ne, as shown in Figure 9.10. This leads to the outputs (which we saw earlier) (9.27) + (high frequency terms which will be cut out in a moment) (9.28) + (high frequency terms which will be cut out in a moment) Channel Filtering and Equalizers cos(Wct) r'(t) rs(t) Figure 9.10 sin(Wct) The first part of an optimal receiver front end where NC (t)=n(t)cos = Wct and NS s(t)=n(t)sinw, To simplify the presentation and keep us from having to bring two different signals with us in the rest of our discussion, we can represent this as a single complex signal, namely (9.29) which leads to (9.30) + (high frequency terms, cut out in a moment) +n'(t) + (high frequency terms, cut out in a moment) (9.31) r'(t)= s'(t) n'(t)+(high frequency terms,cut out in a moment) (9.32) where (t) = (t) + jn (t) and Now, each I = Ae18, is one of a finite number of values (one of M values, to be exact, because I represents the output from the coder, which takes in n bits and outputs one of M=2" = possible I's. See Figure 9.3 for an illustrative example.) That means that is a time function that is one of a finite number of time functions (one of ML possible time functions to be exact). So we have '(t)=s'(t) n (t), where is one of a finite number of time functions. Chapter Nine If you look back to Chapter 5, we saw there that: if you receive r(t) = s(t) + n(t), where s(t) takes on one of a finite number of values, M values, to be exact, then an optimal receiver front end is the one shown in Figure 9.11 (a). So, now that we have v' (t) = s'(t) + n'(t), where s'(t) takes on one of a finite num- ber of values (one of M values), then an optimal receiver front end for r'(t) must be the one shown in Figure 9.11(b). S1 (t) (t) = s (t) + n(t) SM(t) COS Wct S'1 (t) r'(t) = s'(t) + n'(t) sin Wct S'ML(t) Figure 9.11 (a) Optimal receiver front end for received signal of Chapter 5, namely = (b) Corresponding optimal receiver front end for received signal of Chapter 9: (1) first creating r((t) = s (t) + n'(t), = then (2) using receiver front end analogous to (a)