(New page: ==Problem== Compute the energy and power of x(t) = <math>(3t+2)^2</math> ==Energy== <math>E=\int_0^{2}{(3t + 2)^2dt}</math> <math>E = \dfrac{1}{9}(3t+2)^3|_{t=0}^{t=2}</math> E = 56) |
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| Line 3: | Line 3: | ||
==Energy== | ==Energy== | ||
| + | <math>E=\int_{t_1}^{t_2}x(t)dt</math> | ||
| + | |||
<math>E=\int_0^{2}{(3t + 2)^2dt}</math> | <math>E=\int_0^{2}{(3t + 2)^2dt}</math> | ||
| Line 8: | Line 10: | ||
E = 56 | E = 56 | ||
| + | |||
| + | ==Power== | ||
| + | <math>P=\dfrac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}x(t)dt</math> | ||
| + | |||
| + | <math>P = E*.5</math> | ||
| + | |||
| + | P = 28 | ||
Latest revision as of 07:20, 5 September 2008
Problem
Compute the energy and power of x(t) = $ (3t+2)^2 $
Energy
$ E=\int_{t_1}^{t_2}x(t)dt $
$ E=\int_0^{2}{(3t + 2)^2dt} $
$ E = \dfrac{1}{9}(3t+2)^3|_{t=0}^{t=2} $
E = 56
Power
$ P=\dfrac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}x(t)dt $
$ P = E*.5 $
P = 28
