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<math> X [k] = \sum_{k=0}^{N-1} x[n].e^{-j.2\pi k n/N}</math>
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<math> N=3 </math>
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<math>x[n]= e^{-j \frac{2}{3} \pi n}</math>
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<math> X [k] = \sum_{k=0}^{2}e^{-j(n)(\frac{2}{3}\pi)(1+k)}</math>
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<math> X [k] = 1+ e^{-j(1)(\frac{2}{3}\pi)(1+k)} +e^{-j\frac{4}{3}\pi(1+k)}</math>
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?- AJFunche
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*Answer/question
 
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Revision as of 14:43, 18 October 2010

Practice Question 1, ECE438 Fall 2010, Prof. Boutin

On Computing the DFT of a discrete-time periodic signal


Compute the discrete Fourier transform of the discrete-time signal

$ x[n]= e^{-j \frac{2}{3} \pi n} $.

How does your answer related to the Fourier series coefficients of x[n]?


Post Your answer/questions below.

$ X [k] = \sum_{k=0}^{N-1} x[n].e^{-j.2\pi k n/N} $

$ N=3 $

$ x[n]= e^{-j \frac{2}{3} \pi n} $

$ X [k] = \sum_{k=0}^{2}e^{-j(n)(\frac{2}{3}\pi)(1+k)} $

$ X [k] = 1+ e^{-j(1)(\frac{2}{3}\pi)(1+k)} +e^{-j\frac{4}{3}\pi(1+k)} $

?- AJFunche

  • Answer/question
  • Answer/question
  • Answer/question

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