TRB 2006 - Question 10

A voice graded channel of the telephone network has a Bandwidth of $3.43\; kHz$. The channel capacity of the channel for a SNR of $30dB$ is 

a) $33888.56928 \;bits/sec$

b) $1000\; bits/sec$ 
c) $10\; bits/sec $
d) $10^{-6}\; bps$

Correct Answer: Option A

SNR(dB) = $30$ dB

$10\;log_{10}(SNR) = 30$ 

SNR$=10^{3}=1000$ 

Channel capacity, $$C=B\;log_2(1+SNR)$$ $$=3.4×10^3\; log_2(1+1000)$$$$=3.4×10^3\; log_2(1001)$$ $$=3.4×10^3\; log_2(1001)$$ $$=3.4×10^3\; log_2(1001)$$ $$=3.4×10^3×9.96722625$$ $$=33888.56928\; bits/sec$$ 

If calc is not allowed, round off $1001$ to nearest $2^x$ value. Hence $$C\lt 3.4×10^3\; log_2(1024)$$ $$C\lt 3.4×10^3\; log_2(2^{10})$$ $$C\lt 34000\;bps$$