Question

JU Q1. Sketch the following SIGNALS and determine for each whether it is: Periodic or aperiodic. If periodic, • Even, odd, or neither. If it is neither, specify To. decompose it into even part Ev{x} Energy signal, power signal, or and odd part Od{x} and sketch them. neither. Find the total energy E. and average power PC- 12 -1<n<1 (a) xa n] = lo otherwise (b) xo(t) = xy(t) + x2(t), where x,(t) and x2(t) are the signals shown below: xi(t) -4 -3 -2 -1 0 1 2 3 4 5 6 7 8 9 X2() - -4 -1 -3 -2 0 1 2 3 4 5 9 6 7 8 .. TATT n*e- x[n] and the average power is Hint: for discrete-time signals, the total energy is E. = P = lim - 1 **-n | x[n] 12. 1

Answer 1

Given's las Ka[n] =42;-15 asi otherwise - 100 2 2 0 -00 from Figure, this signal xa [n] is Z o r z 3 aperiodic. Energy: EE ! Pa60212 E = { 1212 [E = 4] p=lim ! È IRQ[n] 12 P= lim ayanti 2 (2) 2 p= lim 2+1 (4) no1 Power- 102N+Incent ir E is finite value and P is zero. then xa [n] is Energy signal. + Ra[n] = $-0,2, , 2,0-$ - n=0 xa [n] = {..0, 2, 2, 2,0.-} D=0 Hence, Ha [n] = ka[n] req [n] is even signal.

Given by eght) = x, t) +22) 72,4) -4-3-2- ol ! 2 3 4 -4-3-2-1 0 1 2 3 4 5 6 - Peñad Ti=4 Periodic signal Period T2 =6 a periodic signal 30. To = Lema CT1, Tz] - To = Lem [4,6] T. 212) ľ M, Ct). 2264 dt = 0, Hence this signels are Orthogonal. 80, Energy = e =E, TE2 and power p =P, TP2 Energy 942465:- Energy of Rally: E = pix, es pat E = pa ,12,4077* det E2= E2= 1 EL = 1 Power of x, et Power of Nally a P2 = 1 {Tole neurat - P2 Topola P, = to 2014 P, 2 * уч- P2 = $ $ 1 mzury jest P2 = 1 2 3 , 2dt P = 4 P2=t

80, E = 3, + EZ ET+1=2 P= P, + P2 =Ę + -0.4167 Thus, xoets is neither Energy nor power signal. -> Now, 4 -3 -2 lo 2. 3 Rent- 1 - +- 26-4):- 1 1 - O I 2 3 4 $'t Ratt -3 -2 -1 2 3 Y s t -5-4-3-2-1 i 2 3 4 %, CH 7 x,(-t) Kzlt) & R21-7) 2, 7 -275) 82. (t) 7-821-1) Hence, Molt is neither even mor od R$60:- Tables -Y -3 -2 -1 1 2 3 Even ! Even par- Evi RJ - Rble + Ult creado 15) --W(-) Odd part del = OdC9