Line 5: Line 5:
 
Frequency Response:
 
Frequency Response:
  
<math>y(t) = \int^{\infty}_{-\infty} h(t) * x(t) dt\,</math> where <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math>
+
<math>H(s) = \int^{\infty}_{-\infty} h(\tao)e^{-j\omega_0 r} d\tao</math> by definition
 
+
<math>y(t) = \int^{\infty}_{-\infty} K \delta(t) * (1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}))</math>
+

Revision as of 11:20, 25 September 2008

LTI System: $ y(t) = Kx(t)\, $ where K is a constant

Unit Impulse Response: $ h(t) = K \delta(t)\, $

Frequency Response:

$ H(s) = \int^{\infty}_{-\infty} h(\tao)e^{-j\omega_0 r} d\tao $ by definition

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood