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| author | hardythe1 | 2015-06-03 15:27:17 +0530 |
|---|---|---|
| committer | hardythe1 | 2015-06-03 15:27:17 +0530 |
| commit | 47d7279a724246ef7aa0f5359cf417992ed04449 (patch) | |
| tree | c613e5e4813d846d24d67f46507a6a69d1a42d87 /Analog_and_digital_communication/Chapter11.ipynb | |
| parent | 435840cef00c596d9e608f9eb2d96f522ea8505a (diff) | |
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diff --git a/Analog_and_digital_communication/Chapter11.ipynb b/Analog_and_digital_communication/Chapter11.ipynb new file mode 100755 index 00000000..642532b4 --- /dev/null +++ b/Analog_and_digital_communication/Chapter11.ipynb @@ -0,0 +1,1653 @@ +{
+ "metadata": {
+ "name": "",
+ "signature": "sha256:d35f66e241c83fce49bf0f825461c9bef7321fa8529d41a1200b86c1371dbb65"
+ },
+ "nbformat": 3,
+ "nbformat_minor": 0,
+ "worksheets": [
+ {
+ "cells": [
+ {
+ "cell_type": "heading",
+ "level": 1,
+ "metadata": {},
+ "source": [
+ "Chapter 11 - Information theory"
+ ]
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 1 - pg 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the information content in all symbols\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "Px_1=1./2;#probability 1\n",
+ "Px_2=1./4;#probability 2\n",
+ "Px_3=1./8;#probability 3\n",
+ "Px_4=1./8;#probability 4\n",
+ "\n",
+ "#calculations\n",
+ "Ix_1 = log2(1/(Px_1))#information content in first probability\n",
+ "Ix_2 = log2(1/(Px_2))#information content in first probability\n",
+ "Ix_3 = log2(1/(Px_3))#information content in first probability\n",
+ "Ix_4 = log2(1/(Px_3))#information content in first probability\n",
+ "\n",
+ "#results\n",
+ "print \"i. Information content of first symbol (bits) = \",Ix_1\n",
+ "print \"ii. Information content of second symbol (bits) = \",Ix_2\n",
+ "print \"iii. Information content of third symbol (bits) = \",Ix_3\n",
+ "print \"iV. Information content of fourth symbol (bits)\",Ix_4\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i. Information content of first symbol (bits) = 1.0\n",
+ "ii. Information content of second symbol (bits) = 2.0\n",
+ "iii. Information content of third symbol (bits) = 3.0\n",
+ "iV. Information content of fourth symbol (bits) 3.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 1
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 2 - pg 488"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the amount of information\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "Px_i = 1./4#probability of a symbol\n",
+ "\n",
+ "#calculation\n",
+ "Ix_i = (log(1/Px_i))/log(2)#formula for amount of information of a symbol\n",
+ "\n",
+ "#result\n",
+ "print \"i. Amount of information (bits) = \",Ix_i\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i. Amount of information (bits) = 2.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 2
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 3 - pg 489"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the amount of information\n",
+ "#given\n",
+ "import math\n",
+ "from math import log\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#since there are only two binary levels i.e. 1 or 0. Since, these two binary levels occur with equal likelihood of occurrence will be\n",
+ "Px_1 = 1./2#probability of zero level\n",
+ "Px_2 = 1./2#probability of first level\n",
+ "\n",
+ "#calculations\n",
+ "Ix_1 = log2(1/Px_1)#amount of information of zero level with base 2\n",
+ "Ix_2 = log2(1/Px_2)#amount of information of first level with base 2\n",
+ "Ix_1= log(1/Px_1)/log(2)#amount of information content with base 10\n",
+ "Ix_2 = Ix_1 \n",
+ "\n",
+ "#result\n",
+ "print \"i.Amount of information content wrt binary PCM 0 (bit) = \",Ix_1\n",
+ "print \"ii.Amount of information content wrt binary PCM 1 (bit) = \",Ix_2\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Amount of information content wrt binary PCM 0 (bit) = 1.0\n",
+ "ii.Amount of information content wrt binary PCM 1 (bit) = 1.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 3
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 4 - pg 489"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the amount of information\n",
+ "import math\n",
+ "from math import log\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#given\n",
+ "Px_1 = 1./4#probability wrt to binary PCM '0'\n",
+ "Px_2 = 3./4#probability wrt to binary PCM '1'\n",
+ "\n",
+ "#calculations\n",
+ "Ix_1 = log2(1/Px_1)#amount of information of zero level with base 2\n",
+ "Ix_2 = log2(1/Px_2)#amount of information of first level with base 2\n",
+ "Ix_1= log(1/Px_1)/log(2)#amount of information content with base 10\n",
