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== Part A ==
 
== Part A ==
 
  
 
<math>h(t) = 3\delta(t-1)+\delta(t+3)-\delta(t)\!</math>
 
<math>h(t) = 3\delta(t-1)+\delta(t+3)-\delta(t)\!</math>
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<math>H(s) = 3e^{-s} + e^{3s} - 1\!</math>
 
<math>H(s) = 3e^{-s} + e^{3s} - 1\!</math>
 +
 +
== Part B ==

Revision as of 17:40, 26 September 2008

CT LTI signal:

$ y(t) = 3x(t-1)+x(t+3)-x(t)\! $


Part A

$ h(t) = 3\delta(t-1)+\delta(t+3)-\delta(t)\! $

$ H(j\omega) = \int_{-\infty}^{\infty}3\delta(t-1)+\delta(t+3)-\delta(t)\,dt\! $

$ H(s) = 3e^{-s} + e^{3s} - 1\! $

Part B

Alumni Liaison

Questions/answers with a recent ECE grad

Ryne Rayburn