Hopefully you are familiar with the notion of the eigenvectors of a "matrix system," if not they do a quick review of eigen-stuff. We can develop the same ideas for LTI systems acting on signals. A linear time invariant (LTI) system $\mathscr{H}$ operating on a continuous input $f\left(t\right)$ to produce continuous time output $y\left(t\right)$

$$\mathscr{H}\left[f\left(t\right)\right]=y\left(t\right)$$

1

is mathematically analogous to an $N$x$N$ matrix $A$ operating on a vector $x\in {\u2102}^{N}$ to produce another vector $b\in {\u2102}^{N}$ (see Matrices and LTI Systems for an overview).

$$Ax=b$$

2

Just as an eigenvector of $A$ is a $v\in {\u2102}^{N}$ such that $Av=\lambda v$, $\lambda \in \mathbb{C}$,

we can define an eigenfunction (or eigensignal) of an LTI system $\mathscr{H}$ to be a signal $f\left(t\right)$ such that
$$\forall \lambda ,\lambda \in \mathbb{C}:\left(\mathscr{H}\left[f\left(t\right)\right]=\lambda f\left(t\right)\right)$$

3

Eigenfunctions are the **simplest** possible
signals for
$\mathscr{H}$
to operate on: to calculate the output, we simply multiply the
input by a complex number
$\lambda $.

The class of LTI systems has a set of eigenfunctions in common:
the
complex exponentials
${e}^{st}$,
$s\in \mathbb{C}$
are eigenfunctions for **all** LTI systems.

$$\mathscr{H}\left[{e}^{st}\right]={\lambda}_{s}{e}^{st}$$

4

Note:

While
$\left\{\forall s,s\in \mathbb{C}:\left({e}^{st}\right)\right\}$
are always eigenfunctions of an LTI system, they are not
necessarily the We can prove Equation 4 by expressing the output as a convolution of the input ${e}^{st}$ and the impulse response $h\left(t\right)$ of $\mathscr{H}$:

$$\begin{array}{rcl}\hfill \mathscr{H}\left[{e}^{st}\right]& \hfill =\hfill & {\int}_{-\infty}^{\infty}h\left(\tau \right){e}^{s(t-\tau )}d\tau \hfill \\ \hfill & \hfill =\hfill & {\int}_{-\infty}^{\infty}h\left(\tau \right){e}^{st}{e}^{-(s\tau )}d\tau \hfill \\ \hfill & \hfill =\hfill & {e}^{st}{\int}_{-\infty}^{\infty}h\left(\tau \right){e}^{-(s\tau )}d\tau \hfill \end{array}$$

5

$$\mathscr{H}\left[{e}^{st}\right]={\lambda}_{s}{e}^{st}$$

6

Since the action of an LTI operator on its eigenfunctions ${e}^{st}$ is easy to calculate and interpret, it is convenient to represent an arbitrary signal $f\left(t\right)$ as a linear combination of complex exponentials. The Fourier series gives us this representation for periodic continuous time signals, while the (slightly more complicated) Fourier transform lets us expand arbitrary continuous time signals.