zl=100 #in ohms\n",
+ "vg=4 #in volts\n",
+ "l=60 #in m\n",
+ "c=3*10**8 #speed of light in m/s\n",
+ "u=0.1*c #in m/s\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "gammag=(zg-zo)/(zg+zo)\n",
+ "gammal=(zl-zo)/(zl+zo)\n",
+ "Vo=zo*vg/(zo+zg) #in V\n",
+ "t1=l/u #in micro sec\n",
+ "Io=vg/(zo+zg) #in mA\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "t1=[0,4,5,8,9,12,13,15]\n",
+ "I1=[40,31.43,-8.571,-7.959,0.6123,0.5685,-0.0438,-0.438]\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.step(t1,I1,where='post')\n",
+ "ax.set_xlabel('Time (micro s)')\n",
+ "ax.set_ylabel(r'I(0,t) in mA')\n",
+ "plt.show()\n",
+ "\n",
+ "t2=[0,2,6,7,10,11,14]\n",
+ "I2=[0,34.3,31.9,-2.46,-2.28,0.176,0.176]\n",
+ "fig = plt.figure()\n",
+ "ax = fig.add_subplot(111)\n",
+ "ax.step(t2,I2,where='post')\n",
+ "ax.set_xlabel('Time (micro s)')\n",
+ "ax.set_ylabel(r'I(l,t) in mA')\n",
+ "plt.show()\n",
+ "\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x5f89c90>"
+ ]
+ },
+ {
+ "output_type": "display_data",
+ "png": 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+ "text": [
+ "<matplotlib.figure.Figure at 0x6342070>"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.10, Page number: 527<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A certain microstrip line has fused quartz (Er=3.8) as a substrate. \n",
+ "If the ratio of line width to substrate thickness is w/h = 4.5, determine \n",
+ "\n",
+ "(a) The effective relative permittivity of the substrate \n",
+ "(b) The characteristic impedance of the line \n",
+ "(c) The wavelength of the line at 10 GHz '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Er=3.8 #relative permittivity\n",
+ "c=3*10**8 #speed of wave in m/s\n",
+ "r=4.5 #ratio of line width to substrate thickness\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Eeff=((Er+1)/2)+((Er-1)/(2*(1+12/r)**0.5))\n",
+ "Zo=(120*scipy.pi)/((r+1.393+(0.667*scipy.log(r+1.444)))*((Eeff)**0.5))\n",
+ "f=10**10\n",
+ "l=c/(f*scipy.sqrt(Eeff))\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The effective relative permittivity of the substrate =',round(Eeff,3)\n",
+ "print 'The characteristic impedance of the line =',round(Zo,2),'ohms'\n",
+ "print 'The wavelength of the line at 10 GHz =',round(l*1000,2),'mm'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The effective relative permittivity of the substrate = 3.131\n",
+ "The characteristic impedance of the line = 30.08 ohms\n",
+ "The wavelength of the line at 10 GHz = 16.95 mm\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 11.11, Page number: 527<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "At 10 GHz, a microstrip line has the following parameters: \n",
+ "h = 1 mm \n",
+ "w = 0.8 mm \n",
+ "Er = 6.6 \n",
+ "tan(theta) = 10^-4 \n",
+ "sigma_r = 5.8 X 10^7 S/m \n",
+ "Calculate the attenuation due to conduction loss and dielectric loss. '''\n",
+ "\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "h=1 #in mm\n",
+ "w=0.8 #in mm\n",
+ "Er=6.6 #relative permittivity\n",
+ "P=scipy.arctan(0.0001) \n",
+ "c=5.8*10**7 #conductivity in S/m\n",
+ "f=10**10 #frequency in Hz\n",
+ "mu=4*scipy.pi*10**-7 #permeability of free space\n",
+ "C=3*10**8 #speed of wave in m/s\n",
+ "r=w/h\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Ee=((Er+1)/2.0)+((Er-1)/(2.0*(1+12/r)**0.5))\n",
+ "Zo=(120.0*scipy.pi)/((r+1.393+(0.667*scipy.log(r+1.444)))*((Ee)**0.5))\n",
+ "Rs=scipy.sqrt((scipy.pi*f*mu)/c)\n",
+ "ac=8.686*Rs/(w*(10**-3)*Zo)\n",
+ "l=C/(f*(Ee)**0.5)\n",
+ "ad=27.3*(Ee-1)*Er*scipy.tan(P)/((Er-1)*Ee*l)\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'attenuation due to conduction loss =',round(ac,2),'dB/m'\n",
+ "print 'attenuation due to dielectric loss =',round(ad,3),'dB/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "attenuation due to conduction loss = 4.35 dB/m\n",
+ "attenuation due to dielectric loss = 0.177 dB/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_12.ipynb b/Elements_of_Electromagnetics/chapter_12.ipynb new file mode 100644 index 00000000..b6ded6a6 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_12.ipynb @@ -0,0 +1,452 @@ +{
+ "metadata": {
+ "name": "chapter_12.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 12: Waveguides<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 12.1, Page number: 557<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "from __future__ import division\n",
+ "'''\n",
+ "A rectangular waveguide with dimensions a = 2.5 cm, b = 1 cm is to \n",
+ "operate below 15.1 GHz. How many TE and TM modes can the waveguide transmit\n",
+ "if the guide is filled with a medium characterized by sigma = 0, epsilon = 4\n",
+ "epsilon_o,mu_r = 1? Calculate the cutoff frequencies of the modes. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "a=2.5*10**-2 #in m\n",
+ "b=1*10**-2 #in m\n",
+ "c=0\n",
+ "Ur=1 #relative permeability\n",
+ "Er=4 #relative permittivity\n",
+ "C=3*10**8 #speed of wave in m/s\n",
+ "fc=0\n",
+ "m=0\n",
+ "n=0\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "while (fc*10**-9 < 15.1) :\n",
+ " fc = (C/(4*a))*scipy.sqrt(m**2+(a*n/b)**2)\n",
+ " if (( fc*10**-9) < 15.1) :\n",
+ " n=n+1\n",
+ " else:\n",
+ " print 'Maximum value of n is ',n-1\n",
+ "\n",
+ "nmax=n-1 \n",
+ "fc=0\n",
+ "m=0\n",
+ "n=0\n",
+ "while(fc*10**-9 < 15.1):\n",
+ " fc =(C/(4*a))*scipy.sqrt(m**2+(a*n/b)**2)\n",
+ " if((fc*10**-9) < 15.1):\n",
+ " m=m+1\n",
+ " else:\n",
+ " print 'Maximum value of m is ',m-1 \n",
+ "\n",
+ "mmax=m-1\n",
+ "m=0\n",
+ "while(m<mmax+1):\n",
+ " n=0\n",
+ " while(n<nmax+1):\n",
+ " p=(C/(4*a))*scipy.sqrt(m**2+(a*n/b)**2)\n",
+ " if((p*10**-9) < 15.1) :\n",
+ " print m,n,'transmission mode is possible'\n",
+ " print 'frequency is',round(p*10**-9,2),'GHz'\n",
+ " else:\n",
+ " print m,n,'transmission mode is not possible'\n",
+ " n=n+1\n",
+ " \n",
+ " m=m+1\n",
+ " "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Maximum value of n is 2\n",
+ "Maximum value of m is 5\n",
+ "0 0 transmission mode is possible\n",
+ "frequency is 0.0 GHz\n",
+ "0 1 transmission mode is possible\n",
+ "frequency is 7.5 GHz\n",
+ "0 2 transmission mode is possible\n",
+ "frequency is 15.0 GHz\n",
+ "1 0 transmission mode is possible\n",
+ "frequency is 3.0 GHz\n",
+ "1 1 transmission mode is possible\n",
+ "frequency is 8.08 GHz\n",
+ "1 2 transmission mode is not possible\n",
+ "2 0 transmission mode is possible\n",
+ "frequency is 6.0 GHz\n",
+ "2 1 transmission mode is possible\n",
+ "frequency is 9.6 GHz\n",
+ "2 2 transmission mode is not possible\n",
+ "3 0 transmission mode is possible\n",
+ "frequency is 9.0 GHz\n",
+ "3 1 transmission mode is possible\n",
+ "frequency is 11.72 GHz\n",
+ "3 2 transmission mode is not possible\n",
+ "4 0 transmission mode is possible\n",
+ "frequency is 12.0 GHz\n",
+ "4 1 transmission mode is possible\n",
+ "frequency is 14.15 GHz\n",
+ "4 2 transmission mode is not possible\n",
+ "5 0 transmission mode is possible\n",
+ "frequency is 15.0 GHz\n",
+ "5 1 transmission mode is not possible\n",
+ "5 2 transmission mode is not possible\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 12.3, Page number: 561<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "In a rectangular waveguide for which a = 1.5 cm, b = 0.8 cm, sigma = 0, \n",
+ "mu = mu_o and epsilon = 4epsilon_o,\n",
+ "\n",
+ "Hx=2sin(pi x/a) cos(3pi y/b)sin(pi X 10^11t - Bz) A/m\n",
+ "\n",
+ "Determine \n",
+ "(a) The mode of operation \n",
+ "(b) The cutoff frequency \n",
+ "(c) The phase constant B \n",
+ "(d) The propagation constant gamma\n",
+ "(e) The intrinsic wave impedance eta '''\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "a=1.5*10**-2 #in m\n",
+ "b=0.8*10**-2 #in m\n",
+ "c=0\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "Ur=1 #relative permeability\n",
+ "Eo=10**-9/(36*scipy.pi) #permittivity of free space\n",
+ "Er=4 #relative permittivity\n",
+ "C=3*10**8 #speed of light in m/s\n",
+ "w=scipy.pi*10**11 #omega in rad/s\n",
+ "m=1\n",
+ "n=3\n",
+ "u=C/2 #speed of wave in m/s\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "f=w/(2*scipy.pi) #frequency of wave in Hz\n",
+ "fc=u*((m*m)/(a*a)+(n*n)/(b*b))**0.5/2 #cutoff frequency in Hz\n",
+ "B=w*scipy.sqrt(1-(fc/f)**2)/u #phase constant in rad/m\n",
+ "eta=377/scipy.sqrt(Er)*scipy.sqrt(1-(fc/f)**2) #intrinsic wave impedance in ohm\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The cutoff frequency =',round(fc*10**-9,2),'GHz'\n",
+ "print 'The phase constant =',round(B,2),'rad/m'\n",
+ "print 'The propagation constant =',round(B,2),'j /m'\n",
+ "print 'The intrinsic wave impedance =',round(eta,1),'ohms'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The cutoff frequency = 28.57 GHz\n",
+ "The phase constant = 1718.93 rad/m\n",
+ "The propagation constant = 1718.93 j /m\n",
+ "The intrinsic wave impedance = 154.7 ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 12.4, Page number: 565<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A standard air-filled rectangular waveguide with dimensions a = 8.636 cm,\n",
+ "b = 4.318 cm is fed by a 4-GHz carrier from a coaxial cable. Determine if a \n",
+ "TE_10 mode will be propagated. If so, calculate the phase velocity and \n",
+ "the group velocity. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "a=8.636*10**-2 #in m\n",
+ "b=4.318*10**-2 #in m\n",
+ "f=4*10**9 #in Hz\n",
+ "u=3*10**8 #speed of wave in m/s\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "fc=u/(2*a)\n",
+ "if(f>fc):\n",
+ " print 'As f>fc, TE10 mode will propogate'\n",
+ "else:\n",
+ " print 'It will not propogate'\n",
+ "\n",
+ "Up=u/scipy.sqrt(1-(fc/f)**2) #phase velocity in m/s\n",
+ "Ug=u*u/Up #group velocity in m/s\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Phase velocity =',round(Up*10**-6,0),'Mm/s'\n",
+ "print 'Group velocity =',round(Ug*10**-6,1),'Mm/s'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "As f>fc, TE10 mode will propogate\n",
+ "Phase velocity = 333.0 Mm/s\n",
+ "Group velocity = 270.2 Mm/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 12.5, Page number: 570<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "An air-filled rectangular waveguide of dimensions a = 4 cm, b = 2 cm \n",
+ "transports energy in the dominant mode at a rate of 2 mW. If the frequency of\n",
+ "operation is lO GHz. Determine the peak value of the electric field \n",
+ "in the waveguide. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "f=10*10**9 #frequency of operation in Hz\n",
+ "a=4*10**-2 #in m\n",
+ "b=2*10**-2 #in m\n",
+ "u=3*10**8 #velocity in m/s\n",
+ "Pavg=2*10**-3 #average power in W\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "fc=u/(2*a) #cutoff frequency in Hz\n",
+ "n=377/scipy.sqrt(1-(fc/f)**2) #intrinsic wave impedance in ohms\n",
+ "E=scipy.sqrt(4*n*Pavg/(a*b)) #peak value of electric field in V/m\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'Peak value of electric field =',round(E,2),'V/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Peak value of electric field = 63.77 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 12.6, Page number: 571<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A copper-plated waveguide (sigma_e = 5.8 X 10 7 S/m) operating at 4.8 GHz \n",
+ "is supposed to deliver a minimum power of 1.2 kW to an antenna. If the guide\n",
+ "is fillcd with polystyrene (sigma = 10^17 S/m, epsilon = 2.55 epsilon_o) and\n",
+ "its dimensions are a = 4.2 cm, b = 2.6 cm, calculate the power dissipated in \n",
+ "a length 60 cm of the guide in the TE_10 mode. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable declaration\n",
+ "\n",
+ "cc=5.8*10**7 #in S/m\n",
+ "f=4.8*10**9 #in Hz\n",
+ "c=10**-17 #in S/m\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "Eo=10**-9/(36*scipy.pi) #permittivity of free space\n",
