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Aug 22, 2023 at 1:58 comment added mhdadk (+1) this is an excellent answer, and the equation $H[e^{\jmath \omega t}] = A e^{\jmath \phi} e^{\jmath \omega t}$ reminds me a lot of the eigenvalue-eigenvector equation $Qv = \lambda v$. We can see that $Q$, the linear operator, is analogous to $H$, $v$, the eigenvector, is analogous to $e^{j\omega t}$, and $\lambda$, the eigenvalue, is analogous to $A e^{\jmath \phi}$.
Apr 13, 2017 at 12:47 history edited CommunityBot
replaced http://dsp.stackexchange.com/ with https://dsp.stackexchange.com/
Feb 15, 2016 at 20:11 comment added Fat32 @Peter K. I think that following the philosophy of choice on the (academic) correctness over "popularity" of an answer, your answer should be be integrated into the above answer provideded by Lorem Ipsum, which despite being selected as the answer with 96 points by the users, lacks this very important point of view.
Dec 8, 2015 at 18:02 history edited Peter K. CC BY-SA 3.0
Changed user name!
Aug 25, 2011 at 20:38 history answered Peter K. CC BY-SA 3.0