As I understand, when the input signal is discrete in time and we want to find the coefficients of Fourier transform, DTFT is used and the coefficients in frequency domain are periodic, but I can't understand why? If DTFT coefficients are periodic, why the coefficients of a continuous signal in frequency domain are not periodic?


3 Answers 3


The result of a DTFT is periodic, because any discrete-time signal has a continuous spectrum. This can be e.g. explained by the following:

Let $x(t)$ be a time-continuous signal. Now, making it discrete corresponds to multiplying it with a Dirac-Train:

$$ x[n] = x(t)\sum_{n\in\mathbb{Z}} \delta(t-nT) $$

Considering the Convolution theorem of the (continuous) Fourier transform and noting that

$$ F\{\sum_{n\in\mathbb{Z}}\delta(t-nT)\} = \sum_{n\in\mathbb{Z}}\delta(f-n/T) $$

we have

$$ X_d(f)=F\{x(t)\sum_{n\in\mathbb{Z}}\delta(t-nT)\}=X(f)*\sum_{n\in\mathbb{Z}}\delta(f-n/T) $$

Now, we can directly express the convolution as:

$$ X_d(f)=\sum_{n\in\mathbb{Z}}X(f-n/T) $$

Hence, the spectrum of the discrete signal is periodic with the period 1/T (i.e. the sampling frequency).

  • $\begingroup$ Your intuition is right, but the first equation is not. The LHS is just a discrete signal (or a time-series) while the RHS is an impulse train. $\endgroup$
    – msm
    Nov 12, 2016 at 13:34

The explanation by @Maximilian Matthé is a standard and formal approach for this question. But I think it is not intuitive and easy to understand the reason. In the following, I will try to explain from another aspect.

First of all, periodicity means infinity. For a signal in the temporal/spatial domain, the periodicity is somewhat easy for realization, such as extending it periodically and infinitely. But for the frequency domain, it is quite different. Different position means different component with different changing speed. Therefore, if periodicity is expected in the frequency domain, infinitely rapid changing is expected in the temporal/spatial domain. In the real world of our life, most signals changes smoothly, thus periodicity is impossible in the frequency domain. While for discrete-time signals, the model is based on the $\delta(t)$ function, for which the changing is infinitely sharp and rapid, thus periodicity becomes possible in this case.

  • $\begingroup$ Why periodicity in freq. domain means infinitely rapid changes in temporal/spatial domain? $\endgroup$
    – Yola
    Feb 3, 2019 at 12:43

The spectrum of any discrete signal has a period of 2*pi. Thus, the fourier coefficients occur periodically at interval of 2*pi. While , in case of continuous time signal the spectrum has no such definite period.

I hope this solves your query.


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