+ "Er=2.55 #relative permittivity\n",
+ "z=60*10**-2 #in m\n",
+ "l=4.2*10**-2 #in m\n",
+ "b=2.6*10**-2 #in m\n",
+ "P=1.2*10**3 #in W\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "n=377/scipy.sqrt(Er)\n",
+ "u=3*10**8/scipy.sqrt(Er)\n",
+ "fc=u/(2*l)\n",
+ "ad=c*n/(2*scipy.sqrt(1-(fc/f)**2))\n",
+ "Rs=scipy.sqrt(scipy.pi*f*Uo/cc)\n",
+ "ac=2*Rs*(0.5+(b/l)*(fc/f)**2)/(b*n*scipy.sqrt(1-(fc/f)**2))\n",
+ "a=ac\n",
+ "Pd=P*(scipy.e**(2*a*z)-1)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'power dissipated =',round(Pd,3),'W'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "power dissipated = 6.096 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 12.8, Page number: 579<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "An air-filled resonant cavity with dimensions a = 5 cm, b = 4 cm, and \n",
+ "c = 10 cm is made of copper (sigma_e = 5.8 X 10^7 mhos/m). Find \n",
+ "(a) The five lowest order modes \n",
+ "(b) The quality factor for TE_101 mode '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ " \n",
+ "a=5*10**-2 #in m\n",
+ "b=4*10**-2 #in m\n",
+ "c=10*10**-2 #in m\n",
+ "C=5.8*10**7 #in mhos/m\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "u=3*10**8 #speed of wave in m/s\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "def f(m,n,p):\n",
+ " fr=scipy.sqrt((m/a)**2+(n/b)**2+(p/c)**2)*u/2 #resonant frequency in Hz\n",
+ " print round(fr*10**-9,3)\n",
+ " \n",
+ "\n",
+ "f101=3.35*10**9\n",
+ "d=scipy.sqrt(1/(scipy.pi*f101*Uo*C))\n",
+ "Q=(a*a+c*c)*a*b*c/(d*(2*b*(a**3+c**3)+a*c*(a*a+c*c))) #quality factor\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Thus the five lowest order modes in ascending order are '\n",
+ "print 'TE101, frequency in GHz ='\n",
+ "f(1,0,1)\n",
+ "print 'TE011, frequency in GHz ='\n",
+ "f(0,1,1)\n",
+ "print 'TE102, frequency in GHz ='\n",
+ "f(1,0,2)\n",
+ "print 'TE110, frequency in GHz ='\n",
+ "f(1,1,0)\n",
+ "print 'TE111 or TM111, frequency in GHz ='\n",
+ "f(1,1,1)\n",
+ "print 'Quality factor =',round(Q,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Thus the five lowest order modes in ascending order are \n",
+ "TE101, frequency in GHz =\n",
+ "3.354\n",
+ "TE011, frequency in GHz =\n",
+ "4.039\n",
+ "TE102, frequency in GHz =\n",
+ "4.243\n",
+ "TE110, frequency in GHz =\n",
+ "4.802\n",
+ "TE111 or TM111, frequency in GHz =\n",
+ "5.031\n",
+ "Quality factor = 14358.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_13.ipynb b/Elements_of_Electromagnetics/chapter_13.ipynb new file mode 100644 index 00000000..0aaf4534 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_13.ipynb @@ -0,0 +1,428 @@ +{
+ "metadata": {
+ "name": "chapter_13.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 13: Antennas<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 13.1, Page number: 601<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A magnetic field strength of 5 muA/m is required at a point on theta = pi/2,\n",
+ "2 km from an antenna in air. Neglecting ohmic loss, how much power must \n",
+ "the antenna transmit if it is \n",
+ "\n",
+ "(a) A Hertzian dipole of length lambda/25?\n",
+ "(b) A half-wave dipole? \n",
+ "(c) A quarter-wave monopole? \n",
+ "(d) A 10-turn loop antenna of radius Po = lambda/20? '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "H=5*10**-6 #magnetic field strentgh in A/m\n",
+ "theta=scipy.pi/2 \n",
+ "r=2*10**3 #distance in m\n",
+ "Bdl=2*scipy.pi/25\n",
+ "N=10 #number of turns\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Ia=4*scipy.pi*r*H/(Bdl*scipy.sin(theta)) #current for part (a) in A\n",
+ "Pa=40*scipy.pi**2*(1/25.0)**2*Ia**2 #power for part (a) in W\n",
+ "def pow(Io,Rrad):\n",
+ " P=0.5*Io**2*Rrad\n",
+ " print round(P*10**3,0),'mW'\n",
+ "\n",
+ "denom=scipy.cos(scipy.pi*scipy.cos(theta)/2) \n",
+ "Ib=H*2*scipy.pi*r*scipy.sin(theta)/denom #current for part (b) in A\n",
+ "Rradb=73 #wave impedance in ohms for (b)\n",
+ "Ic=Ib #current for part (c) in A\n",
+ "Rradc=36.56 #wave impedance in ohms for (c)\n",
+ "Id=H*r*400/(10*scipy.pi**2) #current for part (d) in A\n",
+ "Rradd=320*scipy.pi**6*N**2/20**4 #wave impedance in ohms for (d)\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The power transmitted in mW if antenna is ;'\n",
+ "print '(a) A Hertzian dipole of length lambda/25 =','\\n',round(Pa*10**3,0),'mW'\n",
+ "print '(b) A half-wave dipole ='\n",
+ "pow(Ib,Rradb)\n",
+ "print '(c) A quarter-wave monopole ='\n",
+ "pow(Ic,Rradc)\n",
+ "print '(d) A 10-turn loop antenna of radius Po = lambda/20 ='\n",
+ "pow(Id,Rradd)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The power transmitted in mW if antenna is ;\n",
+ "(a) A Hertzian dipole of length lambda/25 = \n",
+ "158.0 mW\n",
+ "(b) A half-wave dipole =\n",
+ "144.0 mW\n",
+ "(c) A quarter-wave monopole =\n",
+ "72.0 mW\n",
+ "(d) A 10-turn loop antenna of radius Po = lambda/20 =\n",
+ "158.0 mW\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 13.2, Page number: 603<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "An electric field strength of 10 micro V/m is to be measured at an observation \n",
+ "point theta=pi/2, 500 km from a half-wave (resonant) dipole antenna operating\n",
+ "in air at 50 MHz. \n",
+ "(a) What is the length of the dipole? \n",
+ "(b) Calculate the current that must be fed to the antenna. \n",
+ "(c) Find the average power radiated by the antenna. \n",
+ "(d) If a transmission line with Zo = 75 ohms is connected to the antenna,\n",
+ "determine the standing wave ratio. '''\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "c=3*10**8 #speed of wave in m/s\n",
+ "f=50*10**6 #frequency in Hz\n",
+ "E=10*10**-6 #field strength in V/m\n",
+ "theta=scipy.pi/2\n",
+ "r=500*10**3 #distance in m\n",
+ "eta=120*scipy.pi #wave impedance in ohms\n",
+ "Rrad=73 #in ohms\n",
+ "Zo=75 #in ohms\n",
+ "Zl=73+42.5j\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "l=c/(2*f)\n",
+ "I=E*2*r*scipy.pi*sin(theta)/(eta*(cos((scipy.pi/2)*cos(theta))))\n",
+ "P=0.5*I**2*Rrad\n",
+ "T=(Zl-Zo)/(Zl+Zo)\n",
+ "s=(1+abs(T))/(1-abs(T))\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The length of the dipole =',l,'m'\n",
+ "print 'The current that must be fed to the antenna =',round(I*10**3,2),'mA'\n",
+ "print 'The average power radiated by the antenna =',round(P*10**3,1),'mW'\n",
+ "print 'The standing wave ratio =',round(s,4)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The length of the dipole = 3 m\n",
+ "The current that must be fed to the antenna = 83.33 mA\n",
+ "The average power radiated by the antenna = 253.5 mW\n",
+ "The standing wave ratio = 1.7636\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 13.4, Page number: 610<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Determine the electric field intensity at a distance of 10 km from an antenna \n",
+ "having a directive gain of 5 dB and radiating a total power of 20 kW. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "G=5\n",
+ "r=10*10**3 #in m\n",
+ "P=20*10**3 #power in W\n",
+ "n=120*scipy.pi #wave impedance in ohms\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Gd=10**(G/10.0)\n",
+ "E=scipy.sqrt(n*Gd*P/(2*scipy.pi*r*r)) #field intensity in V/m\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'electric field intensity =',round(E,4),'V/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "electric field intensity = 0.1948 V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 13.5, Page number: 611<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The radiation intensity of a certain antenna is \n",
+ "\n",
+ "U(theta,phi) = 2sin(theta) sin^3(phi) , 0<theta<pi,0<phi<pi\n",
+ " = 0, elsewhere\n",
+ "\n",
+ "Determine the directivity of the antenna. '''\n",
+ "\n",
+ "import scipy\n",
+ "import scipy.integrate\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Umax=2.0\n",
+ "\n",
+ "def U(phi,theta):\n",
+ " s=2*scipy.sin(theta)*(scipy.sin(phi))**3/(4.0*scipy.pi)\n",
+ " return s\n",
+ " \n",
+ "#Calculations\n",
+ "\n",
+ "if __name__ == '__main__':\n",
+ " \n",
+ " Uav,er=scipy.integrate.dblquad(lambda theta,phi:U(phi,theta)*scipy.sin(theta), \n",
+ " 0, scipy.pi, lambda theta: 0, lambda theta: scipy.pi)\n",
+ "\n",
+ "D=Umax/Uav #Directivity\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'directivity of the antenna =',D"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "directivity of the antenna = 6.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 13.8, Page number: 624<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Find the maximum effective area of a lamba/2 wire dipole operating at 30 MHz.\n",
+ "How much power is received with an incident plane wave of strength 2 mV/m.'''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "c=3*10**8 #speed of wave in m/s\n",
+ "f=30*10**6 #frequency in Hz\n",
+ "E=2*10**-3 #field strength in V/m\n",
+ "n=120*scipy.pi\n",
+ "R=73 \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "l=c/f #wavelength in m\n",
+ "Gdmax=round(n/(scipy.pi*R),2) \n",
+ "Amax=(l**2/(4*scipy.pi))*Gdmax #maximum effective area in m^2\n",
+ "Pr=(E*E*Amax)/(2*n) #power received in W\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'maximum effective area =',round(Amax,2),'m^2'\n",
+ "print 'power received =',round(Pr*10**9,2),'nW'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "maximum effective area = 13.05 m^2\n",
+ "power received = 69.24 nW\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 13.9, Page number: 624<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The transmitting and receiving antennas are separated by a distance of \n",
+ "200 lambda and have directive gains of 25 and 18 dB, respectively. If 5 mW\n",
+ "of power is to be received, calculate the minimum transmitted power. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Gt=25 #in dB\n",
+ "Gr=18 #in dB\n",
+ "r=200 #in units of lambda\n",
+ "Pr=5*10**-3 #power received in W\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Gdt=10**(Gt/10.0) \n",
+ "Gdr=10**(Gr/10.0)\n",
+ "Pt=Pr*(4*scipy.pi*r)**2/(Gdr*Gdt)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'minimum transmitted power =',round(Pt,3),'W'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "minimum transmitted power = 1.583 W\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 13.10, Page number: 627<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "An S-band radar transmitting at 3 GHz radiates 200 kW. Determine the signal\n",
+ "power density at ranges 100 and 400 nautical miles if the effective area of \n",
+ "the radar antenna is 9 m^2 . With a 20-m^2 target at 300 nautical miles, \n",
+ "calculate the power of the reflected signal at the radar. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "c=3*(10)**8 #speed of wave in m/s\n",
+ "f=3.0*(10)**9 #frequency in Hz\n",
+ "Aet=9 #effective area in m^2\n",
+ "r1=1.852*(10)**5 #distance in m\n",
+ "r2=4*r1 #distance in m\n",
+ "r3=5.556*10**5 #distance in m\n",
+ "Pr=200*(10)**3 #in W\n",
+ "a=20 #target area in m^2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "l=c/f #wavelength in m\n",
+ "Gdt=4*scipy.pi*Aet/(l*l)\n",
+ "P1=Gdt*Pr/(4*scipy.pi*r1*r1) #power at 100 nmiles in W/m^2\n",
+ "P2=Gdt*Pr/(4*scipy.pi*r2*r2) #power at 400 nmiles in W/m^2\n",
+ "Pr=Aet*a*Gdt*Pr/(4*scipy.pi*r3*r3)**2 #power of reflected signal in W\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Signal power density at 100 nautical miles =',round(P1*1000,3),'mW/m^2'\n",
+ "print 'Signal power density at 400 nautical miles =',round(P2*1000,3),'mW/m^2'\n",
+ "print 'Power of reflected signal =',round(Pr*10**12,5),'pico W'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Signal power density at 100 nautical miles = 5.248 mW/m^2\n",
