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h(t) = <math>e^{-6|t|}</math> | h(t) = <math>e^{-6|t|}</math> | ||
| − | Since h(t) <math>\neq</math> | + | Since h(t) <math>\neq</math> 0 for t < 0 so the system is not causal. |
<math>\int_{-\infty}^\infty e^{-6|t|} = 2\int_0^\infty e^{-6t} < \infty</math>. This system is stable. | <math>\int_{-\infty}^\infty e^{-6|t|} = 2\int_0^\infty e^{-6t} < \infty</math>. This system is stable. | ||
This system is stable but not causal. | This system is stable but not causal. | ||
Latest revision as of 20:10, 22 June 2008
Find if each system is stable and causal.
A
h(t) = $ e^{-4t} u(t-2) $
u(t-2) = 1 for t >= 2 making h(t) = 0 for t < 2. The system is causal.
$ \int_{-\infty}^\infty e^{-4t} u(t-2) = /int_2^\infty e^{-4t} < \infty $. Therefore the system is stable.
This system is stable and causal.
B
h(t) = $ e^{-6t} u(3-t) $
u(3-t) = 1 for t<=3, making h(t) $ \neq $ for t < 0. The system is not causal.
$ \int_{-\infty}^\infty e^{-6t} u(3-t) = \int_{-\infty}^3 e^{-6t} = \infty $, therefore the system is not stable.
This system is neither causal or stable.
E
h(t) = $ e^{-6|t|} $
Since h(t) $ \neq $ 0 for t < 0 so the system is not causal.
$ \int_{-\infty}^\infty e^{-6|t|} = 2\int_0^\infty e^{-6t} < \infty $. This system is stable.
This system is stable but not causal.
