NIELIT 2017 OCT Scientific Assistant A (CS) - Section B: 26

Question

If the channel is band limited to $6\;kHz$ and signal to noise ratio is $16,$ what would be the capacity of channel?

• A. $16.15$ kbps
• B. $23.24$ kbps
• C. $40.12$ kbps
• D. $24.74$ kbps

Answer 1

To determine the theoretical highest data rate for a noisy channel we use the Shannon capacity formula.

$\text{Capacity=BW*$\log_2$(1+SNR)}$

where; 1) BW=Bandwidth of the channel

             2) SNR= signal to noise ratio

             3) Capacity= Capacity of the channel in bit per second

$\therefore \text{Capacity=6*$\log_2$(1+16)}$

$\text{Capacity=6*$\log_2$(17) $\approx$24.74 kbps}$

So option $D$ is correct here.

Answer 2

B=6,s/n=16

C=B*log(1+s/n)

    = 6*log(17)

    = 6*4.08

     =24.74