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Chapter 2 (a 2.1-1 Let us denote the signal in question by ¢(t) and its energy by Ey. For parts (a) and (b) ae tp oa pe e,~ [ sintrar= 3 f dt- f cos 2tdt=r4+0S% o “Jo 2 oO an ye tf f sntrur= 3 f dt-s cos Midt= r+ 0a 2s 2 2 an {c) Ey an 1p 4p (d) Ey= (sin 1)? dt = 4] 5 dt-= cos 2dt| = 4{7 +O] = 47 2 2 0 2 Jo fo Sign change and time shift do not uffect the signal energy. Doubling the signal quadrupies its energy. In the same way we can show that the energy of kg(t) is k7Ey. a flare 2 By= Crtdet fi(-1)%ae Exoy = fo (Q)Pdt = 4. Ex-y ¥ of (2Pdr=4 Therefore Far, = Ez ~ Ey. - 20 ai? * a2 20 (DIEs = apars f (-1)?dt = 2m, Ey -f ayers captars f ayars f (-1)2dt = on 0 * 0 dnja + Jano 1/2 an /2 an Fea f carters f ova f (-1)2dt = 49 0 wid an/2 Similasly. we can show that Ex_, = 4" Therefore Exay = Ex + Ey. We are tempted to conclude that Lesy = E, ~ E, in general. Let us see. id " (©) es [ aytas f (-1)?dt = af (dt = 0 a4 lo 4 a a4 Easy = [ (2yrett +f WPdt=aw Bsy= [ (0) at +f (+2)%dt = 30 0 ata 0 ne Therefore. in general Eray # Ex + Ey 21-2 (a) Ff; 21-3 1 e Ras C7 cos*(wot + 0) dt = "a + 00s (2wof + 24)] ut Ts Jy 2To To Ty a c = on sf dt + f wncons 2 = Fy et = TT 2.1-4 This problem is identical 10 Example 2.2b. except that wu ¢ wa. In this case. the third integral in Py (see p. 19 is not zero. This integral is given by 7:2 | cas (at + 61) cos (wit + 2) dt Ts? Cac; Tie TA = lim 2 [f cos() — 82) dt +f cost Quit + + Be) df T™~ Tj2 “172 =n SAE? tr cos(hs ~ Ma)] +0 = CrC2 costs ~ #2) 
