path: root/Analog_and_digital_communication/Chapter8.ipynb

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 "worksheets": [
  {
   "cells": [
    {
     "cell_type": "heading",
     "level": 1,
     "metadata": {},
     "source": [
      "Chapter 8 - Waveform coding techniques"
     ]
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 2 - pg 386"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the transmission bandwidth and final bit rate\n",
      "import math\n",
      "from math import log10\n",
      "#given\n",
      "f_m = 4.2*10**6#bandwidth of television signal\n",
      "q = 512. #quantization levels\n",
      "\n",
      "#calculations\n",
      "#number of bits and quantization levels are related in binary PCM as q = 2^v \n",
      "#where v is code word length\n",
      "v = (log10(q)/log10(2));#code word length\n",
      "BW = v*f_m#transmission channel bandwidth which is greater than or equal to obtained value\n",
      "f_s = 2*f_m#sampling frequency which is greater than or equal to obtained value\n",
      "r = v*f_s#signaling rate of final bit rate\n",
      "SbyN_dB = 4.8 + 6*v#output signal to noise ratio which is less than or equal to obtained value\n",
      "\n",
      "#results\n",
      "print \"i.  Code word length (bits) = \",v\n",
      "print \"ii. Transmission bandwidth (MHz) = \",round(BW/10**6,1)\n",
      "print \"iii.Final bit rate (bits/sec) = \",r\n",
      "print \"iv.Output signal to quantization noise ratio (dB) = \",SbyN_dB\n",
      "print \"Note:There is misprint in the question i.e TV signal bandwidth \"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.  Code word length (bits) =  9.0\n",
        "ii. Transmission bandwidth (MHz) =  37.8\n",
        "iii.Final bit rate (bits/sec) =  75600000.0\n",
        "iv.Output signal to quantization noise ratio (dB) =  58.8\n",
        "Note:There is misprint in the question i.e TV signal bandwidth \n"
       ]
      }
     ],
     "prompt_number": 1
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 3 - pg 387"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the number of bits required, bandwidth required and signalling rate\n",
      "import math\n",
      "#given\n",
      "f_m = 4.*10**3#maximum frequency or bans\n",
      "x_max = 3.8#maximun input signal\n",
      "P = 30.*10**-3#average power of signal\n",
      "SbyN_dB= 20.#signal to noise ratio in db\n",
      "\n",
      "#calculations\n",
      "SbyN = math.exp((SbyN_dB/10)*math.log(10));\n",
      "v = round((math.log10((SbyN*(x_max)**2)/(3*P))/math.log10(2.)/2.));#number of bits required per sample\n",
      "BW = 30*v*f_m#transmission channel bandwidth which is greater than or equal to obtained value\n",
      "r=BW*2#wkt signalling rate is two times the transmission bandwidth\n",
      "\n",
      "#resulta\n",
      "print \"i.Number of bits required (bits) = \",round(v,2)\n",
      "print \"ii.Bandwidth required for 30 PCM coders (kHz) = \",round(BW/1000.,0)\n",
      "print \"iii.Signalling rate (bitspersecond) = \",round(r/1000.,0)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.Number of bits required (bits) =  7.0\n",
        "ii.Bandwidth required for 30 PCM coders (kHz) =  840.0\n",
        "iii.Signalling rate (bitspersecond) =  1680.0\n"
       ]
      }
     ],
     "prompt_number": 2
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 4 - pg 388"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the minumum sampling rate and minumum bit rate required\n",
      "import math\n",
      "#given\n",
      "e_max = .001#maximum quantization error\n",
      "x_max = 10.#maximum amplitude\n",
      "x_min = -10.#minumum amplitude\n",
      "f_m = 100.#bandwidth of ;input signal\n",
      "\n",
      "#calculations\n",
      "delta = 2*e_max#step size\n",
      "q = (2*x_max)/delta#quantization levels\n",
      "f_s = 2*f_m#sampling frequency\n",
      "v = math.ceil(math.log10(q) /math.log10(2));#number of bits in the PCM word\n",
      "r = v * f_s#bit rate required in the PCM signal which is greater than or equal to obtained value\n",
      "BW = .5*r#transmission channel bandwidth which is greater than or equal to obtained value\n",
      "\n",
      "#results\n",
      "print \"i.Minimum sampling rate required (Hz) = \",f_s\n",
      "print \"ii.Number of bits in each PCM word (bits) = \",round(v,2)\n",
      "print \"iii.Minimum bit rate required in the PCM signal (bits/sec) = \",round(r,0)\n",
      "print \"iv.Transmission bandwidth (Hz) = \",round(BW,0)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.Minimum sampling rate required (Hz) =  200.0\n",
        "ii.Number of bits in each PCM word (bits) =  14.0\n",
