Question

Problem 4: Evaluation of the convolution integral too y(t) = (f * h)(t) = f(t)h(t – 7)dt is greatly simplified when either the input f(t) or impulse response h(t) is the sum of weighted impulse functions. This fact will be used later in the semester when we study the operation of communication systems using Fourier analysis methods. a) Use the convolution integral to prove that f(t) *8(t – T) = f(t – T) and 8(t – T) *h(t) = h(t – T), where T is a constant. This is an important result; convolution of an arbitrary function f(t) with an impulse function located at t = T returns the original function f shifted by T seconds. b) Using the result in part a, for each of the pairs of waveforms in the figure below, find the zero- state response of the system with impulse response h(t) and input f(t). Express your answers graphically.

Answer 1

og 4@. fck) *564-1) = ? : 864-7) *hce) a hét-T) ? le have: 2,64) * X,Fe) = fx,(mazlt-podr * {(t-1)xh(t): face-T) hc4-mdr + f(t) *6(4-7) = 5fCW.864-N-1)dr - - j* 6CP-1) h (4 -T) de if we see, for at as variable here, SPT) exists only at N=T the delta Cimpulse) function "St-N-T)". exist at Mat-t, and is zero else where and by Sampling property $Cr-1). h64-F_) = 6CN-T)h(t-T Now, by "sampling property of impulse function : fcro S(t-P-1) = f (-7). &lt=1=1) - h64-T) 76(1-1)dy ahce-T).U. o ffice">6cf-P-1)dt = SfCt-13864--7) dr +($64-1)*5149 = hct-T)) = $4-7958ce-r-ter] = $(t-1) cn S60o dx = 1) 4. fc* $(t-1) = f(t-1)]

@ fcel hrt) isi f)= -2864-1) ht) = $(t-1) + 8(t) +8C++) ->yet) = -2804-1) ¥h(t). ... | nyet) = f(4)*suck) -2h(t-) , = f(e) x[sc4 =10+ 544) 4 864 +1 = f(t) x 5c4+) + f(t) *6c4 ) + f64) *8 (4+1) y(t) = -2h(t-1) fylgte f(t-1) + f 6 ) +(4+1) Ihm f(t-1) (+) = f(t) = 2864) Lyce sa fct) xht} -> output: yce) = f (4) *hct) - 46t) = 2 964) xh (4) ya = 2h(t) 7:6(t-1) *h(4) yet a Dahce-T) D) + I here to fatti - > 1 fff+1) yit)