Difference between revisions of "HW1.5 Nicholas Browdues - Signal Power and Energy ECE301Fall2008mboutin" - Rhea

Latest revision as of 14:19, 5 September 2008

Consider the signal $$ x(t)=cos(t) $$ over the interval 0 to $4\pi$

$Avg. Power = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt$

$Avg. Power = {1\over(4\pi-0)}\int_{0}^{4\pi}\!|cos(t)|^2 dt$

$Avg. Power = {1\over(4\pi)}{1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt$

$Avg. Power = {1\over8\pi}(4\pi+{1\over2}sin(8\pi))$

$Avg. Power = {1\over2}$

$E = \int_{t_1}^{t_2}\!|x(t)|^2\ dt$

$E = \int_{0}^{4\pi}\!|cos(t)|^2\ dt$

$E = {1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt$

$E = {1\over2}(4\pi+{1\over2}sin(8\pi))$

$E = 2\pi$

@@ Line 1: / Line 1: @@
-== Energy ==
-<font size="5">
+<font size="3">
 Consider the signal
 <math>x(t)=cos(t)</math> over the interval 0 to <math>4\pi</math>
+== Average Power ==
+<math>Avg. Power = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt</math>
-<math>E = {1\over(t2-t1)}\int_{t_1}^{t_2}\!|x(t)|^2 dt</math>
+<math>Avg. Power = {1\over(4\pi-0)}\int_{0}^{4\pi}\!|cos(t)|^2 dt</math>
-<math>E = {1\over(4\pi-0)}\int_{0}^{4\pi}\!|cos(t)|^2 dt</math>
+<math>Avg. Power = {1\over(4\pi)}{1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt</math>
-<math>E = {1\over(4\pi)}{1\over4}\int_{0}^{4\pi}\!(1+cos(2t)) dt</math>
+<math>Avg. Power = {1\over8\pi}(4\pi+{1\over2}sin(8\pi)) </math>
-<math>E = {1\over\pi}(4\pi+{1\over2}sin(4\pi)) dt</math>
+<math>Avg. Power = {1\over2}</math>
-<math>E = {2}</math>
 </font>
-== Power ==
+== Energy ==
 <font size="4">
-<math>f(t)=2cos(t)</math>
+<math>E = \int_{t_1}^{t_2}\!|x(t)|^2\ dt</math>
-<math>P = \int_{t_1}^{t_2}\!|f(t)|^2\ dt</math>
+<math>E = \int_{0}^{4\pi}\!|cos(t)|^2\ dt</math>
-<math>P = \int_{0}^{2\pi}\!|2cos(t)|^2\ dt</math>
+<math>E = {1\over2}\int_{0}^{4\pi}\!(1+cos(2t)) dt</math>
-<math>P = (4){1\over2}\int_{0}^{2\pi}\!1+cos(2t) dt</math>
+<math>E = {1\over2}(4\pi+{1\over2}sin(8\pi))</math>
-<math>P = (4){1\over2}(2\pi+{1\over2}sin(2*2\pi)</math>
+<math>E = 2\pi</math>
-<math>P = 4\pi</math>
 </font>