t),s,(t)}, SO the orthonormal basis will be made up of less than or equal to two Q;(t)'s. You can go ahead and do that algorithm I gave you in Section 5.3.1, and you'll get the orthonormal basis (actually, it's good practice, SO take a moment and try it here). OK, here's what you'll find (if you didn't, check your math). The {Q1(t), Qv(t)} for BPSK is just the one-element set {Q1(t)}, where That's right, there's only one element in the orthonormal basis. This means you can write the PSK output signals as so(t)=soiq(t) (5.7) s,(t)=sup) (5.8) The Coefficients: Next, do the integral I showed you (Equation (5.6)) to get the S01 and S11 (Go ahead. Again, it's good practice). Here's what you should get: for so(t), , the S01 = A and for s,(t), the S11 -A. A Plot: One nice thing about an orthonormal basis is that it lets you represent the BPSK signals as a point in space. Once you've got the {Q (t), QN(t)} (in this case the {Q|(t)}), and you've got the coefficients (in this case the So1=A~ 1/2 and S11 = - -A, then you can plot so (t) and s,(t). Figure 5.21 shows you what I mean. It tells you how much Q (t) you need to get so(t), and how many Q1(t) you need to get s,(t). And that's it. represents S1 (t) represents so(t) S11=-AV(T/2) So1=AV(T/2) Figure 5.21 Plot of BPSK signals so(t) and s,(t) Chapter Five 4-PSK Here we'll be asking and answering the following question: "How many Q;(t)'s (or- thonormal basis functions) does it take to represent the 4-PSK signals (and what are those Q;(t)'s)?" The answer is really quite easy. The Orthonormal Basis: If we're going to represent the 4-PSK signals on an orthonormal basis, we've first got to remember what they look like. Take a quick peek at Table 5.4 where you can see the 4-PSK signals (s,(t),s(t),sa(t),s;(t) once again. Now, look at Table 5.6. There, I've written the 4-PSK signals as a sum of a sine and a cosine (using a simple trig identity). I've done this because in fact it suggests a particu- lar orthonormal basis, as I'll show you next. If you carry out the algorithm of S