Question

Q1. A DSB-SC signal is transmitted over a noisy channel with power spectral density of noise given as, 0, otherwise with B-200 kHz and N 10W Hz. The message bandwidth is 10 kHz and its average power is 10 W. Assume A,-1, 100kHz, and the channel attenuates signal power by a factor of 1000 (-30 dB). Assume that a suitable band-pass filter is used at the receiver to limit out-of-band noise. (a) Determine the average signal power at the receiver input? (b) Determine the average noise power at the receiver input? (c) Determine the signal-to-noise ratio at the receiver output?

Answer 1

Given Average signal power at transmitter s_i = 10 w = 10 dB channel attenuation = -30 dB Average signal power at Receiver i/p = S_i - 30 dB S_r - -20 dB S_r = 0.01 w \r\n Given Noise PSD S_N (f) = {N_0 (1 - (f)/B), Vertical-bar f Vertical-bar < B 0 otherwise Given B = 200 KHz N_0 = 10^-9 W/Hz msg signal DSB - SC signal rightarrow effective Noise PSD Noise power S_N = 2 (0.45 N_0 times 20k) + 1/2 (0.1 N_0) (20 k) S_N = 19 times 10^-6 w \r\n SNR at Receiver i/p = S_r/S_N = 0.01/19 times 10^-6 SNR = 526.31 SNR = 27.21 dB