{
 "cells": [
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "# Chapter 7 : Digital Communication"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 1 : pg 285"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 1,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "maximum data rate for four-level code in the available bandwidth 12.8 kb/s\n",
      "maximum data rate for four-level code in the available bandwidth 37.21 kb/s\n"
     ]
    }
   ],
   "source": [
    " \n",
    "#page no 285\n",
    "#prob no 7.1\n",
    "#calculate the max data rate in both cases\n",
    "from math import log\n",
    "# In the given problem a signal is transmitted using a four level code\n",
    "#given\n",
    "M=4.;\n",
    "B=3.2;# in KKz\n",
    "SNR=35.;#in dB\n",
    "#calculations and results\n",
    "#By using Shannon-Hartley theorem, ignoring noise we have\n",
    "c=2*B*log(M) / log(2);\n",
    "print 'maximum data rate for four-level code in the available bandwidth',c,'kb/s'\n",
    "#Now we have to use Shannon limit to find the maximum data rate for any code\n",
    "#SNR in power ratio is \n",
    "SNR1=10**(35./10.);\n",
    "C=B*log(1+SNR1) /log(2);\n",
    "print 'maximum data rate for four-level code in the available bandwidth',round(C,2),'kb/s'\n",
    "# Both results are maxima, we have to choose lesser of the two.\n",
    "# Therefore we choose c=12.8kp/s"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 2 : pg 289"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 2,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The audible frequency is 14.1 KHz\n"
     ]
    }
   ],
   "source": [
    " \n",
    "#page no. 289\n",
    "# prob no. 7.2\n",
    "# In the given problem\n",
    "#calculate the audible frequency \n",
    "#given\n",
    "fm = 30.# in KHz\n",
    "fs = 44.1#sampling rate in KHz\n",
    "#calculations\n",
    "fa = fs - fm# audible frequency\n",
    "#results\n",
    "print 'The audible frequency is',fa,'KHz'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 3 : pg 291"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 3,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "a) The number of levels with m=8 are 256.0 levels\n",
      "b) The number of levels with m=16 are 65536 levels\n"
     ]
    }
   ],
   "source": [
    " \n",
    "#page no 291\n",
    "#prob no 7.3\n",
    "#calculate the number of levels in both cases\n",
    "#part a: no of samples,\n",
    "#given\n",
    "m=8.;\n",
    "#calculations and results\n",
    "N=2**m;# the number of levels\n",
    "print 'a) The number of levels with m=8 are',N,'levels'\n",
    "# part b:\n",
    "m=16;\n",
    "N=2**m;# the number of levels\n",
    "print 'b) The number of levels with m=16 are',N,'levels'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 4 : pg 292"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 4,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "Dynamic range for a linear PCM 98.08 dB\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 292\n",
    "# prob no 7.4\n",
    "#calculate the dynamic range\n",
    "#In the given problem\n",
    "#given\n",
    "m=16.;\n",
    "#calculations\n",
    "DR=1.76 +6.02*m ; #Dynamic range for a linear PCM in dB\n",
    "#results\n",
    "print 'Dynamic range for a linear PCM',DR,'dB'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 5 : pg 295"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 5,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The minimum data rate needed to transmit audio is  560.0 Kb/s\n"
     ]
    }
   ],
   "source": [
    " \n",
    "#page no 295\n",
    "# prob no 7.5\n",
    "#calculate the min data rate required\n",
    "# in the given problem\n",
    "#given\n",
    "fs=40.; m=14;\n",
    "#calculations\n",
    "# the minimum data rate needed to transmit audio is given by\n",
    "D=fs*m;\n",
    "#results\n",
    "print 'The minimum data rate needed to transmit audio is ',D,'Kb/s'"
   ]
  },
  {
   "cell_type": "markdown",
   "metadata": {},
   "source": [
    "## Example 6 : pg 294"
   ]
  },
  {
   "cell_type": "code",
   "execution_count": 6,
   "metadata": {
    "collapsed": false
   },
   "outputs": [
    {
     "name": "stdout",
     "output_type": "stream",
     "text": [
      "The maximum output voltage produced is 0.876 volts\n"
     ]
    }
   ],
   "source": [
    " \n",
    "# page no 294\n",
    "# prob no 7.6\n",
    "#calculate the max output voltage required\n",
    "# In the given problem, input to a mu-law compresser is +ve,\n",
    "# with its voltage one-half the max value\n",
    "from math import log\n",
    "#given\n",
    "u=255.;\n",
    "Vi=1.;#maximum input value is considered as unity volts\n",
    "vi=0.5;\n",
    "V0=1.;#consider maximum output voltage as unity volts\n",
    "#calculations\n",
    "vo=V0* log(1+u*vi/Vi)/log(1+u);\n",
    "#results\n",
    "print 'The maximum output voltage produced is',round(vo,3),'volts'"
   ]
  }
 ],
 "metadata": {
  "kernelspec": {
   "display_name": "Python 2",
   "language": "python",
   "name": "python2"
  },
  "language_info": {
   "codemirror_mode": {
    "name": "ipython",
    "version": 2
   },
   "file_extension": ".py",
   "mimetype": "text/x-python",
   "name": "python",
   "nbconvert_exporter": "python",
   "pygments_lexer": "ipython2",
   "version": "2.7.11"
  }
 },
 "nbformat": 4,
 "nbformat_minor": 0
}