Revision as of 06:00, 25 September 2008 by Nbrowdue (Talk)

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, $

Alumni Liaison

Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett