 Determine the orthonormal basis for the two signals shown in Figure E5.2. Solution: We'll use the orthogonalization procedure, which tells us (E5.1) (E5.2) (E5.3) (E5.4) S1 (t) S2 (t) Figure E5.2 Two signals This first element of the orthonormal basis is drawn in Figure E5.3. Return- ing to the orthogonalization procedure to get the second element, we discover (E5.5) Getting It from Here to There: Modulators and Demodulators (E5.6) (E5.7) (E5.8) (E5.9) Q1 (t) Figure E5.3 The Q (t) To help get a better feeling of what the numerator looks like, you'll find it drawn in Figure E5.4. Using this, we'll finish solving for the second orthonormal basis element like this: (E5.10) (E5.11) Chapter Five (E5.12) =02(t) (E5.13) The two elements of the orthonormal basis are drawn in Figure E5.5. Figure E5.4 A plot of Q1 (t) Q2 (t) Figure E5.5 The orthonormal basis 5.3.2 Representing Modulated Signals Next we'll see some simple examples of what was discussed above. If you totally understood what I said in the last section, then this part will be a piece of cake. If you don't fully understand what I said earlier, then this section will help you nail it. First, we'll look at the signals that leave the PSK modulator, then look at the signals that leave the ASK modulator, and finally look at the signals that leave the QAM modulator. I'm going to tell you about the orthonormal basis (the Q (t) 's) and the coefficients (sij's) corresponding to these signals. Before I do this, let me empha- size that the reason I'm doing this has nothing to do with a great love for orthonormal basis, but it actually turns out to be important when we look at demodulators. Getting It from Here to There: Modulators and Demodulators Let's look at the signals that leave the BPSK modulator. You can see these two signals, called so(t) and s((t), at the top of Table 5.4. The Orthonormal Basis: Given the two-signal set {so(t),s,(t)}, we'll start with the orthonormal basis {Q1(t), , Pn(t)} for these two signals. First, you know you've been given two signals {so(