(New page: == A CT-LTI System == <math>y(t) = 10x(t)\,</math> == Unit Impulse == <math>x(t) = \delta(t)\,</math> <math>h(t) = 10 \delta(t)\,</math> == Frequency Response == <math>y(t) = \int^{\in...) |
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Revision as of 18:50, 24 September 2008
Contents
A CT-LTI System
$ y(t) = 10x(t)\, $
Unit Impulse
$ x(t) = \delta(t)\, $
$ h(t) = 10 \delta(t)\, $
Frequency Response
$ y(t) = \int^{\infty}_{-\infty} h(t) * x(t) dt\, $ where $ x(t) = e^{jwt} \, $
$ y(t) = \int^{\infty}_{-\infty} 10 \delta(t) * e^{jwt} dt\, $
$ y(t) = \int^{\infty}_{-\infty} 10 \delta(r) e^{jw(t-r} dr\, $
$ y(t) = e^{jwt} \int^{\infty}_{-\infty} 10 \delta(r) e^{-jwr} dr\, $
$ H(s) = \int^{\infty}_{-\infty} 10 \delta(r) e^{-jwr} dr\, $
$ H(s) = 10 e^{-jw0}\, $
$ H(s) = 10\, $
Response of the CT system defiend in Q1
CT Periodic Signal : $ x(t) = \cos(3\pi t) + \sin(4\pi t)\, $
$ x(t) = \cos(3\pi t) + \sin(4\pi t)\, $
$ x(t) = \frac{e^{3\pi jt} + e^{-3\pi jt} }{2} + \frac{e^{4\pi jt} - e^{-4\pi jt} }{2j}\, $
$ x(t) = \frac{10e^{3\pi jt} + 10e^{-3\pi jt} }{2} + \frac{10e^{4\pi jt} - 10e^{-4\pi jt} }{2j}\, $
$ x(t) = 10\frac{e^{3\pi jt} + e^{-3\pi jt} }{2} + 10\frac{e^{4\pi jt} - e^{-4\pi jt} }{2j}\, $
$ x(t) = 10\cos(3\pi t) + 10\sin(4\pi t)\, $