+ "Ix_2= log(1/Px_2)/log(2)#amount of information content with base 10\n",
+ "\n",
+ "#results\n",
+ "print \"i.Amount of information carried wrt to binary PCM 0 (bits) = \",Ix_1\n",
+ "print \"ii.Amount of information carried wrt to binary PCM 1 (bits) = \",round(Ix_2,3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Amount of information carried wrt to binary PCM 0 (bits) = 2.0\n",
+ "ii.Amount of information carried wrt to binary PCM 1 (bits) = 0.415\n"
+ ]
+ }
+ ],
+ "prompt_number": 4
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 9 - pg 492"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the entropy and Amount of information\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "\n",
+ "Px_1 = .4#probability of first symbol\n",
+ "Px_2 = .3#probability of second symbol\n",
+ "Px_3 = .2#probability of third symbol\n",
+ "Px_4 = .1#probability of fourth symbol\n",
+ "\n",
+ "#calculations\n",
+ "H_X = -Px_1*log2(Px_1)-Px_2*log2(Px_2)-Px_3*log2(Px_3)-Px_4*log2(Px_4);#entropy\n",
+ "Px1x2x1x3 = Px_1*Px_2*Px_1*Px_3;#product of probabilities\n",
+ "Ix1x2x1x3 =-log2(Px1x2x1x3);#information of four symbols\n",
+ "Px4x3x3x2 = Px_4*Px_3*Px_3*Px_2;#product of probabilities\n",
+ "Ix4x3x3x2 = -log2(Px4x3x3x2);#information of four symbols\n",
+ "\n",
+ "#results\n",
+ "print \"i.Entropy (bits/symbol) = \",round(H_X,2)\n",
+ "print \"ii.Amount of information contained in x1x2x1x3 (bits/symbol) = \",round(Ix1x2x1x3,2)\n",
+ "print \"Thus,Ix1x2x1x3 < 7.4[=4*H_X]bits/symbol\"\n",
+ "print \"iii.Amount of information contained in x4x3x3x2 (bits/symbol) = \",round(Ix4x3x3x2,2)\n",
+ "print \"\\nThus we conclude that\\nIx4x3x3x2 > 7.4[=4*H_X]bits/symbol\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Entropy (bits/symbol) = 1.85\n",
+ "ii.Amount of information contained in x1x2x1x3 (bits/symbol) = 6.7\n",
+ "Thus,Ix1x2x1x3 < 7.4[=4*H_X]bits/symbol\n",
+ "iii.Amount of information contained in x4x3x3x2 (bits/symbol) = 9.7\n",
+ "\n",
+ "Thus we conclude that\n",
+ "Ix4x3x3x2 > 7.4[=4*H_X]bits/symbol\n"
+ ]
+ }
+ ],
+ "prompt_number": 5
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 12 - pg 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the average rate of information convyed\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "\n",
+ "n = 2.*10**6#elements od black and white TV picture\n",
+ "m = 16.#brightness levels of black and white TV picture\n",
+ "o = 32.#repeated rate of pictures per second \n",
+ "\n",
+ "#calculations\n",
+ "Px_i = 1./m#probability of brightness levels of picture\n",
+ "H_X = 0;\n",
+ "for i in range (0,15):\n",
+ "\tH_Xi = (-1./(1./Px_i))*log2(1./(1./Px_i));\n",
+ "\tH_X = H_X +H_Xi\n",
+ "\n",
+ "r = n*o#rate of symbols generated \n",
+ "R = r*math.ceil(H_X)#average rate of information convyed \n",
+ " \n",
+ "#results\n",
+ "print \"i. Average rate of information convyed (Mbits/seconds) = \",R/10**6\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i. Average rate of information convyed (Mbits/seconds) = 256.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 6
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 13 - pg 495"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the average information rate\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "\n",
+ "t_dot = .2#duration of dot symbol\n",
+ "t_dash = .6#duration of dash symbol\n",
+ "t_space = .2#time between the symbols\n",
+ "#wkt sum of the probability is 1 i.e P_dot + P_dash = 1 hence\n",
+ "#P_dot = 2*P_dash weget \n",
+ "P_dot = 2./3#probality of dot symbol\n",
+ "P_dash = 1./3#probality of dash symbol\n",
+ "\n",
+ "#calculations \n",
+ "H_X = -P_dot*log2(P_dot)-P_dash*log2(P_dash);#entropy\n",
+ "T_s = P_dot*t_dot + P_dash*t_dash +t_space;#average time per symbol\n",
+ "r = 1/T_s;#average symbol rate \n",
+ "R = r*H_X;#average information rate of the telegraph sourece\n",
+ "\n",
+ "#result\n",
+ "print \"i.The average information rate of the telegraph source (bits/seconds) = \",round(R,3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.The average information rate of the telegraph source (bits/seconds) = 1.722\n"
+ ]
+ }
+ ],
+ "prompt_number": 7
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 14 - pg 496"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the entropy and information rate\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#given symbols are equally likely all the symbols the probabilities are same\n",
+ "Px_1 = 1./8;#probability of first symbol\n",
+ "Px_2 = 1./8;#probability of second symbol\n",
+ "Px_3 = 3./8;#probability of third symbol\n",
+ "Px_4 = 3./8;#probability of fourth symbol\n",
+ "\n",
+ "r = 2#average symbol rate from problem 11.14\n",
+ "\n",
+ "#calculaitons\n",
