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:<math>\cos x = \dfrac{e^{i x}+e^{-i x}}{2}</math> | :<math>\cos x = \dfrac{e^{i x}+e^{-i x}}{2}</math> | ||
Revision as of 07:09, 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}\, $
- $ \cos x = \dfrac{e^{i x}+e^{-i x}}{2} $
- $ \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
- $ \dfrac{t e^{-2 i x} + t e^{2 i x}}{2} = t\cos 2t $
