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Capacity and Nyquist levels, 3 to 4 MHz, 24 dB

Short answer

Bandwidth B = 4-3 = 1 MHz. Shannon capacity with SNR = 24 dB (S/N ≈ 251) is C = 1×10⁶ · log₂(252.19) ≈ 7.98 Mbps. For Nyquist, log₂L = 7.98/2 ≈ 3.99, so L ≈ 16 levels (4 bits per sample).

State Nyquist's and Shannon's channel capacity formula. The spectrum of a channel is in between 3 MHz and 4 MHz. The SNR = 24 dB. Calculate the bandwidth, the maximum channel capacity using Shannon's formula. Then based on Nyquist's formula, find the number of signal levels required to reach the maximum channel capacity.

Formulas

  • Nyquist (noiseless): C = 2B log₂L
  • Shannon (noisy): C = B log₂(1 + S/N)

Given: spectrum 3 MHz to 4 MHz, SNR = 24 dB.

  1. Bandwidth B = 4 - 3 = 1 MHz.
  2. S/N = 10^(24/10) = 251.19
  3. Maximum capacity C = 10⁶ × log₂(1 + 251.19) = 10⁶ × log₂(252.19) = 7.98 Mbps (≈ 8 Mbps).
  4. Number of levels: C = 2B log₂L, so log₂L = 7.98×10⁶ / (2×10⁶) = 3.99, and L = 2^3.99 = 15.9 ≈ 16 levels.

Answer: B = 1 MHz, C ≈ 7.98 Mbps, L ≈ 16 signal levels (4 bits per level).

✓ Answer checked by Claude Opus 5.5 (high effort)

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