Question

Let the relevant bandwidth \((B)\) of a digital communication system be 1 MHz and \(kT=-174\) dBm/Hz, where \(k\) is Boltzmann's constant and \(T\) is the equivalent noise temperature of the receiver. The power \((S)\) of signal received through an additive Gaussian channel is \(-80\) dBm.
Which of the following options is/are TRUE about Shannon capacity \((C)\) of the channel?

• A. \(C=B\)
• B. \(C=2B\)
• C. \(C>3B\)
• D. \(C<B\)

Hint:

Convert the given dBm values into a linear SNR before applying the Shannon capacity formula.

Answer 1

The Correct Option is C

Step 1: Recall the Shannon capacity formula.
For a channel with bandwidth \(B\) and signal-to-noise ratio \(SNR\), the Shannon capacity is
\[ C=B\log_2(1+SNR) \]

Step 2: Find the noise power in the given bandwidth.
The noise power spectral density is \(kT=-174\ dBm/Hz\). Over a bandwidth \(B=1\ MHz=10^6\ Hz\), the total noise power in dB is
\[ N=kT+10\log_{10}(B)=-174+10\log_{10}(10^6)=-174+60=-114\ dBm \]

Step 3: Find the SNR in dB, then convert to a plain ratio.
\[ SNR(dB)=S-N=-80-(-114)=34\ dB \]
\[ SNR=10^{34/10}=10^{3.4}\approx2512 \]

Step 4: Compute the Shannon capacity.
\[ C=B\log_2(1+SNR)=10^6\times\log_2(2513) \]
Since \(2^{11}=2048\) and \(2^{12}=4096\), \(\log_2(2513)\) sits between \(11\) and \(12\), close to \(11.3\):
\[ C\approx10^6\times11.3=11.3\times10^6\ \text{bits/s}=11.3\ B \]

Step 5: Analyze the options.

(A) \(C=B\): Would mean an SNR of only \(1\) (\(0\) dB), far below the \(34\ dB\) actually present. Incorrect.

(B) \(C=2B\): Would need \(SNR=3\), also far too small. Incorrect.

(C) \(C>3B\): Since \(C\approx11.3B\), which is indeed greater than \(3B\), this statement holds. Correct.

(D) \(C<B\): This would need a negative or zero SNR, which contradicts the large positive SNR here. Incorrect.

Step 6: Final conclusion.
\[ \boxed{C\approx11.3B,\ \text{so}\ C>3B\ \text{is the TRUE statement}} \]