t a good gut, you could use your intuition, using the fact that every 8-PSK signal can be written as a sum of sines and cosines. Either way, you can come to this result: the 8-PSK signals can be represented on the same orthonormal basis as 4-PSK, namely {Q,(t),Q2(t)} where All that's left are the values (501,S02), then (511,S12), and SO on up to (S71,S72) You can get this by using the integral of Equation (5.6), or if you prefer, you can again use that old gut of yours, turn on the intuition, and looking at Table 5.7, see how much of Q1(t) and how much Q2 2(t) you need to create s;(t) (this gives you (Sil,Si2)). Either way, you'll come to the same values, which you'll find in Table 5.8. Input bits Output waveforms 8-PSK s((t) = A COS = S2(t)=Acos(wt+90).(t-iT) )) = = S4(t) = = -A + )1( = = + Table 5.7 8-PSK written as a sum of sines and cosines Getting It from Here to There: Modulators and Demodulators Output Output waveform represented on waveform orthonormal basis 8-PSK So(t) s,(t) 52(t) 53(t) 54(t) 55(t) S6(t) 57(t) Table 5.8 8-PSK written as a sum of sines and cosines Good news. ASK is easy. Look at Tables 5.1 to 5.3. There, you'll see the ASK output signals, and you'll quickly notice one thing: all ASK output signals are simply terms; the only difference is the amplitudes. To get the orthonormal basis for either BASK, 4-ASK, or 8-ASK, you can perform the algorithm of Section 5.3.1 using the ASK symbols {so(t), SM(t)} or, you can simply realize this: If I consider the one-element orthonormal basis {Q1(t)} where I can represent all ASK signals as a constant times this(t)-sothisfellow serves as my orthonormal basis. Next, we turn our attention to computing the coefficients S01, then S11, and SO on up to SM1 You can perform the integral of Equation (5.6), or you can use your intu- ition, looking at Q(t), then looking at S (t), and you can figure out how many Q|(t)'s you need to get s;(t). Either way, you'll get the values shown in Table 5.9. Chapter Five Output Output waveform represented on waveform orthonormal