(New page: ==Problem== Compute the energy and power of x(t) = <math>(79t+43)^2</math> ==Energy== <math>E=\int_{t_1}^{t_2}x(t)dt</math> <math>E=\int_0^{2}{(3t + 2)^2dt}</math> <math>E = \dfrac{1}{9...) |
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==Problem== | ==Problem== | ||
| − | Compute the energy and power of x(t) = <math>( | + | Compute the energy and power of x(t) = <math>(t+1/2)^2</math> |
==Energy== | ==Energy== | ||
<math>E=\int_{t_1}^{t_2}x(t)dt</math> | <math>E=\int_{t_1}^{t_2}x(t)dt</math> | ||
| − | <math>E=\int_0^{ | + | <math>E=\int_0^{5}{(t + 1/2)^2dt}</math> |
| − | <math>E = \dfrac{1}{ | + | <math>E = \dfrac{1}{3}(t+1/2)^3|_{t=0}^{t=5}</math> |
| − | E = | + | E = 1331/24-1/24=665/12 |
==Power== | ==Power== | ||
| Line 16: | Line 16: | ||
<math>P = E*.5</math> | <math>P = E*.5</math> | ||
| − | P = | + | P = 665/6 |
Latest revision as of 10:02, 5 September 2008
Problem
Compute the energy and power of x(t) = $ (t+1/2)^2 $
Energy
$ E=\int_{t_1}^{t_2}x(t)dt $
$ E=\int_0^{5}{(t + 1/2)^2dt} $
$ E = \dfrac{1}{3}(t+1/2)^3|_{t=0}^{t=5} $
E = 1331/24-1/24=665/12
Power
$ P=\dfrac{1}{{t_2}-{t_1}}\int_{t_1}^{t_2}x(t)dt $
$ P = E*.5 $
P = 665/6
