# DTFT and Eigenvalues in frequency domain

Consider an LTI system with impulse response $$h[k]$$. Does the frequency response $$H(e^{j\Omega})$$ equal the eigenvalue corresponding to an eigensignal of frequency $$\Omega$$?

So if I convolve an eigensignal $$x[k]$$ of the system with the impulse response $$h[k]$$, I get the $$H(e^{j\Omega})$$ (=Eigenvalue?) scaled eigensignal $$x[k]$$ as output $$y[k]$$, what does this mean for frequency response plots?

Does this mean, that graphs of the frequency response just show where the eigenvalues lay?

Say I have a maximum at $$\Omega = 0$$ with value 3 and one at $$\Omega = \pi$$ with value 2, are 3 and 2 eigenvalues of the system? Because the magnitude of the eigensignals itself should always be 1, it just gets scaled by the eigenvalue (=$$H(e^{j\Omega})$$). So are frequency response plots just basically "eigenvalue over frequency" plots? This seems to be the case.

Does this sentence hold true in general and for all (LTI) systems? Is the frequency response always "eigenvalues of a system over frequency"?

• Could you please define (1) eigenvalue of a signal; and (2) eigensignal? – MBaz Jun 1 at 17:20
• An eigensignal was defined in the lecture as a signal x[k] as input convolved with the impulse response gives a signal y[k] = a x[k] as output, with a as arbitrary constant. But now when i think about it, shouldnt a be the eigenvalue? – Eigenbrödler Jun 1 at 17:25
• OK, so just to clarify, the terminology comes from linear algebra. If you have a matrix $A$, vector $x \neq 0$, and scalar $\lambda$, then if $Ax=\lambda x$, $x$ is an eigenvector of $A$ and $\lambda$ is the associated eigenvalue. Important: $x$ is not an eigenvector by itself; it is an eigenvector of $A$. Likewise, systems have eigensignals; if $x(t) \ast h(t) = \lambda x(t)$, then $x(t)$ is an eigensignal of the system $h(t)$, and $\lambda$ is the associated eigenvalue. It does not make sense to say "$x(t)$ is an eigensignal" without specifying the system it is an eigensignal of. – MBaz Jun 1 at 17:36
• Given this standard terminology, please edit your question to clarify what you mean. – MBaz Jun 1 at 17:38
• Thank you, I (and someone else, thx) edited the question – Eigenbrödler Jun 1 at 18:09