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| + | <math>y(t)=x(t)*h(t)=\int_{-\infty}^\infty u(\tau)e^{-3(t-\tau)}u(t-\tau)d\tau=e^{-3t}\int_0^\infty e^{3\tau}u(t-\tau)d\tau=\Bigg(e^{-3t}\int_0^t e^{3\tau}d\tau\Bigg)u(t)=\Bigg(\frac{1}{3}e^{-3t}\bigg(e^{3t}-1\bigg)\Bigg)u(t)</math> | ||
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| + | <math>y(t)=\Bigg(\frac{1}{3}-\frac{e^{-3t}}{3}\Bigg)u(t)</math> | ||
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| + | --[[User:Cmcmican|Cmcmican]] 21:00, 4 February 2011 (UTC) | ||
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===Answer 2=== | ===Answer 2=== | ||
Write it here. | Write it here. | ||
Revision as of 17:00, 4 February 2011
Contents
Practice Question on Computing the Output of an LTI system by Convolution
The unit impulse response h(t) of a DT LTI system is
$ h(t)= e^{-3t }u(t). \ $
Use convolution to compute the system's response to the input
$ x(t)= u(t). \ $
You will receive feedback from your instructor and TA directly on this page. Other students are welcome to comment/discuss/point out mistakes/ask questions too!
Answer 1
$ y(t)=x(t)*h(t)=\int_{-\infty}^\infty u(\tau)e^{-3(t-\tau)}u(t-\tau)d\tau=e^{-3t}\int_0^\infty e^{3\tau}u(t-\tau)d\tau=\Bigg(e^{-3t}\int_0^t e^{3\tau}d\tau\Bigg)u(t)=\Bigg(\frac{1}{3}e^{-3t}\bigg(e^{3t}-1\bigg)\Bigg)u(t) $
$ y(t)=\Bigg(\frac{1}{3}-\frac{e^{-3t}}{3}\Bigg)u(t) $
--Cmcmican 21:00, 4 February 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.