+ "H_X = Px_1*log2(1/Px_1) + Px_2*log2(1/Px_2) + Px_3*log2(1/Px_3) + Px_4*log2(1/Px_4);#entropy\n",
+ "R = H_X*r;#information rate\n",
+ "\n",
+ "#results\n",
+ "print \"i.Entropy (bits/symbol) = \",round(H_X,1)\n",
+ "print \"ii.The information rate of all symbols (f_m bits/seconds) = \", round(R,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Entropy (bits/symbol) = 1.8\n",
+ "ii.The information rate of all symbols (f_m bits/seconds) = 3.6\n"
+ ]
+ }
+ ],
+ "prompt_number": 8
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 15 - pg 497"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the information rate\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#given symbols are equally likely all the symbols the probabilities are same\n",
+ "Px_1 = 1./4;#probability of first symbol\n",
+ "Px_2 = 1./4;#probability of second symbol\n",
+ "Px_3 = 1./4;#probability of third symbol\n",
+ "Px_4 = 1./4;#probability of fourth symbol\n",
+ "r = 2#average symbol rate from problem 11.14\n",
+ "\n",
+ "#calculaitons\n",
+ "H_X = Px_1*log2(1/Px_1) + Px_2*log2(1/Px_2) + Px_3*log2(1/Px_3) + Px_4*log2(1/Px_4);#entropy\n",
+ "R = H_X*r;#information rate\n",
+ "\n",
+ "#results\n",
+ "print \"i.The information rate of all symbols (f_m bits/seconds) = \", R\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.The information rate of all symbols (f_m bits/seconds) = 4.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 9
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 16 - pg 498"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Entropy and rate of information\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "Px_1 = 1./2;#probability of first symbol\n",
+ "Px_2 = 1./4;#probability of second symbol\n",
+ "Px_3 = 1./8;#probability of third symbol\n",
+ "Px_4 = 1./16;#probability of fourth symbol\n",
+ "Px_4 = 1./16;#probability of fifth symbol\n",
+ "T_b = 1*10**-3#time required for emittion of each symbol\n",
+ "r = 1./(T_b)#symbol rate\n",
+ "\n",
+ "#calculations\n",
+ "H_X = Px_1*log2(1/Px_1) + Px_2*log2(1/Px_2) + Px_3*log2(1/Px_3) + Px_4*log2(1/Px_4) + Px_4*log2(1/Px_4);\n",
+ "R = r*H_X;#information rate\n",
+ "\n",
+ "#results\n",
+ "print \"i.Entropy of five symbols (bits/symbol) = \",H_X\n",
+ "print \"ii.Rate of information (bits/sec) = \",R\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Entropy of five symbols (bits/symbol) = 1.875\n",
+ "ii.Rate of information (bits/sec) = 1875.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 10
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 17 - pg 498"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the rate of information\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "Px_1 = 1./2;#probability of first symbol\n",
+ "Px_2 = 1./4;#probability of second symbol\n",
+ "Px_3 = 1./8;#probability of third symbol\n",
+ "Px_4 = 1./16;#probability of fourth symbol\n",
+ "Px_5 = 1./16;#probability of fifth symbol\n",
+ "r = 16#outcomes per second\n",
+ "\n",
+ "#calculations\n",
+ "H_X = Px_1*log2(1/Px_1) + Px_2*log2(1/Px_2) + Px_3*log2(1/Px_3) + Px_4*log2(1/Px_4) + Px_5*log2(1/Px_5);\n",
+ "R = r*H_X;#information rate\n",
+ "\n",
+ "#result\n",
+ "print \"Rate of information (bits/sec) = \",R\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of information (bits/sec) = 30.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 11
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 18 - pg 499"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the rate of information\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "Px_1 = 1./4;#probability of first symbol\n",
+ "Px_2 = 1./5;#probability of second symbol\n",
+ "Px_3 = 1./5;#probability of third symbol\n",
+ "Px_4 = 1./10;#probability of fourth symbol\n",
+ "Px_5 = 1./10;#probability of fifth symbol\n",
+ "Px_6 = 1./20;#probability of sixth symbol\n",
+ "Px_7 = 1./20;#probability of seventh symbol\n",
+ "Px_8 = 1./20;#probability of eigith symbol\n",
+ "f_m = 10*10**3#freuency of tranamitting symbol\n",
+ "\n",
+ "#calculations\n",
+ "H_X = Px_1*log2(1/Px_1) + Px_2*log2(1/Px_2) + Px_3*log2(1/Px_3) + Px_4*log2(1/Px_4) + Px_5*log2(1/Px_5) + Px_6*log2(1/Px_6)+ Px_7*log2(1/Px_7)+ Px_8*log2(1/Px_8);#entropy\n",
+ "f_s = 2*f_m#sampling frequency\n",
+ "r = f_s#sampling frequency equal to rate of transmission\n",
+ "R = r*H_X;#information rate\n",
+ "\n",
+ "#result\n",
+ "print \"Rate of information (bits/sec) = \",round(R,0)\n",
+ "print \"Note:Their mistake in calculation of H_X in textbook\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Rate of information (bits/sec) = 54829.0\n",
+ "Note:Their mistake in calculation of H_X in textbook\n"
+ ]
+ }
+ ],
+ "prompt_number": 12