        "iii.Minimum bit rate required in the PCM signal (bits/sec) =  2800.0\n",
        "iv.Transmission bandwidth (Hz) =  1400.0\n"
       ]
      }
     ],
     "prompt_number": 3
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 5 - pg 389"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the transmission bandwidth and sampling frequency\n",
      "#given\n",
      "f_m = 3.4*10**3#maximum frequency in the signal\n",
      "N =24.#number of voice signals\n",
      "r = 1.5*10**6#signaling rate\n",
      "v  = 8.#bits  of encoder\n",
      "\n",
      "#calculations\n",
      "BW = N * f_m#transmission bandwidth\n",
      "r_1 = r/N#bit rate for one channel\n",
      "f_s = r_1/v#sampling frquency\n",
      "\n",
      "#results\n",
      "print \"i.Transmission bandwidth (kHz) = \",BW/1000.\n",
      "print \"ii.Sampling frequency (Hz or samples per second) = \",f_s\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.Transmission bandwidth (kHz) =  81.6\n",
        "ii.Sampling frequency (Hz or samples per second) =  7812.5\n"
       ]
      }
     ],
     "prompt_number": 4
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 6 - pg 389"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the maximum message bandwidth and signal to noise ratio\n",
      "\n",
      "#given\n",
      "v = 7.#bits  of encoder\n",
      "r = 50.*10**6#bit rate of the system\n",
      "\n",
      "#calculations\n",
      "f_m = r/(2*v)#maximum message bandwidth which is less than or equal to obtained value\n",
      "SbyN_dB = 1.8 + 6*v#signal to noise ratio in dB\n",
      "\n",
      "#results\n",
      "print \"i.Maximum message bandwidth (Hz) = \",round(f_m/10**6,2)\n",
      "print \"ii.Signal to noise ratio when modulating frquency is 1MHz applied (dB) = \",SbyN_dB\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.Maximum message bandwidth (Hz) =  3.57\n",
        "ii.Signal to noise ratio when modulating frquency is 1MHz applied (dB) =  43.8\n"
       ]
      }
     ],
     "prompt_number": 5
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 7 - pg 390"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the Minimum sampling rate and bit transmission rate\n",
      "import math\n",
      "from math import log\n",
      "#given\n",
      "f_m = 3*10**3#maximum frequency\n",
      "M = 16.#number of quantization levels\n",
      "q = M #number of quantization levels\n",
      "\n",
      "#calculations\n",
      "v = log(q)/log(2);#number of bits\n",
      "f_s = 2*f_m#sampling frequency or rate which is greater than or equal to obtained value\n",
      "r = v*f_s#bit transmission rate which is greater than or equal to obtained value\n",
      "\n",
      "#results\n",
      "print \"i.Number of bits in a codeword (bits) = \",v\n",
      "print \"ii.Minimum sampling rate (kHz) =  \",f_s/1000.\n",
      "print \"iii.Bit transmission rate (bits/sec) = \",r\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.Number of bits in a codeword (bits) =  4.0\n",
        "ii.Minimum sampling rate (kHz) =   6.0\n",
        "iii.Bit transmission rate (bits/sec) =  24000.0\n"
       ]
      }
     ],
     "prompt_number": 6
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 8 - pg 391"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the signal to noise ratio\n",
      "import math\n",
      "from math import log10\n",
      "#given\n",
      "f_m = 3.5*10**3#maximum frequency\n",
      "r = 50*10**3#bit rate \n",
      "v_rms = .2#rms value of input signal\n",
      "R = 1#resistance\n",
      "x_max = 2#maximum peak voltage\n",
      "\n",
      "#calculations\n",
      "f_s = 2*f_m;#sampling frequency\n",
      "v = math.ceil(r/f_s);#number of bits\n",
      "P = v_rms**2 / R#Normalized signal power\n",
      "SbyN = ((3*P) * 2**(2*v)) /(x_max**2);#signal to noise ratio\n",
      "SbyN_dB = 10*log10(SbyN)#signal to noise ratio in dB\n",
      "\n",
      "#results\n",
      "print \"i.Signal to noise ratio in dB = \",round(SbyN_dB,0)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.Signal to noise ratio in dB =  33.0\n"
       ]
      }
     ],
     "prompt_number": 7
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 10 - pg 392"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the signal to noise ratio and number of bits needed\n",
      "import math\n",
      "#given\n",
      "#x(t) = 3*cos(500*%pi*t)\n",
      "v = 10.#number of bits\n",
      "A_m = 3.#peak voltage\n",
      "SbyN_2 = 40.#signal to noise to noise ratio in second condition\n",
      "\n",
      "#calculations\n",
      "SbyN = 1.8 +6*v#signal to noise ratio in dB\n",
      "v_2 = (40 - 1.8)/6#number of bits needed for SbyN = 40\n",
