| Line 10: | Line 10: | ||
<math>H(j\omega) = \int_{-\infty}^{\infty}3\delta(t-1)+\delta(t+3)-\delta(t)\,dt\!</math> | <math>H(j\omega) = \int_{-\infty}^{\infty}3\delta(t-1)+\delta(t+3)-\delta(t)\,dt\!</math> | ||
| + | |||
| + | <math>H(s) = 3e^{-s} + e^{3s} - 1\!</math> | ||
Revision as of 17:40, 26 September 2008
CT LTI signal:
$ y(t) = 3x(t-1)+x(t+3)-x(t)\! $
Part A
$ h(t) = 3\delta(t-1)+\delta(t+3)-\delta(t)\! $
$ H(j\omega) = \int_{-\infty}^{\infty}3\delta(t-1)+\delta(t+3)-\delta(t)\,dt\! $
$ H(s) = 3e^{-s} + e^{3s} - 1\! $
