If a binary signal is sent over a 3-kHz channel whose signal-to-niose ration is 20dB what is the maximum achievable data rate?

Answer 1

The Nyquist Limit can be disregarded as this is not a noiseless channel (faster signal = more noise, this channel's s/n ratio is provided as 20dB)

thus we use Shannon's result which says the maximum data rate of a noisy channel is X = H Log2 (1 + S/N) bps using 10Log10 S/N as our standard quality

2 = Log10 S/N --> S/N = 102 --> S/N = 100

X = 3000 Log2 (1 + 100) bps which gives you x = 19,974.63bps as your final answer.

~ Mike

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ALTERNATIVE APPROACH

The formula -

Shannon Capacity = Bandwidth * log2 ( 1 + Signal Power / Noise Power )

gets approximated to -

Shannon Capacity = Bandwidth * ( Signal to Noise ratio in dB ) / 3

assuming the fact that ....

Signal to Noise ratio in dB = 10 log10 ( Signal Power /Noise Power )

and also assuming

1 is much much less than Signal Power/Noise Power

So in the present case the approximate answer works out to

Shannon Capacity = Bandwidth * ( Signal to Noise ratio in dB ) / 3 ...

= 3KHz * 20 dB / 3

= 3 * 103 * 20 / 3 bits per second

= 20000 bits per second

~

ANIRUDDHA GHOSH

JADAVPUR UNIVERSITY

BSc Mathematics - 2004 - 2010

MCA - 2007 - 2010

~

Answer 2

C=log 2(1+S/N)=3000*log2(1+100)=19.975 bps