(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 t*cos(-2t).
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If a linear system has a response to <math>e^{2jt}\!</math> of <math>t*e^{-2jt}\!</math> and a response to <math>e^{-2jt}\!</math> of <math>t*e^{2jt}\!</math>, then it's response to <math>cos(2t)\!</math> must be <math>t*cos(2t)\!</math>
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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>
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Based on the given information, we know that the system must be linear. Since it is linear (and has the output shown above in line 1), we can conclude that it must have the output:
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<math> \frac{1}{2}(te^{-2jt}+te^{2jt})\!</math>. 
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Upon converting the output (above) back into a cosine function, we get the output:
  
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<math>t*cos(2t)\!</math>
  
Explanation:
 
  
The system seems to have the following characteristics:
 
x(t) --> SYSTEM --> t*y(-t)
 
  
Therefore, if we put the change the input from x(t) (above) to cos(2t), then it should output the signal t*cos(-2t).
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Jayanth's comments: The answer is correct. Do not use this symbol <math>*</math> for multiplication, it can be misunderstood for convolution.

Latest revision as of 15:07, 19 September 2008

Linearity

If a linear system has a response to $ e^{2jt}\! $ of $ t*e^{-2jt}\! $ and a response to $ e^{-2jt}\! $ of $ t*e^{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 above in line 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)\! $


Jayanth's comments: The answer is correct. Do not use this symbol $ * $ for multiplication, it can be misunderstood for convolution.

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