If we are given a signal of the form $$x(t) = \sum_{k = -\infty}^{+\infty} a_k e^{j k \omega_0 t},$$ can we call it a Fourier Series representation of $x(t)$ right away?

Suppose we are given the signal $x(t) = e^{j2\Omega t} + e^{j4\Omega t}$, this signal can be expressed as $$x(t) = \sum_{k = -\infty}^{+\infty} a_k e^{jk\Omega t},$$ however, in this case, $2\Omega$ is the fundamental frequency of the signal, not $\Omega$ as the equation above might suggest.

I ask this because in the second edition of Signals and Systems by Alan Oppenheim, he derives the Fourier transform of a periodic signal by considering the impulse train $$X(j\omega) = \sum_{k = -\infty}^{+\infty} 2\pi a_k\delta (\omega - k \omega_0)$$ and applying the inverse Fourier transform to obtain $$x(t) = \sum_{k = -\infty}^{+\infty} a_k e^{j k \omega_0 t},$$ which he says is the Fourier series representation of the signal without further discussion. But how do we know that in this case $\omega_0$ is the fundamental frequency of the signal? Couldn't it be the case that it is a signal of the form given above, namely, $x(t) = e^{j2\omega_0 t} + e^{j4\omega_0 t}$, in which case the fundamental frequenct is $2\omega_0$? Thank you in advance.


The Fourier series, if it exists, is unique, apart from the fact that we can always use a multiple of the fundamental period $T$ as the period, or, equivalently, a fraction of the fundamental frequency $\omega_0$. The only consequence of this is that we introduce Fourier coefficients with value zero in between the non-zero coefficients. The values of the non-zero coefficients remain unchanged:

$$x(t)=\sum_{k=-\infty}^{\infty}a_ke^{jk\omega_0}=\sum_{k=-\infty}^{\infty}\tilde{a}_ke^{jk\omega_0/N},\quad N\in\mathbb{Z}^+$$

$$\tilde{a}_k=\begin{cases}a_{k/N},\quad k=lN,\quad l\in\mathbb{Z}\\0,\quad\text{otherwise}\end{cases}$$

  • $\begingroup$ So it is the Fourier series representation of $x(t)$ but there is no way for us to tell whether $a_k$ and $\omega_0$ represent the original coefficients and the fundamental frequency respectively beforehand? $\endgroup$
    – user33568
    Feb 22 '18 at 17:24
  • $\begingroup$ i disagree with your very first statement, @MattL.: "The Fourier series, if it exists, is unique,..." $$ $$ of course that's not true. you can always bump $\omega_0$ to $1/N$ of its previous value (where $N \in \mathbb{Z}$) call that your new $\omega_0$ and fill in zeroes for the $N-1$ intermediate terms between the potentially non-zero coefficients spaced every $N$ terms. i know you're saying essentially this also, but it contradicts your opening premise. $\endgroup$ Feb 22 '18 at 20:37
  • $\begingroup$ of course if $$ x(t+T) = x(t) \qquad \qquad \forall \ t \in \mathbb{R} $$, then it is also always true that $$ x(t+2T) = x(t) \qquad \qquad \forall \ t \in \mathbb{R} $$ or $$ x(t+NT) = x(t) \qquad \qquad \forall \ t \in \mathbb{R}, \ N \in \mathbb{Z} $$ and sometimes it's true that $$ x(t+\tfrac12 T) = x(t) \qquad \qquad \forall \ t \in \mathbb{R} $$ or $$ x(t+\tfrac1N T) = x(t) \qquad \qquad \forall \ t \in \mathbb{R}, \ N \in \mathbb{Z} $$ but if it's not true for any submultiples of $T$, then you have a unique Fourier series. $\endgroup$ Feb 22 '18 at 20:41
  • $\begingroup$ @robertbristow-johnson: well, that's what I say in the following clause of my first sentence, but since that is quite trivial I didn't want to say that it's NOT unique, because using a multiple of $T$ as the period just changes the index of the coefficients, not the coefficients themselves, and that's a triviality. $\endgroup$
    – Matt L.
    Feb 22 '18 at 20:43
  • $\begingroup$ @robertbristow-johnson: I'll reformulate to make this entirely clear ... $\endgroup$
    – Matt L.
    Feb 22 '18 at 20:44

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