(New page: Suppose we have a LTI CT signal y(t)=2x(t) ==Unit Impulse Response h(t) and System Function H(s)== <math>y(t)=2x(t)=>h(t)=2\delta(t)</math> <math>H(j\omega)=\int_{-\infty}^\infty h(\tau)...) |
(→Unit Impulse Response h(t) and System Function H(s)) |
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==Unit Impulse Response h(t) and System Function H(s)== | ==Unit Impulse Response h(t) and System Function H(s)== | ||
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| − | <math> | + | i) |
| + | <math>y(t)=2x(t)=> h(t)=2\delta(t)</math> | ||
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| + | ii) | ||
| + | <math>H(s)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau</math> | ||
| + | <math>\=\int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau</math> | ||
Revision as of 15:55, 26 September 2008
Suppose we have a LTI CT signal y(t)=2x(t)
Unit Impulse Response h(t) and System Function H(s)
i) $ y(t)=2x(t)=> h(t)=2\delta(t) $
ii) $ H(s)=\int_{-\infty}^\infty h(\tau)e^{-j\omega\tau}d\tau $ $ \=\int_{-\infty}^\infty h(\tau)e^{-s\tau}d\tau $
