(Linearity)
(Linearity)
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== Linearity ==
 
== Linearity ==
  
If a linear system has a response to exp(2jt) of t*exp(-2jt) and a response to exp(-2jt) of t*exp(2jt), then it's response to cos(2t) must be <math>t*cos(2t)\!</math>
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If a linear system has a response to <math>e^{2jt}\!</math> of t*exp(-2jt) and a response to exp(-2jt) of t*exp(2jt), then it's response to cos(2t) must be <math>t*cos(2t)\!</math>
  
 
To look at this in more detail, we must first understand that <math>cos(2t)\!</math> can be expressed as follows: <math> \frac{1}{2}(e^{-2jt}+e^{2jt})\!</math>
 
To look at this in more detail, we must first understand that <math>cos(2t)\!</math> can be expressed as follows: <math> \frac{1}{2}(e^{-2jt}+e^{2jt})\!</math>

Revision as of 15:13, 18 September 2008

Linearity

If a linear system has a response to $ e^{2jt}\! $ of t*exp(-2jt) and a response to exp(-2jt) of t*exp(2jt), then it's response to cos(2t) must be $ t*cos(2t)\! $

To look at this in more detail, we must first understand that $ cos(2t)\! $ can be expressed as follows: $ \frac{1}{2}(e^{-2jt}+e^{2jt})\! $


Based on the given information, we know that the system must be linear. Since it is linear (and has the output shown in paragraph 1), we can conclude that it must have the output:

$ \frac{1}{2}(te^{-2jt}+te^{2jt})\! $.

Upon converting the output (above) back into a cosine function, we get the output:

$ t*cos(2t)\! $

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Abstract algebra continues the conceptual developments of linear algebra, on an even grander scale.

Dr. Paul Garrett