Difference between revisions of "Hw4.3 Scott Erdbruegger ECE301Fall2008mboutin" - Rhea

Latest revision as of 06:30, 25 September 2008

CT LTI system

The system is:

$ y(t)=10x(t)+x(t-1) $

unit impulse response

Obtain the unit impulse response h(t) and the system function H(s) of your system. :

$ d (t) => System =>10 d (t) + d(t-1)\, $

$ h(t)=10d(t) +d(t-1)\, $

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

$ H(s)=\int_{-\infty}^{\infty} (10d(t) +d(t-1))e^{-s t}dt $

Using the shifting property,

$ H(s)=10 e^{0 s} + e^{-1 s} \, $

$ H(s)=10 + e^{- s} \, $, where s =jw

Part B

Compute the response of your system to the signal you defined in Question 1 using H(s) and the Fourier series coefficients of your signal.

$ x(t)=e^{.5 j t \pi}-e^{-.5 j t \pi}+3 \, $

$ Response = H(s)*x(t) \, $

$ x(t)=(e^{.5 j t \pi}-e^{-.5 j t \pi}+3)(10+e^{-j.5\pi}) \, $

$ x(t)=10e^{.5 j t \pi}-10e^{-.5 j t \pi}+30 +e^{.5 j t \pi}e^{-j.5\pi}-e^{-.5\pi j t }e^{-j.5\pi}+3e^{-j.5\pi} \, $

$ x(t)=10e^{.5 j t \pi}-10e^{-.5 j t \pi}+30 +e^{.5 j (t-1) \pi}-e^{-.5 j (t-1) \pi}+3e^{-j.5\pi} \, $