Revision as of 08:35, 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}\, $
The Signal is Linear
Since the system is linear you can split the signal in two parts
$ \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 equal to
$ t\cos 2t \, $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett