Question

For the LTI system S, the iput xie) and ies / freguamey respamse Hu) frnd its outpat e) shown helow, Please given are as hee HOu)=J+2 2 =eue) 2. Fined the Founer transform Yejw) of of Xejw) gct) = X6t). SinC wet) i terms For the LTIsystem S, what is its outputt) 3. /t cos (2 72 000t) + sin (2n. 2ooot) ? when t) Sketch yie) hce (+2R Hiju) = 7[hie) >f (Ha) frequeney 3000 2000 -3e0s -200o

This is really easy questions. but needs curiosity in showing the steps. can I get some one to solve this questions. it's a 15 minutes 3 questions
thanks thumbs up is awaiting!

Answer 1

x(t) = e^-t u(t) F.T x(t) = X(J omega) = F e^-t u(t) F e^-at u(t) = 1/a + J omega = 1/1 + J omega H(J omega) = Y(J omega)/X(J omega) Y(J omega) = H(J omega) times X(J omega) = 0/J omega + 2 times 1/1 + J omega = By partial fraction Y(J omega) can be written as. Y(J omega) = -2[1/J omega + 2 - 1/1 + J omega] = 2/1 + J omega - 2/2 + J omega F^-1(Y(J omega)) = y(t) = F^-1 (2/1 + J omega) - F^-1 2/2 + J omega = 2 e^-t u(t) - 2 e^-2t u(t) y(t) = 2 u(t) (e^-t - e^-2t) y(t) = x(t) middot sin(omega_c t) X(J omega) F sin omega_c t = F sin c(t/1/omega_c) = 1/omega_c pi (omega/2 1/omega_c) \r\n F X(t) = X(J omega) Y(t) = x(t) middot sin omega_c t Taking fourier on both sides F y(t) = F(x(t) middot sin omega_c