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| − | Now, | + | Now, we take the product of the each component of the input function h(t) with the system function H(s), which give us the final answer: |
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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> | |
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| − | <math>f(t) = (3+j)\frac{e^{2jt}}{2} | + | |
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Latest revision as of 08:31, 26 September 2008
Contents
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), which give us the final answer:
$ 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})\! $