      "\n",
      "#results\n",
      "print \"i.Signal to noise to ratio in dB \",SbyN\n",
      "print \"ii.Number of bits needed for noise ratio 40 (bits) = \",math.ceil(v_2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.Signal to noise to ratio in dB  61.8\n",
        "ii.Number of bits needed for noise ratio 40 (bits) =  7.0\n"
       ]
      }
     ],
     "prompt_number": 9
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 11 - pg 393"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the maximum frequency\n",
      "\n",
      "#given\n",
      "v = 7.#number of bits\n",
      "r = 56*10**3#signaling rate\n",
      "\n",
      "#calculations\n",
      "SbyN = 1.8 +6*v#signal to noise ratio in dB\n",
      "f_s = r/v#sampling frequency\n",
      "f_m = f_s/2#maximum frequency which is less than or equal to obtained value\n",
      "\n",
      "#results\n",
      "print \"Maximum frequency (Hz) = \",f_m\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum frequency (Hz) =  4000.0\n"
       ]
      }
     ],
     "prompt_number": 11
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 13 - pg 404"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the maximum amplitude\n",
      "import math\n",
      "#given\n",
      "f_m = 3.*10**3#bandwidth or maximum frequency\n",
      "n = 5.#system operation times\n",
      "delta = 250.*10**-3#step size in volts\n",
      "f_m1 = 2.*10**3#given maximum frequency to calculate amplitude\n",
      "\n",
      "#calculations\n",
      "NR = 2 * f_m#nyquist rate\n",
      "f_s = n * NR#sampling frequency\n",
      "T_s = 1/f_s#sampling interval\n",
      "A_m =(delta/(2 * math.pi * f_m1* T_s))#Maximum amplitude\n",
      "\n",
      "#result\n",
      "print \"Maximum amplitude for 2KHz input sinusoid (V) = \",round(A_m,1)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Maximum amplitude for 2KHz input sinusoid (V) =  0.6\n"
       ]
      }
     ],
     "prompt_number": 12
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 14 - pg 406"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the signalling rate\n",
      "import math\n",
      "from math import log\n",
      "#given\n",
      "f_s = 8*10**3#sampling rate\n",
      "q = 64.#quantization levels\n",
      "delta = 31.25#step size\n",
      "\n",
      "#calculations\n",
      "v = log(q)/log(2);#no fo bits in the PC\n",
      "f_s= (2*math.pi*3*10**3)/delta#signalling rate which should be greater than the obtaining value\n",
      "\n",
      "#results\n",
      "print \"Signalling rate (kHz) = \",round(f_s,2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Signalling rate (kHz) =  603.19\n"
       ]
      }
     ],
     "prompt_number": 13
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 15 - pg 407"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the signal to noise ratio\n",
      "import math\n",
      "#given\n",
      "f_m = 2.*10**3#maximum frequency\n",
      "f_s = 64.*10**3#sampling frequency\n",
      "f_M = 4.*10**3#cut off frequency of low pass filter\n",
      "\n",
      "#calculation\n",
      "SNR_0 = (3 * f_s**3) /(8 * math.pi**2 * f_m**2 * f_M);#signal to noise ratio of linear delta modulation system\n",
      "SNR_dB = 10*math.log10(SNR_0);#SNR in dB\n",
      "\n",
      "#result\n",
      "print \"Signal to noise ratio of linear delta modulation system (dB) = \",round(SNR_dB,2)\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Signal to noise ratio of linear delta modulation system (dB) =  27.94\n"
       ]
      }
     ],
     "prompt_number": 15
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 16 - pg 407"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the signal to noise ratio\n",
      "\n",
      "#given\n",
      "r = 64.*10**3#data rate\n",
      "f_s = 8.*10**3#sampling frequency\n",
      "N = 8.#number of samples\n",
      "\n",
      "#calcualtion\n",
      "SNR_q = 1.8 + 6*N#signal to noise ratio\n",
      "\n",
      "#result\n",
      "print \"Signla to noise ratio (dB) = \",SNR_q\n",
      "print \"The SNR of a DM system is 27.94dB which is too poor as \\ncompared to 49.8db of an 8 bit PCM system. Thus, for all\\n the simplicity of Dm,it cannot perform as well as an\\n 8 bit PCM\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Signla to noise ratio (dB) =  49.8\n",
        "The SNR of a DM system is 27.94dB which is too poor as \n",
        "compared to 49.8db of an 8 bit PCM system. Thus, for all\n",
        " the simplicity of Dm,it cannot perform as well as an\n",
        " 8 bit PCM\n"
       ]
      }
     ],
     "prompt_number": 16
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 17 - pg 413"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the sampling frequency and quantizing level\n",
