Question

As the bandwidth approaches infinity, the channel capacity becomes (Referring to Shannon-Hartley theorem: \(C = B \log_2(1 + S/N)\))

• A. 0
• B. \( 1.44 \frac{S}{\eta} \)
• C. \( 0.5 \frac{S}{\eta} \)
• D. \(\infty\)

Hint:

Shannon-Hartley Theorem: \(C = B \log_2(1 + S/N)\).
Noise power \(N = N_0 B\), where \(N_0\) is noise power spectral density.
Limit for infinite bandwidth: \(C_{\infty} = \frac{S}{N_0 \ln 2} = \frac{S}{N_0} \log_2 e \approx 1.443 \frac{S}{N_0}\).
Use approximation \(\ln(1+x) \approx x\) for small \(x\).

Answer 1

The Correct Option is B

The Shannon-Hartley theorem for the capacity C of an AWGN channel is: \[ C = B \log_2(1 + \frac{S}{N}) \text{ bits/sec} \] where B is the bandwidth, S is the average received signal power, and N is the average noise power. The noise power N can be expressed as \(N = N_0 B\), where \(N_0\) is the one-sided power spectral density of the white noise (often denoted as \(\eta\) in some texts). So, \( C = B \log_2(1 + \frac{S}{N_0 B}) \). We need to find the limit of C as \(B \to \infty\): \[ C_{\infty} = \lim_{B \to \infty} B \log_2(1 + \frac{S}{N_0 B}) \] Let \(x = \frac{S}{N_0 B}\). As \(B \to \infty\), \(x \to 0\). The expression is \( \lim_{x \to 0} \frac{S}{N_0 x} \log_2(1+x) \). We use the approximation \(\log_2(1+x) = \frac{\ln(1+x)}{\ln 2}\). For small \(x\), \(\ln(1+x) \approx x\). So, \(\log_2(1+x) \approx \frac{x}{\ln 2}\). Substitute this into the limit expression: \[ C_{\infty} = \lim_{B \to \infty} B \left( \frac{S/N_0 B}{\ln 2} \right) \] (using \(\log_2(1+x) \approx x/\ln 2\)) \[ C_{\infty} = \lim_{B \to \infty} B \cdot \frac{S}{N_0 B \ln 2} = \frac{S}{N_0 \ln 2} \] Since \(\ln 2 \approx 0.693\), then \(1/\ln 2 \approx 1/0.693 \approx 1.443\). So, \[ C_{\infty} = \frac{S}{N_0} \log_2 e \approx 1.443 \frac{S}{N_0} \] If \(\eta\) in the options represents \(N_0\), then the limit is \(1.44 \frac{S}{\eta}\) (approximately). This matches option (b). \[ \boxed{1.44 \frac{S}{\eta}} \]