Line 1: Line 1:
 
LTI System: <math>y(t) = Kx(t)\,</math> where K is a constant
 
LTI System: <math>y(t) = Kx(t)\,</math> where K is a constant
  
Unit Impulse Response: <math>h(t) = K \delta(t)\,</math>
+
Unit Impulse Response: <math>h(t) = K \delta(t)</math>
  
 
Frequency Response:
 
Frequency Response:
Line 9: Line 9:
 
then <math>y(t)=\sum^{\infty}_{k = -\infty}a_k*(h(t)*e^{jj\omega_0 t})</math>
 
then <math>y(t)=\sum^{\infty}_{k = -\infty}a_k*(h(t)*e^{jj\omega_0 t})</math>
  
<math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 r} dt</math> by definition
+
<math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 t} dt</math> by definition
 +
 
 +
<math>H(s) = \int^{\infty}_{-\infty} K \delta(r) e^{-jwr} dr</math>
 +
 
 +
<math>H(s) = K e^{-jw0}</math>
 +
 
 +
<math>H(s) = K</math>

Revision as of 11:27, 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\pi t}\, $

then $ y(t)=\sum^{\infty}_{k = -\infty}a_k*(h(t)*e^{jj\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(r) e^{-jwr} dr $

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

$ H(s) = K $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman