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=== Answer 1  ===
 
=== Answer 1  ===
Write it here.
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<math>\boldsymbol{\chi}(\omega)=\int_{-\infty}^\infty x(t)e^{-j\omega t}dt=\int_{-\infty}^\infty e^{-t} u(t)e^{-j\omega t}dt=\int_{0}^\infty e^{-t(j\omega+1)}dt=\frac{e^{-t(j\omega+1)}}{-(j\omega+1)}\Bigg|_0^\infty=\frac{e^{-\infty}}{-(j\omega+1)}-\frac{e^{0}}{-(j\omega+1)}=\frac{1}{j\omega+1}</math>
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<math>\boldsymbol{\chi}(\omega)=\frac{1}{j\omega+1}</math>
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--[[User:Cmcmican|Cmcmican]] 02:31, 19 February 2011 (UTC)
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=== Answer 2  ===
 
=== Answer 2  ===
 
Write it here.
 
Write it here.

Revision as of 22:31, 18 February 2011

Practice Question on Computing the Fourier Transform of a Continuous-time Signal

Compute the Fourier transform of the signal

$ x(t) = e^{-t} u(t).\ $


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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

$ \boldsymbol{\chi}(\omega)=\int_{-\infty}^\infty x(t)e^{-j\omega t}dt=\int_{-\infty}^\infty e^{-t} u(t)e^{-j\omega t}dt=\int_{0}^\infty e^{-t(j\omega+1)}dt=\frac{e^{-t(j\omega+1)}}{-(j\omega+1)}\Bigg|_0^\infty=\frac{e^{-\infty}}{-(j\omega+1)}-\frac{e^{0}}{-(j\omega+1)}=\frac{1}{j\omega+1} $

$ \boldsymbol{\chi}(\omega)=\frac{1}{j\omega+1} $

--Cmcmican 02:31, 19 February 2011 (UTC)

Answer 2

Write it here.

Answer 3

Write it here.


Back to ECE301 Spring 2011 Prof. Boutin

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva