# Are all exponential functions eigensignals of LTI systems?

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!

• What is your guess? Complex exponential functions do not include real ones (except the constants ones) – Laurent Duval Nov 17 '20 at 18:11
• They don't? My guess was that real exponentials would be a special case of complex exponentials. If not, how can I proof that they are not eigensignals? – Phobos Nov 17 '20 at 18:35
• yes if $2^t$ is an input to LTI system with impulse response $h(t)$ then the output is $y(t) = H(\ln(2)) 2^t$, where $H(s)$ is the Laplace transform of $h(t)$. – Fat32 Nov 17 '20 at 18:38
• Though, this is a mathematical convenience, as the signal $2^t$ (or any other exponential whose magnitude is not unity) will be an unstable signal (more porperly infinite energy signal). But that does not concern the answer to the question, whether it obeys the eigen property or not. – Fat32 Nov 17 '20 at 18:41
• @Jazzmaniac: 1. good to have you back, at least participating in the comment section. 2. it seems that at least in the engineering literature it is commonly accepted, rightly or not, that general complex exponentials $e^{st}$, $s\in\mathbb{C}$, are considered as eigenfunctions of (all) LTI systems. You might not be inclined to write up an answer, but it would be good to have another, more mathematical viewpoint, explained in a way such that the ambitious engineer is able to understand. – Matt L. Nov 18 '20 at 15:31

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.

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.

• Well that would lead to the same solution just without the imaginary number in the exponential. So I would conclude it is also an eigensignal? The input is then scaled with the integral, given that h(t) is chosen such that the integral converges? – Phobos Nov 17 '20 at 19:35
• @Phinie: That sounds right. – Matt L. Nov 17 '20 at 20:18
• Yet, I am troubled by the possibility to use the Fourier transform convolution-to-product theorems for full exponentials – Laurent Duval Nov 18 '20 at 15:29
• @LaurentDuval: Sure, you can't, but for example, $\lambda=\int_{-\infty}^{\infty} h(t)c^{-t}dt$ exists for causal and stable $h(t)$ and $c>0$, and hence the output of the corresponding LTI system excited by $x(t)=c^t$ is $y(t)=\lambda c^t$. – Matt L. Nov 18 '20 at 15:37
• No problem with well-chosen" filters! – Laurent Duval Nov 18 '20 at 15:42

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

• Concerning your last sentence, if $\lambda=\int_{-\infty}^{\infty}h(t)c^{-t}dt$ exists, we would get $\mathcal{T}\{c^t\}=\lambda c^t$. Why wouldn't that make sense? – Matt L. Nov 18 '20 at 7:51
• I don't know if it's evident, I'm taking a naïve engineering viewpoint here. We always depend on the existence of that integral, even with the complex exponentials. Yet I do remember a discussion on this site where it was claimed that it is only complex exponentials, not real ones, that are eigenfunctions of LTI systems. – Matt L. Nov 18 '20 at 9:30
• There is a problem in the textbook "Digital Communication" by Lee and Messerschmitt (problem 2.16 in the admittedly dated 2nd edition) where we were asked to show that for any $z\in\mathbb{C}$, the function $z^t$ is an eigenfunction of any continuous-time LTI system. This claim would include the special case $z\in\mathbb{R}$. – Matt L. Nov 18 '20 at 12:49
• @MattL. There is a problem in Oppenheim's book(problem 3.61) which states "Although some LTI systems may have additional eigenfunctions, the complex exponentials are the only signals that are eigenfunctions of every LTI system." Complex exponentials, in this book, are defined as $e^{st}$ where $s \in \mathbb{C}$. So I think we can consider $e^{at}, \ a \in \mathbb{R}$ as eigenfunctions. – S.H.W Nov 18 '20 at 13:49
• @S.H.W: Yes, so also Oppenheim thinks that we don't require that the magnitude of the eigenfunction be equal to unity, i.e., he considers a broader class than just $e^{j\omega_0t}$, because $s$ can have a non-zero real-part, and it can even have a vanishing imaginary part, which leads to eigenfunctions of the form $c^t$ with $c\in\mathbb{R}$. – Matt L. Nov 18 '20 at 15:05