Let the relevant bandwidth (B) of a digital communication system be 1 MHz and kT = −174 dBm/Hz. The power (S) of signal received is −80 dBm. Which of the following options is/are TRUE about Shannon capacity (C) of the channel?

The correct option is c > 3B.

To determine the Shannon capacity of the channel, we use Shannon's Channel Capacity formula:
$$C=B{\log }_2\left(1+\mathrm{SNR}\right)$$
where:
- $C$ is the channel capacity in bits per second (bps).
- $B$ is the channel bandwidth in Hertz (Hz).
- $\mathrm{SNR}$ is the linear Signal-to-Noise Ratio.

Step 1: Calculate the Noise Power Spectral Density ($N$)
The noise power spectral density is given as:
$$N_0=kT=-174\text{dBm/Hz}$$
The bandwidth is:
$$B=1\text{MHz}={10}^6\text{Hz}$$
We can express the bandwidth in dB-Hz:
$$B_{\text{dB-Hz}}=10{\log }_{10}\left({10}^6\right)=60\text{dB-Hz}$$
Thus, the total noise power $N$ in dBm is:
$$N\left(\text{dBm}\right)=kT\left(\text{dBm/Hz}\right)+B_{\text{dB-Hz}}$$
$$N\left(\text{dBm}\right)=-174+60=-114\text{dBm}$$

Step 2: Calculate the Signal-to-Noise Ratio (SNR)
The received signal power $S$ is given as:
$$S=-80\text{dBm}$$
Calculating the SNR in decibels (dB):
$${\mathrm{SNR}}_{\text{dB}}=S\left(\text{dBm}\right)-N\left(\text{dBm}\right)$$
$${\mathrm{SNR}}_{\text{dB}}=-80-\left(-114\right)=34\text{dB}$$
Converting the SNR from dB to a linear scale:
$$\mathrm{SNR}={10}^{3.4}\approx 2511.89$$

Step 3: Calculate the Shannon Capacity ($C$)
Now we substitute the linear SNR into the capacity formula:
$$C=B{\log }_2\left(1+2511.89\right)$$
$$C=B{\log }_2\left(2512.89\right)$$
Since $2^{11}=2048$ and $2^{12}=4096$, we have:
$${\log }_2\left(2512.89\right)\approx 11.295$$
Therefore:
$$C\approx 11.295B$$
Since $11.295B$ is significantly greater than $3B$, the relation $C>3B$ (or $c>3B$) is indeed true.