(RESPONSE OF SYSTEM TO SIGNAL DEFINED IN QUESTION 1)
(RESPONSE OF SYSTEM TO SIGNAL DEFINED IN QUESTION 1)
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Now, we take the product of the each component of the input function h(t) with the system function H(s):
 
Now, we take the product of the each component of the input function h(t) with the system function H(s):
 
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<math>f(t) = (3+j)\frac{e^{2jt}}{2}(\frac{7}{3} + \frac{9e^{-16j}}{3}) + (3+j)\frac{e^{-2jt}}{2}(\frac{7}{3} + \frac{9e^{16j}}{3}) + (10+j)\frac{e^{7jt}}{2j}(\frac{7}{3} + \frac{9e^{-56j}}{3}) - (10+j)\frac{e^{-7jt}}{2j}(\frac{7}{3} + \frac{9e^{56j}}{3})\!</math>
 
<math>f(t) = (3+j)\frac{e^{2jt}}{2}(\frac{7}{3} + \frac{9e^{-16j}}{3}) + (3+j)\frac{e^{-2jt}}{2}(\frac{7}{3} + \frac{9e^{16j}}{3}) + (10+j)\frac{e^{7jt}}{2j}(\frac{7}{3} + \frac{9e^{-56j}}{3}) - (10+j)\frac{e^{-7jt}}{2j}(\frac{7}{3} + \frac{9e^{56j}}{3})\!</math>
 
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Revision as of 11:56, 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, we take the product of the each component of the input function h(t) with the system function H(s):
$ f(t) = (3+j)\frac{e^{2jt}}{2}(\frac{7}{3} + \frac{9e^{-16j}}{3}) + (3+j)\frac{e^{-2jt}}{2}(\frac{7}{3} + \frac{9e^{16j}}{3}) + (10+j)\frac{e^{7jt}}{2j}(\frac{7}{3} + \frac{9e^{-56j}}{3}) - (10+j)\frac{e^{-7jt}}{2j}(\frac{7}{3} + \frac{9e^{56j}}{3})\! $


When the above expression is simplified, we obtain:


UNDER CONSTRUCTION

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood