Find the channel capacity of the noiseless discrete channel, with
A. C = log<sub>2</sub> 4n
B. C = log<sub>2</sub> 2n
C. C = log<sub>2</sub> n<sup>2</sup>
D. C = log<sub>2</sub> n
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Right Answer is: D
SOLUTION
Channel Capacity Per Symbol Cs
The channel capacity per symbol of a discrete memoryless channel (DMC) is defined as:
\({C_s} = \mathop {\max }\limits_{\left\{ {P\left( {{x_i}} \right)} \right\}} I\left( {X;Y} \right)bits/symbol\)
Where,
I(X;Y) = H(x) – H(x|y)
H(x): Entropy of x
Where the maximization is over all possible input probability distribution {P(xi)} on X. Note that the channel capacity Cs is a function of only the channel transition probabilities which define the channel.
Noiseless Channel:
A channel is called noiseless if it is both lossless and deterministic.
A noiseless channel has been shown in the figure below:
The channel matrix has only one element in each row and each column, and this element is unity. Note that the input and output alphabets are of the same size, that is m = n for the noiseless channel.
Since a noiseless channel is both lossless and deterministic, we have
I(X;Y) = H(X) = H(Y)
And the channel capacity per symbol is
Cs = log2m = log2n