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for a noiseless channel bitrate=8kbps
and for a noisy channel data rate will be approx 20kbps....
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A noiseless channel can carry an arbitrarily large amount of information, no matterhow often it is sampled. Just send a lot of data per sample.

now for second case Shannon limit apply here

Shannon's Theorem

Shannon's Theorem gives an upper bound to the capacity of a link, in bits per second (bps), as a function of the available bandwidth and the signal-to-noise ratio of the link.

The Theorem can be stated as:

C = B * log2(1+ S/N)

where C is the achievable channel capacity, B is the bandwidth of the line, S is the average signal power and N is the average noise power.

The signal-to-noise ratio (S/N) is usually expressed in decibels (dB) given by the formula:

10 * log10(S/N)

but S/N is given 30 dB (in decibal)

so

10 * log10(S/N) = 30 dB.

so S/N=1000

given 4khz

so

C = 4000 * log2(1001)=39.86kbps (shannon'theorem )

Theory :    http://www.inf.fu-berlin.de/lehre/WS01/19548-U/shannon.html

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