Revision as of 11:00, 21 November 2008 by Mcwalker (Talk | contribs)

We are given the input to an LTI system along with the system's impulse response and told to find the output y(t). Since the input and impulse response are given, we simply use convolution on x(t) and h(t) to find the system's output.

$ y(t) = h(t) * x(t) = \int_{-\infty}^\infty h(t-\tau)x(t)d\tau $  (COMMUTATIVE PROPERTY)


Plugging in the given x(t) and h(t) values results in:

$ \begin{align} y(t) & = \int_{-\infty}^\infty e^{-(t-\tau)}u(t-\tau)u(\tau-1)d\tau \\ & = \int_1^\infty e^{-(t-\tau)}u(t-\tau)d\tau \\ & = \int_1^{t} e^{-(t-\tau)}d\tau \\ & = e^{-t}\int_1^{t} e^{\tau}d\tau \\ & = e^{-t}(e^{t} - e) \\ & = 1-e^{-(t-1)}\, \mbox{ for } t > 1 \end{align} $


Since x(t) = 0 when t < 1:

$ y(t) = 0\, \mbox{ for } t < 1 $


$ \therefore y(t) = \begin{cases} 1-e^{-(t-1)}, & \mbox{if }t\mbox{ is} > 1 \\ 0, & \mbox{if }t\mbox{ is} < 1 \end{cases} $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood