Which of the following options is/are TRUE about Shannon capacity ($C$) of the channel?
To determine the Shannon capacity (\(C\)) of a digital communication system, we use the Shannon capacity formula:
\(C = B \log_2(1 + \frac{S}{N})\)
where:
- \(B\) is the bandwidth, given as 1 MHz.
- \(S\) is the signal power, given as -80 dBm.
- \(N\) is the noise power, which can be calculated as: \(N = kT \times B\).
- \(kT\) is given as -174 dBm/Hz.
Step 1: Convert dBm values to linear scale (watts)
Signal power \(S = -80 \text{ dBm} = 10^{-8} \text{ watts}\).
Noise power spectral density \(kT = -174 \text{ dBm/Hz} = 10^{-17.4} \text{ watts/Hz}\).
Step 2: Calculate Noise Power (\(N\))
The noise power can be found by multiplying the noise power spectral density by the bandwidth:
\(N = (10^{-17.4}) \times (1 \times 10^6) = 10^{-17.4 + 6} = 10^{-11.4} \text{ watts}\)
Step 3: Calculate Signal-to-Noise Ratio (SNR)
The signal-to-noise ratio is:
\(\frac{S}{N} = \frac{10^{-8}}{10^{-11.4}} = 10^{3.4}\)
Step 4: Calculate Shannon Capacity (\(C\))
Substituting the values into the Shannon capacity formula, we get:
\(C = 1 \times 10^6 \cdot \log_2(1 + 10^{3.4})\)
Approximating the SNR:
\(\log_2(1 + 10^{3.4}) \approx \log_2(10^{3.4}) = 3.4 \cdot \log_2(10)\)
Using \(\log_2(10) \approx 3.32\), we get:
\(C \approx 1 \times 10^6 \cdot 3.4 \cdot 3.32 \approx 11.288 \times 10^6 \text{ bps}\)
Conclusion
The Shannon capacity \(C\) far exceeds three times the bandwidth:
\(C > 3B = 3 \times 1 \times 10^6 = 3 \times 10^6 \text{ bps}\)
Therefore, the correct option is: \(C > 3B\)
Important Questions from Channel Capacity
Noise factor of a system is defined as:
Match List I with List II:
List I List II (A) Shannon's theorem (I) Capacity of Gaussian Noise channel (B) Shannon-Hartley theorem (II) Rate of Information (C) Bayes theorem (III) Energy of a signal (D) Parseval's theorem (IV) Conditional probabilities Choose the correct answer from the options given below:
The information capacity (bits/sec) of a channel with bandwidth C and transmission time T is given by
The capacity of band-limited additive white Gaussian Noise (AWGN) channel is given by \(C = W{\log _2}\left[ {1 + \frac{P}{{{\sigma ^2}w}}} \right]\) bits per second (bps), where W is the channel Bandwidth, P is the average power received and σ2 is the one-sided power spectral density of the AWGN.
For a fixed \(\frac{P}{{{\sigma ^2}}} = 1000\), the channel capacity (in kbps) with infinite Bandwidth (W → ∞) is approximately
Consider an additive white Gaussian noise (AWGN) channel with bandwidth W and noise power spectral density $\frac{N_0}{2}$. Let $P_{av}$ denote the average transmit power constraint. Which one of the following plots illustrates the dependence of the channel capacity C on the bandwidth W (keeping $P_{av}$ and $N_0$ fixed)?
