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diff --git a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb b/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb deleted file mode 100755 index fcac8b6a..00000000 --- a/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint.ipynb +++ /dev/null @@ -1,959 +0,0 @@ -{
- "metadata": {
- "name": ""
- },
- "nbformat": 3,
- "nbformat_minor": 0,
- "worksheets": [
- {
- "cells": [
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h1>Chapter 42: Filter networks</h1>"
- ]
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 1, page no. 799</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the cut-off frequency and the nominal impedance of each of the low-pass filter sections\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L1 = 2*100E-3;# in Henry\n",
- "C1 = 0.2E-6;# in Fareads\n",
- "L2 = 0.4;# in Henry\n",
- "C2 = 2*200E-12;# in Fareads\n",
- "\n",
- "#calculation:\n",
- " #cut-off frequency\n",
- "fc1 = 1/(math.pi*(L1*C1)**0.5)\n",
- " #nominal impedance\n",
- "R01 = (L1/C1)**0.5\n",
- " #cut-off frequency\n",
- "fc2 = 1/(math.pi*(L2*C2)**0.5)\n",
- " #nominal impedance\n",
- "R02 = (L2/C2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n cut-off frequency \",round(fc1,2),\" Hz and the nominal impedance is \",round( R01,2),\" ohm \"\n",
- "print \"\\n cut-off frequency \",round(fc2,2),\" Hz and the nominal impedance is \",round( R02,2),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " cut-off frequency 1591.55 Hz and the nominal impedance is 1000.0 ohm \n",
- "\n",
- " cut-off frequency 25164.61 Hz and the nominal impedance is 31622.78 ohm "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 2, page no. 801</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a low-pass T section filter, and (b) a low-pass \u0003 section filter to meet these requirements.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 5E6;# in Hz\n",
- "\n",
- " #calculation:\n",
- " #capacitance\n",
- "C = 1/(math.pi*R0*fc)\n",
- " #inductance\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"A low-pass T section filter capcitance is \",round(C*1E12,2),\"pfarad and inductance is\",round( L/2*1E6,2),\"uHenry\"\n",
- "print \"A low-pass pi section filter capcitance is \",round(C/2*1E12,2),\"pfarad and inductance is\",round( L*1E6,2),\"uHenry\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "A low-pass T section filter capcitance is 106.1 pfarad and inductance is 19.1 uHenry\n",
- "A low-pass pi section filter capcitance is 53.05 pfarad and inductance is 38.2 uHenry\n"
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 3, page no. 805</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine (a) the value of the characteristic impedance of the section at a frequency of 90 kHz, and \n",
- "#(b) the value of the characteristic impedance of the equivalent low-pass T section filter.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 500;# in ohm\n",
- "fc = 100000;# in Hz\n",
- "f = 90000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #characteristic impedance of the pi section\n",
- "Zpi = R0/(1 - (f/fc)**2)**0.5\n",
- " #characteristic impedance of the T section\n",
- "Zt = R0*(1 - (f/fc)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\ncharacteristic impedance of the pi section is \",round(Zpi,2),\" ohm\"\n",
- "print \"\\ncharacteristic impedance of the T section is \",round(Zt,2),\" ohm\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "characteristic impedance of the pi section is 1147.08 ohm\n",
- "\n",
- "characteristic impedance of the T section is 217.94 ohm"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 4, page no. 806</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the frequency at which the characteristic impedance of the section is (a) 600ohm (b) 1 kohm(c) 10 kohm\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 2E6;# in Hz\n",
- "Z1 = 600;# in ohm\n",
- "Z2 = 1000;# in ohm\n",
- "Z3 = 10000;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #frequency\n",
- "f1 = fc*(1 - (R0/Z1)**2)**0.5\n",
- "f2 = fc*(1 - (R0/Z2)**2)**0.5\n",
