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for $\sin(2 \pi t)$: Apparently that it's not an eigenfunction real-valued impulse response $h(t)$ but it's a eigenfunction for real-valued and even impulse response $h(t)$

What gives?

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Here's what gives: the actual eigenfunctions are those of the form $e^{i\omega t}$. In particular, we have $$ h(t)*e^{i\omega t} = \int_{-\infty}^\infty h(\tau)e^{i\omega (t-\tau)}\,d\tau = \left[\int_{-\infty}^\infty h(\tau)e^{-i\omega \tau}\,d\tau\right]e^{i\omega t} := H(\omega)e^{i\omega t} $$ It follows that $$ h(t)*\cos(\omega t) = h(t) * \frac 12 \left[e^{i \omega t} + e^{-i \omega t} \right] = \frac12 \left[H(\omega)e^{i \omega t} + H(-\omega)e^{-i \omega t}\right] $$ This will be a multiple of the original function exactly when $H(\omega) = H(-\omega)$. So, as long as $H(2 \pi) = H(- 2 \pi)$, $\cos(2 \pi t)$ will be an eigenfunction.

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