# When is the Fourier transform of a signal periodic?

I mean not the time-domain signal being periodic, but the Fourier transform being periodic.

• The Fourier transform is essentially its own inverse. And the Fourier transform of a periodic function is essentially the Fourier series of the signal, times Dirac delta. – Yves Daoust Sep 18 '15 at 10:17

I'm new to this exchange and I'm not sure how mathy you all get. I think the answer below is cool because it shows that in some sense the continuous-time Fourier transform is never periodic but that in another sense there are lots of ways to get periodic transforms.

For the continuous-time Fourier transform on $\mathbb{R}$, both CMDoolittle's and Robert Bristow-Johnson's answers are correct, though for slightly-different definitions of "Fourier Transform". In the usual way we talk about the continuous Fourier transform, transforms are never periodic. If we're willing to transform distributions (of which the Dirac delta is an example), we can have transforms which are periodic.

Usually, the theory of the continuous Fourier transform is developed on $L^2(\mathbb{R})$. The Fourier transform is a bijection of $L^2(\mathbb{R})$ back onto itself; this means that $L^2(\mathbb{R})$ is also the space of all possible Fourier transforms. However, the zero function is the only periodic function in $L^2(\mathbb{R})$, so we can conclude that continuous Fourier transforms of non-zero functions are never periodic. This is CMDoolittle's answer.

However, we also have a Fourier theory for distributions. Specifically, we define distributions to be the set of linear functionals on the space of so-called test functions. Test functions, in turn, are $C^\infty$ functions which are required to decay rapidly (alternately, sometimes test functions are $C^\infty$ functions with compact support). Some of these distributions can be identified with periodic functions on $\mathbb{R}$. For example, if $f$ is a test function then the mapping $$S : f \mapsto \int_{-\infty}^\infty \sin(t) f(t)\ dt$$ is a distribution which can be identified with the sine function. As we can define the Fourier transform of a distribution, so we have a way of defining the Fourier transform of periodic functions. As it turns out, the Fourier transform is also a bijection of the space of distributions back onto itself. Going back to the distribution $S$, this means that we can write $S$ as the Fourier transform of some other distribution $T$: $S = \hat{T}$; in other words, we conclude that there has to be a distribution $T$ whose transform is the distribution $S$ which is identified with the periodic function sine. Clearly, we could pull this same trick with, say, any bounded periodic function. Ultimately then, we can extend Robert Bristow-Johnson's answer: if we're talking about the Fourier transform of distributions, there are lots of distributions whose Fourier tranforms are periodic.

For a discussion of the Fourier theory of distributions, see Chapter 7 of Rudin's Functional Analysis.

• welcome to dsp.se, Austin. admittedly i play fast-and-loose with $\delta(t)$. i consider it to have integral of 1: $$\int\limits_{-\infty}^{+\infty} \delta(t) \ dt \ = \ 1$$ and to have one Planck Time in width. then i don't have to concern myself with the distribution stuff. (i know that's a lame evasion of mathematical responsibility, i guess i'm sorta irresponsible).  :-) – robert bristow-johnson Sep 20 '15 at 1:50
• Haha! You know, I get it: everything these days is discrete anyway so it's hard to justify worrying about some of the more esoteric continuous-time theory. In this case, I think it's just cool that if you do take this theory into account, you can construct all sorts of functions whose transforms are all the old, familiar, elementary functions. (Sort of, anyway). In my opinion, it's very pleasing that this is true; it's like running into old friends in an unexpected place. I'm certainly not trying to accuse anyone of mathematical irresponsibility! – Austin A. Sep 20 '15 at 16:00
• i've gotten into a few little tiffs here and at the comp.dsp USENET group regarding "naked" $\delta(t)$ functions. the way i like to do the sampling theorem is to treat this: $$T\sum\limits_{n=-\infty}^{+\infty} \delta(t - nT) = \sum\limits_{k=-\infty}^{+\infty} e^{j 2 \pi \frac{k}{T} t}$$ as a regular-old function and apply the regular-old Fourier Transform theorems to it. it's fast and loose, i'll admit. – robert bristow-johnson Sep 21 '15 at 0:37

The question "When is the Fourier Transform of a signal periodic?" is best considered using the notion of duality.

A (finite-power) periodic time-domain signal $x(t)$ of period $T_0$ is uniquely represented in the frequency domain by its Fourier series. We have that $$x(t) = \sum_{n=-\infty}^\infty X_n \exp\left( \frac{j2\pi n t}{T_0}\right)\tag{1}$$ In the context of Fourier Transforms, the Fourier transform of $x(t)$ is $$X(f) = \sum_{n=-\infty}^\infty X_n \delta\left(f - \frac{n}{T_0}\right)\tag{2}$$ which is a sequence of impulses that are at multiples of $\frac{1}{T_0}$ Hz in the frequency domain. Note that mere spacing $\frac 1T$ Hz apart (e.g. at $f=\sqrt{2}+ 100n$ Hz which are spaced $100$ Hz apart) will not do; the impulses must occur at multiples of the spacing. Applying the inverse Fourier Transform integral to $(2)$, the sifting property of impulses gives us that \begin{align} \int_{-\infty}^\infty X(f)\exp(j2\pi tf)\,\mathrm df &= \int_{-\infty}^\infty \sum_{n=-\infty}^\infty X_n \delta\left(f - \frac{n}{T_0}\right) \exp(j2\pi tf)\,\mathrm df\\ &= \sum_{n=-\infty}^\infty X_n \int_{-\infty}^\infty \delta\left(f - \frac{n}{T_0}\right) \exp(j2\pi tf)\,\mathrm df\\ &= \sum_{n=-\infty}^\infty X_n \exp\left( \frac{j2\pi n t}{T_0}\right)\\ &= x(t). \end{align}

