Periodicity of a constant signal!

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.