Difficulty: Medium
Correct Answer: 64 Kbps
Explanation:
Introduction:
Telephony Pulse Code Modulation (PCM) encodes analog voice (nominal 300–3400 Hz) by sampling and quantization. The classic public network standard samples at 8000 samples/s (every 125 microseconds) and uses 8 bits per sample, yielding a familiar channel bit rate. This question tests your ability to connect sampling parameters to bit rate, and to recognize the relationship to CCITT (now ITU-T) 2.048 Mb/s E1 framing.
Given Data / Assumptions:
Concept / Approach:
Bit rate for PCM = samples per second * bits per sample. For a single encoded voice channel, this is independent of framing overhead; framing becomes relevant when multiplexing many channels (e.g., 32 time slots in an E1 line at 2.048 Mb/s).
Step-by-Step Solution:
Verification / Alternative check:
Nyquist sampling for a ~3.4 kHz voice passband leads to a standard 8 kHz sampling rate, which aligns with the 64 Kbps channel rate widely used in telephony.
Why Other Options Are Wrong:
Common Pitfalls:
Confusing the single-channel rate (64 Kbps) with the aggregate E1 rate (2.048 Mb/s); mixing PCM with ADPCM/vocoder assumptions.
Final Answer:
64 Kbps.
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