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Is there an example of an eigenfunction of a linear time invariant (LTI) system that is not a complex exponential? Justin Romberg's Eigenfunctions of LTI Systems says such eigenfuctions do exist, but I am not able to find one.

Rodrigo de Azevedo
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Vinod
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4 Answers4

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All eigenfunctions of an LTI system can be described in terms of complex exponentials, and complex exponentials form a complete basis of the signal space. However, if you have a system that is degenerate, meaning you have eigensubspaces of dimension >1, then the eigenvectors to the corresponding eigenvalue are all linear combination of vectors from the subspace. And linear combinations of complex exponentials of different frequencies are not complex exponentials anymore.

Very simple example: The identity operator 1 as an LTI system has the whole signal space as eigensubspace with eigenvalue 1. That implies ALL functions are eigenfunctions.

Jazzmaniac
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For any arbitrary LTI sytem, the complex exponential is, to the best of my knowledge, the only known eigensignal. On the other hand, consider the ideal LPF. The $\operatorname{sinc}$ function: $$\operatorname{sinc}(t) \triangleq \frac{\sin(\pi t)}{\pi t}$$ can easily be seen to be an eigen signal. This points to the existence of LTI systems (such as the ideal LPF) having signals other than complex exponentials as eigen signals ($\frac{\sin(\pi t)}{\pi t}$ in this case).

