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==Answer 2==
 
==Answer 2==
Write it here
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<math>x[n]= e^{-j \frac{1}{5} \pi n}</math>.
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<math>period = 2*pi / (pi/5) = 10</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 06:04, 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]?

Share your answers below

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 $. Back to ECE438 Fall 2011 Prof. Boutin

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