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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
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 +
----
  
 
Unit Impulse Response: <math>h(t) = K \delta(t)</math>
 
Unit Impulse Response: <math>h(t) = K \delta(t)</math>
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 +
----
  
 
Frequency Response:
 
Frequency Response:
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<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
  
<math>H(s) = \int^{\infty}_{-\infty} K \delta(r) e^{-j\omega_0 t} dt</math>
+
<math>H(s) = \int^{\infty}_{-\infty} K \delta(t) e^{-j\omega_0 t} dt</math>
  
 
<math>H(s) = K e^{-jw0}</math>
 
<math>H(s) = K e^{-jw0}</math>
  
 
<math>H(s) = K</math>
 
<math>H(s) = K</math>
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----
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Response of the CT LTI system in 4.1:
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<math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math>
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<math>x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}</math>
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<math>y(t) = \sum^{\infty}_{k = -\infty} a_k H(s) e^{jk\pi t}</math>
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<math>y(t) = \sum^{\infty}_{k = -\infty} a_k (10) e^{jk\pi t}</math>
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<math>y(t) = K\sum^{\infty}_{k = -\infty} a_k  e^{jk\pi t}</math>
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<math>y(t) = K+K\sin \omega_0 t + K\cos(2\omega_0 t+ \frac{\pi}{4})</math>

Revision as of 11:30, 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^{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\pi t} $

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

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

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

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

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn