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Whenever I try to find the derivation of eigen signal of a LTI system, they always start with assumption that the input signal is complex exponential. Is there any other way to prove that this is the case starting from the basic equation of LTI system $$\int_{-\infty}^{\infty}h(\tau)x(t-\tau)d\tau = \lambda x(t)$$ where $\lambda$ is the eigenvalue of the eigen signal

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    $\begingroup$ Any complex exponential is an eigenfunction of every LTI system. What exactly do you want to prove? $\endgroup$ – Rodrigo de Azevedo Feb 27 at 16:24
  • $\begingroup$ I think the question asks how can we arrive to the conclusion that any complex exponential is an eigenfunction of every LTI system without knowing it a priori. $\endgroup$ – anpar Feb 28 at 14:51
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I think the question asks how can we arrive to the conclusion that any complex exponential is an eigenfunction of every LTI system without knowing it a priori, so here is an attempt starting from the basic equation of LTI systems, as requested.


Let's apply the convolution theorem to $$\int_{-\infty}^{\infty} h(\tau)x(t-\tau) \mathrm{d}\tau = \lambda x(t)$$ to obtain $$H(\omega) X(\omega) = \lambda X(\omega)$$ where $H(\omega)$ and $X(\omega)$ are the Fourier transforms of $h(t)$ and $x(t)$ respectively and $\lambda \neq 0$ is a constant.

This last equation is supposed to be valid for all $\omega$ and for every LTI system, that is for any $H(\omega)$. The question is then: for which $X(\omega)$ is this equation valid for all $\omega$ and for any $H(\omega)$ ?

To answer that, let's first partition the set of all possible values of $\omega$ in two disjoint sets: $\Omega_0 = \{w~|~X(\omega) = 0 \}$ and $\Omega_1 = \{w~|~X(\omega) \neq 0 \}$. Clearly, the above equation is trivially satisfied for all $\omega \in \Omega_0$ (as in this case, the above equation leads to $0 = 0$). Next, for all $\omega \in \Omega_1$, we can write $$H(\omega) = \lambda.$$ Now let's assume $\Omega_1$ contains two distinct frequencies $\omega_1 \neq \omega_2$, the last equation would imply that $H(\omega_1) = \lambda$ and $H(\omega_2) = \lambda$, or $H(\omega_1) = H(\omega_2$) which clearly can only be true for any $H(\omega)$ if $\omega_1 = \omega_2$. It thus means that $\Omega_1$ can only contains a single frequency $w_0$, that is, $X(\omega)$ equals 0 everywhere except in $w_0$, or $$X(\omega) = 2\pi A\delta(w - w_0)$$ for any $A \neq 0$. Going back in the time domain, this corresponds to $$x(t) = Ae^{j\omega_0 t}.$$ We also have that the eigenvalue $\lambda$ is given by $H(\omega_0)$.

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You can use the Convolution Theorem to show that $ \int_{-\infty}^{\infty} h \left( \tau \right) x \left( t - \tau \right) d \tau $ is equivalent to Element Wise multiplication in the Frequency Domain.

Hence for the case $ x \left( t \right) = A {e}^{s t} $ where $ s \in \mathbb{C} $ the output is just a multiplication by scalar of the input (The value of the System Transfer Function at the Delta point of the input).

See Eigen Functions of LTI Systems.

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