0
$\begingroup$

The given sequence $$x[k]=\delta[k-1]-3 \delta[k] + \delta[k-2]$$ represents a period of some discrete periodic signal $e[k]$ with period N=3. Find the Fourier transform of $e[k]$.

This is an exercise in my textbook and i don't get it. Is $x[k]$ the first period of $e[k]$? If so, does that mean that the DFT of $x[k]$ is the same as the DFT of $e[k]$?

$\endgroup$
0
$\begingroup$

The question is to compute the Discrete-time Fourier transform (DTFT) of the $N$-periodic sequence $e[n]$, which I'll call $y[n]$ from now on in order to avoid confusion with Euler's number $e$. $y[n]$ is the periodic continuation of the length $N=3$ sequence $x[n]$, hence it can be written as

$$y[n]=\left(\sum_{k=-\infty}^{\infty}\delta[n-kN]\right)\star x[n]\tag{1}$$

where $\star$ denotes convolution. Since convolution corresponds to multiplication in the frequency domain, the DTFT of $(1)$ is given by

$$\begin{align}Y(\omega)&=\frac{1}{N}\sum_{k=0}^{N-1}\delta(\omega-2\pi k/N)\cdot X(\omega)\\&=\frac{1}{N}\sum_{k=0}^{N-1}X(2\pi k/N)\delta(\omega-2\pi k/N),\quad 0\le\omega<2\pi\tag{2}\end{align}$$

Note that I use $\delta[n]$ to denote the discrete-time unit impulse in $(1)$, and $\delta(\omega)$ to denote the Dirac delta impulse in $(2)$. Also note that $Y(\omega)$ is $2\pi$-periodic and $(2)$ specifies one period. $X(\omega)$ is the DTFT of the sequence $x[n]$. Since $x[n]$ is of finite length $N$, the sampled DTFT $X(2\pi k/N)$ equals the discrete Fourier transform (DFT) of $x[n]$, which I call $\tilde{X}[k]$:

$$X(2\pi k/n)=\tilde{X}[k]=\sum_{n=0}^{N-1}x[n]e^{-jnk2\pi /N}\tag{3}$$

Using $(3)$, the DTFT of $y[n]$ can be written as

$$\frac{1}{N}\sum_{k=0}^{N-1}\tilde{X}[k]\delta(\omega-2\pi k/N),\quad 0\le\omega<2\pi\tag{4}$$

It remains to find the DFT of $x[n]$, which is a straightforward exercise. The result is $\tilde{X}[k]=[-1,-4,-4]$. It can also be obtained by the Matlab/Octave command fft([-3,1,1]). With this result and with $(4)$, the DTFT of $y[n]$ can finally be written as

$$Y(\omega)=-\frac13\left[\delta(\omega)+4\delta(\omega-2\pi/3)+4\delta(\omega-4\pi/3)\right],\quad 0\le\omega<2\pi\tag{5}$$

$\endgroup$
0
$\begingroup$

For a discrete periodic signal you normally compute the Fourier series coefficients. They can be obtained from the DFT by dividing the output of the DFT by N, where number of samples in a period.

The DFT assumes the signal is N-sample periodic anyway.

In Matlab you will implement this as

x=[-3 1 1];
ck=fft(x)/3;
$\endgroup$
  • $\begingroup$ From the question, $x$ should be x=[-3,1,1]. You're right that the Fourier series coefficients of $e[k]$ equal (up to a constant) the DFT of $x[k]$, but the question is to compute the (discrete) Fourier transform of $e[k]$, which exists as a weighted sum of Dirac impulses. The weights are of course the DFT/Fourier series coefficients. See my answer. $\endgroup$ – Matt L. Aug 29 '15 at 20:57
  • $\begingroup$ Matt, in no DSP textbook i am familiar with does "$\delta[n]$" (with brackets "$[\cdot]$", not parenths) mean a Dirac delta, but is instead the Kronecker delta which is defined for only integer $n$. there is no difference, non whatsoever, between the Discrete Fourier Transform and the Discrete Fourier Series. they simply map one discrete and periodic function in one domain to another discrete and periodic function in the reciprocal domain. $\endgroup$ – robert bristow-johnson Aug 29 '15 at 22:55
  • $\begingroup$ @robertbristow-johnson: I haven't seen it either, also not in my answer. I wrote that $\delta[n]$ is the discrete-time unit impulse, whereas $\delta(\omega)$ is the Dirac impulse. You must have misread it. And I also agree with you about DFT/DFS, depending on the definition there might be just a scaling factor between them. $\endgroup$ – Matt L. Aug 30 '15 at 5:38
  • $\begingroup$ yer right on all counts, Matt. i have to read through more carefully. we all know that there is the $\frac{1}{N}$, or sometimes $\sqrt{\frac{1}{N}}$, scaling issue in the definitions of the DFT and its inverse. boy, i would really start ragging on someone who uses the $\frac{1}{N}$ in their definition of DFT and not in the DFS. because some would claim it to be a "difference" between the two concepts when the difference is only a trivial and arbitrary scaling factor. $\endgroup$ – robert bristow-johnson Aug 30 '15 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.