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After reading a few posts, like this. I know that arbitrary LTI systems always have complex exponential eigenfunctions. And that for specific LTI systems you can also have other types of eigenfunctions. But for arbitrary LTI systems are complex exponentials the only eigenfunction that always exists? Preferably with a proof.

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  • $\begingroup$ What exactly is it that you miss in the answers to the question you linked to? All LTI systems have complex exponentials as eigenfunctions, and specific LTI systems can have other eigenfunctions too. $\endgroup$
    – Matt L.
    Nov 9, 2023 at 20:47
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    $\begingroup$ Besides complex exponentials. Is there any other function that is a eigenfunction for arbitrary LTI systems? $\endgroup$
    – roobee
    Nov 9, 2023 at 21:17
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    $\begingroup$ In the most general case you only have complex exponentials as eigenfunctions. Any other eigenfunctions are specific to certain types of LTI systems. $\endgroup$
    – Matt L.
    Nov 9, 2023 at 21:39
  • $\begingroup$ Is there a proof? Or can you explain why informally? $\endgroup$
    – roobee
    Nov 9, 2023 at 21:52

1 Answer 1

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This is not a rigorous proof but just to make it plausible that complex exponentials are the only eigenfunctions for general LTI systems.

We know that for LTI systems the input-output relation is described by convolution. In continuous time we have

$$y(t)=\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau\tag{1}$$

where $y(t)$ is the output, $x(t)$ is the input, and $h(t)$ is the system's impulse response. Let $x(t)$ be a complex exponential:

$$x(t)=e^{s_0t},\qquad s_0\in\mathbb{C}$$

Then, from $(1)$

\begin{align*} y(t) &= \int_{-\infty}^{\infty}h(\tau)e^{s_0(t-\tau)}d\tau\\ &= e^{s_0t}\int_{-\infty}^{\infty}h(\tau)e^{-s_0\tau}d\tau \\ &= e^{s_0t}H(s_0)\tag{2} \end{align*}

where $H(s)$ is the Laplace transform of $h(t)$, i.e., the system's transfer function. [We assume that $s_0$ is inside the region of convergence.]

Eq. $(2)$ shows that $e^{s_0t}$ is an eigenfunction with eigenvalue $H(s_0)$.

Note that in the derivation of $(2)$ we made use of the fact that the input $x(t)=e^{s_0t}$ satisfies $x(t-\tau)=x(t)x(-\tau)$ for all $t$ and $\tau$, otherwise $x(t)$ wouldn't be an eigenfunction for arbitrary $h(t)$. In this math.SE post it is shown that the exponential function is the only continuous function that satisfies the required property $x(t-\tau)=x(t)x(-\tau)$.

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    $\begingroup$ Also worth noting that exponential functions are the only continuous functions that satisfy $x(t-\tau) = x(t)x(-\tau)$ (with some qualifications). Related question from Math.SE: A function with a property $f(x+y)=f(x)f(y)$ $\endgroup$ Nov 10, 2023 at 14:32
  • $\begingroup$ Thanks! I tried to find whether exponential function was the only one that satisfied f(x+y) = f(x)f(y) previously but I couldn't find it. I'll edit the comment into the answer and accept it $\endgroup$
    – roobee
    Nov 10, 2023 at 16:28
  • $\begingroup$ If you consider a trivial system with an impulse response of a scaled delta function. Then every input function is an eigenfunction, since the output is simply a scaled version of the input. $\endgroup$
    – David
    Nov 10, 2023 at 18:29
  • $\begingroup$ @David: That's of course true, but from the question it appears that the OP is aware that specific LTI systems can have many more eigenfunctions than just the complex exponentials. The way I understood the question is if complex exponentials are the only eigenfunctions in the most general case, i.e. for all possible LTI systems. $\endgroup$
    – Matt L.
    Nov 11, 2023 at 11:52

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