A 1 KW signal is transmitted using a communication channel which provides attenuation at the rate of – 2dB per km. If the communication channel has a total length of 5 km, the power of the signal received is [gain in dB = 10 log P0/P1] (a) 900 W (b) 100 W (c) 990 W (d) 1010 W
Correct option is B) Pi=1 kW = 1000W Rate of attenuation of signal = -2dB/km Length of total path = 5 km Gain in attenuation = 5(-2) dB = -10dB Gain in dB =10log(P1P0) −10=10log(1000P0) +1=−loge1000P0 loge10=logeP01000 Taking antilog P01000=10 P0=100 Watt