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 19 - pg 502"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the channel matrix\n",
+ "import numpy\n",
+ "#given\n",
+ "#from fig\n",
+ "P_X = numpy.matrix([.5, .5])#x matrix\n",
+ "P_Xd = numpy.matrix([[.5, 0],[0, .5]])#diagonal x matrix\n",
+ "#calculations\n",
+ "P_YX = numpy.matrix([[.9, .1],[.2, .8]]);#yx matrix representation of given fig\n",
+ "P_Y = P_X*P_YX#y matrix\n",
+ "P_XY = P_Xd * P_YX#xy matrix\n",
+ "\n",
+ "#results\n",
+ "print \"i.Channel matrix of the channelP_YX \"\n",
+ "print(P_YX)\n",
+ "print \"ii.a.P(y1) =\",P_Y[0,0]\n",
+ "print \" b.P(y2) = \",P_Y[0,1]\n",
+ "print \"iii.a.P(x1,y2) = \",P_XY[0,1]\n",
+ "print \" b.P(x2,y1) = \",P_XY[1,0]\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Channel matrix of the channelP_YX \n",
+ "[[ 0.9 0.1]\n",
+ " [ 0.2 0.8]]\n",
+ "ii.a.P(y1) = 0.55\n",
+ " b.P(y2) = 0.45\n",
+ "iii.a.P(x1,y2) = 0.05\n",
+ " b.P(x2,y1) = 0.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 14
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 20 - pg 503"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Channel matrix\n",
+ "import numpy\n",
+ "#given\n",
+ "P_X = numpy.matrix([.5, .5])#x matrix\n",
+ "#calculations\n",
+ "P_YX = numpy.matrix([[.9, .1],[.2, .8]]);#yx matrix representation of given fig\\\n",
+ "P_ZY = numpy.matrix([[.9, .1],[.2, .8]]);#yx matrix representation of given fig\n",
+ "P_Y = P_X *P_YX#y matrix\n",
+ "P_ZX = P_YX * P_ZY#zx matrix\n",
+ "P_Z = P_X *P_ZX#z matrix\n",
+ "\n",
+ "\n",
+ "#results\n",
+ "print \"i.Channel matrix of the channelP_ZX \"\n",
+ "print(P_ZX)\n",
+ "print(\"Matrix P(Z)\")\n",
+ "print(P_Z)\n",
+ "print \"a.P(Z1) = \",P_Z[0,0]\n",
+ "print \"b.P(Z2) = \",P_Z[0,1]\n",
+ "\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Channel matrix of the channelP_ZX \n",
+ "[[ 0.83 0.17]\n",
+ " [ 0.34 0.66]]\n",
+ "Matrix P(Z)\n",
+ "[[ 0.585 0.415]]\n",
+ "a.P(Z1) = 0.585\n",
+ "b.P(Z2) = 0.415\n"
+ ]
+ }
+ ],
+ "prompt_number": 15
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 21 - pg 504"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the probability associated with channel outputs\n",
+ "#given\n",
+ "import numpy\n",
+ "P_X = numpy.matrix([.5, .5])#x matrix\n",
+ "P_YX = numpy.matrix([[.8, .2, 0], [0, .2, .8]]);#yx matrix representation of given fig\n",
+ "\n",
+ "#calculations\n",
+ "P_Y = P_X*P_YX;\n",
+ "\n",
+ "#results\n",
+ "print \"probability associated with the channel outputs for p=.2 is\"\n",
+ "print(P_Y)\n",
+ "print \"a.P(Y1) = \",P_Y[0,0]\n",
+ "print \"b.P(Y2) = \",P_Y[0,1]\n",
+ "print \"C.P(Y3) = \",P_Y[0,2]\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "probability associated with the channel outputs for p=.2 is\n",
+ "[[ 0.4 0.2 0.4]]\n",
+ "a.P(Y1) = 0.4\n",
+ "b.P(Y2) = 0.2\n",
+ "C.P(Y3) = 0.4\n"
+ ]
+ }
+ ],
+ "prompt_number": 16
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 28 - pg 504"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the I_XY value\n",
+ "import numpy\n",
+ "import math\n",
+ "from math import log\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#given\n",
+ "#wkt P_Y = P_X*P_YX from previous problems\n",
+ "alfa = .5\n",
+ "P_1 = .1#probability for first case\n",
+ "P_2 = .5#probability for second case\n",
+ "\n",
+ "#calculations\n",
+ "P_X = numpy.matrix([alfa, alfa]);\n",
+ "#first case\n",
+ "P_YX = ([[1-P_1, P_1],[P_1, 1-P_1]]);\n",
+ "P_Y1 = P_X*P_YX;\n",
+ "H_Y1 = -P_Y1[0,0]*log2(P_Y1[0,0])-P_Y1[0,1]*log2(P_Y1[0,1]);\n",
+ "Q_1 = P_1*log2(P_1) + (1-P_1)*log2(1-P_1)#from proof\n",
+ "I_XY1 = 1 + Q_1;\n",
+ "#second case\n",
+ "P_YX = ([[1-P_2, P_2],[P_2, 1-P_2]]);\n",
+ "P_Y2 = P_X*P_YX;\n",
+ "H_Y2 = -P_Y2[0,0]*log2(P_Y2[0,0])-P_Y2[0,1]*log2(P_Y2[0,1]);\n",
+ "Q_2 = P_2*log2(P_2) + (1-P_2)*log2(1-P_2)#from proof \n",
+ "I_XY2 = 1 + Q_2;\n",
+ "\n",
+ "#results\n",
+ "print \"I_XY for the first case = \",round(I_XY1,3)\n",
+ "print \"I_XY for the second case = \",round(I_XY2,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "I_XY for the first case = 0.531\n",
+ "I_XY for the second case = 0.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 17
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 32 - pg 518"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Entropy in all cases\n",
+ "import math\n",
+ "from math import log\n",
+ "\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#given\n",
+ "a1 = 1.\n",
+ "a2 = 2.\n",
+ "a3 = .5\n",
+ "\n",
+ "#calculations\n",
+ "H_X1 = log2(a1);#Entropy for first case\n",
+ "H_X2 = log2(a2);#Entropy for second case\n",
+ "H_X3 = log2(a3);#Entropy for third case\n",
+ "\n",
+ "#results\n",
