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LTI System: <math>y(t) = Kx(t)\,</math> where K is a constant | LTI System: <math>y(t) = Kx(t)\,</math> where K is a constant | ||
| − | Unit Impulse Response: <math>h(t) = K \delta(t) | + | Unit Impulse Response: <math>h(t) = K \delta(t)</math> |
Frequency Response: | Frequency Response: | ||
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then <math>y(t)=\sum^{\infty}_{k = -\infty}a_k*(h(t)*e^{jj\omega_0 t})</math> | then <math>y(t)=\sum^{\infty}_{k = -\infty}a_k*(h(t)*e^{jj\omega_0 t})</math> | ||
| − | <math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 | + | <math>H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 t} dt</math> by definition |
| + | |||
| + | <math>H(s) = \int^{\infty}_{-\infty} K \delta(r) e^{-jwr} dr</math> | ||
| + | |||
| + | <math>H(s) = K e^{-jw0}</math> | ||
| + | |||
| + | <math>H(s) = K</math> | ||
Revision as of 12:27, 25 September 2008
LTI System: $ y(t) = Kx(t)\, $ where K is a constant
Unit Impulse Response: $ h(t) = K \delta(t) $
Frequency Response:
$ x(t) = \sum^{\infty}_{k = -\infty} a_k e^{jk\pi t}\, $
then $ y(t)=\sum^{\infty}_{k = -\infty}a_k*(h(t)*e^{jj\omega_0 t}) $
$ H(s) = \int^{\infty}_{-\infty} h(t)e^{-j\omega_0 t} dt $ by definition
$ H(s) = \int^{\infty}_{-\infty} K \delta(r) e^{-jwr} dr $
$ H(s) = K e^{-jw0} $
$ H(s) = K $
