(→Computing the Impulse Response and System Function) |
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The System Function is defined by: | The System Function is defined by: | ||
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| + | <math>H(s)=\int_{-\infty}^{\infty} h(t)e^{-st}\,dt\,</math> | ||
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| + | Now computing the actual response: | ||
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| + | <math>H(s)=\int_{-\infty}^{\infty} 0.5 \delta(t-5) u(t) e^{-st}\,dt\,</math> | ||
| + | |||
| + | which is turns into: | ||
| + | |||
| + | |||
| + | <math>H(s)=\int_{0}^{\infty} 0.5 \delta(t-5) e^{-st}\,dt\,</math> | ||
| + | |||
| + | Now using the sifting property of the delta function we obtain: | ||
| + | |||
| + | <math>\ H(s)= 0.5 e^{-5s} </math> | ||
Revision as of 16:14, 26 September 2008
Defining an LTI System
For an input x(t), let the LTI system be defined as:
$ \ y(t)=0.5 x(t-5) u(t) $
Computing the Impulse Response and System Function
Inputting a delta into the system yields:
$ \ y(t)=h(t)=0.5 \delta(t-5) u(t) $
The System Function is defined by:
$ H(s)=\int_{-\infty}^{\infty} h(t)e^{-st}\,dt\, $
Now computing the actual response:
$ H(s)=\int_{-\infty}^{\infty} 0.5 \delta(t-5) u(t) e^{-st}\,dt\, $
which is turns into:
$ H(s)=\int_{0}^{\infty} 0.5 \delta(t-5) e^{-st}\,dt\, $
Now using the sifting property of the delta function we obtain:
$ \ H(s)= 0.5 e^{-5s} $
