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<math>x[n] = \frac{1}{3}\sum_{k=0}^{2} X[k]e^{j\frac{2\pi}{3}kn}</math> | <math>x[n] = \frac{1}{3}\sum_{k=0}^{2} X[k]e^{j\frac{2\pi}{3}kn}</math> | ||
| − | <math>x[n] = \frac{1}{3} \cdot (X[0] + X[1]e^{j\frac{2\pi}{3}n} + X[2]e^{j\frac{2\pi}{3}(2)n})</math> | + | <math>x[n] = \frac{1}{3} \cdot (X[0] + X[1]e^{j\frac{2\pi}{3}n} + X[2]e^{j\frac{2\pi}{3}(2)n})</math> <span style="color:green"> Notice that all the powers of e in this expression are positive, but the signal x[n] is expressed as a negative power of e, so you cannot compare just yet. -pm </span> |
Therefore, | Therefore, | ||
Revision as of 16:35, 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
$ x[n] = \frac{1}{N}\sum_{k=0}^{N-1} X[k]e^{j2\pi k\frac{n}{N}} $
$ x[n] = \frac{1}{3}\sum_{k=0}^{2} X[k]e^{j\frac{2\pi}{3}kn} $
$ x[n] = \frac{1}{3} \cdot (X[0] + X[1]e^{j\frac{2\pi}{3}n} + X[2]e^{j\frac{2\pi}{3}(2)n}) $ Notice that all the powers of e in this expression are positive, but the signal x[n] is expressed as a negative power of e, so you cannot compare just yet. -pm
Therefore,
X[0] = 0
X[1] = 3
X[2] = 0
- Answer/question
- Answer/question
