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− | == cos(t-2) == | + | ==Question == |
+ | Compute the energy and the power of the function | ||
+ | |||
+ | <math>f(t)=\cos \left( t-2 \right).</math> | ||
+ | ---- | ||
+ | ==Answer== | ||
A time shift should not effect the energy or power of periodic function over one period (0 to 2<math>\pi</math> in this case). | A time shift should not effect the energy or power of periodic function over one period (0 to 2<math>\pi</math> in this case). | ||
− | I used [[ | + | I used [[HW1.5_Ben_Laskowski_-_Signal_Power_and_Energy_ECE301Fall2008mboutin|this]] as the original function. |
<math>u = (t-2)</math> | <math>u = (t-2)</math> |
Revision as of 18:16, 5 November 2010
Contents
Question
Compute the energy and the power of the function
$ f(t)=\cos \left( t-2 \right). $
Answer
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} $