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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.

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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.

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    $\begingroup$ Except the null function of course :) Just kidding $\endgroup$ – Laurent Duval Aug 24 '17 at 19:12
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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

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  • $\begingroup$ 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. $\endgroup$ – Jazzmaniac Aug 2 '15 at 13:27
  • $\begingroup$ 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. $\endgroup$ – Jazzmaniac Aug 2 '15 at 13:31
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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).

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    $\begingroup$ 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. $\endgroup$ – Jazzmaniac Aug 1 '15 at 12:04
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    $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Aug 1 '15 at 21:23
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    $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Aug 1 '15 at 21:31
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    $\begingroup$ 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. $\endgroup$ – robert bristow-johnson Aug 2 '15 at 19:22
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    $\begingroup$ @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 (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. $\endgroup$ – Jazzmaniac Aug 25 '17 at 12:02
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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

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