Question

An ideal AWGN. channel has a bandwidth of 16 kHz and the two-sided noise power spectral density of 2.5 nW/Hz. If the average signal power of the binary data transmitting through . the channel is 2.48 mW, then the maximum rate of the binary data that can be allowed through the channel without any error will be_____ kbps.

• A. 80

Answer 1

The correct option is A 80
Noise power in the channel
$N=\frac{N_0}{2}\left(2B\right)=2.5\times 2\times 16=80\mu W$

Given that,
The signal power, s = 2.48 mW

$\frac{S}{N}=\frac{2.48\times {10}^{-3}}{80\times {10}^{-6}}=31$

The maximum data rate that can be transmitted will be the capacity of the channel, which acan be given by

$R_{b\left(max\right)}=C=Blo{g}_2(1+\frac{S}{N})$

= $16lo{g}_2(1+31)$ kbps = 80 kbps