Florent
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CSR
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    It's rather the opposite: The rule is that LTI systems do have degenerate eigensubspaces and therefore eigenvectors that are not complex exponentials. Consider a system with real output. Then $H(\omega)=H^*(-\omega)$, which means that if $H(\omega)$ is real and $\omega\neq 0$, then you already have a two dimensional eigensubspace and the real sine is an eigenvector. That means any LTI system that has a phase response that becomes a multiple of $\pi$ for $\omega\neq 0$ qualifies. That is the rule rather than the exception. – Jazzmaniac Aug 01 '15 at 12:04
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    actually, any pure exponential is an eigenfunction to an LTI system. if you don't mind dealing with quantities rapidly approaching $\infty$, then there is no theoretical requirement for the exponential to be complex or real. – robert bristow-johnson Aug 01 '15 at 21:23
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    i know i edited your answer (to make it more clear and more correct with the semantics), but your answer is mistaken. $$\operatorname{sinc}(t) \triangleq \frac{\sin(\pi t)}{\pi t}$$ is not a general eigenfunction to a general LTI system. it is an eigenfunction for specific LTIs that have $H(f)=1 \quad \forall |f|<\frac{1}{2}$ but not for others. – robert bristow-johnson Aug 01 '15 at 21:31
  • @robertbristow-johnson, ordinary exponentials are not eigenfunctions, because they are not contained in the signal space that is usually considered. That spaced would be the rigged Hilbert space of square integrable functions. The complex exponentials exist in the dual so that they make valid basis functions and can still be considered. The pure exponentials do not have this property. – Jazzmaniac Aug 01 '15 at 22:15
  • Also CSR did clearly state that he considered an ideal low pass filter as the system for the sinc eigenfunction. So your comment about him being mistaken is rather odd. – Jazzmaniac Aug 01 '15 at 22:19
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    evidently "if you don't mind dealing with quantities rapidly approaching ∞" is not the same as "the signal space that is usually considered ... the rigged Hilbert space of square integrable functions." all's i'm saying is that if $$x(t) = e^{st}$$ is your input, then $$y(t) = H(s) \cdot x(t)$$ is your output (where $H(s)$ is the Laplace transform of the LTI impulse response $h(t)$). looks like an eigenfunction to me. but you're right about CSR's specification. – robert bristow-johnson Aug 02 '15 at 19:22
  • @robertbristow-johnson just to make it clear; complex exponentials $e^{st}$ and $z^n$ are eigenfunctions of LTI systems... Just as the Laplace (or $\mathcal{Z}$) transforms exist for signals like $u(t)$, $r(t)$, $\sin(wt)u(t)$, $t^n u(t)$ etc, that are not integrable and/or not bounded. According to Jazzmaniac's criterion then we should discard all these functions, which are not Hilbert L2, from our toolboxes? Yet the engineering books on control theory, communications, circuit theory and signal processing are all based on them? – Fat32 Aug 24 '17 at 20:13
  • @Fat32, you know that Jazz and i don't always agree on things. the claim i made above is dependent only on the convolution integral for continuous-time LTI or the convolution summation for discrete-time LTI systems. and that relationship can be derived from just the fundamental definition of LTI. – robert bristow-johnson Aug 25 '17 at 00:52
  • @robertbristow-johnson It' more than an egreement issue I think. Jazz unnecessarily imposes the constraint on the signals to be absolutely (or square) integrable (members of some L2 space, stable systems and signals) for eigen analysis. Eventhough stable systems and signals are definetely an important class (just as causal ones), however the definition of an eigenfunction does not rely on L2 (stability), it can be any function space. $e^{st}$ and $z^n$ are eigenfunctions. – Fat32 Aug 25 '17 at 08:55
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    @Fat32, demanding a well behaved function space is not about stability and it's far from unnecessary or arbitrary. Most of the useful results in signal processing theory rely on well behaved signal spaces. Especially useful is the spectral theorem (https://en.wikipedia.org/wiki/Spectral_theorem), and this theorem requires certain function spaces, of which $L^2$ is a possible choice. If you want to apply this mathematical framework (and trust me, you want to), then you cannot accept the signals you propose as eigensignals. – Jazzmaniac Aug 25 '17 at 12:02
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    Oh hi @Jazzmaniac ! stability is a very useful and necessary property not only from a practical point of view (unstable systems are generally useless), but also from a rigorous math point of view (as convergent integrals are necessary for analysis etc). But the point is you don't need to imply L2 spaces to show that $e^{st}$ for any $s = \sigma + j \omega$ is an Efunction of LTI systems. You're right if you consider the subspace of L2 functions. Then what if you consider the subspace of causal functions? Then even $e^{jwt}$ will not be an eigenfunction of LTI systems? – Fat32 Aug 25 '17 at 13:46
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    @Fat32, you're missing the point. This is not about in the least about how useful stability is or what it means practically or theoretically. This is about mathematical theorems that rely on the structure of an operation in a certain space. The spectral theorem is the foundation of many important theorems in signal processing. Without it, you can't properly use Fourier analysis or even the convolution theorem. LTI don't diagonalise properly in the space you're proposing. So this is not a choice that you're free to make, it's one imposed onto you by the mathematical tools you want to use! – Jazzmaniac Aug 25 '17 at 14:00
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    @Jazzmaniac ok you are right. I'm probably missing a point. But please keep in mind that in many engineering textbooks on signal processing, (including those of Oppenheim, Proakis, Papoulis, Lathi, Haykin, Lim, etc...) the signal $e^{st}$ is shown to be an eigensignal of LTI systems without a mention of L2 spaces etc. Then probably they are also missing the point you mention, and I'm quite proud of missing the same point with those legends :-) ... – Fat32 Aug 25 '17 at 14:08
  • @Fat32, you'll have to provide actual references with this claim. I've quickly checked Oppenheim's Discrete-Time Signal Processing, and he certainly does not. Neither does the Proakis text I have available, or Brandenburg. In any case, of course you can find texts that make such a statement if they (implicitly) assume a sufficiently large function space. But then nothing useful follows from that statement. All useful consequences of a signal being an Eigensignal require the spectral theorem or a close relative. That's why nobody in signal theory would even start with those function spaces. – Jazzmaniac Aug 25 '17 at 14:21
  • @Jazzmaniac Please look at Oppenheim's Signals and Systems 2ed Ch3 sec 3.2 "Response of LTI systems to complex exponentials" page 182 for $e^{st}$ and $z^n$ being eigenfunctions . I will be looking for other references for explicit referencing. (Oppenheim's Discrete Time signal processing will not involve $e^{st}$ understandably) – Fat32 Aug 25 '17 at 14:34
  • @Jazzmaniac another reference: Signals and Systems_Bern Girod_Wiley ch3, section 3.2.2 "Eigenfunctions of LTI systems" shows $e^{st}$ as an eigenfunction. Btw Haykin's book Signals and Systems only deals with $e^{jwt}$ as eigensignals (I will scan his other books). I'll continue scanning my signal processing and communication books for the claim. – Fat32 Aug 25 '17 at 14:50
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    @Jazzmaniac for your convenience Oppenheim's Discrete-Time Signal Processing: ch1, problem 2.13 states $\alpha^n$ as eigenfunction of LTI systems. Also 2D Signal and Image Processing_JaeS Lim ch1, page 6 defines $\alpha^{n_1} \beta^{n_2}$ as a 2D eigenfunction of 2D LSI systems. Papoulis Signal Analysis ch1 page 10 eq 1-16 defines $L{z^n} = z^n H(z)$ implying that $z^n$ is an eigenfunction. However I could not find a reference from Proakis and Haykin or Lathi, they all use the sibling $e^{j w n}$ for defining the eigenfunctions. Anyway, I hope this much of reference is ok. – Fat32 Aug 25 '17 at 22:30
  • @Jazzmaniac I agree with you on the fact that practically useful point of view do not concern the signals of the form $e^{st}$ or $\alpha^n$... Therefore there is no need to discuss it further. However they are (possibly useless) eigenfunctions of LTI systems as they satisfy the relationship on their related spaces. – Fat32 Aug 25 '17 at 22:35
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I thought I had worded my response clearly---apparently not :-). The original question was, "Are there eigensignals besides the complex exponential for an LTI system?". The answer is, if one is given the fact that the system is LTI but nothing else is known, then the only confirmed eignensignal is the complex exponential. In specific cases, the system may have additional eigensignals as well. The example I gave was the ideal LPF with sinc being such an eigensignal. Note that the sinc function is not an eigensignal of an arbitrary LTI system. I gave the LPF and the sinc as an example to point a non-trivial case---x(t) = y(t) will satisfy a mathematician but not an engineer :->. I am sure one can come up with other specific non-trivial examples that have other signals as eigensignals besides the complex exponential. But these other eigensignals will work for those specific examples only.

Also, cos and sin are not, in general, eigensignals. If cos(wt) is applied and the output is A cos(wt + theta), then this output cannot be expressed as a constant times the input (except when theta is 0 or pi, or A=0), which is the condition needed for a signal to be an eigensignal. There may be conditions under which cos and sin are eigensignals, but they are special cases and not general.

CSR

CSR
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    Are you sure you understood my comment to your other answer? The point is that for real LTI systems it is expected to have a real sine as eigensignal. That doesn't mean that all sines of all frequencies are eigensignals. I specifically gave the precise condition for which they are such, and explained why that condition is met by most LTI systems. – Jazzmaniac Aug 02 '15 at 13:27
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    Also, don't forget that you edited your answer to change the meaning quite a bit. The step from "For a rational transfer function there are no other eigensignals" to "For arbitrary systems there are no general eigen signals besides .." is quite big. So putting it like people did not understand your response correctly is a bit much. – Jazzmaniac Aug 02 '15 at 13:31
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Maybe spatially invariant multidimensional objects like lenses with circular symmetry. It is called the Fourier Bessel expansion. There is no T for time but the convolution frequency domain relations hold