+ "print \"i.Entropy for first case = \",H_X1\n",
+ "print \"ii.Entropy for second case = \",H_X2\n",
+ "print \"iii.Entropy for third case = \",H_X3\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Entropy for first case = 0.0\n",
+ "ii.Entropy for second case = 1.0\n",
+ "iii.Entropy for third case = -1.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 18
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 35 - pg 520"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the capacity of channel\n",
+ "\n",
+ "import math\n",
+ "from math import log\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#given\n",
+ "B = 4000.#bandwidth of AWGN channel\n",
+ "S = .1*10**-3#power of signal\n",
+ "neta = 2*10**-12#spectral dencity \n",
+ "N = neta*B;#power \n",
+ "\n",
+ "#calculations \n",
+ "C = B * log2(1 + (S/N));#capacity of channel\n",
+ "\n",
+ "#result\n",
+ "print \"Capacity of channel (b/s) = \",round(C,0)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Capacity of channel (b/s) = 54439.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 19
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 36 - pg 521"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the information rate, bandwidth and S/N ratio\n",
+ "import math\n",
+ "from math import log\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#given\n",
+ "fm=4000. #Hz\n",
+ "tim=1.25\n",
+ "Pi=1/256.\n",
+ "SN=100.\n",
+ "#calculations \n",
+ "fs=2*fm\n",
+ "r=fs*tim\n",
+ "H=log2(1/Pi)\n",
+ "R=r*H\n",
+ "C = r* log2(1 + SN);#capacity of channel\n",
+ "SN2=(1/Pi-1)\n",
+ "B2=R/log2(1+SN)\n",
+ "#result\n",
+ "print \"information rate of source (kb/s) = \",R/1000.\n",
+ "print \"Capacity of channel (b/s) = \",round(C,0)\n",
+ "print \"Final S/N ratio = \",SN2\n",
+ "print \"bandwidth required (kHz) = \",round(B2/1000.,0)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "information rate of source (kb/s) = 80.0\n",
+ "Capacity of channel (b/s) = 66582.0\n",
+ "Final S/N ratio = 255.0\n",
+ "bandwidth required (kHz) = 12.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 20
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 37 - pg 524"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the efficiency of code and code redundancy\n",
+ "\n",
+ "import math\n",
+ "from math import log\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "#given\n",
+ "Px_1 = 0.9#probability of first symbol\n",
+ "Px_2 = 0.1#probability of second symbol\n",
+ "n1 = 1.#length of the code for x_1\n",
+ "n2 =1.#length of code for x_2\n",
+ "\n",
+ "#calculations\n",
+ "#we know that the average code length L per symbol\n",
+ "L = Px_1*n1 + Px_2*n2#code length\n",
+ "H_X = -Px_1*log2(Px_1) - Px_2*log2(Px_2) #entropy\n",
+ "neta = H_X/L#efficiency \n",
+ "neta1 = neta*100#neta in percentage\n",
+ "gama = 1 - neta#redundancy\n",
+ "gama1 = gama*100#gama in percentage\n",
+ "\n",
+ "#results\n",
+ "print \"i.Efficiency of code (percent) = \",round(neta1,1)\n",
+ "print \"ii.Code redundancy (percent) = \",round(gama1,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Efficiency of code (percent) = 46.9\n",
+ "ii.Code redundancy (percent) = 53.1\n"
+ ]
+ }
+ ],
+ "prompt_number": 21
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 38 - pg 524"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the efficiency of code and code redundancy\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "Px_1 = 0.81#probability of first symbol\n",
+ "Px_2 = .09#probability of second symbol\n",
+ "Px_3 = .09#probability of third symbol\n",
+ "Px_4 = 0.01#probability of forth symbol\n",
+ "n1 = 1.#length of code for a_1\n",
+ "n2 =2.#length of code for a_2\n",
+ "n3 = 3.#length of code for a_3\n",
+ "n4 = 3.#length of code for a_4\n",
+ "\n",
+ "#calculations\n",
+ "#we know that the average code length L per symbol\n",
+ "L = Px_1*n1 + Px_2*n2 + Px_3*n3 + Px_4*n4 #code length\n",
+ "H_X = -Px_1*log2(Px_1) - Px_2*log2(Px_2) - Px_3*log2(Px_3) - Px_4*log2(Px_4)#entropy \n",
+ "neta = H_X/L#efficiency \n",
+ "neta1 = neta*100#neta in percentage\n",
+ "gama = 1 - neta#redundancy\n",
+ "gama1 = gama*100#gama in percentage\n",
+ "\n",
+ "#results\n",
+ "print \"i.Efficiency of code (percent) = \",round(neta1,1)\n",
+ "print \"ii.Code redundancy (percent) = \",round(gama1,1)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "i.Efficiency of code (percent) = 72.7\n",
+ "ii.Code redundancy (percent) = 27.3\n"
+ ]
+ }
+ ],
+ "prompt_number": 22
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 44 - pg 529"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the efficiency \n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "P_x1 = 1./2#probability of first symbol\n",
