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
$ X [0] = 1+ e^{-j(\frac{2}{3}\pi)(1+0)} +e^{-j\frac{4}{3}\pi(1+0)} $
$ X [0] = 1+ e^{-j(\frac{2}{3}\pi)} +e^{-j(\frac{4}{3}\pi)} $
complex result: Noting $ -2/3\pi $ and $ -4/3\pi $ are conjugates cancel the imaginary component.
$ 1+cos(-2/3\pi) +cos(-4/3\pi) = X[0] = 0 $
$ X [1] = 1+ e^{-j(\frac{2}{3}\pi)(1+1)} +e^{-j\frac{4}{3}\pi(1+1)} $
$ X [1] = 1+ e^{-j(\frac{4}{3}\pi)} +e^{-j\frac{8}{3}\pi} $
complex result: Noting $ -4/3\pi $ and $ -8/3\pi $ are conjugates cancel the imaginary component.
$ 1+cos(-4/3\pi) +cos(-8/3\pi) = X[1] = 0 $
$ X [2] = 1+ e^{-j(\frac{2}{3}\pi)(1+2)} +e^{-j\frac{4}{3}\pi(1+2)} $
$ X [2] = 1+ e^{-j(2\pi)} +e^{-j(4\pi)} $
$ X [2] = 1+ 1 + 1 = 3 $
- 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}(1)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 n j}=1 $ to rewrite x[n] as a positive power of e. (Just add $ 2 \pi n j $ to the exponent of e). -pm
$ \begin{align} x[n]&= e^{-j \frac{2}{3} \pi n} \\ &= e^{-j \frac{2}{3} \pi n} e^{j 2 \pi n} \text{ (since this is the same as multiplying by one, for any integer n)}\\ &= e^{-j \frac{2}{3} \pi n +j 2 \pi n } \\ & = e^{j \frac{4}{3} \pi n} \\ & = e^{j 2 \frac{2\pi n }{3} } \end{align} $
Now compare with the inverse DFT formula.
$ e^{j 2 \frac{2\pi n }{3} } \ \ compare \ with \ \ \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)}) $
X[0] = 0
X[1] = 0
X[2] = 3
This could easily be shown that due to the modulation property that this is a shifted delta in the DFT world. But to do a little mathematical proof.... $ \begin{align} X [k] &= \sum_{n=0}^{N-1} x[n]e^{\frac{-j2\pi k n}{N}} \\ &= \sum_{n=0}^{N-1} e^{-j\frac{2}{3}\pi n}e^{\frac{-j2\pi k n}{N}} \text{ where N=3 because of periodicity of the signal} \\ &= \sum_{n=0}^{2} e^{j\frac{4}{3}\pi n}e^{-j \frac{2}{3} \pi n k} \\ &= \sum_{n=0}^{2} e^{-j\frac{2}{3}\pi n (k-2)} \end{align} $
Now to get the final result you must compare this equation to the IDFT formula and you get that
$ \begin{align} X [k] &= \sum_{n=0}^{N-1} x[n]e^{\frac{-j2\pi k n}{N}} &= 3 \delta (k-2) \end{align} $
In class we compared the IDFT and the Fourier Series expansion and the Fourier coefficients can be expressed (if I remember correctly) as
$ A_{k}= \frac{X[k]}{N} $
- This is a different and interesting way of looking at the problem. -pm----
- Answer/question
