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<math>y(t)=\frac{-j}{2}(H(j3\pi)e^{j3\pi t}-H(-j3\pi)e^{-j3\pi t})</math> | <math>y(t)=\frac{-j}{2}(H(j3\pi)e^{j3\pi t}-H(-j3\pi)e^{-j3\pi t})</math> | ||
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The systems response to this signal by direct computation would be... | The systems response to this signal by direct computation would be... | ||
Latest revision as of 07:23, 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}\, $
System Response to Q.1
The input in Q1 was not CT, but if it was it would have been:
$ x(t)=sin(3\pi t)\, $
The systems response to this signal would be...
$ x(t)=sin(3\pi t)\,=\frac{-j}{2}(e^{j3\pi t}-e^{-j3\pi t}) $
$ y(t)=\frac{-j}{2}(H(j3\pi)e^{j3\pi t}-H(-j3\pi)e^{-j3\pi t}) $
The systems response to this signal by direct computation would be...
$ y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}x(\tau)\,d\tau\, $
$ y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}sin(3\pi\tau)\,d\tau\, $ $ y(t)=\int_{-\infty}^{t} e^{-t}e^{\tau}sin(3\pi\tau)\,d\tau\, $
$ y(t)=e^{-t}\int_{-\infty}^{t}e^{\tau}sin(3\pi\tau)\,d\tau\, $
from the table of Integrals...
$ y(t)=e^{-t}[\frac{e^\tau}{1^2+(3\pi)^2}(sin(3\pi\tau)-3\pi cos(3\pi\tau))]_{-\infty}^t\, $
$ y(t)=\frac{1}{1+9\pi^2}(sin(3\pi t)-3\pi cos(3\pi t))\, $