+ "Signal power density at 400 nautical miles = 0.328 mW/m^2\n",
+ "Power of reflected signal = 0.02706 pico W\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_14.ipynb b/Elements_of_Electromagnetics/chapter_14.ipynb new file mode 100644 index 00000000..1a5d389b --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_14.ipynb @@ -0,0 +1,172 @@ +{
+ "metadata": {
+ "name": "chapter_14.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 14: Modern Topics<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 14.1, Page number: 643<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The following S-parameters are obtained for a microwave transistor operating \n",
+ "at 2.5 GHz: S11 = 0.85 / 30\u00b0 ,S12 = 0.07 /56\u00b0 , S21 = 1.68 /120\u00b0 , \n",
+ "S22 = O.85 /-40 . Determine the input reflection coefficient when ZL=Zo=75 ohm. '''\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "S11=0.85*scipy.e**(-30j*scipy.pi/180)\n",
+ "S12=0.07*scipy.e**(56j*scipy.pi/180)\n",
+ "S21=1.68*scipy.e**(120j*scipy.pi/180)\n",
+ "S22=0.85*scipy.e**(-40j*scipy.pi/180)\n",
+ "Zl=75 \n",
+ "Zo=75\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Tl=(Zl-Zo)/(Zl+Zo)\n",
+ "Ti=S11+(S12*S21*Tl)/(1-S22*Tl) #reflection coefficient\n",
+ "Timod=abs(Ti) #mod of Ti\n",
+ "Tiang=scipy.arctan(Ti.imag/Ti.real)*180/scipy.pi #argument of Ti in degrees\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'input reflection coefficient =',Timod,'/',Tiang,'degrees'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "input reflection coefficient = 0.85 / -30.0 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 14.2, Page number: 654<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A step-index fiber has a core diameter of 80 micro m, a core refractive index \n",
+ "of 1.62, and a numerical aperture of 0.21. Calculate: \n",
+ "(a) the acceptance angle, (b) the refractive index that the fiber can \n",
+ "propagate at a wavelength of 0.8 micro m, (c) the number of modes that the\n",
+ "fiber can propagate at a wavelength of 0.8 micro m.'''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "d=80*(10)**-6 #diameter in m\n",
+ "n1=1.62 #core refractive index\n",
+ "NA=0.21 #numerical aperture\n",
+ "L=8*(10)**-7 #wavelength in m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "P=scipy.arcsin(NA)*180/scipy.pi #acceptance angle\n",
+ "n2=scipy.sqrt(n1**2-NA**2) #refractive index\n",
+ "V=(scipy.pi*d/L)*scipy.sqrt(n1**2-n2**2)\n",
+ "N=V**2/2 #number of modes\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Acceptance angle =',round(P,2),'degrees'\n",
+ "print 'Refractive index =',round(n2,3)\n",
+ "print 'No. of modes =',round(N,0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Acceptance angle = 12.12 degrees\n",
+ "Refractive index = 1.606\n",
+ "No. of modes = 2176.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 14.3, Page number: 655<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Light pulses propagate through a fiber cable with an attenuation of 0.25 dB/km.\n",
+ "Determine the distance through which the power of pulses is reduced by 40%. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "a=0.25 #in dB/km\n",
+ "P=1-0.4 #strength of pulse im %\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "l=(10/a)*scipy.log(1/P)/scipy.log(10) #distance in km\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'distance through which the power is reduced by 40% =',round(l,3),'km'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "distance through which the power is reduced by 40% = 8.874 km\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_2.ipynb b/Elements_of_Electromagnetics/chapter_2.ipynb new file mode 100644 index 00000000..042db1c8 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_2.ipynb @@ -0,0 +1,318 @@ +{
+ "metadata": {
+ "name": "chapter_2.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 2: Coordinate Systems and Transformation<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 2.1, Page number: 36<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Given point P(-2,6,3) and vector A = y a_x + (x+z) a_y ,\n",
+ "express P and A in cylindrical and spherical coordinates.\n",
+ "Evaluate A at P in the Cartesian, cylindrical, and spherical systems. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "x=-2\n",
+ "y=6\n",
+ "z=3 \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "r=scipy.sqrt(x**2+y**2)\n",
+ "phi=scipy.arctan(-y/x) #phi in radians in 1st quadrant\n",
+ "phid=180-(phi*180/scipy.pi) #phi in degrees in second quadrant\n",
+ "phic=scipy.pi*phid/180.0 #phi in radians in second quadrant\n",
+ "R=scipy.sqrt(x**2+y**2+z**2) \n",
+ "theta=scipy.arctan(r/z) \n",
+ "\n",
+ " #P in cylindrical coordinates\n",
+ " \n",
+ "Pcyl=array([round(r,2),round(phid,2),z]) \n",
+ "\n",
+ " #P in spherical coordinates\n",
+ " \n",
+ "Psph=array([round(R,2),round(theta*180/scipy.pi,2),\n",
+ " round(phid,2)]) \n",
+ "\n",
+ " #Vector A in cylindrical coordinate system\n",
+ "\n",
+ "Xc=r*scipy.cos(phic)\n",
+ "Yc=r*scipy.sin(phic)\n",
+ "Zc=z \n",
+ "Ar=Yc*scipy.cos(phic)+(Xc+Zc)*scipy.sin(phic)\n",
+ "Aphi=-Yc*scipy.sin(phic)+(Xc+Zc)*scipy.cos(phic)\n",
+ "Az=0\n",
+ "Acyl=array([round(Ar,4),round(Aphi,3),Az])\n",
+ "\n",
+ " #Vector A in spherical coordinate system\n",
+ "\n",
+ "Xs=R*scipy.cos(phic)*scipy.sin(theta)\n",
+ "Ys=R*scipy.sin(phic)*scipy.sin(theta)\n",
+ "Zs=R*scipy.cos(theta)\n",
+ "AR=Ys*scipy.sin(theta)*scipy.cos(phic)+(Xs+Zs)*scipy.sin(theta)*scipy.sin(phic) \n",
+ "Ath=Ys*scipy.cos(theta)*scipy.cos(phic)+(Xs+Zs)*scipy.cos(theta)*scipy.sin(phic) \n",
+ "Aph=-Ys*scipy.sin(phic)+(Xs+Zs)*scipy.cos(phic)\n",
+ "Asph=array([round(AR,4),round(Ath,4),round(Aph,3)])\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'P in cylindrical coordinates =',Pcyl\n",
+ "print 'P in spherical coordinates =',Psph\n",
+ "print 'A in cylindrical coordinates =',Acyl\n",
+ "print 'A in spherical coordinates =',Asph"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "P in cylindrical coordinates = [ 6.32 108.43 3. ]\n",
+ "P in spherical coordinates = [ 7. 64.62 108.43]\n",
+ "A in cylindrical coordinates = [-0.9487 -6.008 0. ]\n",
+ "A in spherical coordinates = [-0.8571 -0.4066 -6.008 ]\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 2.2, Page number: 39<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Express vector B = 10/R a_r+ r cos(theta) a_(theta) + a(phi)\n",
+ "in Cartesian and cylindrical coordinates. \n",
+ "Find B (- 3, 4, 0) and B (5, pi/2, - 2). '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ " \n",
+ "#Variable Declaration\n",
+ "\n",
+ "x=-3\n",
+ "y=4\n",
+ "z=0\n",
+ "p=5\n",
+ "phi=scipy.pi/2 \n",
+ "Zc=-2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ " #B in cartesian coordinates\n",
+ "\n",
+ "R=scipy.sqrt(x**2+y**2+z**2)\n",
+ "r=scipy.sqrt(x**2+y**2) \n",
+ "P=scipy.arcsin(r/R) #in radians\n",
+ "Q=scipy.arccos(x/r) #in radians \n",
+ "f=10/R \n",
+ "Bx=f*scipy.sin(P)*scipy.cos(Q)+R*(scipy.cos(P))**2*scipy.cos(Q)-scipy.sin(Q) \n",
+ "By=f*scipy.sin(P)*scipy.sin(Q)+R*(scipy.cos(P))**2*scipy.sin(Q)+scipy.cos(Q) \n",
+ "Bz=f*scipy.cos(P)-R*scipy.cos(P)*scipy.sin(P) \n",
+ "Bcart=array([round(Bx,0),round(By,0),round(-Bz,0)])\n",
+ "\n",
+ " #B in cylindrical coordinates\n",
+ " \n",
+ "Rc=sqrt(p**2+Zc**2) \n",
+ "Pc=scipy.arccos(Zc/Rc) #in radians\n",
+ "Br=(10/Rc)*scipy.sin(Pc)+Rc*(scipy.cos(Pc))**2 \n",
+ "Bp=1 \n",
+ "Bzc=(10/Rc)*scipy.cos(Pc)-Rc*scipy.cos(Pc)*scipy.sin(Pc) \n",
+ "Bcyl=array([round(Br,3),Bp,round(Bzc,3)])\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'B(-3,4,0) in cartesian coordinates is',Bcart\n",
+ "print 'B(5,pi/2,-2) in cylindrical coordinates is',Bcyl"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "B(-3,4,0) in cartesian coordinates is [-2. 1. 0.]\n",
+ "B(5,pi/2,-2) in cylindrical coordinates is [ 2.467 1. 1.167]\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 2.3, Page number: 44<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Two uniform vector fields are gIven by E=-5a_p+10a_phi+3a_z and\n",
+ "F=a_p+2a_phi-6a_z. Calculate \n",
+ "(a) |E X F|\n",
+ "(b) The vector component of E at P(5, pi/2, 3) parallel to the line x=2, z=3 \n",
+ "(c) The angle E makes with the surface z = 3 at P '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "E=array([-5,10,3]) #in cylindrical coordinates\n",
+ "F=array([1,2,-6]) #in cylindrical coordinates\n",
+ "P=array([5,scipy.pi/2,3]) #in cylindrical coordinates\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "exf=cross(E,F)\n",
+ "ansa=scipy.sqrt(dot(exf,exf)) #|EXF|\n",
+ "ay=array([round(scipy.sin(scipy.pi/2),0),\n",
+ " round(scipy.cos(scipy.pi/2),0),0])\n",
+ "ansb=dot(E,ay)*ay\n",
+ "modE=scipy.sqrt(dot(E,E))\n",
+ "az=array([0,0,1])\n",
+ "thetaEz=(180/scipy.pi)*arccos(dot(E,az)/modE) #in degrees\n",
+ "ansc=90-thetaEz #in degrees\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print '|EXF| =',round(ansa,2)\n",
+ "print 'The vector component of E at P parallel to the line x=2,z=3 =',ansb,','\n",
+ "print 'in spherical coordinates'\n",
+ "print 'The angle E makes with the surface z = 3 at P =',round(ansc,2),'degrees'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "|EXF| = 74.06\n",
+ "The vector component of E at P parallel to the line x=2,z=3 = [-5. -0. -0.] ,\n",
+ "in spherical coordinates\n",
+ "The angle E makes with the surface z = 3 at P = 15.02 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 2.4, Page number: 45<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ " '''\n",
+ "Given a vector field \n",
+ "D = r*sin(phi)a_r-(1/r)*sin(theta)*cos(phi)a_theta+ r^2*a_phi, determine\n",
+ "(a) D at P(10,150\u00b0,330\u00b0) \n",
+ "(b) The component of D tangential to the spherical surface r = 10 at P \n",
+ "(c) A unit vector at P perpendicular to D and tangential to the cone 0 = 150\u00b0 '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "aR=array([1,0,0]) #Unit vector along radial direction\n",
+ "ath=array([0,1,0]) #Unit vector along theta direction\n",
+ "aph=array([0,0,1]) #Unit vector along phi direction\n",
+ "P=array([10,scipy.pi*150/180,scipy.pi*330/180])\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "r=dot(P,aR)\n",
+ "q=dot(P,aph)\n",
+ "p=dot(P,ath)\n",
+ "R=r*scipy.sin(q)\n",
+ "Ph=-scipy.sin(p)*scipy.cos(q)/r\n",
+ "Q=r*r\n",
+ "D=array([R,Ph,Q]) #D at P(10,150\u00b0,330\u00b0)\n",
+ "rDr=round(dot(aR,D),0) #radial component of D\n",
+ "rDth=round(dot(-ath,D),3) #theta component of D\n",
+ "rDph=round(dot(aph,D),0) #phi component of D\n",
+ "\n",
+ "Dn=array([r*scipy.sin(q),0,0]) #Component of D normal to surface r=10\n",
+ "Dt=D-Dn #Component of D tangential to surface r=10\n",
+ "Dtr=round(dot(aR,Dt),0) #radial component of Dt\n",
+ "Dtth=round(dot(-ath,Dt),3) #theta component of Dt\n",
+ "Dtph=round(dot(aph,Dt),0) #phi component of Dt\n",
+ "rDt=array([Dtr,Dtth,Dtph])\n",
+ "\n",
+ " #Unit vector normal to D and tangential to cone theta=45 degrees\n",
+ "\n",
+ "U=cross(D,ath)\n",
+ "u=U/scipy.sqrt(dot(U,U)) \n",
+ "ru=array([round(dot(aR,u),4),round(dot(ath,u),4),round(dot(aph,u),4)])\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'D at P(10,150\u00b0,330\u00b0) = [',rDr,' ',rDth,' ',rDph,']'\n",
+ "print 'The component of D tangential to the spherical surface r = 10 at P ='\n",
+ "print '[',Dtr,' ',Dtth,' ',Dtph,']'\n",
+ "print 'A unit vector at P perpendicular to D and tangential to cone 0 = 150\u00b0 ='\n",
+ "print ru"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "D at P(10,150\u00b0,330\u00b0) = [ -5.0 0.043 100.0 ]\n",
+ "The component of D tangential to the spherical surface r = 10 at P =\n",
+ "[ 0.0 0.043 100.0 ]\n",
+ "A unit vector at P perpendicular to D and tangential to cone 0 = 150\u00b0 =\n",
