CT LTI signal:

$ y(t) = 3x(t-1)+x(t+3)-x(t)\! $


Part A

$ h(t) = 3\delta(t-1)+\delta(t+3)-\delta(t)\! $

$ H(j\omega) = \int_{-\infty}^{\infty}3\delta(t-1)+\delta(t+3)-\delta(t)\,dt\! $

$ H(s) = 3e^{-s} + e^{3s} - 1\! $

Part B

Signal from Question 1

$ x(t)=cos(\frac{\pi}{2}t)u(t) = \frac{1}{2}(e^{j0.5\pi t}+e^{-j0.5\pi t})u(t)\! $

$ y(t) = (3e^{-j0.5\pi} + e^{j1.5\pi} - 1)\times\frac{1}{2}[e^{j0.5\pi t}]u(t)+(3e^{-j0.5\pi} + e^{j1.5\pi} - 1)\times\frac{1}{2}[e^{-j0.5\pi t}]u(t)\! $

$ y(t)=\frac{1}{2}[3e^{j0.5\pi(t-1)}+e^{j0.5\pi(t+3)}-e^{j0.5\pi t}+3e^{-j0.5\pi(t+1)}+e^{j0.5\pi(3-t)}-e^{-j0.5\pi t}]u(t)\! $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood