(New page: Since <math>e^{2jt}\rightarrow t e^{-2jt}</math> and <math>e^{-2jt}\rightarrow t e^{2jt}</math>, we can conclude that the system performs the function <math>x(t)\rightarrow t x(-t)</m...) |
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| + | ==The Obvious Answer== | ||
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Since | Since | ||
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since cosine is an even function. | since cosine is an even function. | ||
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| + | ==The Correct Answer== | ||
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| + | It was pointed out by those commenting on my answer that my original method for solving the problem works in only a subset of cases, something I did not think about when answering this question. A better way to approach this problem is to write the cosine function as a sum of complex exponentials. | ||
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| + | <math>cos(x)=\frac{1}{2}\left[e^{jx}+e^{-jx}\right]</math> | ||
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| + | The response to each of these complex exponential terms was given in the problem statement, so we can apply linearity to arrive at the response. | ||
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| + | <math>e^{2jt}\rightarrow t e^{-2jt}</math> and | ||
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| + | <math>e^{-2jt}\rightarrow t e^{2jt}</math>. | ||
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| + | The system is given to be linear, so we can take the coefficient of each term and apply it to the corresponding term in the output. Then we have | ||
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| + | <math>cos(t)=\frac{1}{2}\left[e^{jt}+e^{-jt}\right] \rightarrow [system] \rightarrow \frac{1}{2} \left[t e^{-2jt}+ t e^{2jt} \right] = \frac{1}{2}t \left[ e^{2jt}+e^{-2jt} \right] = t cos(2t)</math> | ||
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| + | We arrive at the same answer using both methods, but in general, we would not. The most correct way to solve a problem of this type is to use complex exponentials. | ||
Latest revision as of 05:12, 18 September 2008
The Obvious Answer
Since
$ e^{2jt}\rightarrow t e^{-2jt} $
and
$ e^{-2jt}\rightarrow t e^{2jt} $,
we can conclude that the system performs the function
$ x(t)\rightarrow t x(-t) $.
Therefore, we assume the response to our input is
$ cos(2t)\rightarrow t cos(-2t) = t cos(2t) $,
since cosine is an even function.
The Correct Answer
It was pointed out by those commenting on my answer that my original method for solving the problem works in only a subset of cases, something I did not think about when answering this question. A better way to approach this problem is to write the cosine function as a sum of complex exponentials.
$ cos(x)=\frac{1}{2}\left[e^{jx}+e^{-jx}\right] $
The response to each of these complex exponential terms was given in the problem statement, so we can apply linearity to arrive at the response.
$ e^{2jt}\rightarrow t e^{-2jt} $ and
$ e^{-2jt}\rightarrow t e^{2jt} $.
The system is given to be linear, so we can take the coefficient of each term and apply it to the corresponding term in the output. Then we have
$ cos(t)=\frac{1}{2}\left[e^{jt}+e^{-jt}\right] \rightarrow [system] \rightarrow \frac{1}{2} \left[t e^{-2jt}+ t e^{2jt} \right] = \frac{1}{2}t \left[ e^{2jt}+e^{-2jt} \right] = t cos(2t) $
We arrive at the same answer using both methods, but in general, we would not. The most correct way to solve a problem of this type is to use complex exponentials.
