Topic: Digital Communications
A communication channel has a bandwidth of 3 kHz and Signal-to-Noise Ratio (SNR) of 30 dB. What is the channel capacity?
- A. ~30 kbps
- B. ~20 kbps
- C. ~2.1 kbps
- D. ~3.3 kbps
Answer: A. ~30 kbps
Explanation: First, convert the SNR from dB to linear scale: $SNR_{linear} = 10^{30/10} = 10^3 = 1000$.
Using Shannon's Channel Capacity theorem: $C = B \log_2(1 + SNR)$.
$C = 3000 \times \log_2(1 + 1000) = 3000 \times \log_2(1001)$.
Since $\log_2(1024) = 10$, $\log_2(1001)$ is approximately 9.967.
$C \approx 3000 \times 9.967 = 29901$ bps, which is roughly 30 kbps.