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Chapter 6 The Impulse Response and Convolutione At – 0.5e – t + 1.5e – 3t – 2e – t + 2e – 3t=3--e – t 3 3t– --e –1.5e – t –8 8 0.5e – 3tThe impulse response is obtained from (6.5), Page 6−1, that is,then,ht () = xt () = e At bu 0 () tht () = xt ()x – 0.5e – t + 1.5e – 3t – 2e – t + 2e – 3t= 1=x32 --e – t 3 43t– --e –1.5e – t –8 8 0.5e – 3t 0 u 0()t =– 2e t +–3--e – t 3 3t–2 2 --e –– 6e 3tu 0 () t(6.10)In Example 5.10, Chapter 5, Page 5−22, we definedandThen,orx 1x 2==i Lv Cht () = x 2 = v C () t = 1.5e – t – 1.5e – 3tht () = v C () t = 1.5( e – t – e – 3t )(6.11)Of course, this answer is not the same as that of Example 5.10, because the inputs and initial conditionswere defined differently.6.2 Even and Odd Functions of TimeA functionft ()is an even function of time if the following relation holds.f( – t) = ft ()(6.12)that is, if in an even function we replace t with – t, the function ft () does not change. Thus, polynomialswith even exponents only, and with or without constants, are even functions. Forinstance, the cosine function is an even function because it can be written as the power seriescost= 1 –Other examples of even functions are shown in Figure 6.4.t 2----2!t 4+ ----4!t 6– ---- + …6!6−4Signals and Systems with MATLAB ® Computing and Simulink ® Modeling, Fourth EditionCopyright © Orchard Publications