Question
Compare the maximum data rate of a noiseless $4-\mathrm{kHz}$ channel using(a) Analog encoding (e.g., QPSK) with 2 bits per sample.(b) The T1 PCM system.
Step 1
g., QPSK) with 2 bits per sample: In QPSK (Quadrature Phase Shift Keying), each symbol represents 2 bits, and there are 4 possible symbols. The maximum data rate can be calculated using the Nyquist formula: Maximum Data Rate = 2 * Bandwidth * log2(Number of Show more…
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