- 2071 Shrawan · 3 marks
Channel capacity for 6 kHz, 6 dB
Short answer
Given B = 6 kHz, SNR = 6 dB → S/N ≈ 3.98, Shannon capacity C = B log2(1 + S/N) = 6000 × log2(4.98) = 6000 × 2.316 ≈ 13.9 kbps.
Calculate the channel capacity having bandwidth and SNR of 6 kHz and 6 dB respectively.
Given: B = 6 kHz, SNR = 6 dB. Use Shannon's formula C = B log₂(1 + S/N).
- S/N = 10^(6/10) = 3.981
- 1 + S/N = 4.981
- log₂(4.981) = 2.316
- C = 6000 × 2.316 = 13,899 bps
Answer: C ≈ 13.9 kbps.
✓ Answer checked by Claude Opus 5.5 (high effort)
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