- "f3 = fc*(1 - (R0/Z3)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"frequency at which the characteristic impedance of the section is 600 ohm is \",f1,\" Hz \"\n",
- "print \"and 1000 Ohm is \",f2*1E-3,\"kHz and 10000 ohm is \",round(f3*1E-3,2),\"kHz \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "frequency at which the characteristic impedance of the section is 600 ohm is 0.0 Hz \n",
- "and 1000 Ohm is 1600.0 kHz and 10000 ohm is 1996.4 kHz "
- ]
- }
- ],
- "prompt_number": 4
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 5, page no. 809</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each of the high-pass filter sections (i) the cut-off frequency, and (ii) the nominal impedance\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L1 = 100*1E-3;# in Henry\n",
- "C1 = 0.2*1E-6;# in Fareads\n",
- "L2 = 200*1E-6;# in Henry\n",
- "C2 = 4000*1E-12;# in Fareads\n",
- "\n",
- "#calculation:\n",
- " #cut-off frequency\n",
- "fc1 = 1/(4*math.pi*(L1*C1/2)**0.5)\n",
- " #nominal impedance\n",
- "R01 = (L1*2/C1)**0.5\n",
- " #cut-off frequency\n",
- "fc2 = 1/(4*math.pi*(L2*C2/2)**0.5)\n",
- " #nominal impedance\n",
- "R02 = (L2/(C2*2))**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n cut-off frequency \",round(fc1,0),\" Hz and the nominal impedance is \",round( R01,0),\" ohm\"\n",
- "print \"\\n cut-off frequency \",round(fc2/1000,0),\"KHz and the nominal impedance is \",round( R02,0),\" ohm \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " cut-off frequency 796.0 Hz and the nominal impedance is 1000.0 ohm\n",
- "\n",
- " cut-off frequency 126.0 KHz and the nominal impedance is 158.0 ohm "
- ]
- }
- ],
- "prompt_number": 1
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 6, page no. 811</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a high-pass T section filter and (b) a high-pass pi-section filter to meet these requirements\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 25000;# in Hz\n",
- "\n",
- "#calculation:\n",
- " #capacitance\n",
- "C1 = 2/(4*math.pi*R0*fc)\n",
- " #inductance\n",
- "L1 = R0/(4*math.pi*fc)\n",
- " #capacitance\n",
- "C2 = 1/(4*math.pi*R0*fc)\n",
- " #inductance\n",
- "L2 = 2*R0/(4*math.pi*fc)\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n A low-pass T section filter capcitance is \",round(C1*1E9,2),\"nfarad and inductance is\",round(L1*1E3,2),\"mHenry\"\n",
- "print \"\\n A high-pass pi section filter capcitance is \",round(C2*1E9,3),\"nfarad and inductance is\",round(L2*1E3,2),\"mHenry\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " A low-pass T section filter capcitance is 10.61 nfarad and inductance is 1.91 mHenry\n",
- "\n",
- " A high-pass pi section filter capcitance is 5.305 nfarad and inductance is 3.82 mHenry"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 8, page no. 814</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the frequency at which the characteristic impedance of the section is (a) zero, (b) 300 ohm, (c) 590 ohm\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 500;# in Hz\n",
- "Z1 = 0;# in ohm\n",
- "Z2 = 300;# in ohm\n",
- "Z3 = 590;# in ohm\n",
- "\n",
- " #calculation:\n",
- " #frequency\n",
- "f1 = fc\n",
- "f2 = fc/(1 - (Z2/R0)**2)**0.5\n",
- "f3 = fc/(1 - (Z3/R0)**2)**0.5\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"requency at which the characteristic impedance of the section is 0 ohm is \",f1,\" Hz \"\n",
- "print \"and 300 Ohm is \",round(f2,2),\" Hz and 590 ohm is \",round(f3,2),\" Hz \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "requency at which the characteristic impedance of the section is 0 ohm is 500 Hz \n",
- "and 300 Ohm is 577.35 Hz and 590 ohm is 2750.1 Hz "
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 9, page no. 817</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for each (i) the attenuation coefficient, and (ii) the phase shift coefficient.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "r1 = 1.25 + 0.52j;# propagation coefficients \n",
- "rr = 1.794;# propagation coefficients \n",
- "thetar = -39.4;# in ddegrees\n",
- "\n",
- "#calculation:\n",
- " #r\n",
- "r2 = rr*math.cos(thetar*math.pi/180) + 1j*rr*math.sin(thetar*math.pi/180)\n",
- " #attenuation coefficient\n",
