Line 9: Line 9:
 
Frequency Response:
 
Frequency Response:
  
<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\,</math>
+
<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\,</math>
  
then <math>y(t)=\sum^{\infty}_{k = -\infty}a_k*(h(t)*e^{j\omega_0 t})</math>
+
then <math>y(t)=\sum^{\infty}_{k = -\infty}a_k*[h(t)*e^{j\omega_0 t}]</math>
  
 
<math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 t} dt</math> by definition
 
<math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 t} dt</math> by definition
Line 28: Line 28:
 
<math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math>
 
<math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math>
  
<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}</math>
+
<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}</math>
  
<math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}</math>
+
<math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\omega_0 t}</math>
  
<math>y(t) = \sum^{\infty}_{k = -\infty} a_k (10) e^{jk\pi t}</math>
+
<math>y(t) = \sum^{\infty}_{k = -\infty} a_k (10) e^{jk\omega_0 t}</math>
  
<math>y(t) = K\sum^{\infty}_{k = -\infty} a_k  e^{jk\pi t}</math>
+
<math>y(t) = K\sum^{\infty}_{k = -\infty} a_k  e^{jk\omega_0 t}</math>
  
 
<math>y(t) = K+K\sin \omega_0 t + K\cos(2\omega_0 t+ \frac{\pi}{4})</math>
 
<math>y(t) = K+K\sin \omega_0 t + K\cos(2\omega_0 t+ \frac{\pi}{4})</math>

Latest revision as of 11:33, 25 September 2008

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


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


Frequency Response:

$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t}\, $

then $ y(t)=\sum^{\infty}_{k = -\infty}a_k*[h(t)*e^{j\omega_0 t}] $

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

$ H(s) = \int^{\infty}_{-\infty} K \delta(t) e^{-j\omega_0 t} dt $

$ H(s) = K e^{-jw0} $

$ H(s) = K $



Response of the CT LTI system in 4.1:

$ x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}) $

$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t} $

$ y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\omega_0 t} $

$ y(t) = \sum^{\infty}_{k = -\infty} a_k (10) e^{jk\omega_0 t} $

$ y(t) = K\sum^{\infty}_{k = -\infty} a_k e^{jk\omega_0 t} $

$ y(t) = K+K\sin \omega_0 t + K\cos(2\omega_0 t+ \frac{\pi}{4}) $

Alumni Liaison

Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood