Latest revision as of 18:20, 5 November 2010

HW1, ECE301, Fall 2008, Prof. Boutin

Compute the energy and the power of the function

$f(t)=\cos \left( t-2 \right).$

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.

$$ u = (t-2) $$

$E=\int_{-2}^{2\pi-2}{|cos(u)|^2du}$

$E=\frac{1}{2}\int_{-2}^{2\pi-2}(1+cos(2(u)))du$

$E=\frac{1}{2}((u+\frac{1}{2}sin(2(u)))|_{u=-2}^{u=2\pi-2}$

$E=\frac{1}{2}(2\pi-2 + .378 -(-2 - .378))$

$E=\pi$

$P=\frac{1}{2\pi-0}\int_{-2}^{2\pi-2}{|cos(u)|^2du}$

$P=\frac{1}{2\pi-0} *{\frac{1}{2}}\int_{-2}^{2\pi-2}(1+cos(2u))du$

$P=\frac{1}{4\pi}((u)+\frac{1}{2}sin(2u))|_{u=-2}^{u=2\pi-2}$

$P=\frac{1}{4\pi}(2\pi-2+.378-(-2+.378))$

$P=\frac{1}{2}$

@@ Line 1: / Line 1: @@
-== cos(t-2) ==
+=[[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.
-== Energy ==
 <math>u = (t-2)</math>
-<math>E=\int_0^{2\pi}{|cos(u)|^2du}</math>
+== Energy ==
+<math>E=\int_{-2}^{2\pi-2}{|cos(u)|^2du}</math>
-<math>E=\frac{1}{2}\int_0^{2\pi}(1+cos(2(u)))du</math>
+<math>E=\frac{1}{2}\int_{-2}^{2\pi-2}(1+cos(2(u)))du</math>
@@ Line 21: / Line 34: @@
-<math>E=\frac{1}{2\pi-0}\int_0^{2\pi}{|cos(u)|^2du}</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_0^{2\pi}(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=-\pi}^{u=2\pi-\pi}</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}(\pi+0-(0-\pi))</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>