 Channel Filtering and Equalizers Of course, nobody wants to build the receiver front end shown in Figure 9.11 (b) because there are ML branches, which can be quite a lot. But, this did give some smart engineer an idea-namely, to take a closer look at that Uk coming out of the receiver front end of Figure 9.11(b). That Uk corresponds to: (9.33) (9.34) (9.35) (9.36) where we have r' (for i = 0,1,...,L - 1), we could generate every Uk we want. We don't have to build a receiver front end as big as the one of Figure 9.11(b)-all we have to do is build one that provides the value of r' for i = 0,1,2, ..., L-1. If we can do that, then from this receiver front end we have the values needed to build the receiver front end of Figure 9.11(b) if we want it-or any other important receiver we want to build, for that matter. Take a look at Figure 9.12. From this receiver front end, r(t) is mapped to v (t) by the cosine and sine multiplication; then with v (t) coming into the filter and sampler, v' comes out-like this: Let 0 output of filter and sampler at sample time i; then (9.37) (9.38) (9.39) (9.40) Figure 9.12, then, is our receiver front end. It takes in r(t) and it puts out r' i = =0,1,2,...,L-1. ..., - With these values r', we have all the numbers we need to generate any values we might want in our receiver. Chapter Nine cos wct rc(t) h(-t) r'(t) rs(t) sin(Wct) Figure 9.12 Optimal receiver front end 9.5 Optimal Rest-of-the-Receiver We've gone from the new modulator, through the new channel that put out the new r(t), and through a receiver front end which outputs r' You can see it all together in Figure 9.13. Now let's build the optimal "rest-of-the-receiver," which puts out m', the best possible guess on the transmitted bits m. RECEIVER TRANSMITTER CHANNEL FRONT END COS Wct pulse coder shaper Re {.} h(-t) (bits) sin Wct Figure 9.13 The new modulator (transmitter) with pulse shaping, followed by the new channel (with a channel filter), followed by the optimal receiver front end 9.5.1 The Input Coming into what I'm calling