+ "P_x2 = 1./4#probability of second symbol\n",
+ "P_x3 = 1./8#probability of third symbol\n",
+ "P_x4 = 1./8#probability of fouth symbol\n",
+ "n1 = 1.\n",
+ "n2 = 2.\n",
+ "n3 = 3.\n",
+ "n4 = 3.\n",
+ "\n",
+ "#calculations\n",
+ "I_x1 = -log2(P_x1);\n",
+ "I_x2 = -log2(P_x2);\n",
+ "I_x3 = -log2(P_x3);\n",
+ "I_x4 = -log2(P_x4);\n",
+ "H_x = P_x1*I_x1 + P_x2*I_x2 + P_x3*I_x3 + P_x4*I_x4;\n",
+ "L = P_x1*n1 + P_x2*n2 + P_x3*n3 + P_x4*n4;\n",
+ "neta = H_x/L;\n",
+ "P_neta = neta*100#efficiency in percentage\n",
+ "\n",
+ "#results\n",
+ "print \"Efficiency = \",neta\n",
+ "print \"Efficiency in percentage (percent) = \",P_neta\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency = 1.0\n",
+ "Efficiency in percentage (percent) = 100.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 23
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 45 - pg 532"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the efficiency using shannon code and huffman code\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "P_x1 = .2#probability of first signal\n",
+ "P_x2 = .2#probability of second signal\n",
+ "P_x3 = .2#probability of third signal\n",
+ "P_x4 = .2#probability of fourth signal\n",
+ "P_x5 = .2#probability of fifth signal\n",
+ "n1 = 2.#number of bits in code obtained from table given textbook\n",
+ "n2 = 2.#number of bits in code obtained from table given textbook\n",
+ "n3 = 2.#number of bits in code obtained from table given textbook\n",
+ "n4 = 3.#number of bits in code obtained from table given textbook\n",
+ "n5 = 3.#number of bits in code obtained from table given textbook\n",
+ "\n",
+ "#calculations\n",
+ "I_x1 = -log2(P_x1);\n",
+ "I_x2 = -log2(P_x2);\n",
+ "I_x3 = -log2(P_x3);\n",
+ "I_x4 = -log2(P_x4);\n",
+ "I_x5 = -log2(P_x5);\n",
+ "H_x = P_x1*I_x1 + P_x2*I_x2 + P_x3*I_x3 + P_x4*I_x4 + P_x5*I_x5;#entropy\n",
+ "L1 = P_x1*n1 + P_x2*n2 + P_x3*n3 + P_x4*n4 + P_x5*n5;\n",
+ "neta1 = H_x/L1;\n",
+ "P_neta1 = neta1*100#efficiency in percentage using Shannon Fano code\n",
+ "L2 = P_x1*n1 + P_x2*n2 + P_x3*n3 +P_x4*n4 +P_x5*n5\n",
+ "neta2 = H_x/L2;\n",
+ "P_neta2 = neta2*100#efficiency in percentage using huffman code\n",
+ "\n",
+ "#results\n",
+ "print \"Efficiency in percentage using Shannon Fano code (percent) = \",round(P_neta1,1)\n",
+ "print \"Efficiency in percentage using huffman code (percent) = \",round(P_neta2,1)"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency in percentage using Shannon Fano code (percent) = 96.7\n",
+ "Efficiency in percentage using huffman code (percent) = 96.7\n"
+ ]
+ }
+ ],
+ "prompt_number": 24
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 46 - pg 532"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the efficiency using shannon code and huffman code\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "P_x1 = .4#probability of first signal\n",
+ "P_x2 = .19#probability of second signal\n",
+ "P_x3 = .16#probability of third signal\n",
+ "P_x4 = .15#probability of fourth signal\n",
+ "P_x5 = .1#probability of fifth signal\n",
+ "n1 = 1.#number of bits in code obtained from table given textbook\n",
+ "n2 = 2.#number of bits in code obtained from table given textbook\n",
+ "n3 = 2.#number of bits in code obtained from table given textbook\n",
+ "n4 = 3.#number of bits in code obtained from table given textbook\n",
+ "n5 = 3.#number of bits in code obtained from table given textbook\n",
+ "\n",
+ "#calculations\n",
+ "I_x1 = -log2(P_x1);\n",
+ "I_x2 = -log2(P_x2);\n",
+ "I_x3 = -log2(P_x3);\n",
+ "I_x4 = -log2(P_x4);\n",
+ "I_x5 = -log2(P_x5);\n",
+ "H_x = P_x1*I_x1 + P_x2*I_x2 + P_x3*I_x3 + P_x4*I_x4 + P_x5*I_x5;#entropy\n",
+ "L1 = P_x1*n1 + P_x2*n2 + P_x3*n3 + P_x4*n4 + P_x5*n5;\n",
+ "neta1 = H_x/L1;\n",
+ "P_neta1 = neta1*100#efficiency in percentage using Shannon Fano code\n",
+ "L2 = P_x1*1 + (P_x2 + P_x3 +P_x4 +P_x5 )*3\n",
+ "neta2 = H_x/L2;\n",
+ "P_neta2 = neta2*100#efficiency in percentage using huffman code\n",
+ "\n",
+ "#results\n",
+ "print \"Efficiency in percentage using Shannon Fano code (percent) = \",round(P_neta1,1)\n",
+ "print \"Efficiency in percentage using huffman code (percent) = \",round(P_neta2,1)\n",
+ "print \"Note: There is mistake in the textbook in calculation of L using SHannon Fano code\""
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency in percentage using Shannon Fano code (percent) = 116.2\n",
+ "Efficiency in percentage using huffman code (percent) = 97.7\n",
+ "Note: There is mistake in the textbook in calculation of L using SHannon Fano code\n"
+ ]
+ }
+ ],
+ "prompt_number": 26
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 47 - pg 532"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Efficiency\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "P_x1 = .05#probability of first signal\n",
