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y(t) & = \int_{-\infty}^\infty e^{-(t-\tau)}u(t-\tau)u(\tau-1)d\tau \\ | 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^\infty e^{-(t-\tau)}u(t-\tau)d\tau \\ | ||
| − | & = \int_1^{t} e^{-t-\tau}d\tau \\ | + | & = \int_1^{t} e^{-(t-\tau)}d\tau \\ |
& = e^{-t}\int_1^{t} e^{\tau}d\tau \\ | & = e^{-t}\int_1^{t} e^{\tau}d\tau \\ | ||
& = e^{-t}(e^{t} - e) \\ | & = e^{-t}(e^{t} - e) \\ | ||
Latest revision as of 16:56, 30 June 2008
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} $
