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This can be a very silly question, but I'm quite confused:

  1. If we take the Fourier transform of any constant signal, we get an impulse at zero, which says that its frequency is zero and, hence, it is non-repeating and its period is infinity.

  2. By the definition of a periodic signal, $$F(z+p) = F(z)$$ therefore, for a constant function $$F(z) = c$$ $$F(z+p) = c, \forall p >0$$ Therefore, a constant function is periodic, but its period is undefined or can be defined as anything.

Which of the above arguments are correct and why? Please clarify. Thanks in advance.

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    $\begingroup$ You need to use the least positive $p$, which does not exist for a constant function. See this: en.wikipedia.org/wiki/Periodic_function#cite_note-1 $\endgroup$ – Atul Ingle Oct 19 '17 at 16:57
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    $\begingroup$ i would vote for the constant function being a periodic function that can have any period. $\endgroup$ – robert bristow-johnson Oct 19 '17 at 17:49
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    $\begingroup$ @robertbristow-johnson by that reasoning, all periodic signals can have any period $p' = n\cdot p$ (with integral $n$). That's ugly because it means the period is never really well-defined; hence as Atul Ingle says, the definition of the period is actually the least positive $p$ that fulfills $F(z+p) \equiv F(z)$. $\endgroup$ – leftaroundabout Oct 19 '17 at 20:31
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    $\begingroup$ that is true, Leftie. any periodic signal that is periodic with period $P$ is also periodic with period $2P$ or $3P$. $\endgroup$ – robert bristow-johnson Oct 19 '17 at 22:09
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    $\begingroup$ There are different standards. A similar question was closed on SE.math Is a constant function periodic? $\endgroup$ – Laurent Duval Oct 20 '17 at 16:37
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As you say, the constant function is periodic. A signal $x(t)$ is said to be periodic with period $p$ or to have a period $p$ if there exists a $p>0$ such that $x(t+p)=x(t)$ for all real numbers $t$. Note that I said "a period" instead of "the period" -- periodic signals have an infinite number of periods, maybe even an uncountable infinite of them!

Often, what is referred to as the period of a signal is actually more formally described as the "fundamental period", which is the smallest among all periods.

So, a constant signal is periodic, it has an uncountably infinite number of periods (since any real number $p>0$ is a period), but it does not have a fundamental period.

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  • $\begingroup$ "periodic signals may have many (even an infinite) number of periods." Actually, all periodic functions have an infinite number of periods. $\endgroup$ – leonbloy Oct 19 '17 at 20:52
  • $\begingroup$ @leonbloy You're right, of course -- edited to remark on what might be actually surprising (the number of periods may be an uncountable infinite). $\endgroup$ – MBaz Oct 19 '17 at 21:58
  • $\begingroup$ Edited to add a few words. Please roll back if you don't like the changes I made. I like your answer much better than the accepted one. $\endgroup$ – Dilip Sarwate Oct 20 '17 at 15:34
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    $\begingroup$ "periodic signals may have many (even an infinite) number of periods." Actually, all periodic functions have an infinite number of periods > Well, this assertion depends a lot of where you locate $t$. With a torus shape one might have only a finite number of "different" periods (within a ring structure). Yes, it's a little bit pedantic :) $\endgroup$ – Laurent Duval Oct 20 '17 at 16:32
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    $\begingroup$ The number of possible periods of a constant is aleph one, as opposed to non-constant periodic waveforms which have aleph zero. Two different infinities. $\endgroup$ – hotpaw2 Oct 20 '17 at 21:35
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When you are in doubt, use the limiting approach as an aid in your deductions:

For example, you can consider a constant signal $x_C(t)=1$ as the limit of a periodic sine wave $x_p(t)= \cos(\omega_0 t)$ when the frequency goes to zero; i.e:

$$x_C(t) = 1 = \lim_{\omega_0 \to 0} \cos(\omega_0 t) $$

And you now that the Fourier transform of a sine wave is a frequency impulse located at $\omega = \omega_0$ ; i.e., $$\mathcal{F}\{\cos(\omega_0 t) \} = \pi \delta(\omega - \omega_0) + \pi \delta(\omega + \omega_0) $$

And therefore in the limiting case the signal $x_C(t)$ will have the Fourier transform of $$\mathcal{F}\{1\} = 2\pi \delta(\omega - 0) = 2\pi \delta(\omega) $$

Indeed the two impulse from right and left converge it the middle and add up !

At this point then, you don't have to consider whether the definition of periodicity applies or makes sense for the signal $x_C(t)=1$, since it's a degenerate case. Yet if you wish you can consider that its period is infinity or else zero, which is an ill consideration nevertheless.

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    $\begingroup$ @LaurentDuval That's actually an intuitive limit, rather than a formal topological one. I remember that in our signals and systems book, (OppWillsky) the transition from continuous time Fourier series to th continuous time Furier transform was also based on a similar intuitive limiting process, where one thinks of the CTFT as the limit of CTFS when the period goes to infinity (while the signal being nonzero only for a finite length)... $\endgroup$ – Fat32 Oct 20 '17 at 20:51
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    $\begingroup$ I understood the analogy. However, I just got a paper from Feichtinger, on how LTI system courses can be misleading, and wanted to mention the potential caveats $\endgroup$ – Laurent Duval Oct 20 '17 at 21:15
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    $\begingroup$ dx.doi.org/10.1007/978-3-319-55556-0_3 $\endgroup$ – Laurent Duval Oct 20 '17 at 21:59
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    $\begingroup$ @LaurentDuval Thanks. As an engineer, I'm not convinced that the way we teach should change, but as a professor one should be aware of the mathematical nuances. $\endgroup$ – MBaz Oct 20 '17 at 22:20
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    $\begingroup$ @MBaz just sharing alternative facts :) "Who knows where the road may lead us, only a fool would say. Who knows if we'll meet along the way. " Alan Parsons Project) $\endgroup$ – Laurent Duval Oct 20 '17 at 22:30
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If we take the Fourier transform of any constant signal, we get an impulse at zero, which says that its frequency is zero and, hence, it is non-repeating and its period is infinity.

No, this does not work for a zero signal (Fourier is flat-flat, no impulse). Plus, something over zero is traditionally undefined, and could be any number. And THIS is the very case here.

A periodic function $P$ is such that its values repeat "in some way" at "regular intervals or periods". Trigonometric functions are natural examples. Indeed, you can derive them from the exponential series, which is convergent and its own derivative

This notion of repetition can be multidimensional in the variable $x$. There are multiple avatars for periodicity: the most commonly used are simply periodic, doubly periodic, triply periodic functions.

In the complex plane (the most useful for signal practitioners so far), you can have a doubly periodic function $P$, with two incommensurate least/fundamental complex periods:

$$P(x+x_1)=P(x+x_2)=P(x)$$

with $x_1/x_2$ is not real. Those are called elliptic functions. If you now restrict to univariate single-valued functions, then a theorem by Jacobi states that is is impossible for them to have more than two distinct (least/fundamental) periods.

But a periodic function can have "no least period" at all, and this is the case with constant functions:

The constant function $P(x)=0$ is periodic with any period $x_0$ for all nonzero real numbers $x_0$, so there is no concept analogous to the least period for constant functions.

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    $\begingroup$ ok I understood my mistake and deleted the provokative comment. Now I'm also believing that a constant is a periodic signal with an undefined fundamental period. Now I'm happy :-) $\endgroup$ – Fat32 Oct 20 '17 at 22:25
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    $\begingroup$ A way clearer definition :) I'll borrow it for my lectures to come :) $\endgroup$ – Laurent Duval Oct 20 '17 at 22:32

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