| Line 6: | Line 6: | ||
h(t) = 10delta(t-1) | h(t) = 10delta(t-1) | ||
| + | |||
| + | <math>s = j\omega</math> | ||
<math>H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}</math> | <math>H(s) = \int_{-\infty}^{\infty}h(t)e^{-st}</math> | ||
Revision as of 12:03, 25 September 2008
CT LTI system:
y(t) = 10x(t-1)
plugging in delta(t) into the system we get:
h(t) = 10delta(t-1)
$ s = j\omega $
$ H(s) = \int_{-\infty}^{\infty}h(t)e^{-st} $
$ H(s) = \int_{-\infty}^{\infty}10delta(t-1)e^{-st} $
$ H(s) = 10\times\int_{-\infty}^{\infty}delta(t-1)e^{-st} $
$ H(s) = 10e^{-s}\, $
