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== System Response to Q.1 == | == System Response to Q.1 == | ||
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+ | The input in Q1 was not CT, but if it was it would have been: | ||
+ | |||
+ | <math>x(t)=sin(3\pi t)\,</math> | ||
+ | |||
+ | The systems response to this signal would be... | ||
+ | |||
+ | <math>y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}x(\tau)\,d\tau\,</math> | ||
+ | |||
+ | <math>y(t)=\int_{-\infty}^{t} e^{-(t-\tau)}sin(3\pi\tau)\,d\tau\,</math> | ||
+ | <math>y(t)=\int_{-\infty}^{t} e^{-t}e^{\tau}sin(3\pi\tau)\,d\tau\,</math> | ||
+ | |||
+ | <math>y(t)=e^{-t}\int_{-\infty}^{t}e^{\tau}sin(3\pi\tau)\,d\tau\,</math> | ||
+ | |||
+ | from the table of Integrals... | ||
+ | |||
+ | |||
+ | <math>y(t)=e^{-t}[\frac{e^\tau}{1^2+(3\pi)^2}(sin(3\pi\tau-3\pi cos(3\pi\tau))]_-\infty^t,</math> |
Revision as of 07:00, 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...
$ 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, $