- "a1 = r1.real\n",
- "a2 = r2.real\n",
- " #phase shift coefficient\n",
- "b1 = r1.imag\n",
- "b2 = r2.imag\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient are for (a) is \",a1,\" N and for (b) is \",round(a2,2),\" N \"\n",
- "print \"\\nphase shift coefficient are for (a) is \",b1,\" rad and for (b) is \",round(b2,2),\" rad \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient are for (a) is 1.25 N and for (b) is 1.39 N \n",
- "\n",
- "phase shift coefficient are for (a) is 0.52 rad and for (b) is -1.14 rad "
- ]
- }
- ],
- "prompt_number": 8
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 10, page no. 818</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine for the section (a) the attenuation coefficient, \n",
- "#(b) the phase shift coefficient, and (c) the propagation coefficient. \n",
- "#(d) If five such sections are cascaded determine the output current of the fifth stage and \n",
- "#the overall propagation constant of the network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "ri1 = 0.024;# in amperes\n",
- "ri2 = 0.008;# in amperes\n",
- "thetai1 = 10;# in ddegrees\n",
- "thetai2 = -45;# in ddegrees\n",
- "\n",
- "#calculation:\n",
- " #currents\n",
- "I1 = ri1*math.cos(thetai1*math.pi/180) + 1j*ri1*math.sin(thetai1*math.pi/180)\n",
- "I2 = ri2*math.cos(thetai2*math.pi/180) + 1j*ri2*math.sin(thetai2*math.pi/180)\n",
- " #ir\n",
- "ir = I1/I2\n",
- "irmag = ri1/ri2\n",
- "thetai = thetai1-thetai2\n",
- " #attenuation coefficient\n",
- "a = math.log(irmag)\n",
- " #phase shift coefficient\n",
- "b = thetai*math.pi/180\n",
- " #propagation coefficient \n",
- "r = a + 1j*b\n",
- " #output current of the fifth stage\n",
- "I6 = I1/(ir**5)\n",
- "x = ir**5\n",
- "xmg = abs(x)\n",
- " #overall attenuation coefficient\n",
- "ad = math.log(xmg)\n",
- " #overall phase shift coefficient\n",
- "bd = cmath.phase(complex(x.real,x.imag))\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient is \",round(a,3),\" N \"\n",
- "print \"\\nphase shift coefficient is \",round(b,3),\" rad \"\n",
- "print \"\\npropagation coefficient is \",round(a,3),\" + (\",round(b,3),\")i \"\n",
- "print \"\\nthe output current of the fifth stage is \",round(abs(I6*1E6),1),\"/_\",round(cmath.phase(complex(I6.real,I6.imag))*180/math.pi,2),\"deg mA \"\n",
- "print \"and the overall propagation coefficient is \",round(ad,2),\" + (\",round(bd+(2*math.pi),2),\")i\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient is 1.099 N \n",
- "\n",
- "phase shift coefficient is 0.96 rad \n",
- "\n",
- "propagation coefficient is 1.099 + ( 0.96 )i \n",
- "\n",
- "the output current of the fifth stage is 98.8 /_ 95.0 deg mA \n",
- "and the overall propagation coefficient is 5.49 + ( 4.8 )i\n"
- ]
- }
- ],
- "prompt_number": 2
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 11, page no. 819</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the attenuation coefficient, (b) the phase shift coefficient and (c) the propagation coefficient gamma\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "XL = 5j;# in ohms\n",
- "Xc = -1j*10;# in ohms\n",
- "RL = 12;# in ohms\n",
- "I1 = 1;# in amperes (lets say)\n",
- "\n",
- " #calculation:\n",
- " #current I2\n",
- "I2 = (Xc/(Xc + XL + RL))*I1\n",
- " #current ratio\n",
- "Ir = I1/I2\n",
- "Irmg = abs(Ir)\n",
- " #attenuation coefficient\n",
- "a = math.log(Irmg)\n",
- " #phase shift coefficient\n",
- "b = cmath.phase(complex(Ir.real, Ir.imag))\n",
- " #propagation coefficient \n",
- "r = a + 1j*b\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\nattenuation coefficient is \",round(a,2),\" N \"\n",
- "print \"\\nphase shift coefficient is \",round(b,2),\" rad \"\n",
- "print \"\\npropagation coefficient is \",round(a,2),\" + (\",round(b,2),\")i \""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- "attenuation coefficient is 0.26 N \n",
- "\n",
- "phase shift coefficient is 1.18 rad \n",
- "\n",
- "propagation coefficient is 0.26 + ( 1.18 )i "
- ]
- }
- ],
- "prompt_number": 10
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 12, page no. 823</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#(a) the time delay for the signal to pass through the filter, assuming the phase shift is small, and \n",
