The channel capacity under the Gaussian noise environment for a discrete memoryless channel with a bandwidth of 4 MHz and SNR of 31 is
- A. 20 Mbps
- B. 4 Mbps
- C. 8 Kbps
- D. 4 Kbps
Answer: A. 20 Mbps
Explanation: According to Shannon's channel capacity theorem: $C = B \log_2(1+SNR) = 4 \log_2(1+31) = 4 \log_2(32) = 4 \times 5 = 20$ Mbps.