+ "[-0.9988 0. -0.0499]\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_3.ipynb b/Elements_of_Electromagnetics/chapter_3.ipynb new file mode 100644 index 00000000..88c80683 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_3.ipynb @@ -0,0 +1,389 @@ +{
+ "metadata": {
+ "name": "chapter_3.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 3: Vector Calculus<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 3.1, Page number: 58<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Consider the object shown in Figure 3.7. Calculate \n",
+ "(a) The distance BC \n",
+ "(b) The distance CD \n",
+ "(c) The surface area ABCD \n",
+ "(d) The surface area ABO \n",
+ "(e) The surface area AOFD \n",
+ "(f) The volume ABDCFO '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "import scipy.integrate\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "A=array([5,0,0])\n",
+ "B=array([0,5,0])\n",
+ "C=array([0,5,10])\n",
+ "D=array([5,0,10])\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ " #A,B,C,D in cylindrical coordinates\n",
+ " \n",
+ "A=array([5,0,0])\n",
+ "B=array([5,scipy.pi,0])\n",
+ "C=array([5,scipy.pi,10])\n",
+ "D=array([5,0,10])\n",
+ "\n",
+ "p=5\n",
+ "\n",
+ "def BC(z): \n",
+ " return 1\n",
+ "ansa, erra = scipy.integrate.quad(BC, 0, 10)\n",
+ " \n",
+ "def CD(phi): \n",
+ " return p\n",
+ "ansb, errb = scipy.integrate.quad(CD, 0, scipy.pi/2)\n",
+ "ansbb=ansb/scipy.pi #answer in multiples of pi\n",
+ "\n",
+ "def ABCD(phi,z): \n",
+ " return p\n",
+ "ansc, errc = scipy.integrate.dblquad(lambda z , phi: ABCD(phi,z), \n",
+ " 0, scipy.pi/2, lambda z: 0, lambda z: 10) \n",
+ "anscc=ansc/scipy.pi #answer in multiples of pi\n",
+ " \n",
+ "def ABO(phi,rho): \n",
+ " return rho\n",
+ "ansd, errd = scipy.integrate.dblquad(lambda rho , phi: ABO(phi,rho), \n",
+ " 0, scipy.pi/2, lambda rho: 0, lambda rho: 5)\n",
+ "ansdd=ansd/scipy.pi #answer in multiples of pi\n",
+ "\n",
+ "def AOFD(rho,z): \n",
+ " return 1\n",
+ "anse, erre = scipy.integrate.dblquad(lambda z , rho: AOFD(rho,z), \n",
+ " 0, 10, lambda z: 0, lambda z: 5)\n",
+ " \n",
+ "def ABDCFO(z,phi,rho):\n",
+ " return rho\n",
+ "ansf, errf=scipy.integrate.tplquad(ABDCFO,0,5,lambda rho:0,\n",
+ " lambda rho:scipy.pi/2,lambda rho,phi:0,lambda rho,phi:10)\n",
+ "ansff=ansf/scipy.pi #answer in multiples of pi\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The distance BC =',ansa\n",
+ "print 'The distance CD =',ansbb,'pi'\n",
+ "print 'The surface area ABCD =',anscc,'pi'\n",
+ "print 'The surface area ABO =',ansdd,'pi'\n",
+ "print 'The surface area AOFD =',anse\n",
+ "print 'The volume ABDCFO =',ansff,'pi'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The distance BC = 10.0\n",
+ "The distance CD = 2.5 pi\n",
+ "The surface area ABCD = 25.0 pi\n",
+ "The surface area ABO = 6.25 pi\n",
+ "The surface area AOFD = 50.0\n",
+ "The volume ABDCFO = 62.5 pi\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 3.2, Page number: 61<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Given that F = x^2 a_x - xza_y - y^2 a_z , calculate the \n",
+ "circulation of F around the (closed) path shown in Figure 3.10. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "import scipy.integrate\n",
+ "from fractions import Fraction\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "def C1(x): \n",
+ " return x**2\n",
+ "seg1, err1 = scipy.integrate.quad(C1, 1, 0) #segment 1\n",
+ "Seg1=Fraction(seg1).limit_denominator(100) #converting to fraction\n",
+ "\n",
+ "def C2(y): \n",
+ " return 0\n",
+ "seg2, err2 = scipy.integrate.quad(C2, 0, 1) #segment 2\n",
+ "\n",
+ "def C3(x): \n",
+ " return (x**2-1)\n",
+ "seg3, err3 = scipy.integrate.quad(C3, 0, 1) #segment 3\n",
+ "Seg3=Fraction(seg3).limit_denominator(100) #converting to fraction\n",
+ "\n",
+ "def C4(y): \n",
+ " return (-y-y**2)\n",
+ "seg4, err4 = scipy.integrate.quad(C4, 1, 0) #segment 4\n",
+ "Seg4=Fraction(seg4).limit_denominator(100) #converting to fraction\n",
+ "\n",
+ "seg=Seg1+seg2+Seg3+Seg4 #total circulation around path\n",
+ "Seg=Fraction(seg).limit_denominator(100) #converting to fraction\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'F along segment 1 is',Seg1\n",
+ "print 'F along segment 2 is',seg2\n",
+ "print 'F along segment 3 is',Seg3\n",
+ "print 'F along segment 4 is',Seg4\n",
+ "print 'Circulation of F around the path is',Seg"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "F along segment 1 is -1/3\n",
+ "F along segment 2 is 0.0\n",
+ "F along segment 3 is -2/3\n",
+ "F along segment 4 is 5/6\n",
+ "Circulation of F around the path is -1/6\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.4, Page number: 68"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Given W = x^2 y^2 + xyz, compute VW and the direction derivative dW/dl in the direction \n",
+ "3a_x + 4a_y + 12a_z at (2, -1, 0). '''\n",
+ "\n",
+ "import scipy\n",
+ "from fractions import Fraction\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "Al=array([3,4,12])\n",
+ "x=2\n",
+ "y=-1\n",
+ "z=0\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "gradW=(2*x*y**2+y*z)*ax+(2*x**2*y+x*z)*ay+(x*y)*az\n",
+ "gradWl=Fraction(dot(gradW,Al)/scipy.sqrt(dot(Al,Al))).limit_denominator(1000)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'dW/dl =',gradWl\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "dW/dl = -44/13\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 3.7, Page number: 74"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "If G(r) = lOe^-2z(pa_p + a_z\u001f), determine the flux of G out of the entire surface of the cylinder \n",
+ "p = 1, 0 < Z < 1. Confirm the result using the divergence theorem. '''\n",
+ "\n",
+ "import scipy\n",
+ "import scipy.integrate\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ap=array([1,0,0]) #Unit vector along radial direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "def psi1(phi,p): \n",
+ " return 10*scipy.e**(-2)*p\n",
+ "psit, errt = scipy.integrate.dblquad(lambda p , phi: psi1(phi,p), #flux through top\n",
+ " 0, 2*scipy.pi, lambda p: 0, lambda p: 1) \n",
+ "\n",
+ "def psi2(phi,p): \n",
+ " return -10*p\n",
+ "psib, errb = scipy.integrate.dblquad(lambda p , phi: psi2(phi,p), #flux through bottom\n",
+ " 0, 2*scipy.pi, lambda p: 0, lambda p: 1) \n",
+ "\n",
+ "def psi3(phi,z): \n",
+ " return 10*scipy.exp(-2*z)\n",
+ "psis, errs = scipy.integrate.dblquad(lambda z , phi: psi3(phi,z), #flux through side\n",
+ " 0, scipy.pi*2, lambda z: 0, lambda z: 1) \n",
+ "\n",
+ "psi=psit+psib+psis #total flux through cylinder\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The total flux through the cylinder is',psi\n",
+ "print 'The total flux through cylinder using divergence theorem is also 0'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total flux through the cylinder is 0.0\n",
+ "The total flux through cylinder using divergence theorem is also 0\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 3.9, Page number: 81<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "If A = p cos(phi) a_p + sin(phi)a_phi, evaluate closed integral A\u00b7dl\n",
+ "around the path shown in Figure 3.22. \n",
+ "Confirm this using Stokes's theorem. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "import scipy.integrate\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ap=array([1,0,0]) #Unit vector along radial direction\n",
+ "ath=array([0,1,0]) #Unit vector along theta direction\n",
+ "aph=array([0,0,1]) #Unit vector along phi direction\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ " #segment 1\n",
+ "def ab(phi): \n",
+ " return 2*scipy.sin(phi)\n",
+ "seg1,err1 = scipy.integrate.quad(ab,scipy.pi*60/180,scipy.pi*30/180) \n",
+ "\n",
+ " #segment 2\n",
+ "def bc(p): \n",
+ " return p*scipy.cos(scipy.pi*30/180)\n",
+ "seg2,err2 = scipy.integrate.quad(bc,2,5) \n",
+ "\n",
+ " #segment 3\n",
+ "def cd(phi): \n",
+ " return 5*scipy.sin(phi)\n",
+ "seg3,err3 = scipy.integrate.quad(cd,-scipy.pi*30/180,scipy.pi*60/180)\n",
+ "\n",
+ " #segment 4\n",
+ "def da(p): \n",
+ " return p*scipy.cos(scipy.pi*60/180)\n",
+ "seg4,err4 = scipy.integrate.quad(da,5,2)\n",
+ "\n",
+ "I1=seg1+seg2+seg3+seg4\n",
+ "\n",
+ " #using stoke's theorem\n",
+ "\n",
+ "def curlA(phi,p): \n",
+ " return ((1+p)*scipy.sin(phi))\n",
+ "I2, err = scipy.integrate.dblquad(lambda p , phi: curlA(phi,p), \n",
+ " scipy.pi*30/180, scipy.pi*60/180, lambda p: 2, lambda p: 5)\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The integral calculated segment wise =',round(I1,3)\n",
+ "print 'The integral calculated using Stokes Theorem =',round(I2,3)\n",
+ "print 'Since I1 = I2, Stokes theorem is verified'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The integral calculated segment wise = 4.941\n",
+ "The integral calculated using Stokes Theorem = 4.941\n",
+ "Since I1 = I2, Stokes theorem is verified\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_4.ipynb b/Elements_of_Electromagnetics/chapter_4.ipynb new file mode 100644 index 00000000..14ac4af2 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_4.ipynb @@ -0,0 +1,699 @@ +{
+ "metadata": {
+ "name": "chapter_4.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 4: Electrostatic Fields<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4.1, Page number: 107<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Point charges 1 mC and -2 mC are located at (3,2,-1) and (-1,-1,4),\n",
+ "respectively. Calculate the electric force on a 10 nC charge located\n",
+ "at (0,3,1) and the electric field intensity at that point. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "Q1=1*10**-3 #charge 1 at (-1,-1,4) in C\n",
+ "Q2=-2*10**-3 #charge 2 at (3,2,-1) in C\n",
+ "Q=10*10**-9 #charge 3 at (0,3,1) in C\n",
+ "P1=array([0,3,1])-array([3,2,-1]) #distance vector from charge 3 to 1\n",
+ "P2=array([0,3,1])-array([-1,-1,4]) #distance vector from charge 3 to 2\n",
+ "e=10**-9/(36*scipy.pi) #permittivity in Farad/m \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "modP1=scipy.sqrt(dot(P1,P1))\n",
+ "modP2=scipy.sqrt(dot(P2,P2))\n",
+ "F1=(Q*Q1)*P1*10**3/(4*scipy.pi*e*modP1**3) #force on charge 3 by 1 in mN\n",
+ "F2=(Q*Q2)*P2*10**3/(4*scipy.pi*e*modP2**3) #force on charge 3 by 2 in mN\n",
+ "\n",
+ " #Total force on charge 3\n",
+ " \n",
+ "Fx=round(dot(F1+F2,ax),3)\n",
+ "Fy=round(dot(F1+F2,ay),3)\n",
+ "Fz=round(dot(F1+F2,az),3)\n",
+ "F=array([Fx,Fy,Fz]) #Total force in mN\n",
+ "E=(10**-6)*(F/Q) #Electric field in kV/m\n",
+ "\n",
+ "#Results \n",
+ "\n",
+ "print 'Total force on charge at (0,3,1) =',F,'in mN'\n",
+ "print 'Electric field at (0,3,1) =',E,'kV/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total force on charge at (0,3,1) = [-6.512 -3.713 7.509] in mN\n",
+ "Electric field at (0,3,1) = [-651.2 -371.3 750.9] kV/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.3, Page number: 109"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A practical application of electrostatics is in electrostatic separation of solids. For example, \n",
+ "Florida phosphate ore, consisting of small panicles of quartz and phosphate rock, can be \n",
+ "separated into its components by applying a uniform electric field as in Figure 4.4. Assum- \n",
+ "ing zero initial velocity and displacement, determine the separation between the particles \n",
+ "after falling 80 cm. Take E = 500 kV/m and Q/m = 9 microC/kg for both positively and neg- \n",
+ "atively charged particles. '''\n",
+ "\n",
+ "import scipy\n",
+ "#Variable Declaration\n",
+ "\n",
+ "E=500*10**3 #electric field in V/m\n",
+ "Qm=9*10**-6 #Q/m in C/kg\n",
+ "y=0.8 #distance fallen in m\n",
+ "g=9.8 #acceleration due to gravity in m/s^2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "t=scipy.sqrt(2*y/g) #time taken to fall in seconds\n",
