Eigenfunctions of discrete-time LTI systems

Given a discrete-time signal x[n], how does one determine if it is an eigenfunction of a stable, discrete-time LTI system?

For example, consider

1. $5^nu[n]$
2. $2^nu[-n-1]$

Intuitively, second one seems to be an eigenfunction of an LTI system while first one does not. Since the second sequence is absolutely summable while first one is not, is this the criteria for deciding?

• @Jazzmaniac how can you pick a system "freely" and still consider it be "LTI" ? – Fat32 Mar 8 '16 at 22:09
• @Jazzmaniac All stable LTI systems are the "same" from the point of view of eigen-analysis. I dont think so a signal not an eigenfunction for one LTI can be so for another. But I'm not sure. Can you give a simple example ? – Fat32 Mar 8 '16 at 22:14
• @Jazzmaniac. That is a nice example, even though it doesn't yield a method for the generality of LTI systems. Nevertheless eigen analysis rely on the generality of stable LTI definition as it must? apply to all members and not a specific subset. However as you pointed out there can be (that I didn't really care until now) subsets of LTI system for which a signal can be eigenfunction while it is not for the general definition. Hence it can be quite useless although being correct. – Fat32 Mar 8 '16 at 22:26
• @Jazzmaniac What is the use of it? In the sense that complex exponential $e^{j\omega}$ is used to describe the output of "any" stable LTI by expanding "any" input into it. ("any" here refers to practically existing) I really don't know a use of it. May be only pure mathematics? – Fat32 Mar 8 '16 at 22:38
• @Fat32, trying to take "only" pure mathematics not personally ;). But seriously, there are many applications in signal theory. For example a Hilbert transformer has analytical and anti-analytical signals as eigenvectors. Or a perfect band/low-pass has bandlimited signals as eigensignals. There is a very rich and powerful theory behind this all, and it makes many arguments a lot simpler. – Jazzmaniac Mar 8 '16 at 22:45

i think the only family of functions that are eigenfunctions to LTI systems the sole exponential.

for continuous-time, if it's LTI:

$$y(t) = \int\limits_{-\infty}^{+\infty} h(u) \ x(t-u) \ du$$

the eigenfunction is $x(t) = A e^{st}$ and the output is

$$y(t) = H(s) \ x(t)$$

where

$$H(s) = \int\limits_{-\infty}^{+\infty} h(t) \ e^{-st} \ dt$$

for discrete-time LTI systems, it is similar

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

the eigenfunction is $x[n] = A z^n$ and the output is

$$y[n] = H(z) \ x[n]$$

where

$$H(z) = \sum\limits_{n=-\infty}^{+\infty} h[n] \ z^{-n}$$

often $s=j\Omega$ or $z=e^{j\omega}$ but it wouldn't have to be that.

• Write "to all" LTI systems to be specific. Without a quantification the statement at least vague. But even then you are giving a statement that does not answer the OP's question. Given a sequence like $1+(-1)^n$, is that an eigenfunction to a (even non-trivial) LTI system? Yes, it is, and it's not hard to construct such a system. But the function is not a complex exponential. – Jazzmaniac Mar 9 '16 at 11:29
• it's the sum of two systems, each with specific exponential driving function that passes with gain = 1 and that have zero response to the other's driving exponential. big deal. – robert bristow-johnson Mar 10 '16 at 0:16
• Its probably a typo but y(t)=H(s) x(t) seems incorrect mixing functions in x and s. Should be y(s)=H(s) x(s) in s domain, or y(t)=h(t)*x(t) in time domain. – bcperth Jul 20 at 10:32
• @bcperth , actually it's not a typo. if $x(t) = A e^{s_0 t}$ is input to a system with impulse response $h(t)$ and transfer function $$H(s) = \int\limits_{-\infty}^{+\infty} h(t) \ e^{-st} \ dt$$ then the output is $$y(t) = H(s_0) \ x(t)$$ $s_0 = \sigma_0 + j \omega_0$ is just a symbol of a complex number. – robert bristow-johnson Jul 21 at 3:20
• Yes I agree. The equations holds for a specific value of H(s) at s0 which evaluates to a single complex number. So y(t)=H(s0) x(t) holds where x(t) is an Eigenfunction. But obviously y(t)=H(s) x(t) does not hold for all general signals f(t) where f(t) is a linear combination of eigenfunctions. Your original post talks of s not s0, and I apologise for being pedantic, but the reflection helped my own understanding! – bcperth Jul 22 at 2:21

