0 votes 0 votes The channel capacity of a noise free channel having $\text{M}$ symbols is given by : $\text{M}$ $2^{\text{M}}$ $\log \text{M}$ None of these Others nielit2021dec-scientistb + – soujanyareddy13 asked Dec 7, 2021 • edited Dec 14, 2021 by soujanyareddy13 soujanyareddy13 528 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes C. log M … Noise free channel is both lossless and deterministic, i.e. Noise free channel = Loseless + deterministic …. For loseless channel, H(x|y) = 0 .. I(x ; y) = H(x) So, Cs = max H(x) = log2 m .. Where, m = numbers of symbols in X ... For deterministic channel H(y|x) = 0.. I(x ; y) = H(y) So, Cs = max H(y) = log2 n .. Where, n = number of symbols in Y Hence, for Noiseless channel, we can write: I(x ; y) = H(x) = H(y), i.e. Cs = log2 m = log2 n … 33 answered Mar 13, 2022 33 comment Share Follow See all 0 reply Please log in or register to add a comment.