Revision as of 08:07, 18 September 2008 by Serdbrue (Talk)

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 $

Alumni Liaison

BSEE 2004, current Ph.D. student researching signal and image processing.

Landis Huffman