t/T): (a) in time and (b) in frequency )=RCa(t) a=0(-) a = 0.5 (--) x = 1 (....) x(f) = F{RCa = (t)} 1/2T(1+a) -1/2T (1+a) Figure 9.8 x(t) = RCa(t): = (a) in time and (b) in frequency Chapter Nine 9.3.3 The Proposed Receiver: Problems and Usefulness Let's look at the receiver that we've been discussing SO far. It's shown in Figure 9.6, and we figured out that the v value is: (9.22) . We also saw that if we chose f(t) and g(t) carefully, we could make sure that the x(t) was such that x(0) = 1 and X (kT - iT) = 0, which would lead to rk=1k+nk (9.23) This looks great. We've now been able to get the receiver to produce a kth sample that has in it the kth information symbol Ik and some noise Nk. But there are problems with the noise Nk. This noise corresponds to . Now, f(t) was chosen to make x(t) satisfy the Nyquist criteria; we made no effort to consider how f(t) would affect the noise. Indeed, f(t) could cause dramatic noise amplification, making the information I lost in the sea of loud noise Nk. So, generally speaking, this receiver of Figure 9.6 is not used in its current form, except it does have one really nice use: when the channel filter c(t) is flat over the transmission frequency-that is, when c(t) has a frequency response C(f) such as that shown in Figure 9.9. In this case, the channel C(f) = 1 for all intents and purposes, as long as we make sure all the transmit frequencies fall in the flat part of C(f). For all practical purposes, the channel can be modeled as cg(t)=8(t). = C(f) = frequency response of channel filter flat over range of transmission Figure 9.9 A possible channel filter c(t) shown in the frequency domain Channel Filtering and Equalizers In this case, for the modulator and receiver we're considering, we have an V equal to (9.24) where this time X x(t)=g(t)*cp(t)*f(t)=g(t)*f(t) and, as before, = =n"(kT) . If we want to try to make V as close as possible to I k , then we'll make sure that (9.25) If we do this and also make sure g(t) = f(t), it can be shown (with a bit of statistical wrangli