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=== Answer 2 === | === Answer 2 === | ||
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| + | So it should be like this. | ||
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| + | <math>\mathcal X (\omega) = \sum_{n=-\infty}^\infty (u[n+1]-u[n-2])e^{-j\omega n}=\sum_{n=-1}^1 e^{-j\omega n}=</math> | ||
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| + | <math>\mathcal X (\omega) = e^{j\omega}+1+e^{-j\omega}</math> | ||
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| + | --[[User:Cmcmican|Cmcmican]] 11:57, 2 March 2011 (UTC) | ||
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=== Answer 3 === | === Answer 3 === | ||
Write it here. | Write it here. | ||
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[[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | [[2011_Spring_ECE_301_Boutin|Back to ECE301 Spring 2011 Prof. Boutin]] | ||
Revision as of 07:57, 2 March 2011
Contents
Practice Question on Computing the Fourier Transform of a Discrete-time Signal
Compute the Fourier transform of the signal
$ x[n] = u[n+1]-u[n-2].\ $
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
$ \mathcal X (\omega) = \sum_{n=-\infty}^\infty (u[n+1]-u[n-2])e^{-j\omega n}=\sum_{n=-1}^2 e^{-j\omega n}= $
$ \mathcal X (\omega) = e^{j\omega}+1+e^{-j\omega}+e^{-j2\omega} $
--Cmcmican 19:57, 28 February 2011 (UTC)
- TA's comments: You have a small mistake in that. Note that $ u[n-2] $ starts at $ n=2 $ and not $ n=3 $.
Answer 2
So it should be like this.
$ \mathcal X (\omega) = \sum_{n=-\infty}^\infty (u[n+1]-u[n-2])e^{-j\omega n}=\sum_{n=-1}^1 e^{-j\omega n}= $
$ \mathcal X (\omega) = e^{j\omega}+1+e^{-j\omega} $
--Cmcmican 11:57, 2 March 2011 (UTC)
Answer 3
Write it here.
