Question

Calculate the convolution integral of the following signals. Find the energy and power of the input and output signals. x(t) y(t) x(t) = cos(it)[u(t + 1) – uſt – 3)] h(t) = u(t + 2) – uſt – 1) del mes h(t) de ser LTI System

Answer 1

D itd to t = 3 cut +1) - ult-3) - E so, xft) will be only in 3 interve interval X(t) = cos(TC4) [2(t++) - (4-3) AX(t) h (l) = u(t+2) - Ult-t) | 1 (3) -2 convolution Integral = (x(2) h (t-2) dz o h (2) TX (2)

tilt-z) 2 +t -- ++ convolution is sum of X (12) h (A-2) rie, multifly the X) and h(t-2) ware form and then take their sun. aħ (-2) from Both graph it = 0-3 to will overlaß when it vary t= 4 4 So, convolution = (x(2). hrt-2) dz -3 si catre i da ps con (72) de

din A Sintran) = sinkt T & Sint = 0 + 0 (Sin4 =o) & fin 31=0 - 0 so, output = 0 Energy and power of the į o. output signal shnd x64) - colt:)[(+) – uits] HARMITTAAM1111111111111 s'font/r4) de limit I ['cos rt at THA 27 limit THA I LT t+cos 27t It I limet TH / 2T 2 I f & Cosatt at

to set up (doel Sin (-2TCT) 2 T a bind linit 1 2 T / TA IT In Griven Signal Time period = 21 & wat | So Tim Period = 2 TAA Sinda - Sinh TL = 0 I power = watt itt a Periodic Signal...tt di And we know, Energy of signal Power Signal & infinite .. E=0 h() 1 Plt)=ult +2) - ultt)

Energy = (244) It 5274) 04 monOCAAAAARRRRRRRRR = + [t- (-2) = 3 joule . It is a finite duration signal . It is Energy Signal And we know, Power of Energy Signal become o. P=0.