(→cos(t-2)) |
(→Power) |
||
Line 26: | Line 26: | ||
− | <math> | + | <math>P=\frac{1}{2\pi-0}\int_{-2}^{2\pi-2}{|cos(u)|^2du}</math> |
− | <math>=\frac{1}{2\pi-0} *{\frac{1}{2}}\int_{-2}^{2\pi-2}(1+cos(2u))du</math> | + | <math>P=\frac{1}{2\pi-0} *{\frac{1}{2}}\int_{-2}^{2\pi-2}(1+cos(2u))du</math> |
− | <math>=\frac{1}{4\pi}((u)+\frac{1}{2}sin(2u))|_{u=-2}^{u=2\pi-2}</math> | + | <math>P=\frac{1}{4\pi}((u)+\frac{1}{2}sin(2u))|_{u=-2}^{u=2\pi-2}</math> |
− | <math>=\frac{1}{4\pi}(2\pi-2+.378-(-2+.378))</math> | + | <math>P=\frac{1}{4\pi}(2\pi-2+.378-(-2+.378))</math> |
− | <math>=\frac{1}{2}</math> | + | <math>P=\frac{1}{2}</math> |
Revision as of 05:56, 5 September 2008
cos(t-2)
A time shift should not effect the energy or power of periodic function over one period (0 to 2$ \pi $ in this case).
I used [this] as the original function.
$ u = (t-2) $
Energy
$ E=\int_{-2}^{2\pi-2}{|cos(u)|^2du} $
$ E=\frac{1}{2}\int_{-2}^{2\pi-2}(1+cos(2(u)))du $
$ E=\frac{1}{2}((u+\frac{1}{2}sin(2(u)))|_{u=-2}^{u=2\pi-2} $
$ E=\frac{1}{2}(2\pi-2 + .378 -(-2 - .378)) $
$ E=\pi $
Power
$ P=\frac{1}{2\pi-0}\int_{-2}^{2\pi-2}{|cos(u)|^2du} $
$ P=\frac{1}{2\pi-0} *{\frac{1}{2}}\int_{-2}^{2\pi-2}(1+cos(2u))du $
$ P=\frac{1}{4\pi}((u)+\frac{1}{2}sin(2u))|_{u=-2}^{u=2\pi-2} $
$ P=\frac{1}{4\pi}(2\pi-2+.378-(-2+.378)) $
$ P=\frac{1}{2} $