+ "P_x2 = .15#probability of second signal\n",
+ "P_x3 = .2#probability of third signal\n",
+ "P_x4 = .05#probability of fourth signal\n",
+ "P_x5 = .15#probability of fifth signal\n",
+ "P_x6 = .3#probability of sixth signal\n",
+ "P_x7 = .1#probability of seventh signal\n",
+ "n1 = 4.#number of bits in code obtained from table given textbook\n",
+ "n2 = 3.#number of bits in code obtained from table given textbook\n",
+ "n3 = 2.#number of bits in code obtained from table given textbook\n",
+ "n4 = 4.#number of bits in code obtained from table given textbook\n",
+ "n5 = 3.#number of bits in code obtained from table given textbook\n",
+ "n6 = 2.#number of bits in code obtained from table given textbook\n",
+ "n7 = 3.#number of bits in code obtained from table given textbook\n",
+ "\n",
+ "#calculations\n",
+ "I_x1 = -log2(P_x1);\n",
+ "I_x2 = -log2(P_x2);\n",
+ "I_x3 = -log2(P_x3);\n",
+ "I_x4 = -log2(P_x4);\n",
+ "I_x5 = -log2(P_x5);\n",
+ "I_x6 = -log2(P_x6);\n",
+ "I_x7 = -log2(P_x7);\n",
+ "H_x = P_x1*I_x1 + P_x2*I_x2 + P_x3*I_x3 + P_x4*I_x4 + P_x5*I_x5 + P_x6*I_x6 + P_x7*I_x7;#entropy\n",
+ "L = P_x1*n1 + P_x2*n2 + P_x3*n3 + P_x4*n4 + P_x5*n5 + P_x6*n6 + P_x7*n7;\n",
+ "neta = (H_x*100)/L#Efficiency in percentage\n",
+ "\n",
+ "#results\n",
+ "print \"Efficiency in percentage (percent) = \",round(neta,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Efficiency in percentage (percent) = 98.88\n"
+ ]
+ }
+ ],
+ "prompt_number": 27
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 49 - pg 534"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Variance of codeword length\n",
+ "\n",
+ "#given\n",
+ "P_x1 = .4#probability of first signal\n",
+ "P_x2 = .2#probability of second signal\n",
+ "P_x3 = .8#probability of third signal\n",
+ "P_x4 = .08#probability of fourth signal\n",
+ "P_x5 = .02#probability of fifth signal\n",
+ "n1 = 2.#number of bits in code obtained from table given textbook\n",
+ "n2 = 3.#number of bits in code obtained from table given textbook\n",
+ "n3 = 1.#number of bits in code obtained from table given textbook\n",
+ "n4 = 4.#number of bits in code obtained from table given textbook\n",
+ "n5 = 4.#number of bits in code obtained from table given textbook\n",
+ "\n",
+ "#calculations\n",
+ "L = P_x1*n1 + P_x2*n2 + P_x3*n3 + P_x4*n4 + P_x5*n5;#average codeword length per symbol\n",
+ "#since sigma = sqrt(summation of product of probability and (n- L)**2)\n",
+ "sigmasquare = P_x1*(n1-L)**2 + P_x2*(n2-L)**2 +P_x3*(n3-L)**2 + P_x4*(n4-L)**2 +P_x5*(n5-L)**2;#Variance of codewoed length\n",
+ "\n",
+ "#results\n",
+ "print \"Variance of codeword length = \",sigmasquare\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Variance of codeword length = 2.42\n"
+ ]
+ }
+ ],
+ "prompt_number": 28
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 50 - pg 535"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Entropy and Information rate\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "P_x1 = 1./2#probability of first signal\n",
+ "P_x2 = 1./4#probability of second signal\n",
+ "P_x3 = 1./8#probability of third signal\n",
+ "P_x4 = 1./16#probability of fourth signal\n",
+ "P_x5 = 1./32#probability of fifth signal\n",
+ "P_x6 = 1./32#probability of sixth signal\n",
+ "r = 16#message rate in outcomes per second\n",
+ "\n",
+ "#calculations\n",
+ "I_x1 = -log2(P_x1);\n",
+ "I_x2 = -log2(P_x2);\n",
+ "I_x3 = -log2(P_x3);\n",
+ "I_x4 = -log2(P_x4);\n",
+ "I_x5 = -log2(P_x5);\n",
+ "I_x6 = -log2(P_x6);\n",
+ "H_X = P_x1*I_x1 + P_x2*I_x2 + P_x3*I_x3 + P_x4*I_x4 + P_x5*I_x5 + P_x6*I_x6 #entropy\n",
+ "R = H_X*r#Information rate\n",
+ "\n",
+ "#results\n",
+ "print \"Entropy of the system (bits/message) = \",H_X\n",
+ "print \"Information rate (bits/seconds) = \",R\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Entropy of the system (bits/message) = 1.9375\n",
+ "Information rate (bits/seconds) = 31.0\n"
+ ]
+ }
+ ],
+ "prompt_number": 29
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 51 - pg 535"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Entropy\n",
+ "import numpy\n",
+ "#given\n",
+ "import math\n",
+ "from math import log\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "P_x1 = .3#probability of first signal\n",
+ "P_x2 = .4#probability of second signal\n",
+ "P_x3 = .3#probability of third signal\n",
+ "P_YX = numpy.matrix([[.8, .2, 0],[0, .1, 0],[0, .3, 0.7]])#matrix obtained from the figure \n",
+ "\n",
+ "#calculations\n",
+ "I_x1 = -log2(P_x1);\n",
+ "I_x2 = -log2(P_x2);\n",
+ "I_x3 = -log2(P_x3);\n",
+ "H_X = P_x1*I_x1 + P_x2*I_x2 + P_x3*I_x3 #entropy\n",
