(→Basics of Linearity) |
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:<math>e^{2 x i}=t e^{-2 x i}\, </math> | :<math>e^{2 x i}=t e^{-2 x i}\, </math> | ||
:<math>e^{-2 x i}=t e^{2 x i}\, </math> | :<math>e^{-2 x i}=t e^{2 x i}\, </math> | ||
| + | : The Signal is Linear | ||
| + | |||
| + | :Since the system is linear you can split the signal in two parts | ||
| − | |||
:<math>\cos 2x = \dfrac{e^{2 i x}+e^{-2 i x}}{2}</math> | :<math>\cos 2x = \dfrac{e^{2 i x}+e^{-2 i x}}{2}</math> | ||
| Line 11: | Line 13: | ||
is | is | ||
<math>\ \dfrac{t e^{-2 i x} + t e^{2 i x}}{2} </math> | <math>\ \dfrac{t e^{-2 i x} + t e^{2 i x}}{2} </math> | ||
| + | but | ||
| + | :<math>e^{2 x i}=\cos 2x + i \sin 2x \, </math> and | ||
| + | :<math>e^{-2 x i}=\cos 2x - i \sin 2x \, </math> | ||
| + | :so the response is equal to | ||
| + | |||
| + | :<math>t\cos 2t \, </math> | ||
Latest revision as of 07:35, 18 September 2008
Basics of Linearity
Given
- $ e^{2 x i}=t e^{-2 x i}\, $
- $ e^{-2 x i}=t e^{2 x i}\, $
- The Signal is Linear
- Since the system is linear you can split the signal in two parts
- $ \cos 2x = \dfrac{e^{2 i x}+e^{-2 i x}}{2} $
The Systems response to $ \cos 2x $ is $ \ \dfrac{t e^{-2 i x} + t e^{2 i x}}{2} $ but
- $ e^{2 x i}=\cos 2x + i \sin 2x \, $ and
- $ e^{-2 x i}=\cos 2x - i \sin 2x \, $
- so the response is equal to
- $ t\cos 2t \, $
