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<math>y(t)= \int_{-\infty}^{\infty}h(T)x(t-T)\, dt </math> | <math>y(t)= \int_{-\infty}^{\infty}h(T)x(t-T)\, dt </math> | ||
| − | = <math> \int_{-\infty}^{\infty}h(T){e^ | + | = <math> \int_{-\infty}^{\infty}h(T){e^{s(t-T)}}\, dt </math> |
Revision as of 15:39, 23 September 2008
A Signal for which the output signal is constant times the input is referred as
an eigenfunction of the system,andthe amplitude is called the
system's eigenvalue'
let the input be
$ x(t)= e^{st} $
we can determine the output using convolution integral
$ y(t)= \int_{-\infty}^{\infty}h(T)x(t-T)\, dt $
= $ \int_{-\infty}^{\infty}h(T){e^{s(t-T)}}\, dt $
