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==Answer 1== | ==Answer 1== | ||
| − | + | <math>x[n]= e^{-j \frac{1}{5} \pi n}=cos(\frac{\pi n}{5})+jsin(\frac{\pi n}{5})</math>. | |
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| + | period=10, therefor, by comparing with<math>x[n]=e^{-j2\pi k_0 n/N}</math>. | ||
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| + | we get <math>N=10</math>,<math>k_0=1</math>. | ||
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| + | From DFT transfer pair, <math>X[k]=10\delta[k-1]</math>. repeated with period 10. | ||
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==Answer 2== | ==Answer 2== | ||
Revision as of 07:15, 2 October 2011
Practice Problem
Compute the discrete Fourier transform of the discrete-time signal
$ x[n]= e^{-j \frac{1}{5} \pi n} $.
How does your answer related to the Fourier series coefficients of x[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
$ x[n]= e^{-j \frac{1}{5} \pi n}=cos(\frac{\pi n}{5})+jsin(\frac{\pi n}{5}) $.
period=10, therefor, by comparing with$ x[n]=e^{-j2\pi k_0 n/N} $.
we get $ N=10 $,$ k_0=1 $.
From DFT transfer pair, $ X[k]=10\delta[k-1] $. repeated with period 10.
Answer 2
Write it here
