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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?

stillanoob
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  • Eigenfunctions of Stable LTI systems can be described as follows: if $x[n]$ is an eigenfunction then corresponding output $y[n]$ is given as $K x[n]$ where K is an eigenvalue corresponding to $x[n]$. To see if this holds apply the convolution sum to input to get the output $y[n]=\sum {h[k] x[n-k]}$ and see if you can simplify it into a form of $K x[n]$. Note that even though $K$ is a number, a simple constant, its expression can be complicated expression including summations due to the definition of the convolution. Therefore they are not. – Fat32 Mar 08 '16 at 21:09
  • Any signal can be made an eigenfunction of an LTI system if you can pick the system freely. So your question lacks context or a constraint to be meaningful. – Jazzmaniac Mar 08 '16 at 22:05
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    @Jazzmaniac how can you pick a system "freely" and still consider it be "LTI" ? – Fat32 Mar 08 '16 at 22:09
  • @Fat32, not sure I understand your question. Of course I meant picking an LTI system freely. – Jazzmaniac Mar 08 '16 at 22:10
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    @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 08 '16 at 22:14
  • @Fat32, all LTIs share the same complex exponential eigenbasis. However, you can have additional eigenvectors if you have degenerate subspaces. Within the subspaces you can mix eigenvectors and get a new eigenvector. The simple gain LTI system y[n] = g*x[n] is the extreme case. It's fully degenerate because all eigenvalues are g. Therefore any vector in the space the LTI acts on is an eigenvector with eigenvalue g. – Jazzmaniac Mar 08 '16 at 22:17
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    @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 08 '16 at 22:26
  • @Fat32, oh, it's not so useless. Only because the example was trivial, it doesn't mean that there are not more relevant examples. For LTI systems, the eigenvalues are in the image of the complex frequency response $H(\omega)$. For straight forward topological reasons, discrete systems with continuous frequency response and an a total frequency response phase difference that exceeds $2\pi$ always have at least one degenerate subspace. That means that nearly all relevant LTI systems have a whole subspace of eigenvectors that are not complex exponentials. – Jazzmaniac Mar 08 '16 at 22:34
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    @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 08 '16 at 22:38
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    @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 08 '16 at 22:45
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    @Jazzmaniac of course not take it so ;) I admit It must have applications in signal theory ;) But 1-This discussion adds nothing to the question being asked, whose answer relies on the simplest and pure definition of stable LTI system and not a specific subset. 2-Btw those Hilbert transformers and the ideal low/band pass system are not practically realizable, but I really do not want to go any further as i- it is not useful ii- I don't have the interest :)) – Fat32 Mar 08 '16 at 22:53
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    I think it has to do everything with the question asked! – Jazzmaniac Mar 08 '16 at 22:57
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    @Jazzmaniac so go ahead and please provide your answer based on your theory to show that those two signals are eigenfunctions of "any" stable LTI system. – Fat32 Mar 08 '16 at 23:00
  • @Fat32, where does the question say "any"? – Jazzmaniac Mar 08 '16 at 23:01
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    @Jazzmaniac you mean by your above comment that the question is asking for a "specific" LTI system or a specific "subset" of it ? Then which subset is it, whose description is not given other than a stable LTI? Cannot you see the complete contradiction of that statement ? – Fat32 Mar 08 '16 at 23:04
  • @Fat32, I think you are missing the point. The question does not specify the system, but it also does not require that all (or as you wrote "any") LTI systems match the criterium. That's why I asked for additional context and a clarification. As the question currently stands, both given signals are eigensignals to the simple gain LTI that I gave as an example. That is just to point out that in its current form, the question nearly meaningless. – Jazzmaniac Mar 08 '16 at 23:07
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    @Jazzmaniac given "the question". I say "these two signals are not eigenfunctions of stable LTI systems". Provide your answer please ? Don't change the question please, you are trying to fit the question into your answer ! – Fat32 Mar 08 '16 at 23:10
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    @Fat32, I'm not trying to change the question. The questions says "eigenfunction of a stable LTI", without specifying it. So yes, they are eigenvectors of a stable LTI, namely the gain LTI. They are not of others, but that's not what the OP asks for. – Jazzmaniac Mar 08 '16 at 23:12
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    @Jazzmaniac This is a "classic" question from undergraduate DSP curriculum and the "jargon" here should be read as the "generality of all stable LTI systems" and not a specific subset unless otherwise explicitly stated. You are imposing an unnecessary complexity into it. May be technically you are right, but "aim" of the question is as I described. And indeed would it be ever meaningful to ask such a question if its answer would depend on the choosen specific subset? Provided that no such subset is mentioned ? You are right if he chooses the change the question inline of your investigation... – Fat32 Mar 08 '16 at 23:19
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    @Fat32, I'm trying to answer the question that is asked, not the one that you believe has been asked. I asked for further clarification from the OP because the question as it is asked does not seem to be what the author intended. That's why I wrote a comment pointing out the problem and ask for a more precise formulation or a confirmation of the question as it stands. There is no need to argue about that and I am very surprised that you put so much effort into the attempt to prove my comment wrong instead of just waiting for the needed clarification. – Jazzmaniac Mar 08 '16 at 23:22
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    @Jazzmaniac I'm pretty sure that he will repeat what I said. I'm smelling a simple undergraduate question here. My complete experience argues that "someone who cannot find whether $5^n u[n]$ is an eigenfunction (to what so ever system) cannot possibly be asking for a topological exposition of degenerate subspaces as what you describe as an alternate answer."... I really have nothing to do with your comments... – Fat32 Mar 08 '16 at 23:30
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    @Fat32, that's your own opinion and you are entitled to it. Just like I am entitled to mine. So please, let's wait for the clarification. You are arguing with me just for the sake of not being wrong. I only pointed out that the question was not clear. Call me pedantic, but I prefer to be precise with these things. It's not your position to judge whether the question was meant in one way or another. – Jazzmaniac Mar 08 '16 at 23:36
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    @Jazzmaniac I understand your tendency of "formal mathematical rigor" which is a right tool used on this wrong occasion. You're ignoring the meta meaning of the purpose of asking a question, just like a computer would. Other than a typographical error could be, OP is aiming for an anwer which applies to as general as possible unless otherwise explicitly stated...You are technically right as the answer might change based on the specific subset, but lets see that it will not ;) – Fat32 Mar 08 '16 at 23:51

