{ "metadata": { "name": "" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "markdown", "metadata": {}, "source": [ "

Chapter 42: Filter networks

" ] }, { "cell_type": "markdown", "metadata": {}, "source": [ "

Example 1, page no. 799

" ] }, { "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": [ "

Example 2, page no. 801

" ] }, { "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": [ "

Example 3, page no. 805

" ] }, { "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": [ "

Example 4, page no. 806

" ] }, { "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": [ "

Example 5, page no. 809

" ] }, { "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": [ "

Example 6, page no. 811

" ] }, { "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": [ "

Example 8, page no. 814

" ] }, { "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": [ "

Example 9, page no. 817

" ] }, { "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": [ "

Example 10, page no. 818

" ] }, { "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": [ "

Example 11, page no. 819

" ] }, { "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": [ "

Example 12, page no. 823

" ] }, { "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": [ "

Example 13, page no. 824

" ] }, { "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": [ "

Example 14, page no. 824

" ] }, { "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": [ "

Example 15, page no. 829

" ] }, { "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": [ "

Example 16, page no. 832

" ] }, { "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": [ "

Example 17, page no. 835

" ] }, { "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": {} } ] }