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An analog baseband signal, band-limited to 100 Hz, is sampled at the Nyquist rate. The samples are quantized into four message symbols that occur independently with probabilities p1 = p4 = 0.125 and p2 = p3. The information rate (bits/sec) of the message source is __________

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Concept:

The Entropy of a source, which is also the average information content is given by:

\(H = \sum P\left( {{x_i}} \right){\log _2}\left( {\frac{1}{{P\left( {{x_i}} \right)}}} \right)\)

Where P(xi) = Probability of each codeword.

Application:

We have P1 = P4 = 0.125

Now, P1 + P2 + P3 + P4 = 1

Since P2 = P3, the above equation becomes:

\(0.12 + 2{{\rm{P}}_2} + 0.125 = 1\)

2P2 = 0.75

P2 = P3 = 0.375

Average information will be:

H = - P1log2P1 - P2log2P2 - P3log2P3 - P4log2P4

H = - 0.125.log2 0.125 - 0.375.log2 0.375 - 0.375.log0.375 - 0.125.log0.125

H = 0.375 + 0.531 + 0.531 + 0.375

H = 1.812 bits/symbol.

Sampling rate is the Nyquist rate, i.e.

r = 2 × 100 = 200 samples/sec

Thus, the information rate will be: r × H

= 200 × 1.812

= 362.4 bits/sec.

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