{ "metadata": { "name": "", "signature": "sha256:ee7d80ecb4663168394a1a558769dfa112d29db92a494569a7cd8e5de95a8ba8" }, "nbformat": 3, "nbformat_minor": 0, "worksheets": [ { "cells": [ { "cell_type": "heading", "level": 1, "metadata": {}, "source": [ "CHAPTER 5 - Cellular antenna system design considerations" ] }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.1 - PG NO.133" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no.133\n", "op=15.\n", "l=2.\n", "n=2.\n", "l1=n*l#connector loss\n", "l2=3.#coaxial cable loss\n", "tl=l1+l2#total loss\n", "ip=op-tl#input=output-total loss\n", "print '%s %s %d %s' %('signal level at the i/p of the antenna is =','+',ip,'dBm')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "signal level at the i/p of the antenna is = + 8 dBm\n" ] } ], "prompt_number": 1 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.2 - PG NO.136" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no. 136\n", "import math\n", "ci=18.\n", "CI=10.**((ci)/10.)\n", "q=(6*(CI))**0.25\n", "K=math.ceil(q*q/3)#cluster size\n", "print'%s %d' %('minimum cluster size is K =',K)\n", "k=7\n", "q1=math.sqrt(3*k)\n", "c1i1=q1**4/6\n", "C1I1=10*math.log10(c1i1)\n", "if (C1I1<20):\n", "\tprint('cluster size cannot meet the desired C/I requirement')\n", "\tC2I2=10**(20/10)\n", "\tq2=(6*C2I2)**0.25\n", "\tk1=math.ceil((q2)**2/3)\n", "\tprint'%s %d' %('nearest valid cluster size is K =',k1)\n", "else: \n", "\tprint('cluster size determined is adequate')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "minimum cluster size is K = 7\n", "cluster size cannot meet the desired C/I requirement\n", "nearest valid cluster size is K = 9\n" ] } ], "prompt_number": 2 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.4 - PG NO.139" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no. 139\n", "import math\n", "Y=4.#path loss exponent\n", "N=6.\n", "\n", "K=7.\n", "q=math.sqrt(3.*K)\n", "CI=(2.*(q-1.)**(-Y)+2.*q**(-Y)+2.*(q+1.)**(-Y))**(-1.)#C/I for omnidirectional operating cell\n", "CIdB=10.*math.log10(CI)\n", "print'%s %d %s' %('co-channel interfernce ratio C/I for K=7 is =',CIdB,'dB')\n", "\n", "K1=9.\n", "q1=math.sqrt(3.*K1)\n", "CI1=(2.*(q1-1.)**(-Y)+2.*q1**(-Y)+2.*(q1+1.)**(-Y))**(-1.)\n", "CI1dB=10.*math.log10(CI1)\n", "print'%s %.1f %s' %('co-channel interfernce ratio C/I for K=9 is =',CI1dB,'dB')\n", "\n", "K2=12.\n", "q2=math.sqrt(3.*K2)\n", "CI2=(2.*(q2-1.)**(-Y)+2.*q2**(-Y)+2.*(q2+1.)**(-Y))**(-1.)\n", "CI2dB=10.*math.log10(CI2)\n", "print'%s %.1f %s' %('co-channel interfernce ratio C/I in dB for K=12',CI2dB,'dB')\n", "\n", "\n", "if (CIdB<18) :\n", "\tprint('K=7 is imperfect')\n", "else :\n", "\tprint('K=7 is perfect')\n", "#end\n", "\n", "if (CI1dB<18):\n", "\tprint('K=9 is imperfect')\n", "else: \n", "\tprint('K=9 is perfect')\n", "#end\n", "\n", "if (CI2dB<18) :\n", "\tprint('K=12 is imperfect')\n", "else: \n", "\tprint('K=12 is perfect')\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "co-channel interfernce ratio C/I for K=7 is = 17 dB\n", "co-channel interfernce ratio C/I for K=9 is = 19.8 dB\n", "co-channel interfernce ratio C/I in dB for K=12 22.5 dB\n", "K=7 is imperfect\n", "K=9 is perfect\n", "K=12 is perfect\n" ] } ], "prompt_number": 3 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.5 - PG NO.142" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no.142\n", "import math\n", "N=2.\n", "Y=4.\n", "K=7.\n", "q=math.sqrt(3*K)\n", "CI=((q**(-Y)+(q+0.7)**(-Y)))**(-1)#C/I for 3-sector\n", "CIdB=10*math.log10(CI)\n", "print'%s %.1f %s' %('worst case signal to co-channel interfernce ratio C/I is =',CIdB,'dB')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "worst case signal to co-channel interfernce ratio C/I is = 24.5 dB\n" ] } ], "prompt_number": 4 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.6 - PG NO.143" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no.143\n", "import math\n", "N=2\n", "Y=4\n", "K=4\n", "\n", "q=math.sqrt(3*K)\n", "CI=((q**(-Y)+(q+0.7)**(-Y)))**(-1)#C/I for 