2 Answers2

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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.

robert bristow-johnson
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  • 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 09 '16 at 11:29
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    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 00:16
  • 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 '19 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 '19 at 03:20
  • 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 '19 at 02:21
  • Yes, it does @bcperth . "$s$" is not dependent on $t$. it's just a symbol of a complex number, $s=\sigma + j \omega$ in exactly the manner that $s_0$ is. – robert bristow-johnson Jul 22 '19 at 20:03
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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.

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Matt L.
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  • Ideality of a system is entirely unrelated to stability. It's related to causality. I don't see any reason for excluding ideal systems other than the willingness to give a specific answer. Also, all input signals with a sparse frequency domain representation give rise to a construction of a causal (and stable) LTI system. That's a large enough class to not dismiss, even if you require causality. – Jazzmaniac Mar 09 '16 at 11:26
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    @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 09 '16 at 11:59
  • Frequency selective filters are perfectly stable as a mathematical system. They are just the perfectly well defined projectors in the frequency basis. Numerical approximations of such systems turn out to be unstable, but that's a different story. Also, you don't need perfect frequency selection to create a frequency-contiguous eigensubspace. A frequency response that is constant over an open set is already sufficient. Your "E.g" listed only cases that you believe don't fit the stability criterium, so you were giving an example but dismissing that example in the same statement. – Jazzmaniac Mar 09 '16 at 12:11
  • Your selection of what you accept as a valid system and what not is purely based on your personal choice, which makes your answer strongly biased and incomplete. You don't give any method of testing if a given signal can be an eigenfunction of an LTI system, even if you ignore the, in your eyes, "trivial" case. – Jazzmaniac Mar 09 '16 at 12:14
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    @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 09 '16 at 12:32
  • For continuous systems and distributional spaces, the BIBO criterium is not really very useful. Linear systems have a very close relation to the L^2 norm. Signal theory often neglects discussions of criteria other than BIBO because the mathematics becomes more complicated and you need to understand functional analysis well enough to use them. A trivial example for why BIBO provokes problems is for example that the statement "If a system is stable on a certain space, it is also stable on all of this space's subspaces" is wrong. So my arguments always refer to rigged Hilbert spaces. – Jazzmaniac Mar 09 '16 at 12:39
  • If you dismiss the "trivial case", please say so. Your statement currently is "... in general it is not possible to construct an LTI system for a given [eigen]function ...". That's wrong. Insert "non-trivial" and you're right. However the answer would be much more useful and less biased if you quantified the "in general" instead of just implying that it's an exotic special case that is not interesting to the discussion of this problem. – Jazzmaniac Mar 09 '16 at 12:41
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    @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 09 '16 at 13:24
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    And yes, I still claim that "in general" the statement is true, even though there are infinitely many special cases that are exceptions. If the OP is really interested in those, he should probably ask another question. – Matt L. Mar 09 '16 at 13:25
  • "How do I determine if an integer solves a quadratic equation?" - "In general, integers don't solve quadratic equations. Complex numbers do." Perfectly correct answer, right? – Jazzmaniac Mar 09 '16 at 13:37
  • Still, you don't answer the OP's question "how to determine, if [a given signal] is an eigenfunction to a stable LTI". Or do you take the statement "in general it isn't" is an encouragement to just assume it is not? A good generally applicable test would be: If the frequency domain support of the signal is a proper subset of the set of all frequencies, then there is a non-trivial LTI system to which the signal is an eigensignal. Furthermore, if the complement of the support is not of measure zero, there is an LTI which is also BIBO stable. (L^2 stability is always given) – Jazzmaniac Mar 09 '16 at 14:01
  • Start with a constant frequency response. Then add a a BIBO stable frequency response with frequency support in the open set that complements the eigenband. A triangular response function would be a suitable choice. Any Lipschitz-continuous response will work. There's likely a weaker condition that also works, but I cannot tell without thinking more about it. – Jazzmaniac Mar 09 '16 at 15:24