Exponentials as eigenfunctions in LTI Systems

[Tanja]

I hope I catched the correct forum.

Under http://en.wikipedia.org/wiki/LTI_system_theory
e^{s t} is an eigenvalue.
I don't really understand that the following is an eigenvalue equation:
\quad = e^{s t} \int_{-\infty}^{\infty} h(\tau) \, e^{-s \tau} \, d \tau
\quad = e^{s t} H(s),

where

H(s) = \int_{-\infty}^\infty h(t) e^{-s t} d t .

The integrals H(s) and \int_{-\infty}^{\infty} h(\tau) \, e^{-s \tau} \, d \tau should be unique, isn't it? So there is only one solution for H(s) = \int_{-\infty}^{\infty} h(\tau) \, e^{-s \tau} \, d \tau with one e^{s t}, isn't it?

Thanks
Tanja

 

[EnumaElish]

e^{s t} is an eigenvalue.
I don't really understand that the following is an eigenvalue equation:
\quad = e^{s t} \int_{-\infty}^{\infty} h(\tau) \, e^{-s \tau} \, d \tau
\quad = e^{s t} H(s),

where

H(s) = \int_{-\infty}^\infty h(t) e^{-s t} d t.

The integrals H(s) and \int_{-\infty}^{\infty} h(\tau) \, e^{-s \tau} \, d \tau should be unique, isn't it? So there is only one solution for H(s) = \int_{-\infty}^{\infty} h(\tau) \, e^{-s \tau} \, d \tau with one e^{s t}, isn't it?

F(x) = \int_a^xf(t)dt is unique by the fundamental theorem of calculus.