Are complex exponentials the only eigenfunction for arbitrary LTI systems?

Question

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.

Answer 1

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 $x(t-\tau)=x(t)x(-\tau)$.