Question

Page 3 out of 4 Z10: ZID TONS FOR FULL CREDIT Bive multiple contradictory answers orrect attempts a problem. Scratch off a The AM modulator used k-0.7 and the signal carries a speech signal 8) An AM receiver receives a signal with -103d8m. with bandwidth 8kHz and the power of the speech signal is 100mw. a) (0.4pt) What is the amplitude A, of the carrier received? b) (0.4pt) What is the SNR (in dB) at the output of the AM receiver? (use No 4 102 w/Ha)

Answer 1

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Given AM SNR = -103 dBm P_dBm = 10 log_10(P_mw) P_mw = 10^P_dBm/10 P_mw = 10^-103/10 = 10^-10.3 SNR P_mw = 5.011 times 10^-11 mw received power K_a = 0.7, w = 8 k Hz: N_0 = 4 times 10^-21 w/Hz P = 100 mw we know SNR = A^2_c (1 + k^2_a P)/2 w N_0 5.011 times 10^-14 = A^2_c (1 + (0.7)^2 times 100 times 10^-3)/2 times 8 times 10^3 times 4 times 10^-21 A^2_c = 5.011 times 10^-14 times 64 times 10^-18/1.049 A^2_c = 320.707/1.049 times 10^-32 A_c = squareroot 305 .72 times 10^-32 A_c = 17.48 times 10^-16 J \r\n SNR = A^2_c(1 + k^2_a P)/2 w N_0 = (17.48 times 10^-16)^2 times (1 + (0.7)^2 middot 100 times 10^-3/2 times 8 times 10^3 times 4 times 10^-21 = 305.72 times 10^-32 times (1.049)/64 times 10^-18 = 320.707/64 times 10^-14 SNR = 5.011 times 10^-14 SNR_dB = 10 log(5.011 times 10^-14) SNR_dB = -133.0007 dB