 {.} h(-t) symbol sin Wct NEW OPTIMAL NEW MODULATOR NEW CHANNEL RECEIVER FRONT Decision device LINEAR EQUALIZER Figure 9.20 (a) The modulator, channel, and receiver front end (b) The new rest-of-the-receiver (a linear equalizer followed by a decision device) 9.6.1 Zero Forcing Linear Equalizer To understand this case, let's start with a look at what is coming into the receiver, namely V' (9.59) The intention in the zero forcing linear equalizer is to force the ISI to zero-to get the output rk to have the form rk=1k+nk (9.60) That's easy. All we do in this case is make Ck undo XK . That is, we choose Ck to be the inverse filter for the filter Xk . Mathematically, we choose (using the z-transform domain) C(z) = 1/X(z) (9.61) Channel Filtering and Equalizers There's one problem with building the linear equalizer in this way: it focuses on getting rid of the filtering effect (the X ), but it pays no attention to what happens to the noise. The noise passes through the filter Ck , and it may get much bigger. This phe- nomenon, known as noise amplification, can cause the I to get lost in the much larger noise nk For this reason, this type of linear equalizer is rarely used. Instead, engineers have a fondness for the... 9.6.2 MMSE (Minimum Mean Squared Error) Equalizer With this type of equalizer, in comes (9.62) or, written differently, (9.63) rk=Ikxo+ (9.64) We'd like to choose the equalizer (the filter) Ck SO that what comes out, namely rk=rk*ck is as close to las possible. With that in mind we say, mathematically, that we want to choose Ck to minimize the function: (9.65) where E[x] is the expected value of X. That is, in words: on average, we want the output of the filter to be as close as possible to Ik . After about a page of statistical wrangling, you can show that this requirement is met by choosing Ck to satisfy (in the z-transform domain) C(2)=1/[X(2)+N,] = (9.66) where N /2 is the variance of the noise introduced in the channel and X(2) is the z-transform of Xk. Example 9.5 If a receiver sees the input (E9.23