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<math>H(s)=\frac{-1}{s+1}|e^{-(s+1)t}|_0^{\infty}\,=\frac{1}{s+1}\,</math> | <math>H(s)=\frac{-1}{s+1}|e^{-(s+1)t}|_0^{\infty}\,=\frac{1}{s+1}\,</math> | ||
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| + | == System Response to Q.1 == | ||
Revision as of 06:48, 25 September 2008
Impulse Response
Consider the following CT LTI system defined by:
$ y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}x(\tau)\,d\tau\, $
the impulse response is...
$ h(t)=\int_{-\infty}^{t} e^{-(t-\tau)}\delta(\tau)\,d\tau\, = e^{-(t-\tau)} |_{ \tau=0}= e^{-t} $
but this will diverge when t is less than 0 so...
$ h(t)= e^{-t}u(t)\, $
System Function
The system function is...
$ H(s)=\int_{-\infty}^{\infty} h(t)e^{-st}\,dt\, $
Where $ s=j\omega\, $
for this system....
$ H(s)=\int_{-\infty}^{\infty} e^{-t}u(t)e^{-st}\,dt\, $
$ H(s)=\int_{0}^{\infty} e^{-t}e^{-st}\,dt\, $
$ H(s)=\int_{0}^{\infty} e^{-(s+1)t}\,dt\, $
$ H(s)=\frac{-1}{s+1}|e^{-(s+1)t}|_0^{\infty}\,=\frac{1}{s+1}\, $
