Question

$\text{An AWGN channel has a bandwidth of 4 kHz and a two-sided noise power spectral density of}{10}^{-10}\text{W/Hz. The minimum signal power to be maintained at the input of the receiver is}100\mu \text{W. The capacity of the channel will be}$

• A. 8.4 kbps
• B. 9.6 kbps
• C. 31.88 kbps
• D. 27.9 kbps

Answer 1

The correct option is D 27.9 kbps

Minimum signal-to-noise ratio has to be maintained in the channel is,

$SNR=\frac{S}{N_{0}B}=\frac{S}{\left(\frac{N_0}{2}\right)\left(2B\right)}$

$=\frac{100\times {10}^{-6}}{\left({10}^{-10}\right)(2\times 4\times {10}^3)}=125$

The capacity of the channel can be given by,

$C=B{\log }_2(1+SNR)$

$=4{\log }_2(1+125)\simeq 27.91\text{kbps}$