Question

Problem 3.8 Find and sketch g(t)=x(t)*x(t) for the pair of functions shown below. Did you notice anything interesting? x(t) xz(t)xz(t) x2(t) - 3 3 -1 -1 of xi(t) xz(t) xz(t)xzt) X(t) xz(t) xz(t) * x2(t) xi(t) xz(t) xi(t) = x2(t) Note: Everlasting signal AAA

Answer 1

x₂ (+) a) _ ^ vices B 22 4 6 + 24 X (+-7) Whem -ict co- - 6+t - 4+t -5 h When t <-1 -6+t -htt -5 h = (x₂ (T) x, (t-T) dt -6tt h+t - 2y (t) = 07 = TAB dT When ost<! - AB[=htt)-(-5) = AB (L+t) When 1<t <2 MB - 5 cha -6+t -htt -5 - y (+) = J AB de - Ав -bit tt y (t) = PAB dT 6+t I t< AB(1++) -stco AB octet As AB(2-t) 1<t <2 t> = AB(-4-(-6+t) = AB[2-t) AB yet) y(t) =/ - 1012

X, Lt) = u(t+2) xz(t) = ult+2)- ult-1) -2 - 2 1 Note ult) * u(t) = r(t) ultra) * u(t+b) = r(ta+6) X, (t) * x₂ (t) = u(+2) * (u(t+2) -u (t-1)] 3 = ult+2) * u(t+2) - a (+72) * ult-4) - rlt+4) - alt t) yet) - ret&th) - r(+11) - -

c) 1 X2 CT x (t) X (t-) an 6 2 T -2++ -2xt t Whenort <2 ct SE- +9dt When to O T -2t+1 z 2 Ty(t) = 0 + -2+t t 2 Jo When t > 1 -2t²tt N/ (y(t) = // t- tco octc2 t> 4)

(7) ix 40 10) * (T) X2 (-T) x₂(t-T) yet) = ( x, (2). 4, (tt) de yee Yuce) de y = 0 4t to 4 t=h Area =0 Area=0 TyH) = 0) As

X, (t) * (t-1) -8tt yet) = 1 (0) 3 (4-7) de tot 1x,ce) dt Area=0 (y (+) =0 4 ) Ay [ Area=o 7 e) AM x, (t) - Xz(t) x2 Ct -T) 1 2+t * (T) yet) - "J+, (C) & (t-1)dt 2 = 218 x, (T) at -216 n Lyct-o ] Aus I Area=o