+ "x=Qm*E*t**2/2 #half the separation between particles in m\n",
+ "sep=2*x #separation between particles in m\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The separation between particles is',round(sep*100,2),'cm'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The separation between particles is 73.47 cm\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.5, Page number: 120"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The finite sheet 0 <= x <= 1, 0 <= y <= 1 on the z = 0 plane has a charge density \n",
+ "Ps=xy(x^2 + y^2 + 25)^3/2 nC/m^2 . Find \n",
+ "\u001f\u001d",
+ "(a) The total charge on the sheet \n",
+ "(b) The e1ectric field at (0, 0, 5) \n",
+ "(c) The force experienced by a -1 mC charge located at (0, 0, 5) '''\n",
+ "\n",
+ "import scipy\n",
+ "import scipy.integrate\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Eo=10**-9/(36*scipy.pi) #permittivity of free space\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "q=-1 #charge in mC\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "def charge(x,y): \n",
+ " return x*y*(x**2+y**2+25)**(1.5)\n",
+ "Q, errq = scipy.integrate.dblquad(lambda y , x: charge(x,y), #total charge in nC\n",
+ " 0, 1, lambda y: 0, lambda y: 1) \n",
+ "\n",
+ "d=(4*scipy.pi*Eo*(x**2+y**2+25)**(1.5))\n",
+ "\n",
+ "def elecx(x,y): \n",
+ " return 10**-9*x*y*(x**2+y**2+25)**(1.5)*(-x)/d #x component of electric field\n",
+ "Ex, errx = scipy.integrate.dblquad(lambda y , x: elecx(x,y), \n",
+ " 0, 1, lambda y: 0, lambda y: 1) \n",
+ "\n",
+ "def elecy(x,y): \n",
+ " return 10**-9*x*y*(x**2+y**2+25)**(1.5)*(-y)/d #y component of electric field\n",
+ "Ey, erry = scipy.integrate.dblquad(lambda y , x: elecy(x,y), \n",
+ " 0, 1, lambda y: 0, lambda y: 1) \n",
+ "\n",
+ "def elecz(x,y): \n",
+ " return 10**-9*x*y*(5)/(4*scipy.pi*Eo) #z component of electric field\n",
+ "Ez, errz = scipy.integrate.dblquad(lambda y , x: elecz(x,y), \n",
+ " 0, 1, lambda y: 0, lambda y: 1) \n",
+ "\n",
+ "E=array([round(Ex,1),round(Ey,1),round(Ez,2)]) #electric field in V/m\n",
+ "\n",
+ "F=q*E #force in mN \n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Total charge =',round(Q,2),'nC'\n",
+ "print 'Electric field at (0,0,5) =',E,'V/m'\n",
+ "print 'Force experienced by -1mC =',F,'mN'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total charge = 33.15 nC\n",
+ "Electric field at (0,0,5) = [ -1.5 -1.5 11.25] V/m\n",
+ "Force experienced by -1mC = [ 1.5 1.5 -11.25] mN\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4.6, Page number: 121<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Planes x=2 and y=-3, respectively, carry charges 10 nC/m^2 and 15 nC/m^2 .\n",
+ "If the line x=0, z=2 carries charge 10 pi nC/m, calculate E at (1,1,-1)\n",
+ "due to the three charge distributions. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "ps1=10*10**-9 #Surface charge density of plane 1\n",
+ "ps2=15*10**-9 #Surface charge density of plane 2\n",
+ "pl=10*scipy.pi*10**-9 #charge density of line\n",
+ "e=(10**-9)/(36*scipy.pi) #permittivity of free space in Farad/m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "E1=(ps1/(2*e))*-ax/scipy.pi #field due to plane 1 in multiples of pi in V/m\n",
+ "E2=(ps2/(2*e))*ay/scipy.pi #field due to plane 2 in multiples of pi in V/m\n",
+ "\n",
+ " #field due to line charge in multiples of pi in V/m\n",
+ " \n",
+ "a=(ax-3*az) \n",
+ "moda=scipy.sqrt(dot((ax-3*az),(ax-3*az)))\n",
+ "e3=(pl/(2*scipy.pi*e*moda**2))*a\n",
+ "E3=array([dot(e3,ax)/scipy.pi,0,dot(e3,az)/scipy.pi])\n",
+ "\n",
+ " #total field in multiples of pi in V/m\n",
+ " \n",
+ "E=E1+E2+E3 \n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The total electric field at (1,1,-1) =',E,'Pi V/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The total electric field at (1,1,-1) = [-162. 270. -54.] Pi V/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4.7, Page number: 123<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Determine D at (4, 0, 3) if there is a point charge - 5pi mC\n",
+ "at (4, 0, 0) and a line charge 3pi mC/m along the y-axis. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "Q=-5*scipy.pi*10**-3 #charge at (4,0,0) in C\n",
+ "pl=3*scipy.pi*10**-3 #charge density of line charge in C/m\n",
+ "r=array([4,0,3]) #point where D is to be found \n",
+ "rp=array([4,0,0]) #position of point charge\n",
+ "\n",
+ "#Calculations \n",
+ "\n",
+ "R=r-rp \n",
+ "modR=scipy.sqrt(dot(R,R)) \n",
+ "Dq=(Q*R)/(4*scipy.pi*modR**3) #flux density due to point charge in C/m^2\n",
+ "p=scipy.sqrt(dot(r,r))\n",
+ "ap=r/p \n",
+ "Dl=(pl/(2*scipy.pi*p))*ap #flux density due to line charge in C/m^2\n",
+ "D=(Dq+Dl)*10**6 #total flux density in micro C/m^2\n",
+ "Dz=round(dot(D,az),0)\n",
+ "Dx=round(dot(D,ax),0)\n",
+ "Dy=round(dot(D,ay),0)\n",
+ "Dround=array([Dx,Dy,Dz]) #value of D rounded to 0 decimal points\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'D at (4,0,0) due to point charge and line charge =',Dround,'micro C/m^2'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "D at (4,0,0) due to point charge and line charge = [ 240. 0. 41.] micro C/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4.8, Page number: 130<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Given that D = zp cos^2(phi) a_z C/m^2 , calculate the charge density \n",
+ "at (1,pi/4, 3) and the total charge enclosed by the cylinder of \n",
+ "radius 1 m with -2< Z <2 m. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "import scipy.integrate\n",
+ "from fractions import Fraction\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ap=array([1,0,0]) #Unit vector along rho direction\n",
+ "aph=array([0,1,0]) #Unit vector along phi direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "point=array([1,scipy.pi/4,3])\n",
+ "p1=0\n",
+ "p2=1\n",
+ "ph1=0\n",
+ "ph2=2*scipy.pi\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "pointp=dot(point,ap)\n",
+ "pointph=dot(point,aph)\n",
+ "pv=pointp*scipy.cos(pointph)**2 #charge density at (1,pi/4,3) in C/m^3\n",
+ "\n",
+ "def ctop(phi,p): \n",
+ " return 2*p**2*(scipy.cos(phi)**2)\n",
+ "psya, erra = scipy.integrate.dblquad(lambda p , phi: ctop(phi,p), \n",
+ " ph1, ph2, lambda p: p1, lambda p: p2)\n",
+ "\n",
+ "def cbot(phi,p): \n",
+ " return 2*p**2*(scipy.cos(phi)**2)\n",
+ "psyb, errb = scipy.integrate.dblquad(lambda p , phi: cbot(phi,p), \n",
+ " ph1, ph2, lambda p: p1, lambda p: p2)\n",
+ " \n",
+ "psy=psya+psyb #Charge in C\n",
+ "psyp=psy/scipy.pi #Charge in multiples of Pi in C\n",
+ "psyf=Fraction(psyp).limit_denominator(100) #converting to fraction\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Charge density at (1,pi/4,3) =',pv,'C/m^3'\n",
+ "print 'Total charge enclosed by the cylinder =',psyf,'Pi C'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Charge density at (1,pi/4,3) = 0.5 C/m^3\n",
+ "Total charge enclosed by the cylinder = 4/3 Pi C\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4.10, Page number: 136<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Two point charges -4 micro C and 5 micro C are located at (2, -1, 3) \n",
+ "and (0, 4, -2), respectively. Find the potential at (1, 0, 1) \n",
+ "assuming zero potential at infinity. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Q1=-4 #charge 1 in micro C\n",
+ "Q2=5 #charge 2 in micro C\n",
+ "e=10**-9/(36*scipy.pi) #permittivity of free space in Farad/m \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R1=array([1,0,1])-array([2,-1,3]) #distance vector from (1,0,1) to charge 1\n",
+ "R2=array([1,0,1])-array([0,4,-2]) #distance vector from (1,0,1) to charge 2\n",
+ "modR1=scipy.sqrt(dot(R1,R1))\n",
+ "modR2=scipy.sqrt(dot(R2,R2)) \n",
+ "V=10**-9*((Q1/modR1)+(Q2/modR2))/(4*scipy.pi*e) #potential in kV\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The potential at (1, 0, 1) =',round(V,3),'kV'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The potential at (1, 0, 1) = -5.872 kV\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 4.11, Page number: 136"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A point charge 5 nC is located at (-3, 4, 0) while line y = 1, z = 1 carries uniform charge \n",
+ "2 nC/m. \n",
+ "\u001f\u001d",
+ "(a) If V = 0V at O(0, 0, 0), find V at A(5, 0, ]). \n",
+ "(b) If V = 100V at B(1, 2, 1), find V at C(-2, 5, 3). \n",
+ "(c) If V = -5V at O, find V_BC '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Eo=10**-9/(36*scipy.pi) #permittivity of free space\n",
+ "Vo=0 #potential at O in V\n",
+ "Vb=100 #potential at B in V\n",
+ "po=scipy.sqrt(2)\n",
+ "ro=5\n",
+ "pa=1\n",
+ "ra=9\n",
+ "pb=1\n",
+ "rb=scipy.sqrt(21)\n",
+ "pc=scipy.sqrt(20)\n",
+ "rc=scipy.sqrt(11)\n",
+ "pl=2*10**-9 #charge density of the line in C/m\n",
+ "Q=5*10**-9 #point charge at (-3,4,0) in C\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Va=Vo-(-pl*scipy.log(po/pa)/(2*scipy.pi*Eo)+Q*(ra-ro)/(4*scipy.pi*Eo*ra*ro))\n",
+ "Vc=Vb+(-pl*scipy.log(pc/pb)/(2*scipy.pi*Eo)+Q*(rb-rc)/(4*scipy.pi*Eo*rb*rc))\n",
+ "Vbc=Vc-Vb\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Va =',round(Va,3),'V'\n",
+ "print 'Vc =',round(Vc,3),'V'\n",
+ "print 'Vbc =',round(Vbc,3),'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Va = 8.477 V\n",
+ "Vc = 49.825 V\n",
+ "Vbc = -50.175 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4.12, Page number: 140<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Given the potential V = (10/r**2)sin(theta)cos(phi), \n",
+ "(a) Find the electric flux density D at (2, pi/2, 0). \n",
+ "(b) Calculate the work done in moving a 10 micro C charge from \n",
+ "point A( 1, 30\u00b0, 120\u00b0) to B( 4, 90\u00b0, 60\u00b0). '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ar=array([1,0,0]) #Unit vector along radial direction\n",
+ "ath=array([0,1,0]) #Unit vector along theta direction\n",
+ "aph=array([0,0,1]) #Unit vector along phi direction\n",
+ "e=(10**-9)/(36*scipy.pi) #permittivity of free space in Farad/m\n",
+ "\n",
+ " #The point (2, pi/2, 0)\n",
+ "r=2\n",
+ "th=scipy.pi/2\n",
+ "ph=0\n",
+ " #Point A\n",
+ "ra=1\n",
+ "tha=scipy.pi*30/180\n",
+ "pha=scipy.pi*120/180\n",
+ " #Point B\n",
+ "rb=4\n",
+ "thb=scipy.pi/2\n",
+ "phb=scipy.pi*60/180\n",
+ "\n",
+ "q=10*10**-6 \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Er=(20.0/r**3)*scipy.sin(th)*scipy.cos(ph) #Radial component of E in V/m\n",
+ "Eth=-(10/r**3)*scipy.cos(th)*scipy.cos(ph) #Theta component of E in V/m\n",
+ "Eph=(10/r**3)*scipy.sin(ph) #Phi component of E in V/m\n",
+ "E=array([Er,Eth,Eph])\n",
+ "D=E*e*10**12 #Electric flux density D in pC/m^2\n",
+ "Dr=round(dot(D,ar),1) #Radial component of D in V/m rounded to 1 decimal\n",
+ "Dth=round(dot(D,ath),0) #Theta component of D in pC/m^2 rounded to 0 decimal\n",
+ "Dph=round(dot(D,aph),0) #Phi component of D in pC/m^2 rounded to 0 decimal\n",
+ "Dc=array([Dr,Dth,Dph]) #Rounded D in pC/m^2\n",
+ "\n",
+ "Va=10*scipy.sin(tha)*cos(pha)/ra**2 #potential at point A in V\n",
+ "Vb=10*scipy.sin(thb)*cos(phb)/rb**2 #potential at point B in V\n",
+ "W=q*(Vb-Va)*10**6 #work done in micro J\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The electric flux density D at (2, pi/2, 0) =',Dc,'pC/m^2'\n",
+ "print 'Work done in moving the charge =',W,'micro J'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electric flux density D at (2, pi/2, 0) = [ 22.1 -0. 0. ] pC/m^2\n",
+ "Work done in moving the charge = 28.125 micro J\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4.13, Page number: 145<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Two dipoles with dipole moments -5 a_z nC/m and 9 a_z nC/m are\n",
+ "located at points (0,0,-2) and (0,0,3), respectively.\n",