3-sector\n", "CIdB=10*math.log10(CI)\n", "print'%s %d %s' %('worst case C/I is',round(CIdB),'dB')\n", "if CIdB>18 :\n", "\ta= CIdB-6\n", "\tif a>18 :\n", "\t\tprint('K=4 is adequate system as C/I is still geater than 18dB after considering the practical conditions with reductions of 6dB ')\n", "\n", "\telse :\n", "\t\tprint('K=4 is inadequate system as C/I is smaller than 18dB after considering the practical conditions with reductions of 6dB ')\n", "\t#end\n", "\n", "else: \n", "\tprint('K=4 is inadequate system as C/I is less than the minimum required value of 18dB ')\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "worst case C/I is 20 dB\n", "K=4 is inadequate system as C/I is smaller than 18dB after considering the practical conditions with reductions of 6dB \n" ] } ], "prompt_number": 5 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.7 - PG NO.145" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no.145\n", "import math\n", "N=1.\n", "Y=4.\n", "K=7.\n", "q=math.sqrt(3.*K)\n", "CI=((q+0.7)**(-Y))**(-1.)#C/I for 6-sector\n", "CIdB=10.*math.log10(CI)\n", "print'%s %d %s' %('signal to co-channel interfernce ratio C/I is =',round(CIdB),'dB')\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "signal to co-channel interfernce ratio C/I is = 29 dB\n" ] } ], "prompt_number": 6 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.8 - PG NO.146" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no. 146\n", "import math\n", "N=1.\n", "Y=4.\n", "K=4.\n", "q=math.sqrt(3.*K)\n", "CI=((q+0.7)**(-Y))**(-1)#C/I for 6-sector\n", "CIdB=10.*math.log10(CI)\n", "print'%s %.2f %s' %('signal to co-channel interfernce ratio C/I is =',CIdB,'dB')\n", "\n", "if CIdB>18 :\n", "\ta= CIdB-6\n", "\tif a>18:\n", "\t\tprint('K=4 is adequate system as C/I is still geater than 18dB after considering the practical conditions with reductions of 6dB ')\n", "\n", "\telse :\n", "\t\tprint('K=4 is inadequate system as C/I is smaller than 18dB after considering the practical conditions with reductions of 6dB ')\n", "\t#end\n", "\n", "else: \n", "\tprint('K=4 is inadequate system as C/I is less than the minimum required value of 18dB ')\n", "#end\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "signal to co-channel interfernce ratio C/I is = 24.78 dB\n", "K=4 is adequate system as C/I is still geater than 18dB after considering the practical conditions with reductions of 6dB \n" ] } ], "prompt_number": 7 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.9 - PG NO.146" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no.146\n", "CIdB=15.\n", "CI=10.**(CIdB/10.)\n", "q=(6.*(CI))**0.25\n", "K=q*q/3.\n", "\n", "\n", "if K >4:\n", " K=7.\n", "print'%s %d' %('optimum value of K for an omnidirectional antenna design is K =',K)\n", "q1=(CI**0.25-0.7)\n", "k=q1*q1/3.\n", "\n", "\n", "if k<3: \n", "\tk=3\n", "#end\n", "\tprint'%s %d' %('practical value of K for 6-sector 60deg. directionl antenna design is K =',k)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "optimum value of K for an omnidirectional antenna design is K = 7\n", "practical value of K for 6-sector 60deg. directionl antenna design is K = 3\n" ] } ], "prompt_number": 8 }, { "cell_type": "heading", "level": 2, "metadata": {}, "source": [ "EXAMPLE 5.10 - PG NO.148" ] }, { "cell_type": "code", "collapsed": false, "input": [ "#page no. 148\n", "N=312.\n", "K=7.\n", "Nspc=3.\n", "Ntcpc=N/K\n", "Ntcps=Ntcpc/Nspc#number of traffic channels per sector\n", "print'%s %.f' %('number of traffic channels per sector for System A is =',Ntcps)\n", "\n", "N1=312.\n", "K1=4.\n", "Nspc1=6.\n", "Ntcpc1=N1/K1\n", "Ntcps1=Ntcpc1/Nspc1#number of traffic channels per sector\n", "print'%s %.f' %('number of traffic channels per sector for System B is =',Ntcps1)\n" ], "language": "python", "metadata": {}, "outputs": [ { "output_type": "stream", "stream": "stdout", "text": [ "number of traffic channels per sector for System A is = 15\n", "number of traffic channels per sector for System B is = 13\n" ] } ], "prompt_number": 9 } ], "metadata": {} } ] }