(New page: Let us use the CT LTI system: <math>y(t) = x(3t) + x(t+3)</math> ---- The impulse response, h(t), of this system is computed using the following: <math>x(t) = \delta (t)</math> <math>...) |
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Let us use the CT LTI system: | Let us use the CT LTI system: | ||
− | <math>y(t) = | + | <math>y(t) = 3x(t) + 7x(t+3)</math> |
---- | ---- | ||
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<math>x(t) = \delta (t)</math> | <math>x(t) = \delta (t)</math> | ||
− | <math>h(t) = \delta ( | + | <math>h(t) = 3\delta (t) + 7\delta (t+3)</math> |
+ | |||
+ | The system function, H(s) is: | ||
+ | |||
+ | <math>H(s) = \int^{\infty}_{- \infty} h(t) e^{- s t} dt</math> | ||
+ | |||
+ | <math>H(s) = \int^{\infty}_{- \infty} [3\delta (t) + 7\delta (t+3)]e^{- s t}</math> | ||
+ | |||
+ | <math>H(s) = 3 e^{0} + 7 e^{- 3 s} </math> | ||
+ | |||
+ | <math>H(s) = 3 + 7 e^{- 3 s}</math> | ||
The signal used in question 1: | The signal used in question 1: | ||
<math>x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t}</math> | <math>x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t}</math> |
Revision as of 08:44, 26 September 2008
Let us use the CT LTI system:
$ y(t) = 3x(t) + 7x(t+3) $
The impulse response, h(t), of this system is computed using the following:
$ x(t) = \delta (t) $
$ h(t) = 3\delta (t) + 7\delta (t+3) $
The system function, H(s) is:
$ H(s) = \int^{\infty}_{- \infty} h(t) e^{- s t} dt $
$ H(s) = \int^{\infty}_{- \infty} [3\delta (t) + 7\delta (t+3)]e^{- s t} $
$ H(s) = 3 e^{0} + 7 e^{- 3 s} $
$ H(s) = 3 + 7 e^{- 3 s} $
The signal used in question 1:
$ x(t) = 3cos(4\pi t) + e^{j\frac{2\pi}{5}t} $