I am trying to understand the periodicity of the DFT. How can this property (both in time and frequency domain) be used and can be helpful while developing on a DSP?

It would be good to see some source code or pseudo code, in MATLAB preferably, where this property is exploited/demonstrated.

  • $\begingroup$ As you say right up front, what you really want is MATLAB code. This is definitely off-topic for dsp.SE. If you want to know why the DFT is periodic, then that would be a reasonable question to ask. $\endgroup$ Apr 2 '13 at 13:45

If the expression that defines the DFT is evaluated for all integers $k$ instead of just for $k = 0, \dots, N-1$ , then the resulting infinite sequence is a periodic extension of the DFT, periodic with period $N$.

The periodicity can be shown directly from the definition:

$$ X_{k+N} \stackrel{\mathrm{def}}{=} \ \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} (k+N) n} = \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} k n} \underbrace{e^{-2 \pi i n}}_{1} = \sum_{n=0}^{N-1} x_n e^{-\frac{2\pi i}{N} k n} = X_k.$$

Similarly, it can be shown that the IDFT formula leads to a periodic extension.

Source: Wikipedia.

While the periodic property of DFT isn't very widely utilized, it often causes aliasing problems.

Source: http://www.dspguide.com/ch10/3.htm


In the time domain, the DFT is periodic by definition. While DFT stands for Discrete Fourier Transform, the operation is in fact a discrete fourier series. The signal to be analyzed is assumed to be periodic in the lenght of the signal. This periodic signal is decomposed into a series of periodic sequences. The DFT frequency bins are located at f = 1/T and its integer multiples, where T is the duration of the signal to be analyzed.

In the frequency domain, the DFT is periodic because the time domain signal being analyzed is sampled. Recall that any periodic sequence cannot be uniquely represented for frequencies above fs/2 where fs is the sampling frequency of the sequence (also known as the nyquist frequency). Above fs/2 all signal energy is reflected back into the frequency range 0-fs/2. Between fs/2 and fs, the reflection is in reverse order which gives rise to a DFT period equal to fs.

  • $\begingroup$ 0down votefavorite "I am trying to understand the periodicity of the DFT." ... and this gets a down vote! Go figure. $\endgroup$
    – user2718
    Apr 3 '13 at 11:32
  • $\begingroup$ I didn't downvote your answer, but feel that the downvote is perfectly understandable. If someone is wanting to understand the periodicity of the DFT, saying that "the DFT is periodic by definition" is like a mother answering a child's "Why?" by "Because I say so." What is it in the definition of the DFT that gives it the property of periodicity since the typical definition does not make any mention of periodicity, though it is sometimes discussed right below the definition in a section called "Properties of the DFT"? $\endgroup$ Apr 3 '13 at 12:17
  • $\begingroup$ @Dilip You didn't understand my answer. The reason the DFT is periodic in the time domain is because it is a Fourier series, which by definition describes a periodic function. Do I have to explain why a sine wave is periodic? If you don't 'get' an answer, that isn't reason to down vote it. Just don't choose it as an answer you like. $\endgroup$
    – user2718
    Apr 3 '13 at 14:02
  • $\begingroup$ @DilipSarwate: The only answer that is justified to be down voted (given people take time out of their lives to offer up answers) is one that is incorrect or deliberately misleading. Just my opinion as a working professional. $\endgroup$
    – user2718
    Apr 3 '13 at 14:06
  • 1
    $\begingroup$ Wow - Aother down vote and the answer is essentially the same answer that gets 2 up votes. "The periodicity can be shown directly from the definition:" There must be some very stange visitors to this site. $\endgroup$
    – user2718
    Apr 10 '13 at 15:00

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