Difference between revisions of "HW4.3 Nicholas Browdues ECE301Fall2008mboutin" - Rhea

Latest revision as of 06:23, 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))\, $