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== Linearity == | == Linearity == | ||
| − | If a linear system has a response to <math>e^{2jt}\!</math> of t* | + | 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> |
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> | ||
| − | Based on the given information, we know that the system must be linear. Since it is linear (and has the output shown in | + | 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: |
<math> \frac{1}{2}(te^{-2jt}+te^{2jt})\!</math>. | <math> \frac{1}{2}(te^{-2jt}+te^{2jt})\!</math>. | ||
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<math>t*cos(2t)\!</math> | <math>t*cos(2t)\!</math> | ||
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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.
