Considering the continue, periodic signals $s_1(t)$ and $s_2(t)$, with the period $P_1$ and $P_2$ respectively.

Consider now a new signal $s$ the sum of the two pervious signals.

I come to know that the spectrum of the sum, is not necessarily the sum of the spectrums unless the two signals $s_1(t)$ and $s_2(t)$ have the same period. With spectrum I mean:

$$X\in [\mathbb{Z} \rightarrow \mathbb{C} ]$$

$$X[k] = \langle x(t), \phi_k\rangle = \frac{1}{P} \int_P x(t)e^{-ik\omega_0 t}$$

with $\omega_0 = \frac{2\pi}{P}$. $P$ here represents the period.

Note that $x(t)\in [\mathbb{R} \rightarrow \mathbb{C}]$ is periodic with the period $P$.

I don't understand why? Because at it seems this violates the linearity properties.

  • 1
    $\begingroup$ You define "spectrum" as over one period. Since the periods are different, you have two different integration intervals. Linearity doesn't apply since you have two different definitions of spectrum. If you simply integrate both from $[-\infty,\infty]$ you get two line spectra that do indeed just add and everything is linear again. $\endgroup$ – Hilmar Oct 23 '19 at 12:10
  • $\begingroup$ @Hilmar Never looked at problem this way, makes the other answers much transparent ! But, the signals are power signals so the integration over the $R$ also goes to infinity, that's why I defined them over a finite interval. $\endgroup$ – Sam Farjamirad Oct 23 '19 at 15:38
  • $\begingroup$ The are indeed power signals but the Fourier Transform of a sine/cosine is well defined. It's simply a pair of dirac pulses: see ethz.ch/content/dam/ethz/special-interest/baug/ibk/…, down on page 2 $\endgroup$ – Hilmar Oct 23 '19 at 18:47

the Fourier Series for a single periodic function, having period $P$,

$$ x(t) = x(t+P) \qquad \forall \ -\infty < t < +\infty $$


$$ x(t) = \sum\limits_{k=-\infty}^{\infty} c_k \, e^{j(2\pi k/P)t} $$

each harmonic $c_k$ is labelled with its harmonic number, $k$, in the above. the actual frequency of each harmonic, in Hz or units of 1/time, is $\frac{k}{P}$. the fundamental frequency is the reciprocal $\frac{1}{P}$ of the period. in angular frequency (radians per unit time), multiply both by $2\pi$. but, in the harmonic number $k$ there lacks enough specific information to tell you what the actual frequency is of the harmonic. e.g. the second harmonic of a periodic tone of 220 Hz is the same frequency as the fourth harmonic of a periodic tone of 110 Hz.

so the spectrum of $x(t)$ is about what actual frequencies the energy of a signal is. it doesn't matter whether $x(t)$ is periodic or not (actually it does with the esoteric mathematics, but we won't go there). and we use the Fourier Transform, not the Fourier Series to visualize it.

So each harmonic in the Fourier Series above is attached to a dirac delta function, $\delta(f)$.

$$\begin{align} X(f) &\triangleq \mathscr{F} \Big\{ x(t) \Big\} \\ &= \int\limits_{-\infty}^{\infty} x(t) \, e^{-j2\pi f t} \, \mathrm{d}t \\ &= \mathscr{F} \Big\{ \sum\limits_{k=-\infty}^{\infty} c_k \, e^{j(2\pi k/P)t} \Big\} \\ &= \sum\limits_{k=-\infty}^{\infty} c_k \,\mathscr{F} \Big\{ e^{j(2\pi k/P)t} \Big\} \\ &= \sum\limits_{k=-\infty}^{\infty} c_k \, \delta(f-k\tfrac{1}{P}) \\ \end{align}$$

If you express the spectra of your periodic signals in this form, the spectra sum to a single spectrum.

| improve this answer | |

You cannot sum $X_1[k]$ and $X_2[k]$ (the discrete spectra of the two signals) element-wise because for them $k$ represents different angular frequencies $\frac{k\,2\pi}{P_1}$ and $\frac{k\,2\pi}{P_2}$ in radians. Time domain addition will only translate to frequency domain addition if your frequency variable is equally proportional to angular frequencies for the two spectra.

If you don't want to take an approach with Dirac delta functions as in @robertbristow-johnson's answer, perhaps you can use this: If $\frac{P_1}{P_2}$ is rational, then it is equal to an irreducible fraction $\frac{p}{q}$ and both of the signals will also be $p\,P_2$ -periodic. Using this shortest* shared period you can say that the signals have the same period. It will also make the $k$ represent the same angular frequencies for the two spectra.

*) shortest assuming that $P_1$ and $P_2$ are the fundamental periods.

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  • $\begingroup$ For the future readers: This answer is also correct, and somehow complements the selected answer. $\endgroup$ – Sam Farjamirad Oct 24 '19 at 8:09

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