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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 10 - Digital Multiplexers"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 1 - pg 469"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the minimum value of permissible sampling rate\n",
      "\n",
      "#given\n",
      "X_1 = 4.*10**3#first analog signal in Hz\n",
      "X_2 = 4.5*10**3#second analog signal in Hz\n",
      "\n",
      "#calculation\n",
      "#the highest frequency cmponent of the composite signal consisting among two signal is X_2\n",
      "f_sMIN = 2*X_2;\n",
      "\n",
      "print \"The minimum value of permissible sampling rate (kHz) = \",f_sMIN/1000.\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "The minimum value of permissible sampling rate (kHz) =  9.0\n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2 - pg 469"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the signalling rate and minimum channel bandwidth\n",
      "\n",
      "#given\n",
      "X_1 = 6.*10**3#Nyquist rate in Hz obtained the table\n",
      "X_2 = 2.*10**3#Nyquist rate in Hz obtained the table\n",
      "X_3 = 2.*10**3#Nyquist rate in Hz obtained the table\n",
      "X_4 = 2.*10**3#Nyquist rate in Hz obtained the table\n",
      "\n",
      "#calculations\n",
      "s = 2000#speed of rotation\n",
      "X1 = 3*s#number of samples produced per second for first signal\n",
      "X2 = 1*s#number of samples produced per second for second signal\n",
      "X3 = X2#number of samples produced per second for third signal\n",
      "X3 = X2#number of samples produced per second for fourth signal\n",
      "SR = X1 + 3*X2#signalling rate\n",
      "BW = .5*SR#minimum channel bandwidth \n",
      "\n",
      "#results\n",
      "print \"If the sampling commutator rotates at the rate of 2000 rotations per second the the signals X_1,X_2,X_3,X_4 will be sampled at their Nyquist rate\"\n",
      "print \"Signalling rate (samples per second) = \",SR\n",
      "print \"Minimum channel bandwidth (Hz) = \",BW\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "If the sampling commutator rotates at the rate of 2000 rotations per second the the signals X_1,X_2,X_3,X_4 will be sampled at their Nyquist rate\n",
        "Signalling rate (samples per second) =  12000\n",
        "Minimum channel bandwidth (Hz) =  6000.0\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3 - pg 470"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the sampling rate and Nyquist rate\n",
      "\n",
      "#given\n",
      "SR = 8000.#sampling rate in samples per second\n",
      "T = 1.*10**-6#pulse duration\n",
      "f = 3.4*10**3#highest frequency component\n",
      "\n",
      "#calculations\n",
      "#second case\n",
      "NR = 2.*f#Nyquist rate of sampling\n",
      "T2 = 1./NR#time taken for one rotation of commutator\n",
      "\n",
      "\n",
      "#results\n",
      "print \"sampling rate for first condition = \",SR\n",
      "print \"There are 24 voice signals + 1 synchronizing pulse\"\n",
      "print \"Pulse width of each voice channel and synchronizing pulseis 1 microseconds \"\n",
      "print \"Now, time taken by the commutator for 1 rotation =1/8000 = 125*10**-6 seconds\"\n",
      "print \"Number of pulses produced in one rotation = 24 + 1 = 25\"\n",
      "print \"Therefore, the leading edges of the pulses are at 125/25 = 5*10^-6 seconds distance\"\n",
      "print \"Nyquist rate for second condition (kHz) = \",NR/1000.\n",
      "print \"Time taken for one rotation of commutator (museconds) = \",round(T2*10**6,2)\n",
      "print \"Therefore, 147*10**-6 seconds corresponds to 25 pulses\"\n",
      "print \"therefore, 1 pulse corresponds to 5.88*10^-6 seconds\"\n",
      "print \"As the pulse width of each pulse is 1*10^-6 seconds, the spacing between adjacent pulses will be 4.88*10**-6 seconds\\n and if we assume tou = 0 then the spacing between the adjacent pulses will be 5.88*10**-6 seconds \"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "sampling rate for first condition =  8000.0\n",
        "There are 24 voice signals + 1 synchronizing pulse\n",
        "Pulse width of each voice channel and synchronizing pulseis 1 microseconds \n",
        "Now, time taken by the commutator for 1 rotation =1/8000 = 125*10**-6 seconds\n",
        "Number of pulses produced in one rotation = 24 + 1 = 25\n",
        "Therefore, the leading edges of the pulses are at 125/25 = 5*10^-6 seconds distance\n",
        "Nyquist rate for second condition (kHz) =  6.8\n",
        "Time taken for one rotation of commutator (museconds) =  147.06\n",
        "Therefore, 147*10**-6 seconds corresponds to 25 pulses\n",
        "therefore, 1 pulse corresponds to 5.88*10^-6 seconds\n",
        "As the pulse width of each pulse is 1*10^-6 seconds, the spacing between adjacent pulses will be 4.88*10**-6 seconds\n",
        " and if we assume tou = 0 then the spacing between the adjacent pulses will be 5.88*10**-6 seconds \n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4 - pg 471"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Signaling rate and minimum channel bandwidth\n",
      "\n",
      "#given\n",
      "N = 6.#number of channels\n",
      "f_m = 5.*10**3#bandwidth of each channel\n",
      "\n",
      "#calculations\n",
      "SR1= 2*f_m#minimum sampling rate\n",
      "SR = N*SR1#sampling rate\n",
      "BW =N*f_m#minimum channel bandwidth\n",
      "\n",
      "#results\n",
      "print \"Signaling rate (Kbits per second) = \",SR/1000.\n",
      "print \"Minimum channel bandwidth (kHz) = \",BW/1000.\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Signaling rate (Kbits per second) =  60.0\n",
        "Minimum channel bandwidth (kHz) =  30.0\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6 - pg 476"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the number of framing plus signaling bits per frame\n",
      "\n",
      "#given\n",
      "channel=64*10**3 #kb/s\n",
      "bitrate=2.048*10**6 #bits/sec\n",
      "\n",
      "#calculations\n",
      "fs_min=channel*2\n",
      "x=bitrate/fs_min\n",
      "\n",
      "#results\n",
      "print 'No. of bits per frame = ',x"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "No. of bits per frame =  16.0\n"
       ]
      }
     ],
     "prompt_number": 5
    }
   ],
   "metadata": {}
  }
 ]
}