(New page: == Signal energy == Engery can be found via the following formula: <math>E = \int_{t_1}^{t_2} \! |x(t)|^2\ dt</math> I will solve for the energy for the following function) |
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− | I will solve for the energy for the following function | + | I will solve for the energy for the following function <math>f(x) = 5\!</math> on the interval <math>[1,5]\!</math>. If we go through all of the math, our answer turns out to be <math>5(5) - 5(1)\!</math>, which equal <math>24\!</math>. |
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+ | == Signal power == | ||
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+ | The power can be found using this function: | ||
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+ | <math>P_{avg}=\frac{1}{t_2-t_1} \int_{t_1}^{t_2} |x(t)|^2\ dt \!</math> | ||
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+ | I will solve the same function on the same interval. After all of the math, our answer turns out to be <math>24/4\!</math>, which is <math>6\!</math>. | ||
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+ | Good night, or actually, at this hour, it is more like good morning. |
Revision as of 20:38, 4 September 2008
Signal energy
Engery can be found via the following formula:
$ E = \int_{t_1}^{t_2} \! |x(t)|^2\ dt $
I will solve for the energy for the following function $ f(x) = 5\! $ on the interval $ [1,5]\! $. If we go through all of the math, our answer turns out to be $ 5(5) - 5(1)\! $, which equal $ 24\! $.
Signal power
The power can be found using this function:
$ P_{avg}=\frac{1}{t_2-t_1} \int_{t_1}^{t_2} |x(t)|^2\ dt \! $
I will solve the same function on the same interval. After all of the math, our answer turns out to be $ 24/4\! $, which is $ 6\! $.
Good night, or actually, at this hour, it is more like good morning.