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<math> X [k] = \sum_{k=0}^{N-1} x[n].e^{-j.2\pi k n/N}</math> | <math> X [k] = \sum_{k=0}^{N-1} x[n].e^{-j.2\pi k n/N}</math> | ||
| − | <math> N=3 </math> | + | <math> N=3 </math> <span style="color:green"> That's correct! -pm </span> |
<math>x[n]= e^{-j \frac{2}{3} \pi n}</math> | <math>x[n]= e^{-j \frac{2}{3} \pi n}</math> | ||
| − | <math> X [k] = \sum_{k=0}^{2}e^{-j(n)(\frac{2}{3}\pi)(1+k)}</math> | + | <math> X [k] = \sum_{k=0}^{2}e^{-j(n)(\frac{2}{3}\pi)(1+k)}</math> <span style="color:green"> You are using the long route, instead of the short route. -pm </span> |
| − | <math> X [k] = 1+ e^{-j(1)(\frac{2}{3}\pi)(1+k)} +e^{-j\frac{4}{3}\pi(1+k)}</math> | + | <math> X [k] = 1+ e^{-j(1)(\frac{2}{3}\pi)(1+k)} +e^{-j\frac{4}{3}\pi(1+k)}</math> <span style="color:green"> This gives you a very complicated answer. -pm </span> |
| − | ?- AJFunche | + | ?- AJFunche <span style="color:green"> Nice effort! -pm </span> |
| + | ---- | ||
*Answer/question | *Answer/question | ||
Revision as of 15:09, 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 $ That's correct! -pm
$ x[n]= e^{-j \frac{2}{3} \pi n} $
$ X [k] = \sum_{k=0}^{2}e^{-j(n)(\frac{2}{3}\pi)(1+k)} $ You are using the long route, instead of the short route. -pm
$ X [k] = 1+ e^{-j(1)(\frac{2}{3}\pi)(1+k)} +e^{-j\frac{4}{3}\pi(1+k)} $ This gives you a very complicated answer. -pm
?- AJFunche Nice effort! -pm
- Answer/question
- Answer/question
- Answer/question
