A periodic function can be represented as an infinite sum in many bases. I know at least one other series apart from fourier:

Power series:
\$\sin(t) = \sum\limits_{k=0}^{\infty} \frac{(-1)^kt^{2k}}{(2k)!}\$

Why do linear circuits respond like fourier series and not like a power series?

Is it possible to have circuits like opamps that respond to polynomial functions in an useful manner? For example:
input: \$t^2 + 5t^3+t^{10}+t^{150}\$
low pass filter output: \$t^2+5t^3\$
high pass filter output: \$t^{150}\$

  • 3
    \$\begingroup\$ What do you mean by 'respond linearly'? \$\endgroup\$ – Chu Mar 25 '20 at 15:25
  • 2
    \$\begingroup\$ Components are dumb. They don't calculate the Fourier series and decide how to respond. It's the other way around: components respond in a certain way by (their) nature and scientist like Fourier found how to describe this behaviour. \$\endgroup\$ – Huisman Mar 25 '20 at 15:27
  • 2
    \$\begingroup\$ "Fourier series" is, like you said, a representation of a periodic function. It's something you can write on a piece of paper. What does "respond like fourier series" mean? \$\endgroup\$ – Solomon Slow Mar 25 '20 at 15:27
  • \$\begingroup\$ oh i was thinking of superposition, but linear circuits respond linearly to nonsinusoidal inputs too, so the word "linearly" is redundant. I'll remove that. Thank you :) \$\endgroup\$ – across Mar 25 '20 at 15:28
  • \$\begingroup\$ What do you mean by "respond to"? \$\endgroup\$ – Hearth Mar 25 '20 at 15:31

Short answer: The complex exponential functions are eigenfunctions of the derivative and integral operators.

Some more explanation: The sine and cosine functions are somewhat more cleanly expressed as complex exponentials; that is, functions of the form $$f(x)=e^{i\omega t}$$

using the identities $$\sin \omega t = \frac{e^{i\omega t}-e^{-i\omega t}}{2i}$$ $$\cos \omega t = \frac{e^{i\omega t}+e^{-i\omega t}}{2}$$

And, you're probably aware, when you take the derivative or integral of an exponential function, you get another exponential out, with the same terms in the exponent. (This is what it means to be an eigenfunction of those operators).

So, if you design a circuit that performs linear operations (some combination of scalar multiplication, derivative, and integral, or a sum of several such operations) on an input signal, and you put a sine or cosine signal in, you will get a combination of sines and cosines out, at the same frequency.

Is it possible to have circuits like opamps that respond to polynomial functions?

I don't know if it's impossible, but it would at least be very difficult, for at least a couple of reasons:

  • A polynomial function with a finite number of terms will eventually trend toward infinity beyond some positive and negative limits in time. But no real op-amp can produce a non-bounded output voltage.

  • The response would have to be non-time-invariant. Meaning the circuit would have to keep track of the time and respond differently at different moments in time. It would also have to do this for times before \$t=0\$, meaning it would have to predict the future to know when some event is going to happen that will designate the reference instant of time for the system.

  • \$\begingroup\$ The Photon so you mean if we input an exponential function to a linear cirucit, we just get back the scaled version of the same exponetial function. But this is not possible with any other basis? Ofcourse it is certainly not possible with polynomials. Differentiation and integration mess up the exponents.. \$\endgroup\$ – across Mar 25 '20 at 15:44
  • \$\begingroup\$ I don't know much math but is it possible to have other eigenfunctions to derivative/integral operators? or the exponential functions are THE eigenfunctions? (I don't even know if the terminology I'm using makes sense) \$\endgroup\$ – across Mar 25 '20 at 15:46
  • 1
    \$\begingroup\$ There are other basis functions with similar properties, but they don't normally apply to functions in the time domain. For example, in problems where the variable is position, and with cylindrical symmetry, we often encounter the Bessel function basis for the radial dependence of a "signal" because the Bessel functions are eigenfunctions of the spatial derivatives in cylindrical coordinates. \$\endgroup\$ – The Photon Mar 25 '20 at 15:47
  • 1
    \$\begingroup\$ Yes to your last formula. \$\endgroup\$ – The Photon Mar 25 '20 at 16:03
  • 1
    \$\begingroup\$ @beccaboo It'd probably be helpful to you to do some research into the Laplace transform, as well. Helps with understanding this sort of thing a lot, I've found. \$\endgroup\$ – Hearth Mar 25 '20 at 16:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.