{ "metadata": { "name": "", "signature": "sha256:8bd5af3b2acf26b36843842ce0fbb28501e7f5eae117a0af6c810fe6d60c2891" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "Chapter 11 The Transmission of binary data in communication systems" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.1 Page no 392" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "t=0.0016\n", "No_words=256.0\n", "bits_word = 12.0\n", "\n", "#Calculation\n", "tword= t/No_words\n", "tbit = tword/bits_word\n", "bps =1/tbit\n", "\n", "#Result\n", "print\"(a) The time duration of the word \",tword*10**8,\"microsecond\"\n", "print\"(b) The time duration of the one bit is \",round(tbit*10**8,4),\"microseconds\"\n", "print\"(c) The speed of transmission is \",bps/10**5,\"kbps\"\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a) The time duration of the word 625.0 microsecond\n", "(b) The time duration of the one bit is 52.0833 microseconds\n", "(c) The speed of transmission is 19.2 kbps\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.2 Page no 400" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "B=12.5*10**3\n", "SN_dB= 25\n", "\n", "#Calculation\n", "import math\n", "C_th = 2*B\n", "SN=316.2\n", "C =B*3.32*log10(SN+1)\n", "N= 2**(C/(2.0*B))\n", "\n", "#Result\n", "print\"(a) The maximum theorotical data rate is \",C_th/10**3,\"kbps\"\n", "print\"(b) The maximum theorotical capacity of channel is \",round(C/10**3,1),\"Kbps\"\n", "print\"(c) The number of levels needed to acheive maximum speed are \",round(N,2)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "(a) The maximum theorotical data rate is 25.0 kbps\n", "(b) The maximum theorotical capacity of channel is 103.8 Kbps\n", "(c) The number of levels needed to acheive maximum speed are 17.78\n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "Example 11.3 Page no 430" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#Given\n", "block =512\n", "packets =8\n", "BER = 2*10**-4\n", " \n", "#Calculation\n", "avg_errors = block*packets*8*BER\n", "\n", "print\"Average number of errors are \",avg_errors\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "Average number of errors are 6.5536\n" ] } ], "prompt_number": 3 } ], "metadata": {} } ] }