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Nyquist levels for 3 to 4 MHz channel

Short answer

Bandwidth B = 1 MHz, SNR = 24 dB gives S/N ≈ 251.2, Shannon capacity C = 7.98 Mbps. Using Nyquist, C = 2B log2 L → log2 L = 3.989, so L ≈ 16 discrete signal levels (4 bits/symbol).

How is the Nyquist theorem applied for a noiseless channel? Calculate the number of discrete signal levels in the channel if a channel has a spectrum of 3 to 4 MHz with signal to noise ratio of 24 dB.

Nyquist theorem for a noiseless channel

For a noiseless channel, the maximum bit rate depends only on the bandwidth B and the number of signal levels L:

C = 2B log₂L bps

At most 2B symbols per second can be sent without inter-symbol interference. By using more levels, each symbol carries log₂L bits, so the capacity rises. In a real channel, noise limits L, so Shannon's formula is used to find the practical limit.

Numerical

Channel spectrum 3 MHz to 4 MHz, so B = 4 - 3 = 1 MHz. SNR = 24 dB.

  1. S/N = 10^(24/10) = 251.19
  2. Shannon capacity: C = B log₂(1 + S/N) = 10⁶ × log₂(252.19) = 10⁶ × 7.978 = 7.978 × 10⁶ bps
  3. Nyquist: C = 2B log₂L, so log₂L = C / 2B = 7.978×10⁶ / (2×10⁶) = 3.989 L = 2^3.989 = 15.88 ≈ 16

Answer: about 16 signal levels (4 bits per level); capacity ≈ 7.98 Mbps.

✓ Answer checked by Claude Opus 5.5 (high effort)

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