- "#(b) the time delay for a signal to pass through the section at the cut-off frequency.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "L = 2*0.5;# in Henry\n",
- "C = 2E-9;# in Farad\n",
- "\n",
- "#calculation:\n",
- " #time delay\n",
- "t = (L*C)**0.5\n",
- " #time delay at the cut-off frequency\n",
- "tfc = t*math.pi/2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n time delay is \",round(t*1E6,2),\"usec \"\n",
- "print \"\\ntime delay at the cut-off frequency is \",round(tfc*1E6,2),\"usec\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " time delay is 44.72 usec \n",
- "\n",
- "time delay at the cut-off frequency is 70.25 usec"
- ]
- }
- ],
- "prompt_number": 11
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 13, page no. 824</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine (a) the values of the elements in each section, and (b) the value of n.\n",
- "from __future__ import division\n",
- "import math\n",
- "#initializing the variables:\n",
- "fc = 500000;# in Hz\n",
- "t1 = 9.55E-6;# in secs\n",
- "R0 = 1000;# in ohm\n",
- "\n",
- "#calculation:\n",
- " #for a low-pass filter section, capacitance\n",
- "C = 1/(math.pi*R0*fc)\n",
- " #inductance\n",
- "L = R0/(math.pi*fc)\n",
- " #time delay\n",
- "t2 = (L*C)**0.5\n",
- " #number of cascaded sections required\n",
- "n = t1/t2\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for low-pass T section inductance is \",round(L/2*1E6,2),\"uH and capacitance is \",round(C*1E12,2),\"pF\"\n",
- "print \"\\n for low-pass pi section inductance is \",round(L*1E6,2),\"uH and capacitance is \",round(C/2*1E12,2),\"pF\"\n",
- "print \"\\nnumber of cascaded sections required is \",round(n,2)"
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for low-pass T section inductance is 318.31 uH and capacitance is 636.62 pF\n",
- "\n",
- " for low-pass pi section inductance is 636.62 uH and capacitance is 318.31 pF\n",
- "\n",
- "number of cascaded sections required is 15.0"
- ]
- }
- ],
- "prompt_number": 12
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 14, page no. 824</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#determine the component values for each section if the filter is (a) a low-pass T network, and (b) a high-pass \u0003 network.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "n = 8;# sections in cascade\n",
- "R0 = 1000;# in ohm\n",
- "t1 = 4E-6;# in secs\n",
- "\n",
- "\n",
- "#calculation:\n",
- " #time delay\n",
- "t2 = t1/n\n",
- " #capacitance\n",
- "C = t2/R0\n",
- " #inductance\n",
- "L = t2*R0\n",
- "\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for low-pass T section inductance is \",L/2*1E6,\"uH and capacitance is \",C*1E12,\"pF\"\n",
- "print \"\\n for high-pass pi section inductance is \",2*L*1E6,\"uH and capacitance is \",C*1E12,\"pF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for low-pass T section inductance is 250.0 uH and capacitance is 500.0 pF\n",
- "\n",
- " for high-pass pi section inductance is 1000.0 uH and capacitance is 500.0 pF"
- ]
- }
- ],
- "prompt_number": 13
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 15, page no. 829</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) an appropriate \u2018mderived\u2019 T section, and (b) an appropriate \u2018m-derived\u2019 pi section.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 5000; # in Hz\n",
- "finf = 5500; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (fc/finf)**2)**0.5\n",
- "C = 1/(math.pi*R0*fc)\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "LT = m*L/2\n",
- "CT = m*C\n",
- "Ls = (1- m**2)*L/(4*m)\n",
- "\n",
- "Cpi = m*C/2\n",
- "Lpi = m*L\n",
- "Cp = (1- (m**2))*C/(4*m)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n for mderived T section inductance is \",round(Ls*1000,2),\"mH and capacitance is \",round(CT*1E9,2),\"nF\"\n",
- "print \"\\n for mderived pi section inductance is \",round(Lpi*1000,2),\"mH and capacitance is \",round(Cp*1E9,2),\"nF\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " for mderived T section inductance is 18.94 mH and capacitance is 44.2 nF\n",
