- 2071 Chaitra · 3+2 marks
Required SNR for 40 Mbps over 6 MHz
Short answer
Shannon gives S/N = 2^(40/6) - 1 ≈ 100.6, or 20 dB. For a noiseless channel, Nyquist yields log₂L = C/(2B) = 40/12 ≈ 3.33 bits/sample, L ≈ 10 levels.
Given a channel with an intended capacity of 40 Mbps. The bandwidth of the channel is 6 MHz. What signal-to-noise ratio is required in order to achieve this capacity? Also find number of bits/sample if channel becomes noiseless.
Required SNR (Shannon)
C = B log₂(1 + S/N)
40 × 10⁶ = 6 × 10⁶ × log₂(1 + S/N)
log₂(1 + S/N) = 40/6 = 6.667
1 + S/N = 2^6.667 = 101.59
S/N = 100.59
In dB: SNR = 10 log₁₀(100.59) = 20.03 dB
Answer: S/N ≈ 100.6, that is about 20 dB.
Bits per sample if the channel is noiseless (Nyquist)
C = 2B log₂L, so log₂L = C/2B = 40×10⁶ / (12×10⁶) = 3.33
L = 2^3.33 = 10.08 levels
Answer: about 3.33 bits per sample. Since L must be a whole number in practice, 3 bits (L = 8) would give only 36 Mbps and 4 bits (L = 16) would give 48 Mbps; L ≈ 10 is the exact figure for 40 Mbps.
✓ Answer checked by Claude Opus 5.5 (high effort)
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