A voice-grade line has SNR = 30.1 dB (power ratio ≈ 1023:1) and usable spectrum from 300 Hz to 4300 Hz (bandwidth = 4000 Hz). Using Shannon’s capacity formula, what is the maximum achievable data rate on this line?

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Introduction / Context: Channel capacity places an upper bound on error-free data rate for a given bandwidth and signal-to-noise ratio. For telephony-grade channels, Shannon’s theorem provides the theoretical limit irrespective of modulation specifics. Given Data / Assumptions: SNR = 30.1 dB ≈ power ratio 1023:1. Frequency span: 300 Hz to 4300 Hz → bandwidth B = 4000 Hz. We apply Shannon capacity: C = B * log2(1 + SNR). Concept / Approach: Convert SNR from dB to linear: SNR_linear ≈ 10^(30.1/10) ≈ 1023. Then compute C with base-2 logarithm. Notation: use * for multiplication and / for division. Step-by-Step Solution: 1) SNR_linear ≈ 1023.2) 1 + SNR_linear ≈ 1024.3) log2(1024) = 10.4) C = B * log2(1 + SNR) = 4000 * 10 = 40000 bps. Verification / Alternative check: If one mistakenly uses 3000 Hz bandwidth, C would be 30000 bps, which is lower and does not reflect the given 4000 Hz span. The provided spectrum explicitly yields 40000 bps. Why Other Options Are Wrong: 34,000 bps and 31,000 bps result from using smaller bandwidth or rounding away from the exact log2(1024) = 10.9,600 bps and 28,800 bps are common modem rates subject to practical constraints, not the Shannon upper bound for the given parameters. Common Pitfalls: Mixing Nyquist and Shannon formulas, using dB directly without converting to linear ratio, or assuming 3000 Hz bandwidth by habit. Final Answer: 40,000 bps.

A voice-grade line has SNR = 30.1 dB (power ratio ≈ 1023:1) and usable spectrum from 300 Hz to 4300 Hz (bandwidth = 4000 Hz). Using Shannon’s capacity formula, what is the maximum achievable data rate on this line?

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