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== Part B ==
 
== Part B ==
  
Compute the system's response to
+
Compute the system's response to (from problem 1 [[HW4.1 Jeff Kubascik_ECE301Fall2008mboutin]])
  
 
<math>\,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\,</math>
 
<math>\,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\,</math>

Revision as of 16:07, 25 September 2008

Given the following system

$ \,s(t)=6x(t-2)-5x(t)\, $


Part A

Find the system's unit impulse response $ \,h(t)\, $ and system function $ \,H(s)\, $.


The unit impulse response is simply (plug a $ \,\delta(t)\, $ into the system)

$ \,h(t)=6\delta(t-2)-5\delta(t)\, $


The system function is found using the following formula (for LTI systems)

$ \,H(s)=\int_{-\infty}^{\infty}h(t)e^{-st}dt\, $

$ \,H(s)=\int_{-\infty}^{\infty}(6\delta(t-2)-5\delta(t))e^{-st}dt\, $

$ \,H(s)=6\int_{-\infty}^{\infty}\delta(t-2)e^{-st}dt - 5\int_{-\infty}^{\infty}\delta(t)e^{-st}dt\, $

using the sifting property

$ \,H(s)=6e^{-2s}-5e^{0}\, $

$ \,H(s)=6e^{-2s}-5\, $


Part B

Compute the system's response to (from problem 1 HW4.1 Jeff Kubascik_ECE301Fall2008mboutin)

$ \,x(t)=\frac{3\pi}{2}\cos(\frac{3\pi}{2}t+\pi)\sin(\frac{3\pi}{4}t+\frac{\pi}{2})\, $

Alumni Liaison

has a message for current ECE438 students.

Sean Hu, ECE PhD 2009