(New page: ==CT LTI signal: == <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(j\omega) = \int_{-\infty}^{\infty}3\delta(t...) |
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==CT LTI signal: == | ==CT LTI signal: == | ||
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<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)\!</math> | + | <math>H(j\omega) = \int_{-\infty}^{\infty}3\delta(t-1)+\delta(t+3)-\delta(t)\,dt\!</math> |
'''Part B''' | '''Part B''' | ||
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\! $
Part B
