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| + | <math class="inline">{\mathcal X} (\omega) = \sum_{n=-\infty}^\infty x[n]e^{-j\omega n}=\sum_{n=-\infty}^\infty 3^n u[-n]e^{-j\omega n}=\sum_{n=-\infty}^0 3^n e^{-j\omega n}=\sum_{n=-\infty}^0 \Bigg(\frac{3}{e^{j\omega}}\Bigg)^n</math> Let k=-n | ||
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| + | <math>= \sum_{n=0}^\infty \Bigg(\frac{e^{j\omega}}{3}\Bigg)^k</math> | ||
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| + | <math>\mathcal X (\omega) = \frac{1}{1-\frac{e^{j\omega}}{3}}</math> | ||
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| + | --[[User:Cmcmican|Cmcmican]] 19:42, 28 February 2011 (UTC) | ||
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=== Answer 2 === | === Answer 2 === | ||
Write it here. | Write it here. | ||
Revision as of 14:42, 28 February 2011
Contents
Practice Question on Computing the Fourier Transform of a Discrete-time Signal
Compute the Fourier transform of the signal
$ x[n] = 3^n u[-n].\ $
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 x[n]e^{-j\omega n}=\sum_{n=-\infty}^\infty 3^n u[-n]e^{-j\omega n}=\sum_{n=-\infty}^0 3^n e^{-j\omega n}=\sum_{n=-\infty}^0 \Bigg(\frac{3}{e^{j\omega}}\Bigg)^n $ Let k=-n
$ = \sum_{n=0}^\infty \Bigg(\frac{e^{j\omega}}{3}\Bigg)^k $
$ \mathcal X (\omega) = \frac{1}{1-\frac{e^{j\omega}}{3}} $
--Cmcmican 19:42, 28 February 2011 (UTC)
Answer 2
Write it here.
Answer 3
Write it here.
