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--[[User:Cmcmican|Cmcmican]] 19:42, 28 February 2011 (UTC) | --[[User:Cmcmican|Cmcmican]] 19:42, 28 February 2011 (UTC) | ||
| − | :<span style="color:green">TA's comments: The answer is correct. The geometric series converges because <math class="inline">\color{OliveGreen}{\left|\frac{e^{ | + | :<span style="color:green">TA's comments: The answer is correct. The geometric series converges because <math class="inline">\color{OliveGreen}{\left|\frac{e^{j\omega}}{3}\right|=\frac{1}{3}<1}</math>.</span> |
=== Answer 2 === | === Answer 2 === | ||
Write it here. | Write it here. | ||
Revision as of 15:14, 1 March 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)
- TA's comments: The answer is correct. The geometric series converges because $ \color{OliveGreen}{\left|\frac{e^{j\omega}}{3}\right|=\frac{1}{3}<1} $.
Answer 2
Write it here.
Answer 3
Write it here.
