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  1. 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

    2. A.
    77 K bits/sec
    B.
    92 K bits/sec
    C.
    84.8 K bits/sec
    D.
    82.8 K bits/sec

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    Answer : Option D

    Explanation :

    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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