On studying about Fourier series, I encountered 2 doubts:

  1. How is it that a non-periodic function has a Fourier series?
  2. When expressing a periodic function as summation of sinusoids, why is the fundamental (minimum) frequency taken to be that of the periodic signal? Why cannot one of the sinusoids have a frequency less than the fundamental?

I mean, we think of representing the signal in terms of sinusoids with frequencies greater than and equal to the fundamental frequency, and of course, there is the DC part as well. But, why is there no sinusoid with frequency less than the fundamental as well?

Is there any mathematical logic for the second point?

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    $\begingroup$ Why do you think that non-periodic signals have a Fourier series? $\endgroup$ – MBaz Jan 31 '19 at 16:38
  1. A function that is not periodic doesn't have a Fourier series. What you can do is take a finite length interval and construct a Fourier series which approximates the function in that interval (and continues it periodically outside that interval).
  2. The sum of several sinusoids with one of the sinusoids having (fundamental) period $T$ can never have a period less than $T$, simply because that one sinusoid will destroy periodicity (with period $<T$) due to its larger period.
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  • $\begingroup$ Regarding point 2, why will that sinusoid destroy periodicity? Even a DC part exists. A DC part has the same value at every value of t. It's period is 0 in that sense. Right? $\endgroup$ – Curiosity Jan 31 '19 at 16:45
  • $\begingroup$ @Curiosity: Well, a constant has the same value in every period, so it doesn't affect periodicity. This is different from a periodic function (like a sinusoid), which cannot satisfy $x(t)=x(t+T)$ unless $T$ is an integer multiple of its own period. $\endgroup$ – Matt L. Jan 31 '19 at 16:47
  • $\begingroup$ I understand. Still, from a mathematical standpoint, can it be proved that frequency of any of the sinusoids cannot be less than the fundamental frequency? $\endgroup$ – Curiosity Jan 31 '19 at 16:54
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    $\begingroup$ @Curiosity: Sure, just use the equation in my previous comment. By definition it cannot be satisfied by a sinusoid with period greater than $T$. $\endgroup$ – Matt L. Jan 31 '19 at 17:30
  • $\begingroup$ Matt, regarding #2, how are you defining the "period $T$" of "one of the sinusoids"? on the surface, it doesn't seem true unless you define explicitly that the period $T$ of a periodic function must be the smallest real and positive value $T$, such that $x(t+T) = x(t) \quad \forall t \in \mathbb{R} \quad.$ that's what it seems to me. if you don't make that the explicit definition of the period, then i think that #2 is false. E.g. for $x(t)=\cos(2\pi t)$, this function is periodic and any positive integer can be its period $T$. $\endgroup$ – robert bristow-johnson Feb 1 '19 at 2:30

Answering question 2: Say you have a periodic signal with fundamental period $T_0$ and fundamental frequency $f_0=1/T_0$. You want to write it as the sum $$\sum_k a_k \cos(2\pi k f_1 t + \phi_k),$$ where $f_1 < f_0$.

The problem is that the sum has fundamental period $T_1 = 1/f_1$, which contradicts your assumption that the fundamental period was $T_0$.

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  • $\begingroup$ Yes. But, why does the frequency of the original signal need to be the fundamental? Fourier series even has a DC part with same value for all t. In that sense, it's period should be 0. $\endgroup$ – Curiosity Jan 31 '19 at 16:47
  • $\begingroup$ The period is given by first harmonic. DC provides an offset, it cannot change the periodicity. By the way, a constant signal has no fundamental period, but it has all possible periods. Since a period has to be positive, it's incorrect to say that the period of DC is zero. $\endgroup$ – MBaz Jan 31 '19 at 16:54
  • $\begingroup$ Right. Got it now. But, can it be mathematically proved that the frequency of any of the sinusoids cannot be less than the fundamental frequency? $\endgroup$ – Curiosity Jan 31 '19 at 16:55
  • $\begingroup$ I thought that's what I had done in my answer :) All you need to do is prove that the sum has fundamental $f_1$, which can be seen directly. $\endgroup$ – MBaz Jan 31 '19 at 16:57
  • $\begingroup$ I meany, why cannot be take the fundamental to be something like f/2? Just by dividing the phase in your expression by 2 as a whole? $\endgroup$ – Curiosity Jan 31 '19 at 16:58

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