Now, duality tells us that what is sauce for the goose is sauce for the gander and that one man's Mede is another man's Persian and so

A signal whose Fourier transform is periodic with period $f_0$ is represented in the time domain by a sequence of impulses that occur at multiples of $\frac{1}{f_0}$ seconds, that is, $$x(t) = \sum_{n=-\infty}^\infty x_n \delta\left(t - \frac{n}{f_0}\right)$$ and \begin{align} X(f) &= \int_{-\infty}^\infty x(t) \exp(-j2\pi ft)\,\mathrm dt\\ &= \int_{-\infty}^\infty \sum_{n=-\infty}^\infty x_n \delta\left(t - \frac{n}{f_0}\right) \exp(-j2\pi ft)\,\mathrm dt\\ &= \sum_{n=-\infty}^\infty x_n \int_{-\infty}^\infty \delta\left(t - \frac{n}{f_0}\right) \exp(-j2\pi ft)\,\mathrm dt\\ &= \sum_{n=-\infty}^\infty x_n \exp\left(-\frac{j2\pi nf}{f_0}\right). \tag{3} \end{align} As a function of $f$, $\exp\left(-\frac{j2\pi nf}{f_0}\right)$ has period $\frac{n}{f_0}$, that is, a multiple of $\frac{1}{f_0}$ and so $(3)$ is indeed a periodic function of period $\frac{1}{f_0}$.

When you sample an analog signal to get a discrete-time signal with sampling frequency $f_s$, the spectrum of the latter will be periodic. To be more precise, the spectrum of your analog signal will be shifted by $n\cdot f_s$, for every integer $n$ and the sum of these shifted versions is the spectrum of the discrete-time signal.

The reason you need to have a sampling frequency high enough is to avoid overlapping between the shifted spectrum parts, thus cancelling the opportunity to recreate the original analog signal. Generally, if you choose an $f_s$ higher than twice the signal bandwidth, you are always good to go, but in some cases a lower sampling frequency might be sufficient as well.

(image source: R. G. Lyons: Understanding Digital Signal Processing (2nd Edition))

• What a spectacular drawing! The artist’s mother must be very proud. – Richard Lyons Sep 17 '15 at 11:00
• Looks like it's straight of your book, Rick! :-( – Peter K. Sep 17 '15 at 19:05
• @RichardLyons : I am really sorry for this. What is the policy regarding pictures "from the Internet", that, in this case, turns out to be your own work? – hryghr Sep 17 '15 at 21:24
• Thanks for asking. I don't know StackExchange's formal policy (I'm just barely learning to navigate this site myself), but when you use someone's drawing it's considered good manners to briefly state from where a drawing came. That way no one can accuse you of taking credit for someone else's work. – Richard Lyons Sep 19 '15 at 15:52
• I included the source. Pardon me for forgetting about giving credit properly! – hryghr Sep 19 '15 at 19:09

The Discrete Fourier Transform is always periodic, but we usually focus only on what's inside the Nyquist window and ignore the periodic copies outside. The regular Fourier Transform (function of continuous frequency) cannot be periodic, to my knowledge.

• Delta function - though not real has a periodic FT. I can't think of a real function with a periodic FT either . – hakunamatata Sep 16 '15 at 22:53
• IIRC, an infinite periodic train of impulses in one domain will produce an infinite periodic train of impulses in the other domain (of an FT). – hotpaw2 Sep 17 '15 at 0:38
• That is true, I meant to say a continuous signal which technically can't be real now cause everything is digital and discrete :P. – hakunamatata Sep 17 '15 at 17:20
• You are right, hotpaw2, but any discrete signal can be thought of as a continuous signal multiplied with a periodic train of time domain impulses. This leads to convolution in the frequency domain between the transforms of the continuous signal and the impulse train (which is indeed an impulse train with period equal to sampling frequency). So we're saying the same thing. – CMDoolittle Sep 17 '15 at 18:41
• Oops. Semantics are critical in this discussion. I'm pretty sure the product of a continuous time signal and a continuous train of time-domain impulses results in a continuous signal. Whereas the phrase "discrete signal" typically means a sequence (a list) of individual numbers. As such, a "discrete signal" cannot be continuous. – Richard Lyons Sep 19 '15 at 16:17

when the time-domain signal being Fourier Transformed is uniformly sampled.

the continuous Fourier Transform of

$$x(t) \sum\limits_{n=-\infty}^{+\infty} \delta\left(t - \frac{n}{f_\text{s}}\right) = \sum\limits_{n=-\infty}^{+\infty} x\left(\frac{n}{f_\text{s}}\right) \delta\left(t - \frac{n}{f_\text{s}}\right) = \sum\limits_{n=-\infty}^{+\infty} x[n] \delta\left(t - \frac{n}{f_\text{s}}\right)$$

is

$$f_\text{s} \sum\limits_{k=-\infty}^{+\infty} X\left(f - k \, f_\text{s} \right)$$

which is periodic with period $f_\text{s}$.

The effect of delaying a signal means that the Fourier transform gets multiplied by $e^{-Ts}$, which when using Euler's identity can be split up into a real valued cosine and a complex valued sine function, which are periodic.

A signal whose Fourier transform is constant is the Dirac's delta function. You can combine multiple of such functions and still have a periodic Fourier transform, however the delays have to be integer multiples of each other, such that each they have a common period. This can be seen as constructing a Fourier series in the Fourier domain. And a Fourier series can represent any bounded periodic function. Thus any function, whose Fourier transform is periodic and bounded, has to be of this form.

• Multiplying a Fourier transform by e^−Ts, ...what an unusual thought. Is 's' the complex variable used in Laplace transforms? Your first paragraph certainly does invite meditation. – Richard Lyons Sep 19 '15 at 16:23