+ "Find the potential at the origin. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "p1=-5*10**-9 #dipole moment of dipole 1 in C/m\n",
+ "p2=9*10**-9 #dipole moment of dipole 2 in C/m\n",
+ "z1=2 #z component of position vector of dipole 1\n",
+ "z2=-3 #z component of position vector of dipole 2\n",
+ "e=10**-9/(36*scipy.pi) #permittivity of free space in Farad/m\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "V=(1/(4*scipy.pi*e))*((p1*abs(z1)/z1**3)+(p2*abs(z2)/z2**3))\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'Potential at origin =',V, 'V'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Potential at origin = -20.25 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 4.14, Page number: 148<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Three point charges - 1 nC, 4 nC, and 3 nC are located at (0, 0, 0), \n",
+ "(0, 0, 1), and (1, 0, 0), respectively. Find the energy in the system. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Q1=-1*10**-9 #Charge 1 in C\n",
+ "Q2=4*10**-9 #Charge 2 in C\n",
+ "Q3=3*10**-9 #Charge 3 in C\n",
+ "e=10**-9/(36*scipy.pi) #permittivity of free space in farad/m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "V1=(1/(4*scipy.pi*e)*(Q2+Q3))\n",
+ "V2=(1/(4*scipy.pi*e)*(Q1+Q3/(2**.5)))\n",
+ "V3=(1/(4*scipy.pi*e)*(Q1+Q2/(2**.5)))\n",
+ "W=0.5*((V1*Q1)+(V2*Q2)+(V3*Q3))*10**9 #Energy in nJ\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'Energy in the system =',round(W,2),'nJ'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Energy in the system = 13.37 nJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_5.ipynb b/Elements_of_Electromagnetics/chapter_5.ipynb new file mode 100644 index 00000000..882e9145 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_5.ipynb @@ -0,0 +1,523 @@ +{
+ "metadata": {
+ "name": "chapter_5.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 5: Electric Fields in Material Space<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5.1, Page number: 167<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "If the current density J=1/r^3(2cos(theta) a_r + sin(theta) a_t) Ampere/m^2,\n",
+ "calculate the current passing through\n",
+ "(a) A hemispherical shell of radius 20 cm.\n",
+ "(b) A spherical shell of radius 10 cm. '''\n",
+ "\n",
+ "import scipy\n",
+ "import scipy.integrate\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "r1= 0.2 # radius of hemispherical shell in metres\n",
+ "r2= 0.1 # radius of spherical shell in metres\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Calculation of current through hemispherical shell \n",
+ "\n",
+ "def J1(phi,theta):\n",
+ "\ts1=(1/r1)*(2* scipy.cos(theta)* scipy.sin(theta))\n",
+ "\treturn s1\n",
+ "\n",
+ "if __name__ == '__main__':\n",
+ "\n",
+ " I1, error = scipy.integrate.dblquad(lambda theta , phi: J1(phi,theta), \n",
+ " 0, 2*scipy.pi, lambda theta: 0, lambda theta: scipy.pi/2) \n",
+ "\t \n",
+ "#Calculation of current through spherical shell \n",
+ "\n",
+ "def J2(phi,theta):\n",
+ "\ts2=(1/r2)*(2* scipy.cos(theta)* scipy.sin(theta))\n",
+ "\treturn s2\n",
+ "\n",
+ "if __name__ == '__main__':\n",
+ "\n",
+ " I2, error = scipy.integrate.dblquad(lambda theta , phi: J1(phi,theta), \n",
+ " 0, 2*scipy.pi, lambda theta: 0, lambda theta: scipy.pi) \n",
+ "\t \n",
+ "#Results\n",
+ "\n",
+ "print 'Current through hemispherical shell=',round(I1,1),'A' \n",
+ "print 'Current through spherical shell=',round(I2,0),'A'\n",
+ "\t"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Current through hemispherical shell= 31.4 A\n",
+ "Current through spherical shell= 0.0 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5.2, Page number: 168<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A typical example of convective charge transport is found \n",
+ "in the Van de Graaff generator where charge is transported \n",
+ "on a moving belt from the base to the dome.\n",
+ "If a surface charge density 10^7 C/m^2 is transported at a velocity of 2 m/s, \n",
+ "calculate the charge collected in 5 s. Take the width of the belt as 10 cm. '''\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ps=10**-7 #Surface charge density of the belt in Couloumb/metre^2\n",
+ "u=2 #Speed of the belt in metres/sec\n",
+ "w=0.1 #Width of the belt in metres\n",
+ "t=5 #Time taken in seconds \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "I=ps*u*w #Current in amperes\n",
+ "Q=I*t*10**9 #Charge collected in 5 seconds in nano Coloumbs\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print \"The charge collected in 5 seconds is \",Q,\"nC\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The charge collected in 5 seconds is 100.0 nC\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5.3, Page number: 169<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A wire of diameter 1mm and conductivity 5X10^7 S/m has 10^29 \n",
+ "free electrons/m^3 when an electric field of 10mV /m is applied. Determine \n",
+ "(a) The charge density of free electrons \n",
+ "(b) The current density \n",
+ "(c) The current in the wire \n",
+ "(d) The drift velocity of the electrons. \n",
+ "Take the electronic charge as e = -1.6X10^-19 C. '''\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "n=10**29 #Number density of electrons in m^-3\n",
+ "e=-1.6*10**-19 #Electronic charge in Coloumbs\n",
+ "sigma=5*10**7 #Current density in S/m\n",
+ "E=10**-2 #Electric Field in V/m\n",
+ "S=(3.14*10**-6)/4 #Cross sectional area of the wire in m^2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "pv=n*e #Charge density of free electrons in C/m^3\n",
+ "J=sigma*E*10**-3 #Current density in kA/m^2\n",
+ "I=J*S*10**3 #Current in amperes\n",
+ "u=J*10**3/pv #Drift velocity in m/s\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print \"The charge density is \",pv,\"C/m^3\" \n",
+ "print \"The current density is \",J,\"kA/m^2\" \n",
+ "print \"The current is \",round(I,3), \"A\"\n",
+ "print \"The drift velocity is \",-u,\"m/s\" "
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The charge density is -16000000000.0 C/m^3\n",
+ "The current density is 500.0 kA/m^2\n",
+ "The current is 0.393 A\n",
+ "The drift velocity is 3.125e-05 m/s\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5.4, Page number: 170<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A lead (sigma=5X10^6 S/m) bar of square cross section has a\n",
+ "hole bored along its length of 4 cm so that its cross section \n",
+ "becomes a square of side 3 cm with a hole of diameter 1 cm \n",
+ "drilled through the centre,\n",
+ "Find the resistance between the square ends. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "l=4 #Length of the lead bar in m\n",
+ "d=3 #Width of the lead bar in cm\n",
+ "r=0.5 #Radius of the hole drilled in cm\n",
+ "sigma=5*10**6 #Conductivity of the bar in S/m\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "S=(d**2-(scipy.pi*r**2)) #Cross sectional area in cm^2\n",
+ "R=l/(S*sigma*10**-4) #Resistance in ohms\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The resistance between the square ends is',round(R*10**6),'micro ohms'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The resistance between the square ends is 974.0 micro ohms\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5.6, Page number: 177<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The electric field intensity in polystyrene (epsilon_r= 2.55) filling\n",
+ "the space between the plates of a parallel-plate capacitor is 10 kV/m. \n",
+ "The distance between the plates is 1.5 mm. Calculate: \n",
+ "(a) D \n",
+ "(b) P \n",
+ "(c) The surface charge density of free charge on the plates \n",
+ "(d) The surface density of polarization charge \n",
+ "(e) The potential difference between the plates '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "e0=10**-9/(36*scipy.pi) #permittivity of free space in Farad/m\n",
+ "er=2.55 #relative permittivity (dimensionless)\n",
+ "E=10*10**3 #Electric field in V/m\n",
+ "chi=er-1.0 #Electric susceptibility (dimensionless)\n",
+ "d=1.5 #Distance between plates in mm\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "D=e0*er*E*10**9 #D in nC/m^2\n",
+ "\n",
+ "P=chi*e0*E*10**9 #P in nC/m^2\n",
+ "\n",
+ "ps=D #The surface charge density of \n",
+ " #free charge in nC/m^2\n",
+ " \n",
+ "pps =P #The surface charge density of\n",
+ " #polarization charge in nC/m^2\n",
+ " \n",
+ "V=E*d*10**-3 #The potential difference between \n",
+ " #the plates in volts\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'D =',round(D,2),'nC/m^2'\n",
+ "print 'P =',round(P,0),'nC/m^2'\n",
+ "print 'Surface charge density of free charge =',round(ps,2),'nC/m^2'\n",
+ "print 'Surface charge density of polarization charge =',round(pps,0),'nC/m^2'\n",
+ "print 'The potential difference between the plates =',V,'V'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "D = 225.47 nC/m^2\n",
+ "P = 137.0 nC/m^2\n",
+ "Surface charge density of free charge = 225.47 nC/m^2\n",
+ "Surface charge density of polarization charge = 137.0 nC/m^2\n",
+ "The potential difference between the plates = 15.0 V\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5.7, Page number: 178<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A dielectric sphere (epsilon_r = 5.7) of radius 10 cm \n",
+ "has a point charge 2 pC placed at its center. \n",
+ "Calculate: \n",
+ "(a)The surface density of polarization charge on\n",
+ " the surface of the sphere \n",
+ "(b)The force exerted by the charge on a -4 pC \n",
+ " point charge placed on the sphere '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "e0=10**-9/(36*scipy.pi) #permittivity of free space\n",
+ " #in Farad/m\n",
+ " \n",
+ "er=5.7 #relative permittivity\n",
+ " #(dimensionless)\n",
+ " \n",
+ "chi=er-1 #Electric susceptibility\n",
+ " #(dimensionless)\n",
+ " \n",
+ "r=0.1 #radius of sphere in m\n",
+ "\n",
+ "q1=2 #charge on sphere in pC\n",
+ "\n",
+ "q2=-4 #value of point charge in pC\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "E=q1/(4*scipy.pi*e0*er*r**2) #Electric field on the\n",
+ " #sphere in pV/m\n",
+ " \n",
+ "P=chi*e0*E #Polarisation in pC/m^2\n",
+ "\n",
+ "pps=P #The surface density of polarization \n",
+ " #charge in pC/m^2\n",
+ " \n",
+ "F=(q1*q2*10**-12)/(4*scipy.pi*e0*er*r**2) #Force exerted on point charge in pN\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The surface density of polarization'\n",
+ "print 'charge on the surface of the sphere =',round(pps,2),'pC/m^2'\n",
+ "print 'Force exerted on -4 pC charge =',round(F,3),'pN in the radial direction'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The surface density of polarization\n",
+ "charge on the surface of the sphere = 13.12 pC/m^2\n",
+ "Force exerted on -4 pC charge = -1.263 pN in the radial direction\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5.9, Page number: 188<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Two extensive homogeneous isotropic dielectrics meet on \n",
+ "plane z = O. For z >=0, epsilon_r1 = 4 and for z <= 0, \n",
+ "epsilon_r2 = 3. A uniform electric field E1= 5 a_x - 2 a_y + 3 a_z kV/m \n",
+ "exists for z >= O. Find \n",
+ "(a) E2 for z <= 0 \n",
+ "(b) The angles El and E2 make with the interface \n",
+ "(c) The energy densities in J/m^3 in both dielectrics \n",
+ "(d) The energy within a cube of side 2 m centered at (3,4,-5) '''\n",
+ "\n",
+ "#Variable Declarartion\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "an=array([0,0,1]) #Unit vector normal to the interface\n",
+ "E1=array([5,-2,3]) #Electric field for z >=0 in kV/m\n",
+ "e_r1=4 #Relative permittivity for z >=0 (dimensionless)\n",
+ "e_r2=3 #Relative permittivity for z <=0 (dimensionless)\n",
+ "e0=(10**-9)/(36*scipy.pi) #Permittivity of free space in Farad/m\n",
