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<math>h(t)=K\delta (t-a)</math> | <math>h(t)=K\delta (t-a)</math> | ||
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| + | <math>H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}=\int_{-\infty}^{\infty}\frac{1}{2}u(\tau)e^{-s\tau}</math> | ||
Revision as of 17:23, 26 September 2008
Part A
$ y(t) = K x(t-a) $
if $ x(t)=e^{jwt} $ was inputed to the system
$ y(t) = K e^{jw(t-a)} $
$ = K e^{-jwa}e^{jwt} $
eigen function is $ e^{-jwa} $
$ H(jw)=Ke^{-jwa} $
$ h(t)=K\delta (t-a) $
$ H(s)=\int_{-\infty}^{\infty}h(\tau)e^{-s\tau}=\int_{-\infty}^{\infty}\frac{1}{2}u(\tau)e^{-s\tau} $
