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=== Answer 1 === | === Answer 1 === | ||
| − | + | <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 21:31, 18 February 2011
Contents
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).\ $
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.