+ "V=2*2*2 #Volume of cube placed in region 2 in m^3\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "E1n=array([0,0,dot(E1,an)]) #The normal component of E1 in kV/m\n",
+ "E1t=E1-E1n #Transverse component of E1 in kV/m\n",
+ "E2t=E1t #Transverse component of E2 in kV/m\n",
+ "E2n=e_r1*E1n/e_r2 #Normal Component of E2 in kV/m\n",
+ "E2=E2n+E2t #The total field E2 in kV/m\n",
+ "\n",
+ "theta1= 90- 180*scipy.arccos(dot(E1,an)/ #Angle between E1 and \n",
+ " scipy.sqrt(dot(E1,E1)))/scipy.pi #interface in degrees\n",
+ " \n",
+ "theta2= 90- 180*scipy.arccos(dot(E2,an)/ #Angle between E2 and \n",
+ " scipy.sqrt(dot(E2,E2)))/scipy.pi #interface in degrees\n",
+ "\n",
+ "\n",
+ "We1= 0.5*e0*e_r1*dot(E1,E1)*10**6 # The energy density of E1 in J/m^3\n",
+ "We2= 0.5*e0*e_r2*dot(E2,E2)*10**6 # The energy density of E2 in J/m^3\n",
+ "W= We2*V # The energy within the cube in J\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The electric field for the region z <=0 is',E2,'kV/m'\n",
+ "print 'The angle E1 makes with the boundary is',round(theta1,1),'degrees'\n",
+ "print 'The angle E2 makes with the boundary is',round(theta2,1),'degrees'\n",
+ "print 'The energy density in dielectric 1 is',round(We1*10**6,0),'J/m^3'\n",
+ "print 'The energy density in dielectric 2 is',round(We2*10**6,0),'J/m^3'\n",
+ "print 'The energy within the cube is',round(W*1000,3),'mJ'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The electric field for the region z <=0 is [ 5 -2 4] kV/m\n",
+ "The angle E1 makes with the boundary is 29.1 degrees\n",
+ "The angle E2 makes with the boundary is 36.6 degrees\n",
+ "The energy density in dielectric 1 is 672.0 J/m^3\n",
+ "The energy density in dielectric 2 is 597.0 J/m^3\n",
+ "The energy within the cube is 4.775 mJ\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 5.10, Page number: 190<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Region y <= 0 consists of a perfect conductor while region y >= 0 \n",
+ "is a dielectric medium (epsilon_1r = 2). If there is a surface charge\n",
+ "of 2 nC/m^2 on the conductor, determine E and D at \n",
+ "(a) A(3, -2,2) \n",
+ "(b) B(-4, 1,5) '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "e=(10**-9)/(36*scipy.pi) #Permittivity of free space in Farad/m\n",
+ "er=2 #Relative permittivity (dimensionless)\n",
+ "ps=2 #Surface charge in nC/m^2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Point A is in the region y <=0. Hence E=D=0\n",
+ "#For point B which is in the region y >=0,\n",
+ "\n",
+ "Dn=ps #Displacement current in nC/m^2\n",
+ "En=Dn*10**-9/(e*er) #Electric Field\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'E at point A= 0'\n",
+ "print 'D at point A= 0'\n",
+ "print 'E at point B=',round(En,2),'V/m along positive y direction'\n",
+ "print 'D at point B=',Dn,'nC/m^2 along positive y direction'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "E at point A= 0\n",
+ "D at point A= 0\n",
+ "E at point B= 113.1 V/m along positive y direction\n",
+ "D at point B= 2 nC/m^2 along positive y direction\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_6.ipynb b/Elements_of_Electromagnetics/chapter_6.ipynb new file mode 100644 index 00000000..21f353b6 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_6.ipynb @@ -0,0 +1,80 @@ +{
+ "metadata": {
+ "name": "chapter_6.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 6: Electrostatic Boundary Value Problems<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 6.12, Page number: 238<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Determine the capacitance of each of the capacitors in Figure 6.20.\n",
+ "Take epsilon_r1 = 4, epsilon_r2 = 6, d = 5 mm, S = 30 cm^2 '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "e0=10**-9/(36*scipy.pi) #Permittivity of free space in Farad/m\n",
+ "er1=4 #Relative permittivity of material 1 (dimensionless)\n",
+ "er2=6 #Relative permittivity of material 2 (dimensionless)\n",
+ "d=5*10** -3 #The spacing between the capacitor plates in m\n",
+ "S=30*10** -4 #The surface area of the plates\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "#Calculation for Capacitor 1\n",
+ "\n",
+ "C1=e0*er1*S*2/d #Capacitance in Farads\n",
+ "C2=e0*er2*S*2/d #Capacitance in Farads\n",
+ "Ca=C1*C2/(C1+C2) #Total capacitance in Farads\n",
+ "\n",
+ "#Calculation for Capacitor 2\n",
+ "\n",
+ "C1= e0*er1*S/(2*d) #Capacitance in Farads\n",
+ "C2= e0*er2*S/(2*d) #Capacitance in Farads\n",
+ "Cb= C1+C2 #Total capacitance in Farads\n",
+ "\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'The capacitance of capacitor 1 =',round(Ca*10**12,2),'pF'\n",
+ "print 'The capacitance of capacitor 2 =',round(Cb*10**12,2),'pF'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The capacitance of capacitor 1 = 25.46 pF\n",
+ "The capacitance of capacitor 2 = 26.53 pF\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_7.ipynb b/Elements_of_Electromagnetics/chapter_7.ipynb new file mode 100644 index 00000000..0c96dd8d --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_7.ipynb @@ -0,0 +1,224 @@ +{
+ "metadata": {
+ "name": "chapter_7.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 7: Magnetostatic Fields<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 7.1, Page number: 266<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The conducting triangular loop in Figure 7.6(a) carries a current of lOA.\n",
+ "Find H at (0, 0, 5) due to side 1 of the loop. '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "p=5 #Distance from side 1 of loop to (0,0,5) in m\n",
+ "l=2 #Length of the side in m\n",
+ "I=10 #Current through loop in A\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "a1=scipy.arccos(l/(scipy.sqrt(l**2+p**2))) #Angle in radians\n",
+ "a2=scipy.pi/2 #Angle in radians\n",
+ "H=I*(scipy.cos(a1)-scipy.cos(a2))/(4*scipy.pi*p) #Field Intensity in A/m\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'H=',round(H*1000,1),'mA/m in the negative y direction'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "H= 59.1 mA/m in the negative y direction\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 7.2, Page number: 268<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ " Find H at (-3, 4, 0) due to the current filament shown in Figure 7.7 (a). '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "a1=scipy.arccos(0)\n",
+ "a2=scipy.arccos(1)\n",
+ "b1=scipy.arccos(0.6)\n",
+ "b2=scipy.arccos(1)\n",
+ "p1=5\n",
+ "p2=4\n",
+ "I1=3 #current in A\n",
+ "I2=3 #current in A\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "Hz=I1/(4*scipy.pi*p1)*(cos(a2)-cos(a1))*array([0.8,0.6,0])\n",
+ "Hx=I2/(4*scipy.pi*p2)*(cos(b2)-cos(b1))*array([0,0,1])\n",
+ "Hzcyl=-I1/(4*scipy.pi*p1)*array([0,1,0])\n",
+ "Hzx=round(dot(Hz,ax),4)\n",
+ "Hzy=round(dot(Hz,ay),5)\n",
+ "Hxz=round(dot(Hx,az),5)\n",
+ "Hxr=array([0,0,Hxz])\n",
+ "Hzr=array([Hzx,Hzy,0])\n",
+ "Hzcyly=round(dot(Hzcyl,ay),5)\n",
+ "Hzcylr=array([0,Hzcyly,0])\n",
+ "Hcart=(Hxr+Hzr)*10**3 #H in cartesian coordinates in mA \n",
+ "Hcyl=(Hxr+Hzcylr)*10**3 #H in cylindrical coordinates in mA\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'H at (-3, 4, 0) in cartesian coordnates =',Hcart,'mA/m'\n",
+ "print 'H at (-3, 4, 0) in cylindrical coordnates =',Hcyl,'mA/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "H at (-3, 4, 0) in cartesian coordnates = [ 38.2 28.65 23.87] mA/m\n",
+ "H at (-3, 4, 0) in cylindrical coordnates = [ 0. -47.75 23.87] mA/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 7.5, Page number: 279<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Planes z=0 and z=4 carry current K=-10a_x A/m and K=10a_x A/m, respectively. \n",
+ "Determine H at \n",
+ "(a) (1,1,1) \n",
+ "(b) (0,-3,10) '''\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "i0=-10 #current through plane z=0 in A/m\n",
+ "i4=10 #current through plane z=4 in A/m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "H0a=0.5*i0*-1 #H in positive Y direction in A/m\n",
+ "H4a=0.5*i4*-1*-1 #H in positive Y direction in A/m\n",
+ "Ha=H0a+H4a #H at (1,1,1) in A/m \n",
+ "H0b=0.5*i0*-1 #H in positive Y direction in A/m\n",
+ "H4b=0.5*i4*-1 #H in negative Y direction in A/m\n",
+ "Hb=H0b+H4b #H at (0,-3,10) in A/m\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'H at (1,1,1) =',Ha,'A/m'\n",
+ "print 'H at (0,-3,10) =',Hb,'A/m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "H at (1,1,1) = 10.0 A/m\n",
+ "H at (0,-3,10) = 0.0 A/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 7.7, Page number: 287<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Given the magnetic vector potential A=-p^2/4 a_z Wb/m, calculate \n",
+ "the total magnetic flux crossing the surface phi=pi/2, 1 <= p <= 2 m,\n",
+ "0 <= Z <= 5 m. '''\n",
+ "\n",
+ "import scipy.integrate\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "def B(z,p): \n",
+ " return 0.5*p\n",
+ "psy, err = scipy.integrate.dblquad(lambda p , z: B(z,p), \n",
+ " 0, 5, lambda p: 1, lambda p: 2)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'Total magnetic flux crossing the surface phi=pi/2 is',psy,'Wb'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Total magnetic flux crossing the surface phi=pi/2 is 3.75 Wb\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_8.ipynb b/Elements_of_Electromagnetics/chapter_8.ipynb new file mode 100644 index 00000000..e9adf8d2 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_8.ipynb @@ -0,0 +1,454 @@ +{
+ "metadata": {
+ "name": "chapter_8.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 8: Magnetic Forces, Materials and Devices<h1>"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 8.1, Page number: 308<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A charged particle of mass 2 kg and charge 3 C starts at point (1,-2,0)\n",
+ "with velocity 4a_x+3a_z m/s in an electric field 12a_x+10a_y V/m.\n",
+ "At time t=1s, determine \n",
+ "(a) The acceleration of the particle \n",
+ "(b) Its velocity \n",
+ "(c) Its kinetic energy \n",
+ "(d) Its position '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "m=2 #mass in kg\n",
+ "q=3 #charge in C\n",
+ "v=array([4,0,3]) #initial velocity in m/s\n",
+ "E=array([12,10,0]) #electric field in V/m\n",
+ "t=1 #time in sec\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "a=q*E/m #acceleration in m/s^2 after 1 sec\n",
+ "u=array([18*t+4,15*t,3]) #velocity in m/s after 1 sec\n",
+ "modofu=scipy.sqrt(dot(u,u))\n",
+ "KE=0.5*m*(modofu)**2 #kinetic energy in J at t=1 sec\n",
+ "s=array([9*t**2+4*t+1,7.5*t**2-2,3*t]) #position after 1 sec in m\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'At time t=1 sec,'\n",
+ "print' The acceleration of the particle =',a,'m/s^2'\n",
+ "print 'Its velocity =',u,'m/s' \n",
+ "print 'Its kinetic energy =',KE,'J'\n",
+ "print 'Its position =',s,'m'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "At time t=1 sec,\n",
+ " The acceleration of the particle = [18 15 0] m/s^2\n",
+ "Its velocity = [22 15 3] m/s\n",
+ "Its kinetic energy = 718.0 J\n",
+ "Its position = [ 14. 5.5 3. ] m\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.6, Page number: 322"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A small current loop L1 with magnetic moment 5a_z A/m^2 is located at the origin while \n",
+ "another small loop current I2 with magnetic moment 3a_y A/m^2 is located at (4, - 3, 10). \n",
+ "Determine the torque on L2 . '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ar=array([1,0,0]) #Unit vector along radial direction\n",
+ "ath=array([0,1,0]) #Unit vector along theta direction\n",
+ "aph=array([0,0,1]) #Unit vector along phi direction \n",
+ "x=4\n",
+ "y=-3\n",
+ "z=10\n",
+ "muo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "m1=5 #magnetic moment in A/m^2\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "r=scipy.sqrt(x**2+y**2+z**2)\n",
+ "p=scipy.sqrt(x**2+y**2)\n",
+ "sinphi=y/p\n",
+ "cosphi=x/p\n",
+ "sintheta=1/scipy.sqrt(5)\n",
+ "costheta=2/scipy.sqrt(5)\n",
+ "B1=muo*m1*(2*costheta*ar+sintheta*ath)/(4*scipy.pi*r**3)\n",
+ "m2=3*(sintheta*sinphi*ar+costheta*sinphi*ath+cosphi*aph)\n",
+ "T2=cross(m2,B1)*10**9\n",
+ "T2x=round(dot(T2,ar),3)\n",
+ "T2y=round(dot(T2,ath),3)\n",
+ "T2z=round(dot(T2,aph),3)\n",
+ "T2r=array([T2x,T2y,T2z]) #torque in nNm\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'Torque T2 =',T2r,'nNm'\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Torque T2 = [-0.384 1.536 0.902]\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 8.7, Page number: 330"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Region 0 < Z < 2 m is occupied by an infinite slab of permeable material (mur = 2.5). If \n",
+ "B = IOy ax - 5x ay mWb/m^2 within the slab, determine: (a) J, (b) Jb, (c) M, (d) Kb on \n",
+ "z=0 '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "muo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "mur=2 #relative permeability\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction\n",
+ "\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "J=(-5-10)*10**-6/(4*scipy.pi*10**-7*2.5) #in kA/m^2\n",
+ "Jb=1.5*J #in kA/m^2\n",
+ "MbyB=(1.5)*10**4/(4*scipy.pi*2.5) \n",
+ "Mv=MbyB*10*10**-3*ax+MbyB*5*10**-3*ay\n",
+ "Kb=cross(az,Mv)\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'J =',round(J,3),'kA/m^2'\n",
+ "print 'Jb =',round(Jb,3),'kA/m^2'\n",
+ "print 'M =(',round(dot(Mv,ax),3),'y,',round(dot(Mv,ay),3),'x, 0) kA/m'\n",
+ "print 'Kb =(',round(dot(Kb,ax),3),'x,',round(dot(Kb,ay),3),'y, 0) kA/m' \n",
+ " \n",
+ "\n",
+ " \n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "J = -4.775 kA/m^2\n",
+ "Jb = -7.162 kA/m^2\n",
+ "M =( 4.775 y, 2.387 x, 0) kA/m\n",
+ "Kb =( -2.387 x, 4.775 y, 0) kA/m\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 8.8, Page number: 332<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Given that H=-2a_x + 6a_y + 4 a_z A/m in region y - x - 2 <= 0 where \n",
+ "mu_1=5mu_0 , calculate \n",
+ "(a) M_1 and B_1 \n",
+ "(b) H_2 and B_2 in region y - x - 2 >= 0 where mu_2=2mu_0 '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "ax=array([1,0,0]) #Unit vector along x direction\n",
+ "ay=array([0,1,0]) #Unit vector along y direction\n",
+ "az=array([0,0,1]) #Unit vector along z direction \n",
+ "H1=array([-2,6,4]) #in A/m\n",
+ "mu0=4*scipy.pi*10**-7 #permeability of free space\n",
+ "mur1=5 #relative permeabililty in region 1\n",
+ "mur2=2 #relative permeabililty in region 2\n",
+ "an=array([-1,1,0])/scipy.sqrt(2)\n",
+ "\n",
+ "#Calculatios\n",
+ "\n",
+ "mu1=mu0*mur1\n",
+ "mu2=mu0*mur2\n",
+ "M1=(mur1-1)*H1 # magnetisation in region 1 in A/m\n",
+ "B1=mu1*H1*10**6 # field in micro Wb/m^2\n",
+ "B1x=round(dot(B1,ax),2) # x component of B1\n",
+ "B1y=round(dot(B1,ay),1) # y component of B1\n",
+ "B1z=round(dot(B1,az),2) # z component of B1\n",
+ "B1r=array([B1x,B1y,B1z]) # B1 rounded to 2 decimal places\n",
+ "H1n=dot(H1,an)*an \n",
+ "H1t=H1-H1n\n",
+ "H2t=H1t # using transverse boundary condition\n",
+ "H2n=(mu1/mu2)*H1n # using normal boundary condition\n",
+ "H2=H2t+H2n # in A/m\n",
+ "B2=mu2*H2*10**6 # field in micro Wb/m^2\n",
+ "B2x=round(dot(B2,ax),2) # x component of B2\n",
+ "B2y=round(dot(B2,ay),2) # y component of B2\n",
+ "B2z=round(dot(B2,az),2) # z component of B2\n",
+ "B2r=array([B2x,B2y,B2z]) # B2 rounded to 2 decimal places\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'M1= ',M1,'A/m'\n",
+ "print 'B1= ',B1r,'micro Wb/m^2'\n",
+ "print 'H2= ',H2,'A/m'\n",
+ "print 'B2= ',B2r,'micro Wb/m^2'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "M1= [-8 24 16] A/m\n",
+ "B1= [-12.57 37.7 25.13] micro Wb/m^2\n",
+ "H2= [ -8. 12. 4.] A/m\n",
+ "B2= [-20.11 30.16 10.05] micro Wb/m^2\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 8.14, Page number: 350<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The toroidal core of Figure 8.26(a) has p_o = 10 cm and a circular cross \n",
+ "section with a = 1 cm. If the core is made of steel (mu = 1000mu_o) and \n",
+ "has a coil with 200 turns. calculate the amount of current that will\n",
+ "produce a flux of 0.5 mWb in the core. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable declaration\n",
+ "\n",
+ "p=10*10**-2 #in m\n",
+ "a=1*10**-2 #in m\n",
+ "Ur=1000 #relative permeability\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "n=200 #number of turns\n",
+ "phi=0.5*10**-3 #flux in the core in Wb\n",
+ "U=Uo*Ur #permeability of steel core\n",
+ "\n",
+ "#Calculation\n",
+ "\n",
+ "I=phi*2*scipy.pi*p/(U*n*scipy.pi*a*a) #current in A\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The current that will produce a flux of 0.5 mWb =',round(I,3),'A'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current that will produce a flux of 0.5 mWb = 3.979 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 8.15, Page number: 351<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "In the magnetic circuit of Figure 8.27, calculate the current in the coil \n",
+ "that will produce a magnetic flux density of 1.5 Wb/m^2 in the air gap\n",
+ "assuming that mu_1 = 50 mu_o and that all branches have the same\n",
+ "cross-sectional area of 10 cm^2 . '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "Ur=50 #relative permeability of coil\n",
+ "l1=30*10**-2\n",
+ "s=10*10**-4 \n",
+ "l3=9*10**-2\n",
+ "la=1*10**-2 \n",
+ "B=1.5 #flux density in Wb/m^2\n",
+ "N=400 #number of turns\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "R1=l1/(Uo*Ur*s)\n",
+ "R2=R1\n",
+ "R3=l3/(Uo*Ur*s)\n",
+ "Ra=la/(Uo*s)\n",
+ "R=R1*R2/(R1+R2)\n",
+ "Req=R3+Ra+R\n",
+ "I=B*s*Req/N #current in A\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The current required =',round(I,3),'A'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The current required = 44.165 A\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 8.16, Page number: 353<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A U-shaped electromagnet shown in Figure 8.29 is designed to lift a \n",
+ "400-kg mass (which includes the mass of the keeper). The iron\n",
+ "yoke (mu_r = 3000) has a cross section of 40 cm^2 and mean length of 50 cm,\n",
+ "and the air gaps are each 0.1 mm long. Neglecting the reluctance of \n",
+ "the keeper, calculate the number of turns in the coil when\n",
+ "the excitation current is 1 A. '''\n",
+ "\n",
+ "import scipy\n",
+ "from numpy import *\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "m=400 #mass in kg\n",
+ "g=9.8 #acceleration due to gravity in m/s^2\n",
+ "Ur=3000 #relative permeability of the iron yoke\n",
+ "Uo=4*scipy.pi*10**-7 #permeability of free space\n",
+ "S=40*10**-4 #cross sectional area of iron yoke in m^2\n",
+ "la=1*10**-4 #air gaps in m\n",
+ "li=50*10**-2 #mean length of yoke in m\n",
+ "I=1 #excitation current in A \n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "B=scipy.sqrt(m*g*Uo/S) #field in Wb/m^2\n",
+ "Ra=2*la/(Uo*S) \n",
+ "Ri=li/(Uo*Ur*S) \n",
+ "N=(Ra+Ri)/(Ra*Uo)*B*la #number of turns\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'The nmber of turns in the coil when the excitation current is 1 A ='\n",
+ "print round(N,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "The nmber of turns in the coil when the excitation current is 1 A =\n",
+ "162.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file diff --git a/Elements_of_Electromagnetics/chapter_9.ipynb b/Elements_of_Electromagnetics/chapter_9.ipynb new file mode 100644 index 00000000..02dee160 --- /dev/null +++ b/Elements_of_Electromagnetics/chapter_9.ipynb @@ -0,0 +1,200 @@ +{
+ "metadata": {
+ "name": "chapter_9.ipynb"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h1>Chapter 9: Maxwells Equations<h1>"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 9.1, Page number: 375"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "A conducting bar can slide freely over two conducting rails as shown in Figure 9_6. Calcu- \n",
+ "late the induced voltage in the bar \n",
+ "(a) If the bar is stationed at y = 8 cm and B = 4 cos(10^6t) az mWb/m^2 \n",
+ "(b) If the bar slides at a velocity u = 20ay m/s and B = 4az. m Wb/m 2 \n",
+ "(c) If the bar slides at a velocity u = 20ay m/s and B = 4cos(10^6t - y) az\u001f mWb/m^2 '''\n",
+ "\n",
+ "import scipy\n",
+ "import scipy.integrate\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "u2=20\n",
+ "B2=4\n",
+ "l=0.06\n",
+ "#Calculations\n",
+ "\n",
+ "def ansa(x,y): \n",
+ " return 4*10**3\n",
+ "Va, erra = scipy.integrate.dblquad(lambda y , x: ansa(x,y), #in V \n",
+ " 0, 0.06, lambda y: 0, lambda y: 0.08)\n",
+ "\n",
+ "Vb=-u2*B2*l #in mV\n",
+ "\n",
+ "def ansc(x,y): \n",
+ " return 4\n",
+ "psic, errc = scipy.integrate.dblquad(lambda y , x: ansc(x,y), #in mWb \n",
+ " 0, 0.06, lambda y: 0, lambda y: 1)\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'Va =',Va,'sin(10^6t) V'\n",
+ "print 'Vb =',Vb,'mV'\n",
+ "print 'Vc= ',psic*10**3,'cos(10^6t-y) -',psic*10**3,'cos(10^6t) V'\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Va = 19.2 sin(10^6t) V\n",
+ "Vb = -4.8 mV\n",
+ "Vc= 240.0 cos(10^6t-y) - 240.0 cos(10^6t) V\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 3,
+ "metadata": {},
+ "source": [
+ "Example 9.3, Page number: 379"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "The magnetic circuit of Figure 9.8 has a uniform cross section of 10^-3 m^2 . If the circuit is \n",
+ "energized by a current I1(t) = 3 sin(100pit) A in the coil of N1 = 200 turns, find the emf \n",
+ "induced in the coil of N2 = 100 turns. Assume that mu = 500 muo '''\n",
+ "\n",
+ "import scipy\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "n1=200\n",
+ "n2=100 \n",
+ "S=10**-3 #cross section in m^2\n",
+ "muo=4*scipy.pi*10**-7 #permeabiility of free space\n",
+ "mur=500 #relative permeability\n",
+ "r=10*10**-3 #radius in m\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "psiI=n1*muo*mur*S/(2*scipy.pi*r)\n",
+ "\n",
+ "#Result\n",
+ "\n",
+ "print 'V2 =',psiI*n2*300*scipy.pi,'cos(100pi t) V'\n",
+ "print '= 6Pi cos(100pi t) V'\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "V2 = 188.495559215 cos(100pi t) V\n",
+ "= 6Pi cos(100pi t) V\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "<h3>Example 9.5, Page number: 393<h3>"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "'''\n",
+ "Evaluate the complex numbers \n",
+ " \n",
+ "(a) Zl=(j(3-j4)*)/(-I+j6)(2+j)^2 \n",
+ "(b) Z2=((1+j)/(4-j8))^1/2 '''\n",
+ "\n",
+ "import scipy\n",
+ "import cmath\n",
+ "from numpy import *\n",
+ "\n",
+ "\n",
+ "#Variable Declaration\n",
+ "\n",
+ "z3=1j\n",
+ "z4=3+4j \n",
+ "z5=-1+6j \n",
+ "z6=3+4j\n",
+ "z7=1+1j\n",
+ "z8=4-8j\n",
+ "\n",
+ "#Calculations\n",
+ "\n",
+ "z1=(z3*z4/(z5*z6))\n",
+ "z2=scipy.sqrt(z7/z8)\n",
+ "z1r=round(z1.real,4) #real part of z1 rounded to 4 decimal places\n",
+ "z1i=round(z1.imag,4) #imaginary part of z1 rounded to 4 decimal places\n",
+ "z2r=round(z2.real,4) #real part of z2 rounded to 4 decimal places\n",
+ "z2i=round(z2.imag,4) #imaginary part of z2 rounded to 4 decimal places\n",
+ "\n",
+ "absz2=round(abs(z2),4) #absolute value of z2 rounded to 4 decimal places\n",
+ "\n",
+ "ang=scipy.arctan(z2i/z2r)*180/scipy.pi #in degrees\n",
+ "\n",
+ "#Results\n",
+ "\n",
+ "print 'z1 =',z1r,'+',z1i,'j'\n",
+ "print 'z2 ='\n",
+ "print 'mod =',absz2,'and angle=',round(ang,1),'degrees'"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "z1 = 0.1622 + -0.027 j\n",
+ "z2 =\n",
+ "mod = 0.3976 and angle= 54.2 degrees\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
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