 basis So(t) s,(t) 4-ASK So(t) s((t) 52(t) 53(t) 8-ASK So(t) s((t) S2(t) 53(t) 54(t) 55(t) 56(t) 57(t) Table 5.9 ASK signals represented on orthonormal basis {01(t)} Getting It from Here to There: Modulators and Demodulators The last one! In QAM, the information bits are stuffed into both the phase (0) and the amplitude (A) of the cosine waveform. That is, a typical output waveform for QAM looks like this: (5.9) Now, applying a little bit of trig, we can rewrite this as (5.10) You can plainly see that the (t) can be expressed as a sum of one sine and one cosine term. So it just makes sense that the orthonormal basis consists of a cosine and a sine term. Specifically, one orthonormal basis made up of a sine and a cosine term is this one: {0,(t),Q2(t)} where This will work wonderfully as the orthonormal basis for QAM. Using this orthonormal basis, we can write the QAM output signal Sj (t) above as: (5.11) Now, you can use the integral in Equation (5.6) to get the values S 11 and Sj2, or you can figure these out by comparing the above s ( t ) equations (namely (5.10) and (5.11)) and seeing what these Sj1 and Sj2 must be. Either way, you'll come up with Sj2 A, which means the QAM signal can be written according to (5.12) One nice way to look at this (t) is to plot it as a point in space, which I do in Figure 5.23. O2(t) represents (s)(t) 01(t) Figure 5.23 A plot of a single QAM point Chapter Five Before we head off into the brave 02(t) new world of demodulators, one last thing about QAM. A typical 16-QAM 3A (T/2) constellation looks like Figure 5.24. There, now you've had a chance to see all 16 output signals, and I didn't have to AV(T/2) make a table of all 16 input bit pairs (0000, 0001, 1111) and all the corre- -3A (T/2) (T/2) ((T/2) (T/2) 01(t) sponding output signals. This orthonormal basis stuff sure does come -AV(T/2) in handy sometimes. 5.4 Bring it Home, Baby, or -3AV(T/2) Demodulators What I want to do here is build a de- Figure 5.24 Plot of 4-PSK signals modulator for you, and teach you how to {so(t),s,(t)