| Line 3: | Line 3: | ||
<math>y(t) = 3x(t-1)+x(t+3)-x(t)\!</math> | <math>y(t) = 3x(t-1)+x(t+3)-x(t)\!</math> | ||
| − | + | ||
| + | == Part A == | ||
| + | |||
<math>h(t) = 3\delta(t-1)+\delta(t+3)-\delta(t)\!</math> | <math>h(t) = 3\delta(t-1)+\delta(t+3)-\delta(t)\!</math> | ||
<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> | ||
| − | |||
| − | |||
Revision as of 16:38, 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\! $
