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<math>y(t) = \int^{\infty}_{-\infty} h(t) * x(t) dt\,</math> where <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math>
 
<math>y(t) = \int^{\infty}_{-\infty} h(t) * x(t) dt\,</math> where <math>x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})</math>
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<math>y(t) = \int^{\infty}_{-\infty} K \delta(t) * (1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}))</math>

Revision as of 11:12, 25 September 2008

LTI System: $ y(t) = Kx(t)\, $ where K is a constant

Unit Impulse Response: $ h(t) = K \delta(t)\, $

Frequency Response:

$ y(t) = \int^{\infty}_{-\infty} h(t) * x(t) dt\, $ where $ x(t) = 1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4}) $

$ y(t) = \int^{\infty}_{-\infty} K \delta(t) * (1+\sin \omega_0 t + \cos(2\omega_0 t+ \frac{\pi}{4})) $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang