Are all exponential functions eigensignals of LTI systems?

Question

I know that complex exponential functions are eigensignals to LTI systems. Do these include real exponential functions? E.g. $2^t, e^t, ...$

Thanks for the help!

Answer 1

Mathematically (and theoretically), there is no need for the exponential function to be a complex sinusoid. The math is unchanged. The problem is that practical LTI systems are not boundless nor are they acausal. So setting aside those problems, every LTI system has input/output relationship described by the convolution integral (for continuous-time) or the convolution summation (for discrete-time).

$$ y(t) = \int\limits_{-\infty}^{\infty} h(u) \, x(t-u) \, \mathrm{d}u $$

$$ y[n] = \sum\limits_{m=-\infty}^{\infty} h[m] \, x[n-m] $$

Set the input to be an exponential function for all time, $t$ or $n$.

$$ x(t) \triangleq A e^{st} $$

$$ x[n] \triangleq A z^n $$

and then plug and chug:

$$\begin{align} y(t) &= \int\limits_{-\infty}^{\infty} h(u) \, x(t-u) \, \mathrm{d}u \\ &= \int\limits_{-\infty}^{\infty} h(u) \, A e^{s(t-u)} \, \mathrm{d}u \\ &= A e^{st} \int\limits_{-\infty}^{\infty} h(u) \, e^{-su} \, \mathrm{d}u \\ &= A e^{st} \ H(s) \\ &= x(t) \ H(s) \\ \end{align}$$ .

.

. $$\begin{align} y[n] &= \sum\limits_{m=-\infty}^{\infty} h[m] \, x[n-m] \\ &= \sum\limits_{m=-\infty}^{\infty} h[m] \, A z^{n-m} \\ &= A z^n \sum\limits_{m=-\infty}^{\infty} h[m] \, z^{-m} \\ &= A z^n \ H(z) \\ &= x[n] \ H(z) \\ \end{align}$$

So the eigenvalue is literally the "transfer function", $H(s)$ or $H(z)$, which is the Laplace Transform or the Z Transform of the impulse response $h(t)$ or $h[n]$ of the LTI system.

Answer 2

Complex exponentials are eigenfunctions of LTI systems because they are eigenfunctions of the convolution operator:

$$\begin{align}e^{j\omega_0t}\star h(t)&=\int_{-\infty}^{\infty}h(\tau)e^{j\omega_0(t-\tau)}d\tau\\&=e^{j\omega_0t}\int_{-\infty}^{\infty}h(\tau)e^{-j\omega_0\tau}d\tau\\&=e^{j\omega_0t}H(j\omega_0)\end{align}\tag{1}$$

where $h(t)$ is the system's impulse response and $H(j\omega_0)$ is its frequency response (evaluated at $\omega=\omega_0$).

Now try to prove in the same way that the convolution of an impulse response $h(t)$ with a real-valued exponential results in a scaled version of that exponential. Then draw your conclusions.

Answer 3

Complex exponential functions are most generally defined (up to a constant complex or real factor) as $t\mapsto e^{j\omega_0 t}$, $\omega_0\in \mathbb{R}$. Real exponentials are typically of the form $t\mapsto c^{ t}$, $c>0$. The latter are not a subset of the former: one reason is that complex exponentials have a modulus equal to one. This is not the case for real exponentials. These two families only coincide for $\omega_0 =0$ and $c=1$.

As reals numbers are a genuine subset of complex quantities, this may seem counter-intuitive. One can think about a combination (product) of the above, as a parametrized (by a complex index) family: $f_{a+jb}:t\mapsto e^{(a+jb) t}$. This super family is special: their derivatives are:

$$ f^k_{a+jb}(t) = (a+jb)^kf_{a+jb}(t)\,.$$

Hence, they are eigenfunctions of differential operators, which are linear and time-invariant. Real and complex exponentials can therefore be eigensignals of some LTI systems.

Now, complex exponentials are eigenfunctions of all LTI systems, within a proper definition of convolution. I don't know how to make sense of this for any real exponential. EDIT: I am probably wrong here. Several texts mentioned in the comments consider generic complex exponentials to be eigenfunctions. I still do find this a bit loose, cf. Eigenfunctions of Continuous Time LTI Systems:

Furthermore, the above discussion has been somewhat formally loose as $e^{st}$ may or may not belong to the space on which the system operates