- "\n",
- " for mderived pi section inductance is 15.91 mH and capacitance is 52.62 nF"
- ]
- }
- ],
- "prompt_number": 5
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 16, page no. 832</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Design (a) a suitable \u2018m-derived\u2019 T section, and (b) a suitable \u2018m-derived\u2019 pi section having a cut-off frequency of 20 kHz,\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 500;# in ohm\n",
- "fc = 20000; # in Hz\n",
- "finf = 16000; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (finf/fc)**2)**0.5\n",
- "C = 1/(4*math.pi*R0*fc)\n",
- "L = R0/(4*math.pi*fc)\n",
- "\n",
- "LT = L/m\n",
- "CT = 4*m*C/(1- m**2)\n",
- "Csa = 2*C/m\n",
- "\n",
- "Cpi = C/m\n",
- "Lpi = 4*m*L/(1- m**2)\n",
- "Lsa = 2*L/m\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For an 'm-derived' high-pass T section: series arm contains a capacitance of \",round(Csa*1E9,2),\"nF\"\n",
- "print \"the shunt arm contains an inductance of\",round(LT*1000,3),\" mH in series with a capacitance of\",round(CT*1E9,2),\"nF\"\n",
- "print \"\\n For an 'm-derived' high pass pi section: shunt arms each contain inductance of \",round(Lsa*1000,2),\"mH\"\n",
- "print \"series arm contains a capacitance of \",round(Cpi*1E9,2),\"nF in parallel with an inductance of\",round(Lpi*1E3,3),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For an 'm-derived' high-pass T section: series arm contains a capacitance of 26.53 nF\n",
- "the shunt arm contains an inductance of 3.316 mH in series with a capacitance of 29.84 nF\n",
- "\n",
- " For an 'm-derived' high pass pi section: shunt arms each contain inductance of 6.63 mH\n",
- "series arm contains a capacitance of 13.26 nF in parallel with an inductance of 7.46 mH\n"
- ]
- }
- ],
- "prompt_number": 3
- },
- {
- "cell_type": "markdown",
- "metadata": {},
- "source": [
- "<h3>Example 17, page no. 835</h3>"
- ]
- },
- {
- "cell_type": "code",
- "collapsed": false,
- "input": [
- "#Determine the component values needed if the filter is to comprise a prototype T section, \n",
- "#an \u2018m-derived\u2019 T section and two terminating \u2018m-derived\u2019 halfsections.\n",
- "from __future__ import division\n",
- "import math\n",
- "import cmath\n",
- "#initializing the variables:\n",
- "R0 = 600;# in ohm\n",
- "fc = 10000; # in Hz\n",
- "finf = 11800; #in Hz\n",
- "\n",
- "#calculation:\n",
- "m = (1 - (fc/finf)**2)**0.5\n",
- "C = 1/(math.pi*R0*fc)\n",
- "L = R0/(math.pi*fc)\n",
- "\n",
- "LmT = (1- m**2)*L/(4*m)\n",
- "\n",
- "mH = 0.6\n",
- "LmH = (1- mH**2)*L/(2*mH)\n",
- "\n",
- "#Results\n",
- "print \"\\n\\n Result \\n\\n\"\n",
- "print \"\\n For an Prototype T section: series arm contains a Inductance of \",round(L*1000/2,1),\"mH\"\n",
- "print \"the shunt arm contains an Capacitance of\",round(C*1E6,4),\" uF\"\n",
- "print \"\\n For an 'm-derived' T section: Series arms each contain inductance of \",round(m*L*1000/2,2),\"mH \"\n",
- "print \"Shunt arm contains a capacitance of \",round(m*C*1E6,4),\"uF in Series with an inductance of\",round(LmT*1E3,2),\"mH\"\n",
- "print \"\\n For an 'm-derived' Half section: Series arms each contain inductance of \",round(mH*L*1000/2,1),\"mH\"\n",
- "print \"Shunt arm contains a capacitance of \",round(mH*C*1E6/2,4),\"uF in Series with an inductance of\",round(LmH*1E3,2),\"mH\""
- ],
- "language": "python",
- "metadata": {},
- "outputs": [
- {
- "output_type": "stream",
- "stream": "stdout",
- "text": [
- "\n",
- "\n",
- " Result \n",
- "\n",
- "\n",
- "\n",
- " For an Prototype T section: series arm contains a Inductance of 9.5 mH\n",
- "the shunt arm contains an Capacitance of 0.0531 uF\n",
- "\n",
- " For an 'm-derived' T section: Series arms each contain inductance of 5.07 mH \n",
- "Shunt arm contains a capacitance of 0.0282 uF in Series with an inductance of 6.46 mH\n",
- "\n",
- " For an 'm-derived' Half section: Series arms each contain inductance of 5.7 mH\n",
- "Shunt arm contains a capacitance of 0.0159 uF in Series with an inductance of 10.19 mH\n"
- ]
- }
- ],
- "prompt_number": 5
- }
- ],
- "metadata": {}
- }
- ]
-}
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