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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> <span style="color:green"> This gives you a very complicated answer. -pm </span> | <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 <span style="color:green"> Nice effort! -pm | + | - AJFunche <span style="color:green"> Nice effort! -pm---- |
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<math>x[n] = \frac{1}{N}\sum_{k=0}^{N-1} X[k]e^{j2\pi k\frac{n}{N}}</math> | <math>x[n] = \frac{1}{N}\sum_{k=0}^{N-1} X[k]e^{j2\pi k\frac{n}{N}}</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}\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> <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> | + | <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> | ||
| − | whoops, I was doing the homework. is that correct? - ksoong | + | whoops, I was doing the homework. is that correct? - ksoong <span style="color:green"> |
| + | : Tecnically yes, but not realy useful for computing the DFT. Instead, use the fact that <math>e^{ 2 \pi j}=1</math> to rewrite x[n] as a positive power of e. (Just add <math>2 \pi j</math> to the exponent of e). -pm </span> | ||
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Revision as of 02:34, 19 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
whoops, I was doing the homework. is that correct? - ksoong
- Tecnically yes, but not realy useful for computing the DFT. Instead, use the fact that $ e^{ 2 \pi j}=1 $ to rewrite x[n] as a positive power of e. (Just add $ 2 \pi j $ to the exponent of e). -pm
- Answer/question
- Answer/question
