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Question

A 6-bit PCM system is uniformly quantized with a bit rate of 4.8 kbps. The signal-to-quantization noise ratio that would result when its input is a sine wave with peak amplitude equal to 10 V will be:

A
31.5 dB
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B
37.8 dB
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C
30 dB
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D
27 dB
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Solution

The correct option is B 37.8 dB
n = 6; Am=10

Signal Power = A2m2=50

Quantization Noise Power =Δ212=(2Am2n)2×112

=4(100)212×112

SNQ=50400×212×12=6144

SNQ(dB)=10log10(68.44)=37.8dB

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