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Common Data for Questions 71, 72 and 73: A speech signal, band limited to 4 kHz and peak voltage varying between +5V and -5V, is sampled at the Nyquist rate. Each sample is quantized and represented by 8 bits. If the bits 0 and 1 are transmitted using bipolar pulses, the minimum bandwidth required for distortion free transmission is

1. 64 kHz
2. 32 kHz
3. 8 kHz
4. 4 kHz

Correct answer: 32 kHz

Solution

The minimum bandwidth required for distortion-free transmission of a signal is determined by the Nyquist theorem, which states that the bandwidth must be at least twice the highest frequency of the signal. Since the speech signal is band limited to 4 kHz, the minimum bandwidth needed is 2 times 4 kHz, resulting in 8 kHz. However, when considering the bipolar pulse transmission, the effective bandwidth doubles, leading to a required bandwidth of 32 kHz.

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