+ "P_y1 = P_YX[0,0]*P_x1 + P_YX[0,1]*P_x1 + P_YX[0,2]*P_x1;\n",
+ "P_y2 = P_YX[1,0]*P_x2 + P_YX[1,1]*P_x2 + P_YX[1,2]*P_x2;\n",
+ "P_y3 = P_YX[2,0]*P_x3 + P_YX[2,1]*P_x3 + P_YX[2,2]*P_x3;\n",
+ "I_y1 = -log2(P_y1);\n",
+ "I_y2 = -log2(P_y2);\n",
+ "I_y3 = -log2(P_y3);\n",
+ "H_Y = -P_y1*I_y1 - P_y2*I_y2 - P_y3*I_y3 #entropy\n",
+ "\n",
+ "#results\n",
+ "print \" Entropy H(X) = \",round(H_X,3)\n",
+ "print \"Entropy H(Y) = \",round(H_Y,3)\n",
+ "print \" Note:There is mistake in the calculation of P_y3 in the textbook so their is change in entropy H_Y\"\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ " Entropy H(X) = 1.571\n",
+ "Entropy H(Y) = -1.228\n",
+ " Note:There is mistake in the calculation of P_y3 in the textbook so their is change in entropy H_Y\n"
+ ]
+ }
+ ],
+ "prompt_number": 30
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 52 - pg 536"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Entropy of the second order extension\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "\n",
+ "P_x1 = .7#probability of first signal\n",
+ "P_x2 = .15#probability of second signal\n",
+ "P_x3 = .15#probability of third signal\n",
+ "n = 2#second order extention\n",
+ "\n",
+ "#calculations\n",
+ "I_x1 = -log2(P_x1);\n",
+ "I_x2 = -log2(P_x2);\n",
+ "I_x3 = -log2(P_x3);\n",
+ "H_x = P_x1*I_x1 + P_x2*I_x2 + P_x3*I_x3#entropy\n",
+ "H_x2 = n*H_x#entropy of second order extention\n",
+ "\n",
+ "#results\n",
+ "print \"Entropy of second order extension (bits/symbol) = \",round(H_x2,3)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Entropy of second order extension (bits/symbol) = 2.363\n"
+ ]
+ }
+ ],
+ "prompt_number": 31
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 54 - pg 537"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Entropy of the source\n",
+ "import math\n",
+ "from math import log\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "\n",
+ "#given\n",
+ "P_x1 = 1./3#probability of first signal\n",
+ "P_x2 = 1./6#probability of second signal\n",
+ "P_x3 = 1./4#probability of third signal\n",
+ "P_x4 = 1./4#probability of fourth signal\n",
+ "\n",
+ "#calculations\n",
+ "I_x1 = -log2(P_x1);\n",
+ "I_x2 = -log2(P_x2);\n",
+ "I_x3 = -log2(P_x3);\n",
+ "I_x4 = -log2(P_x4);\n",
+ "H_x = P_x1*I_x1 + P_x2*I_x2 + P_x3*I_x3 + P_x4*I_x4 #entropy\n",
+ "\n",
+ "#results\n",
+ "print \"Entropy of the source (bits/symbol) = \",round(H_x,4)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Entropy of the source (bits/symbol) = 1.9591\n"
+ ]
+ }
+ ],
+ "prompt_number": 32
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 55 - pg 538"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Average number of bits per message\n",
+ "\n",
+ "#given\n",
+ "P_x1 = 1./2#probability of first signal\n",
+ "P_x2 = 1./4#probability of second signal\n",
+ "P_x3 = 1./8#probability of third signal\n",
+ "P_x4 = 1./16#probability of fourth signal\n",
+ "P_x5 = 1./16#probability of fifth signal\n",
+ "n1 = 1.#number of bits in code obtained from table given textbook\n",
+ "n2 = 2.#number of bits in code obtained from table given textbook\n",
+ "n3 = 3.#number of bits in code obtained from table given textbook\n",
+ "n4 = 4.#number of bits in code obtained from table given textbook\n",
+ "n5 = 4.#number of bits in code obtained from table given textbook\n",
+ "\n",
+ "#calculations\n",
+ "L = P_x1*n1 + P_x2*n2 + P_x3*n3 + P_x4*n4 + P_x5*n5;#Average number of bits per message\n",
+ "\n",
+ "#results\n",
+ "print \"Average number of bits per message (bits) = \",L\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Average number of bits per message (bits) = 1.875\n"
+ ]
+ }
+ ],
+ "prompt_number": 33
+ },
+ {
+ "cell_type": "heading",
+ "level": 2,
+ "metadata": {},
+ "source": [
+ "Example 56 - pg 538"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "collapsed": false,
+ "input": [
+ "#calculate the Information capacity of telephone channel\n",
+ "import math\n",
+ "from math import log,exp\n",
+ "#given\n",
+ "def log2(x):\n",
+ "\ty=log(x)/log(2)\n",
+ " \treturn y\n",
+ "B = 3.4*10**3#bandwidth\n",
+ "SbyN = 30#signal to the noise ratio in dB\n",
+ "\n",
+ "\n",
+ "#calculations\n",
+ "SbyN1 = exp((SbyN/10)*log(10))#signal to noise ratio \n",
+ "C = B*log2(1+SbyN1);\n",
+ "\n",
+ "#result\n",
+ "print \"Information capacity of the telephone channel (kbps) = \",round(C/1000.,2)\n"
+ ],
+ "language": "python",
+ "metadata": {},
+ "outputs": [
+ {
+ "output_type": "stream",
+ "stream": "stdout",
+ "text": [
+ "Information capacity of the telephone channel (kbps) = 33.89\n"
+ ]
+ }
+ ],
+ "prompt_number": 34
+ }
+ ],
+ "metadata": {}
+ }
+ ]
+}
\ No newline at end of file |