cities for which a = b corresponds to the binary symmetric channel. Substitution of b=a in Eq. (13.8) yields V = - f(a) and, thus, C = 1 - f(a) - from Eq. (13.7, which is the result obtained earlier in Eq. (13.5). We also obtain from Eq. (13.10) W = 0, and, from Eq. (13.9), q = 1/2, which is the uniform distribution, as expected from the analysis made in Chapter 12. To provide illustrations of channel capacity, we shall consider next the four discrete- channel examples described in Chapter 12. As usual, the input probability distribution is defined by the single parameter = p(x1) = 1 - p(x2). Example 13.1: Z channel The transition matrix is shown in Eq. (12.5), to which the parameters a = 1 and b=1 - E correspond. It is easily obtained from the above definitions that U = 0 and V = W = - f(e)/(1 - E), which gives C = log (1+2 ), (13.12) (13.13) Channel capacity and coding theorem The case E = 0, for which a = b = 1, corresponds to the noiseless channel with identity transition matrix, for which C = 1 and q = 1/2. The limiting case E 1 (a = 1 and b = 0) falls into the category of useless channels for which a + b = 1, with C = 0 and q = 1/2 (the latter only representing a continuity solution,4 4 but not the optimal distribution, as discussed at the end of Appendix K). Example 13.2: Binary erasure channel The transition matrix is shown in Eq. (12.6). One must then calculate the entropies H(Y) and H(Y|X), according to the method illustrated in Appendix K (see Eqs. (K4)-(K5) and (K7)-(K8)). It is left to the reader as an easy exercise to show that H(Y) = f(e) + (1 - E)f(q) and H(Y|X) = f(e). The mutual information is, thus, H(X; Y) = H(Y) - H(Y|X) = (1 E)f(q) and it is maximal for q = 1/2 (df/dq = [log(1 - q)/q]). Substituting this result into H(X; Y) yields the channel capacity C = 1 - E. The number 1 - E = p(y1/x1) = p(y2/x2) corresponds to the fraction of bits that are successfully transmitted through the channel (a fraction E being erased). The conclusion is that the binary-erasure channel capacity is reached with 