(THE SYSTEM FUNCTION)
(RESPONSE OF SYSTEM TO SIGNAL DEFINED IN QUESTION 1)
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<math>f(t) = (3+j)\frac{e^{2jt}}{2} + (3+j)\frac{e^{-2jt}}{2} + (10+j)\frac{e^{7jt}}{2j} - (10+j)\frac{e^{-7jt}}{2j}\!</math>
 
<math>f(t) = (3+j)\frac{e^{2jt}}{2} + (3+j)\frac{e^{-2jt}}{2} + (10+j)\frac{e^{7jt}}{2j} - (10+j)\frac{e^{-7jt}}{2j}\!</math>
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Now, all we have to do is multiply each term of the input signal f(t) by the system function H(s).  By doing so, we obtain the following:
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\frac{7}{3} + \frac{9e^{-8jw}}{3}\!</math>
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<math>f(t) = (3+j)\frac{e^{2jt}}{2}[\frac{7}{3} + \frac{9e^{-8jw}}{3}] + (3+j)\frac{e^{-2jt}}{2}[\frac{7}{3} + \frac{9e^{-8jw}}{3}] + (10+j)\frac{e^{7jt}}{2j}[\frac{7}{3} + \frac{9e^{-8jw}}{3}] - (10+j)\frac{e^{-7jt}}{2j}[\frac{7}{3} + \frac{9e^{-8jw}}{3}]\!</math>
 
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UNDER CONSTRUCTION
 
UNDER CONSTRUCTION

Revision as of 12:44, 25 September 2008

CT LTI SYSTEM

I chose the following continusous-time linear time invariant system:

$ f(t) = \frac{7x(t)}{3} + \frac{9x(t+8)}{2}\! $

UNIT IMPULSE RESPONSE OF SYSTEM

To find the unit impulse response of the system, we set $ x(t) = \delta(t)\! $. Then we obtain the following unit impulse response:


$ h(t) = \frac{7\delta(t)}{3} + \frac{9\delta(t+8)}{2}\! $


THE SYSTEM FUNCTION

In order to compute the system function H(s), we can simply take the laplace transform of the unit impulse response of the system. When we take the laplace transform, we find that $ H(s) = \frac{7}{3} + \frac{9e^{-8jw}}{3}\! $

RESPONSE OF SYSTEM TO SIGNAL DEFINED IN QUESTION 1

Signal used in question 1:
$ f(t) = (3+j)cos(2t) + (10+j)sin(7t)\! $


From question 1, we also know that:
$ f(t) = (3+j)\frac{e^{2jt}}{2} + (3+j)\frac{e^{-2jt}}{2} + (10+j)\frac{e^{7jt}}{2j} - (10+j)\frac{e^{-7jt}}{2j}\! $


Now, all we have to do is multiply each term of the input signal f(t) by the system function H(s). By doing so, we obtain the following:

\frac{7}{3} + \frac{9e^{-8jw}}{3}\!</math>


$ f(t) = (3+j)\frac{e^{2jt}}{2}[\frac{7}{3} + \frac{9e^{-8jw}}{3}] + (3+j)\frac{e^{-2jt}}{2}[\frac{7}{3} + \frac{9e^{-8jw}}{3}] + (10+j)\frac{e^{7jt}}{2j}[\frac{7}{3} + \frac{9e^{-8jw}}{3}] - (10+j)\frac{e^{-7jt}}{2j}[\frac{7}{3} + \frac{9e^{-8jw}}{3}]\! $


UNDER CONSTRUCTION

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva