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From DFT transfer pair, <math>X[k]=10\delta[k-1]</math>. repeated with period 10. | From DFT transfer pair, <math>X[k]=10\delta[k-1]</math>. repeated with period 10. | ||
| − | :Instructor's comment: Why do you need to write the exponential as sine and cosine in order to find the period? Can you find the period directly from the exponential? -pm | + | :<span style="color:purple">Instructor's comment: Why do you need to write the exponential as sine and cosine in order to find the period? Can you find the period directly from the exponential? -pm </span> |
---- | ---- | ||
==Answer 2== | ==Answer 2== | ||
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N= 10 , k0 = 1 | N= 10 , k0 = 1 | ||
| + | :<span style="color:purple"> Instructor's comment: How do you go from here to the answer below? Please justify. -pm </span> | ||
<math>X[k]=10\delta[k-1]</math>. | <math>X[k]=10\delta[k-1]</math>. | ||
| + | ---- | ||
[[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | [[2011_Fall_ECE_438_Boutin|Back to ECE438 Fall 2011 Prof. Boutin]] | ||
Revision as of 08:29, 3 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.
- Instructor's comment: Why do you need to write the exponential as sine and cosine in order to find the period? Can you find the period directly from the exponential? -pm
Answer 2
$ x[n]= e^{-j \frac{1}{5} \pi n} $.
$ period = {2*pi / (pi/5)} = 10 $.
$ x[n]=e^{-j2\pi k_0 n/N} $.
N= 10 , k0 = 1
- Instructor's comment: How do you go from here to the answer below? Please justify. -pm
$ X[k]=10\delta[k-1] $.
