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| − | = | + | =[[Homework_1_ECE301Fall2008mboutin|HW1]], [[ECE301]], Fall 2008, [[user:mboutin|Prof. Boutin]]= |
| − | + | ||
| − | + | ||
| − | == | + | |
| + | ==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). | ||
| + | |||
| + | 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> | ||
| + | |||
| + | == Energy == | ||
| + | |||
<math>E=\int_{-2}^{2\pi-2}{|cos(u)|^2du}</math> | <math>E=\int_{-2}^{2\pi-2}{|cos(u)|^2du}</math> | ||
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| − | <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> |
Latest revision as of 18:20, 5 November 2010
HW1, ECE301, Fall 2008, Prof. Boutin
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} $
