Line 7: Line 7:
 
<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\pi t}\,</math>
  
then y(t)=\sum^{\infty}_{k = -\infty}a_k*(h(t)*e^{jj\omega_0 t})
+
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(\tao)e^{-j\omega_0 r} d\tao</math> by definition
+
<math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 r} dt</math> by definition

Revision as of 12:25, 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 r} dt $ by definition

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn