(New page: 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 convolutio...) |
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| − | + | To convolve two functions we have the following: | |
| − | + | ||
| − | + | ||
<math> | <math> | ||
y(t) = h(t) * x(t) = \int_{-\infty}^\infty x(t)h(t-\tau)d\tau | y(t) = h(t) * x(t) = \int_{-\infty}^\infty x(t)h(t-\tau)d\tau | ||
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= \int_{-\infty}^\infty e^{-\tau}u(\tau)u(t-1-\tau)d\tau | = \int_{-\infty}^\infty e^{-\tau}u(\tau)u(t-1-\tau)d\tau | ||
</math> | </math> | ||
| − | We now change the interval of integration to reflect the step function | + | We now change the interval of integration to reflect the step function <math>u(\tau)</math> |
<math> | <math> | ||
= \int_{0}^\infty e^{-\tau}u(t-1-\tau)d\tau | = \int_{0}^\infty e^{-\tau}u(t-1-\tau)d\tau | ||
</math> | </math> | ||
| − | Finally, by changing the interval of integration again we get: | + | Finally, by changing the interval of integration once again we get: |
<math> | <math> | ||
\begin{align} | \begin{align} | ||
Revision as of 22:10, 1 July 2008
To convolve two functions we have the following:
$ y(t) = h(t) * x(t) = \int_{-\infty}^\infty x(t)h(t-\tau)d\tau $
Plugging in functions for x(t) and h(t) we get:
$ = \int_{-\infty}^\infty e^{-\tau}u(\tau)u(t-1-\tau)d\tau $
We now change the interval of integration to reflect the step function $ u(\tau) $
$ = \int_{0}^\infty e^{-\tau}u(t-1-\tau)d\tau $
Finally, by changing the interval of integration once again we get:
$ \begin{align} &= \int_{0}^{t-1} e^{-\tau}d\tau \\ &= -e^{-(t-1)} - (-e^{0}) \\ &= 1 - e^{-(t-1)} \end{align} $
