to figure out is the orthonormal basis for r(t) = =sm(t)+n(t). In other words, what will that {Q1(t) , N (t)} in Equation (5.16) look like? Let's take a look at r(t) : it is made up of two signals, Sm(t) and n(t), SO we'll take a look at the orthonormal basis for each of these signals, and we'll use that to get an orthonormal basis for r(t). Let's start with Sm(t). We'll pick one out of a hat and say that sm(t) is a 4-PSK signal. That means, from what we saw in the previous section, that the orthonormal basis for Sm(t) is {Q,(t),,22(t)} where (t) = and Getting It from Here to There: Modulators and Demodulators Next, let's look at n(t), a random process described as AWGN. How do you get the orthonormal basis {0,(t),....ON(t)} for a random process like n(t)? Answering that will just introduce a lot of math and stats, which ultimately will lead us to a simple point-so I'm just going to tell you the simple point. For a random process, specifically one described as AWGN, any infinitely long orthonormal basis will do. So we'll choose this one. I pick where {Q,(t),Q2(t)} forms the orthonor- mal basis for Sm(t) (i.e., = and and Q3(t),,44(t), are some other functions which when combined with {Q,(t),Q2(t)} form an infinitely long orthonormal basis (we'll see later that it won't matter what they are). Now we have an orthonormal basis for Sm(t) and an orthonormal basis for n(t), SO you say: what's the orthonormal basis for r(t) = Sm(t)+n(t)? Well, n(t) can be represented on } and Sm(t) can be represented on {Q,(t),Q2(t)} . It follows that r(t) can be represented on since: this basis is adequate to represent Sm(t) (using the first two elements) and adequate to represent n(t) (using all the elements), SO it's got to be enough to repre- sent r(t)=s.,(t)+n(t). Representing r(t) as a vector using the orthonormal basis Now that we've got the orthonormal basis for r(t), let's figure out the in r(t)=r(t)+r2P2(t)+r3P3(t)+... This r is the vector that the receiver front end wants to figure out and hand off to the rest of the receiver.