path: root/Electrical_Circuit_Theory_And_Technology/chapter_42-checkpoint_1.ipynb

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 "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": {}
  }
 ]
}