(New page: ==What we know== The two things that are important when solving this problem is that we know what the output is for <math>e^{j2t}</math> and <math>e^{-j2t}</math> and that the system is l...) |
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<math>e^{-j2t}</math> and that the system is linear. | <math>e^{-j2t}</math> and that the system is linear. | ||
| − | <math>e^{j2t}</math> ---> system ---> <math>te^{-j2t}</math><br> | + | <math>e^{j2t}\!</math> ---> system ---> <math>te^{-j2t}\!</math><br> |
| − | <math>e^{-j2t}</math> ---> system ---> <math>te^{j2t}</math><br> | + | <math>e^{-j2t}\!</math> ---> system ---> <math>te^{j2t}\!</math><br> |
==What we can do with this== | ==What we can do with this== | ||
First step is to take the input <math>cos(2t)</math> and relate it to the two known inputs. It is easy to do this using Euler's Identity <math>e^{jat} = \frac{cos(t) + jsin(t)}{a}</math> | First step is to take the input <math>cos(2t)</math> and relate it to the two known inputs. It is easy to do this using Euler's Identity <math>e^{jat} = \frac{cos(t) + jsin(t)}{a}</math> | ||
| − | <math>cos(2t) = \frac{cos(2t) + jsin(2t)}{2} + \frac{cos(2t) - jsin(2t)}{2}</math> | + | <math>cos(2t) = \frac{cos(2t) + jsin(2t)}{2} + \frac{cos(2t) - jsin(2t)}{2} = \frac{e^{j2t} + e^{-j2t}}{2}</math> |
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| + | So now that we have <math>cos(t)</math> represented as a sum of our two outputs we can run those through the system. | ||
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| + | <math>.5e^{j2t}+.5e^{-j2t}\!</math> ---> system ---> <math>.5te^{j2t}+.5te^{-j2t}\!</math> | ||
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| + | Now, we can use Euler's Identity again to get our answer in terms of cosine. | ||
| + | |||
| + | <math>t\frac{e^{j2t}+e^{-j2t}}{2} = tcos(2t)</math> | ||
Latest revision as of 12:37, 19 September 2008
What we know
The two things that are important when solving this problem is that we know what the output is for $ e^{j2t} $ and $ e^{-j2t} $ and that the system is linear.
$ e^{j2t}\! $ ---> system ---> $ te^{-j2t}\! $
$ e^{-j2t}\! $ ---> system ---> $ te^{j2t}\! $
What we can do with this
First step is to take the input $ cos(2t) $ and relate it to the two known inputs. It is easy to do this using Euler's Identity $ e^{jat} = \frac{cos(t) + jsin(t)}{a} $
$ cos(2t) = \frac{cos(2t) + jsin(2t)}{2} + \frac{cos(2t) - jsin(2t)}{2} = \frac{e^{j2t} + e^{-j2t}}{2} $
So now that we have $ cos(t) $ represented as a sum of our two outputs we can run those through the system.
$ .5e^{j2t}+.5e^{-j2t}\! $ ---> system ---> $ .5te^{j2t}+.5te^{-j2t}\! $
Now, we can use Euler's Identity again to get our answer in terms of cosine.
$ t\frac{e^{j2t}+e^{-j2t}}{2} = tcos(2t) $
