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<math>H(s)= \int_{-\infty}^{\infty}h(T){e^{-sT}}\, dT</math> | <math>H(s)= \int_{-\infty}^{\infty}h(T){e^{-sT}}\, dT</math> | ||
+ | |||
+ | lets assume that: | ||
+ | |||
+ | <math>y(t)= 3x(t)</math> | ||
+ | let <math>x(t)=</math> |
Revision as of 14:12, 26 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 $
= $ e^{st} \int_{-\infty}^{\infty}h(T){e^{-sT}}\, dt $
= $ H(s) e^{st} $
h(t) is the impulse response of the LTI SYSTEM H(s) is the system fuction
$ H(s)= \int_{-\infty}^{\infty}h(T){e^{-sT}}\, dT $
lets assume that:
$ y(t)= 3x(t) $ let $ x(t)= $