t)*cctr)+n(t) (9.7) Chapter Nine pulse channel coder shaper filter Li = Ajej0 new modulator new channel received signal Figure 9.5 New modulator and new channel Here, r(t) is the sent signal passed through the channel filter with a noise added on to it. Plugging in the value for s(t) coming out of the modulator in Equation (9.6) leads us to (9.8) or, using properties of convolution, (9.9) where (t)=g(t)*cg(t) and CE (t) is the baseband filter corresponding to the channel filter c(t) shifted in frequency SO that it is centered around 0 Hz. That is, the channel, through its filter c(t), reshapes the transmitted pulses, changing the shape from g(t) to h(t), and it adds a noise n (t). Example 9.2 Determine the output of a channel that performs a filtering and adds a noise given the channel filter is described according to (E9.8) the modulator used is a BPSK modulator. Solution: Using equation (9.9), we have (E9.9) Channel Filtering and Equalizers where h(t)=g(t)*cg(t) (E9.10) Using the information about the channel in Equation (E9.8) leads us to the result h(1)=g(1)+[8(1)+8(1-1)] (E9.11) h(t)=g(1)+g(1-r) (E9.12) Applying this to equation (E9.9) leads to (E9.13) Knowing that the modulator is BPSK further tells us (from Example 9.1) that I = Aejo, where 01=0° or 180° (E9.14) Equation (E9.13) and (E9.14) together describe the output of the channel filter. 9.3 Receivers: A First Try We now have a general form for the modulators we've been studying. We know if we send the signal s(t) across a channel that does a filtering, we get out (at the receiver side) the r(t) of Equation (9.9). At the receiver side, the job at hand is this: take the r(t) coming in, and figure out m', a best guess of the original bits m that were sent. Alter- natively, if we can come up with I', a good guess on the coder outputs I = (Io,1,, we can figure out m'. So we'll say that the task of the receiver is to come up with I', as good a guess as possible on the value of I. The receiver I'll describe here won't turn out to be the best receiver you can