I'll try to answer your question "how does one determine if it is an eigenfunction of a stable, discrete-time LTI system?" for the type of sequences specified in your question. Let's consider the signal $x[n]=5^nu[n]$ and let $h[n]$ be the system's impulse response. If $x[n]$ is an eigenfunction of the system, then the output signal must be a scaled version of the input signal: $y[n]=c\cdot x[n]$

The convolution sum is

\begin{align}y[n]&=\sum_{k=-\infty}^{\infty}h[k]x[n-k]\\&=\sum_{k=-\infty}^{\infty}h[k]5^{n-k}u[n-k]\\&=5^n\sum_{k=-\infty}^{n}h[k]5^{-k}\tag{1}\end{align}

We can make the sum in $(1)$ zero for $n<0$ by requiring $h[n]$ to be causal, i.e., $h[n]=0$, $n<0$. With this requirement we get

$$y[n]=5^nu[n]\sum_{k=0}^{n}h[k]5^{-k}=x[n]\sum_{k=0}^{n}h[k]5^{-k}\tag{2}$$

This almost looks like the form we're looking for. However, note that the multiplicative term in $(2)$ is not constant but it depends on $n$. So $(2)$ shows that the given $x[n]$ is generally not an eigenfunction of an LTI system. Of course it is an eigenfunction of the trivial system with impulse response $c\cdot \delta[n]$, but any other sequence is so too.

You can use a similar argument to show that also the second signal in your question is not an eigenfunction of any LTI system other than the trivial system mentioned above.

In sum, if you're looking for functions/sequences that are eigenfunctions of all LTI systems, then you'll end up with complex exponentials, as pointed out in RBJ's answer. For special systems there are also other eigenfunctions, such as the ones pointed out by Jazzmaniac in the comments. E.g., any band-limited function is an eigenfunction of an ideally frequency-selective filter with a pass band extending over the frequency range of the input signal. But such ideally frequency-selective systems are not stable, so they're not in the category of systems you are looking for. Note that there are other stable LTI systems with specific eigenfunctions different from complex exponentials. But in general it is not possible to construct an LTI system for a given function such that this function becomes an eigenfunction of the system (if we exclude the trivial system $h[n]=c\cdot \delta[n]$ mentioned above). And this last sentence is also true for the two signals given in your question.

• @Jazzmaniac: I was referring to ideally frequency-selective filters, which are unstable (and also non-causal). Since the OP restricted his question to stable LTI system, I mentioned to him that ideally frequency-selective filters don't fit in his category. There's not much more to it, I don't "dismiss" anything. I did point out that specific systems can have other eigenfunctions, and one example were frequency-selective filters. I never said or meant to say that there are no others, stable or unstable. That's usually the meaning of "E.g., ...". – Matt L. Mar 9 '16 at 11:59
• @Jazzmaniac: I'm not really sure what's wrong here. No, there's nowhere any "personal choice" of mine involved here. The OP dismissed unstable systems, and I told him that one (simple) example of LTI system with other eigenfunctions than complex exponentials are ideally frequency-selective systems, but they're unstable (according to the BIBO-criterion, which every student knows, and which is used exclusively when talking about linear systems). Of course there are other examples, also stable ones (which was implicit, but which I now made explicit if you care to check my edited last paragraph). – Matt L. Mar 9 '16 at 12:26
• @Jazzmaniac: The choice of frequency-selective filters was obviously mine, but the reason was that every beginner knows them. Other more exotic choices may not be that illuminating for a student. And yes, I'm guilty of referring to the system $h[n]=c\cdot \delta[n]$ as "trivial", in the sense that any function is an eigenfunction, which is not very specific. – Matt L. Mar 9 '16 at 12:32
• @Jazzmaniac: I had mentioned the trivial case already, so I hoped that everybody was able to understand what was meant by the sentence starting with "In general ...". Of course you can add disclaimers each and ever time to be 100% correct and clear (?), but this approach has a tendency to clutter text, and make reading a pain rather than a pleasure. Anyway, for the sake of correctness I've added that obvious exception to my statement. – Matt L. Mar 9 '16 at 13:24
• @Jazzmaniac: Why don't you write up an answer discussing a general method? I would like to see how to construct a BIBO-stable LTI system with a perfectly band-limited eigenfunction (bandwidth > 0, not periodic); not that I necessarily doubt that it's possible, I just don't see it right away. – Matt L. Mar 9 '16 at 14:28