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I understand that Fourier series approximate the input signal well and series converge to the original function. If the system is ODE, such as $x''+Ax'+Bx=f(t)$, then $f(t)$ will respond differently to each term of the series according to how close its frequency is to the system natural frequency and thus one of the term will resonates with $f(t)$.

But why can an original input without natural frequency, such as square wave function, can resonate after being transformed? Where is the source of resonance in the original input signal, if it doesn't have in the first place?

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  • $\begingroup$ I can't understand your question. Can you rephrase? What exactly do you mean by resonance? $\endgroup$ – user223391 Jun 30 '16 at 1:01
  • $\begingroup$ @ZacharySelk: en.wikipedia.org/wiki/Resonance $\endgroup$ – Mohamed Mostafa Jun 30 '16 at 1:03
  • $\begingroup$ I can Google/use Wikipedia. I'm asking what YOU mean by resonance. $\endgroup$ – user223391 Jun 30 '16 at 1:04
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    $\begingroup$ the main Fourier series theorem is that $\{e^{2 i \pi n t}\}_{n \in \mathbb{Z}}$ is an orthonormal basis of $L^2([0,2\pi])$, so $\sum_{n=-\infty}^\infty a_n e^{2 i \pi n t} = \sum_{n=-\infty}^\infty c_n e^{2 i \pi n t}$ for every $t \implies a_n = c_n$ for every $n$. finally use that $x(t) = \sum_{n=-\infty}^\infty a_n e^{2 i \pi n t} \implies $ $x''(t)+Ax'(t)+B x(t) = \sum_{n=-\infty}^\infty (-4 \pi^2n^2 + A 2i \pi n+B) a_n e^{2 i \pi n t}$ and $f(t) = \sum_{n=-\infty}^\infty c_n e^{2 i \pi n t} $ to obtain $a_n = \frac{c_n}{-4 \pi^2n^2 + A 2i \pi n+B}$ $\endgroup$ – reuns Jun 30 '16 at 1:21
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    $\begingroup$ overall $$x(t) = \sum_{n=-\infty}^\infty \frac{e^{2 i \pi n t}}{-4 \pi^2n^2 + A 2i \pi n+B} \int_0^{2 \pi} f(x) e^{-2 i \pi n x} dx$$ $\endgroup$ – reuns Jun 30 '16 at 1:24
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The natural frequency is given by $\omega_{0}=\sqrt{B}$.

If a periodic signal $f(t)$ with frequency of $n$-th harmnoic (or $n-1$ overtone) equals to the natural frequency, then the corresponding terms contribute to resonance.

Mathematically, say for example

$$f(t)=\sum_{n=1}^{\infty} a_{n} \sin n\omega t$$

if $\omega_{0}=n\omega$ then the term (partial signal) $a_{n} \sin n\omega t$ contribute to the resonance solution.

In case of a single pulse (or wavepacket), Fourier transform should be used instead.

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  • $\begingroup$ The questing is edited to highlight the exact problem the OP has. Can you come and see? Thank you. $\endgroup$ – Ooker Oct 17 '17 at 9:20
  • $\begingroup$ For periodic wave, there won't be resonance when $\omega_{0} \ne n\omega$. For a single pulse (or wave packet), there's continuous spectrum for all frequencies (with corresponding amplitudes and phases). Say a Gaussian pulse in Fourier space $$e^{-a(\omega-\omega_1)^2t}$$ it's highly resonated when $\omega_1 \approx \omega_0$, otherwise no obvious resonance occurs. $\endgroup$ – Ng Chung Tak Oct 17 '17 at 12:56
  • $\begingroup$ Is your guassian pulse the system itself, or the external force? When you say "no obvious resonance occurs", does it mean that responses do occur, they're just insignificant for all terms? $\endgroup$ – Ooker Oct 19 '17 at 10:22
  • $\begingroup$ No, it's in $\omega$ space. You may do the inverse Fourier transform to get back the signal. $\endgroup$ – Ng Chung Tak Oct 20 '17 at 3:25

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