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| − | <math>H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-4)]e^{st}dt</math> | + | <math>H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-4)]e^{-st}dt</math> |
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| − | <math>=\int_{-\infty}^{+\infty}2\delta(t)e^{st}dt - \int_{-\infty}^{+\infty}3\delta(t-4)e^{st}dt</math> which by the sifting property, | + | <math>=\int_{-\infty}^{+\infty}2\delta(t)e^{-st}dt - \int_{-\infty}^{+\infty}3\delta(t-4)e^{-st}dt</math> which by the sifting property, |
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| − | <math>=2-3e^{4s}\!</math> | + | <math>=2-3e^{-4s}\!</math> |
Revision as of 11:12, 25 September 2008
Define a CT LTI System
$ y(t)=2x(t)-3x(t-4)\! $
Unit Impulse Response
The unit impulse response is simply the systems response to an input $ \delta(t)\! $. Thus, in our case, the unit impulse response is simply $ h(t)=2\delta(t)-3\delta(t-4)\! $
System Function
To find the system function $ H(s)\! $ we use the formula:
$ H(s)=\int_{-\infty}^{+\infty} h(t)e^{st}dt $ where $ s=j\omega\! $.
$ H(s)=\int_{-\infty}^{+\infty}[2\delta(t)-3\delta(t-4)]e^{-st}dt $
$ =\int_{-\infty}^{+\infty}2\delta(t)e^{-st}dt - \int_{-\infty}^{+\infty}3\delta(t-4)e^{-st}dt $ which by the sifting property,
$ =2-3e^{-4s}\! $
