| Line 9: | Line 9: | ||
---- | ---- | ||
Post Your answer/questions below. | Post Your answer/questions below. | ||
| − | + | ||
| + | <math> X [k] = \sum_{k=0}^{N-1} x[n].e^{-j.2\pi k n/N}</math> | ||
| + | |||
| + | <math> N=3 </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] = 1+ e^{-j(1)(\frac{2}{3}\pi)(1+k)} +e^{-j\frac{4}{3}\pi(1+k)}</math> | ||
| + | |||
| + | ?- AJFunche | ||
| + | |||
*Answer/question | *Answer/question | ||
*Answer/question | *Answer/question | ||
Revision as of 15:43, 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
