Topic: Discrete Fourier Transform

(On Computing the DFT of a discrete-time periodic signal.)

## Question

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]?

$X [k] = \sum_{k=0}^{N-1} x[n].e^{-j.2\pi k n/N}$

$N=3$ 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

$X  = 1+ e^{-j(\frac{2}{3}\pi)(1+0)} +e^{-j\frac{4}{3}\pi(1+0)}$

$X  = 1+ e^{-j(\frac{2}{3}\pi)} +e^{-j(\frac{4}{3}\pi)}$

complex result: Noting $-2/3\pi$ and $-4/3\pi$ are conjugates cancel the imaginary component.

$1+cos(-2/3\pi) +cos(-4/3\pi) = X = 0$

$X  = 1+ e^{-j(\frac{2}{3}\pi)(1+1)} +e^{-j\frac{4}{3}\pi(1+1)}$

$X  = 1+ e^{-j(\frac{4}{3}\pi)} +e^{-j\frac{8}{3}\pi}$

complex result: Noting $-4/3\pi$ and $-8/3\pi$ are conjugates cancel the imaginary component.

$1+cos(-4/3\pi) +cos(-8/3\pi) = X = 0$

$X  = 1+ e^{-j(\frac{2}{3}\pi)(1+2)} +e^{-j\frac{4}{3}\pi(1+2)}$

$X  = 1+ e^{-j(2\pi)} +e^{-j(4\pi)}$

$X  = 1+ 1 + 1 = 3$

- AJFunche Nice effort! -pm----

$x[n] = \frac{1}{N}\sum_{k=0}^{N-1} X[k]e^{j2\pi k\frac{n}{N}}$

$x[n] = \frac{1}{3}\sum_{k=0}^{2} X[k]e^{j\frac{2\pi}{3}kn}$

$x[n] = \frac{1}{3} \cdot (X + Xe^{j(\frac{2\pi}{3}(1)n )} + Xe^{j(\frac{2\pi}{3}(2)n)})$

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.

$e^{j 2 \frac{2\pi n }{3} } \ \ compare \ with \ \ \frac{1}{3} \cdot (X + Xe^{j(\frac{2\pi}{3}n)} + Xe^{j(\frac{2\pi}{3}(2)n)})$

X = 0

X = 0

X = 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----
But does that mean my solution is wrong or just unique...? - my