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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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Ooker
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Mohamed Mostafa
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  • I can't understand your question. Can you rephrase? What exactly do you mean by resonance? –  Jun 30 '16 at 01:01
  • @ZacharySelk: https://en.wikipedia.org/wiki/Resonance – Mohamed Mostafa Jun 30 '16 at 01:03
  • I can Google/use Wikipedia. I'm asking what YOU mean by resonance. –  Jun 30 '16 at 01:04
  • Large amplitude of output at input frequencies near the natural frequency. – Mohamed Mostafa Jun 30 '16 at 01:07
  • What's the output you're talking about? A Fourier series approximates a signal. I don't know where output comes into play. –  Jun 30 '16 at 01:10
  • In ODE such as $x''+Ax'+Bx=f(t)$, where $f(t)$ is a periodic function. Is my question clear now ? – Mohamed Mostafa Jun 30 '16 at 01:14
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    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 $a_n = \frac{c_n}{-4 \pi^2n^2 + A 2i \pi n+B}$ – reuns Jun 30 '16 at 01:21
  • So you have an ODE driven by $f$. You Fourier expand the $f$ and you're looking at the solution, $x$. Then the frequencies of $f$ that are closest to the frequency of the solution are the largest in the solution. Then your question is, why does this happen? Do I have your question right? –  Jun 30 '16 at 01:22
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    overall $$x(t) = \sum_{n=-\infty}^\infty \frac{e^{2 i \pi n t}}{-4 \pi^2n^2 + A 2i \pi n+B} \int_0^{2 \pi} f(x) e^{-2 i \pi n x} dx$$ – reuns Jun 30 '16 at 01:24
  • @ZacharySelk that's right. The frequency of the original input signal -without approximating it by Fourier- is far from natural frequency and thus no resonance should happen. But when approximating it using Fourier resonance shows up. How could this happen ? – Mohamed Mostafa Jun 30 '16 at 01:26
  • @user1952009 I'm not asking how to get Fourier series, I'm asking about the behavior of the response. – Mohamed Mostafa Jun 30 '16 at 01:28
  • if you don't say what you don't understand, nobody can help you, and really the answer is in what I wrote – reuns Jun 30 '16 at 01:30
  • I'm not doing engineering, but can square wave resonate in principle? With another squared $f(t)$, or with a sine one? – Ooker Oct 16 '17 at 09:37
  • @reuns I think the question is interesting, and have edited it to highlight the problem. Can you come and see? Thank you – Ooker Oct 16 '17 at 10:01

1 Answers1

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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.

Ng Chung Tak
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  • The questing is edited to highlight the exact problem the OP has. Can you come and see? Thank you. – Ooker Oct 17 '17 at 09:20
  • 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. – Ng Chung Tak Oct 17 '17 at 12:56
  • 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? – Ooker Oct 19 '17 at 10:22
  • No, it's in $\omega$ space. You may do the inverse Fourier transform to get back the signal. – Ng Chung Tak Oct 20 '17 at 03:25