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| − | 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> | + | 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>25(5) - 25(1)\!</math>, which equals <math>100\!</math>. |
== Signal power == | == Signal power == | ||
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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> | + | I will solve the same function on the same interval. After all of the math, our answer turns out to be <math>100/4\!</math>, which is <math>25\!</math>. |
Good night, or actually, at this hour, it is more like good morning. | Good night, or actually, at this hour, it is more like good morning. | ||
Latest revision as of 20:45, 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 $ 25(5) - 25(1)\! $, which equals $ 100\! $.
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 $ 100/4\! $, which is $ 25\! $.
Good night, or actually, at this hour, it is more like good morning.
