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Q1: An audio signal of FM radio station is sampled at a rate of 15% above the Nyquist rate. Signal is sampled in 512 levels and bandwidth of the signal is 4 KHz. The minimum bandwidth required to transmit this signal will be

A 77 K bits/sec

B 92 K bits/sec

C 84.8 K bits/sec

D 82.8 K bits/sec

ANS:D - 82.8 K bits/sec

Nyquist rate = 2 x 4 kHz = 8 kHz Sampling rate = 8 kHz + 1.2 kHz = 9.2 kHz Number of bits required = log2512 = 9 Thus bandwidth = 9 x 9.2 = 82.8 k bits/sec.



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