      "import math\n",
      "#given\n",
      "r = 36000.#bit rate of a channel\n",
      "f_m = 3.2*10**3#maximum frequency\n",
      "\n",
      "#calculations\n",
      "f_s = 2*f_m#sampling frequency\n",
      "v  = math.floor(r/f_s)#number of binary digits\n",
      "q = 2**v#quantizing level\n",
      "f_s2=r/v\n",
      "#results\n",
      "print \"i.Sampling frequency (Hz) = \",f_s\n",
      "print \"ii.Number of binary digits = \",round(v,1)\n",
      "print \"iii.Quantizing level = \",round(q,0)\n",
      "print \"iv. Sampling rate (Hz) = \",f_s2"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.Sampling frequency (Hz) =  6400.0\n",
        "ii.Number of binary digits =  5.0\n",
        "iii.Quantizing level =  32.0\n",
        "iv. Sampling rate (Hz) =  7200.0\n"
       ]
      }
     ],
     "prompt_number": 17
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 20 - pg 415"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the number of required levels and output signal to quantizing noise ratio\n",
      "import math\n",
      "from math import log, exp, sqrt\n",
      "#given\n",
      "SbyN_0dB = 40.#signal to noise ratio in dB\n",
      "SbyN_0 = exp((SbyN_0dB/10)*log(10))#signal to noise ratio\n",
      "q = sqrt((2. / 3) * (SbyN_0));#quantizing level\n",
      "v = math.ceil(log(q)/log(2.))#number of binary bits\n",
      "q_1 = 2**v#number of levels required \n",
      "SbyN_dB1 = 1.76 + 6.02*v#output signal-to-quantizing noise ratio in dB\n",
      "\n",
      "#results\n",
      "print \"Number of required levels = \",v\n",
      "print \"Output signal-to-quantizing noise ratio (dB) = \",SbyN_dB1\n",
      "print \"Note : In the textbook they took number of levels as approximation so we get change\\n in SbyN\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "Number of required levels =  7.0\n",
        "Output signal-to-quantizing noise ratio (dB) =  43.9\n",
        "Note : In the textbook they took number of levels as approximation so we get change\n",
        " in SbyN\n"
       ]
      }
     ],
     "prompt_number": 18
    },
    {
     "cell_type": "heading",
     "level": 2,
     "metadata": {},
     "source": [
      "Example 21 - pg 416"
     ]
    },
    {
     "cell_type": "code",
     "collapsed": false,
     "input": [
      "#calculate the minimum number of quantizing levels and bits, bandwidth required\n",
      "import math\n",
      "from math import log,exp,sqrt\n",
      "#given\n",
      "SbyN_dB = 30.#signal to noise ratio\n",
      "f_s = 8000.#sampling rate\n",
      "\n",
      "#calculations\n",
      "#for Sbyn_dB = 1.76 + 20*logq\n",
      "x1 = (1./ 20)*(SbyN_dB - 1.76)\n",
      "q1 = exp(x1*log(10))#quantizing level for first case\n",
      "v1 = math.ceil(log(q1)/log(2))# number of bits  for first case\n",
      "f_PCM1 = (v1 / 2) * f_s#minimum required bandwidth  for first case\n",
      "#for SbyN = 20logq - 10.1\n",
      "x2 = (1./20) * (SbyN_dB + 10.1)\n",
      "q2 =  exp(x2*log(10))#quantizing level for second case\n",
      "v2 = math.ceil(log(q2)/log(2))# number of bits for second case\n",
      "f_PCM2 =  (v2 / 2) * f_s#minimum required bandwidth for second case\n",
      "\n",
      "#results\n",
      "print \"i.a.Minimum number of quantizing levels for first case = \",round(q1,2)\n",
      "print \"  b.Number of bits for first case = \",v1\n",
      "print \"  c.Minimum system bandwidth required for first case (kHz) = \",f_PCM1/1000.\n",
      "print \"ii.a.Minimum number of quantizing levels for second case = \",round(q2,1)\n",
      "print \"   b.Number of bits for second case = \",v2\n",
      "print \"   c.Minimum system bandwidth required for second case (kHz) = \",f_PCM2/1000.\n",
      "print \"Note:In the text book they took approximation in\\nquantization levels and number bits\"\n"
     ],
     "language": "python",
     "metadata": {},
     "outputs": [
      {
       "output_type": "stream",
       "stream": "stdout",
       "text": [
        "i.a.Minimum number of quantizing levels for first case =  25.82\n",
        "  b.Number of bits for first case =  5.0\n",
        "  c.Minimum system bandwidth required for first case (kHz) =  20.0\n",
        "ii.a.Minimum number of quantizing levels for second case =  101.2\n",
        "   b.Number of bits for second case =  7.0\n",
        "   c.Minimum system bandwidth required for second case (kHz) =  28.0\n",
        "Note:In the text book they took approximation in\n",
        "quantization levels and number bits\n"
       ]
      }
     ],
     "prompt_number": 19
    }
   ],
   "metadata": {}
  }
 ]
}