Question

A channel has \( B = 4 \) KHz, what is the channel capacity having the signal-to-noise ratio of 20 dB?

• A. 24.6 kbits/s
• B. 26.6 kbits/s
• C. 39.8 kbits/s
• D. 20.2 kbits/s

Hint:

- Use Shannon's Capacity Formula: \( C = B \log_2(1 + SNR) \).
- Convert SNR from dB to linear: \( SNR_{linear} = 10^{\frac{SNR_{dB}}{10}} \).

Answer 1

The Correct Option is B

Step 1: Using Shannon's Capacity Formula
Shannon's theorem states that the maximum channel capacity \( C \) is given by: \[ C = B \log_2 (1 + SNR) \] where: - \( B \) is the bandwidth (4 KHz), - \( SNR \) is the signal-to-noise ratio, given in dB as: \[ SNR_{linear} = 10^{\frac{SNR_{dB}}{10}} \]
Step 2: Calculating the Linear SNR Given \( SNR_{dB} = 20 \), we convert it to linear form: \[ SNR_{linear} = 10^{\frac{20}{10}} = 10^2 = 100 \]
Step 3: Computing Channel Capacity \[ C = 4 \times 10^3 \times \log_2(1 + 100) \] \[ = 4 \times 10^3 \times \log_2(101) \] Using logarithm approximation: \[ \log_2(101) \approx 6.658 \] \[ C = 4 \times 6.658 \times 10^3 \] \[ C \approx 26.6 \text{ kbits/s} \]
Step 4: Evaluating the Options
- (A) Incorrect: 24.6 kbits/s is lower than the computed value.
- (B) Correct: 26.6 kbits/s matches the computed value.
- (C) Incorrect: 39.8 kbits/s is much higher than the actual capacity.
- (D) Incorrect: 20.2 kbits/s is lower than the correct answer.