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I know that complex exponentials are eigen functions of LTI systems for example $e^{j2t}, e^{-j5t} , e^{j8t}$ .

If we can define complex exponential as $e^{st}$ where $s$ is a complex number. Can we say that $e^t,e^{2t},e^{(2-j4)t}$ are still complex exponentials and they are eigenfunctions to LTI systems ?

I know that complex number includes real numbers and $1$ , $2$ and $2-j4$ are indeed complex numbers. What I'm curious about is , when $2$ or $2-j4$ is exponent i.e. $e^{2t},e^{(2-j4)t}$ are eigenfunctions to LTI systems?

Thanks for be interested.

HKTS
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  • I think it depends on how you define "eigenfunction". If you're allowed to "preload" the states of the system at some arbitrary point in time, then yes, $h(t) * e^{st} = \lambda e^{s t}$ should be true for all time (note that this is necessary for $s = j \omega$, too -- it's just that if you start with a stable system and a sine wave that starts at $t = -\infty$ then all the transients have died off by the time you get to any finite $t$). What I don't know is if the dodge of "preloading" the states is acceptable to mathematicans. – TimWescott Jan 13 '23 at 20:41
  • I define eigenfunction here as input that produces output as input times a constant number lambda(transfer function for this case) in s domain. $Y(S)=λX(S)$ where X is input and Y is output. I know that when $s=jw$ it is holds. I ask $S=a+jw$ where a has non-zero value , can this still causes constant transfer function ? – HKTS Jan 13 '23 at 20:53
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    Does this answer your question? Are complex exponentials the only eigenfunctions of LTI systems? – Matt L. Jan 13 '23 at 21:57
  • It seems to me that you already answered your question yourself. – Matt L. Jan 13 '23 at 21:59

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