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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. | 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. | ||
| − | <math>y(t) = h(t) * x(t) = \int_{-\infty}^\infty h(\tau)x(t - \tau)d | + | <math>y(t) = h(t) * x(t) = \int_{-\infty}^\infty h(\tau)x(t-\tau)d\tau</math> |
| + | |||
Plugging in the given x(t) and h(t) values results in: | Plugging in the given x(t) and h(t) values results in: | ||
| − | <math>y(t) = \int_{-\infty}^\infty e^{-\tau}u(\tau)u(t - \tau - 1)d(\tau)</math> | + | <math> |
| + | \begin{align} | ||
| + | y(t) & = \int_{-\infty}^\infty e^{-\tau}u(\tau)u(t-\tau-1)d\tau \\ | ||
| + | & = \int_0^\infty e^{-\tau}u(t-\tau-1)d\tau \\ | ||
| + | & = \int_0^{t-1} e^{-\tau}d\tau \\ | ||
| + | & = 1-e^{-(t-1)}\, \mbox{ for } t > 1 | ||
| + | \end{align} | ||
| + | </math> | ||
| − | |||
| − | + | Since x(t) = 0 when t < 1: | |
| − | + | <math>y(t) = 0\, \mbox{ for } t < 1</math> | |
| + | |||
| + | <math>\therefore y(t) = | ||
| + | \begin{cases} | ||
| + | 1-e^{-(t-1)}, & \mbox{if }t\mbox{ is} > 1 \\ | ||
| + | 0, & \mbox{if }t\mbox{ is} < 1 | ||
| + | \end{cases}</math> | ||
| − | + | ==Alternative Solutions== | |
| + | [[Problem 5 - Alternate Solution_OldKiwi]] | ||
| − | + | [[Problem 5 - Graphical Solution_OldKiwi]] | |
Latest revision as of 16:23, 3 July 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(\tau)x(t-\tau)d\tau $
Plugging in the given x(t) and h(t) values results in:
$ \begin{align} y(t) & = \int_{-\infty}^\infty e^{-\tau}u(\tau)u(t-\tau-1)d\tau \\ & = \int_0^\infty e^{-\tau}u(t-\tau-1)d\tau \\ & = \int_0^{t-1} e^{-\tau}d\tau \\ & = 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} $
