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//Chapter 6, Example 6.20, page 277
clc
//Initialisation
pt=20 //transmitter power in watt
Hb=30 //in metre
Hm=3 //in metre
gt=14.2 //trasmitter gain in dB
gr=0.2 //receiver gain in dB
f=450 //frequency in MHz
gm=-2 //in dBd
gr2=-2.2 //in dBi
r1=10
n=20
hb=10
hm=10
//Calculation
gt1=gt+gr2
pr1=-62-38*log10(r1)-20*log10(f*900**-1)+7 //received signal level in suburban
pr2=-64-43*log10(r1)-20*log10(f*900**-1)+7 //received signal level in urban
ao=10*log10(2)+(gr2-6) //in dB (The answer provided in the textbook is wrong)
pr11=-62-38*log10(r1)-20*log10(f*900**-1)+ao //received signal level in rural
pr22=-64-43*log10(r1)-20*log10(f*900**-1)+ao //received signal level in cities
ptd=10*log10(pt*10**3) //in dBm
lp1=ptd-pr11 //Path loss in rural area
lp2=ptd-pr22 //Path loss in cities area
//Results
printf("(2) In the suburban area, Pr = %.1f dBm",pr1)
printf("\n In the urban area, Pr = %.1f dBm",pr2)
printf("\n(3) Path loss in rural area Lp = %.1f dB",lp1) //The answer provided in the textbook is wrong
printf("\n Path loss in cities area Lp = %.1f dB",lp2) //The answer provided in the textbook is wrong
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