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<math>H(s)=\int_{-\infty}^{\infty}3\delta\,\!(T)){e^{-j\omega\,\!T}}\, dT</math> | <math>H(s)=\int_{-\infty}^{\infty}3\delta\,\!(T)){e^{-j\omega\,\!T}}\, dT</math> | ||
| + | |||
| + | <math>H(s)=3{e^{-j\omega\,\!0}}</math> | ||
| + | |||
| + | <math>H(s)=3</math> | ||
Revision as of 14:33, 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)= \delta\,\!(t) $
$ h(t)= 3\delta\,\!(t) $
system function:
$ H(s)=\int_{-\infty}^{\infty}3\delta\,\!(T)){e^{-j\omega\,\!T}}\, dT $
$ H(s)=3{e^{-j\omega\,\!0}} $
$ H(s)=3 $
