Difference between revisions of "HW1.5 Hetong Li - Energy and Average power ECE301Fall2008mboutin" - Rhea

Latest revision as of 16:01, 3 September 2008

$$ f(t)=2cos(t) $$

According to formula of Energy of a singal,we can get:

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

$=\int_0^{2\pi}{|2cos(t)|^2dt}$

$=\int_0^{2\pi}{(2(2cos(t)^2-1)+2)dt}$

$=\int_0^{2\pi}{2+cos(2t))dt}$

$=(2t+sin(2t))|_{t=0}^{t=2\pi}$

$=4\pi+0-0-0$

$=4\pi$

According to formula of Power of a singal,we can get:

$P=\frac{1}{2T}\int_{-T}^{T}\!|f(t)|^2\ dt$

$=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{|2cos(t)|^2dt}$

$=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{(2(2cos(t)^2-1)+2)dt}$

$=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{(2+cos(2t))dt}$

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

$=\frac{1}{4\pi}(4\pi+0-(-4\pi)-0)$

$=\frac{1}{4\pi}(8\pi)$

$$ =2 $$

@@ Line 1: / Line 1: @@
 == Signal ==
 <font size="4">
-<math>y(t)=exp(t)</math>
+<math>f(t)=2cos(t)</math>
 </font>
-== Power ==
+==Energy==
+According to formula of Energy of a singal,we can get:
 <font size="4">
-<math>P = \int_{t_1}^{t_2} \! |y(t)|^2\ dt</math>
+<math>E = \int_{t_1}^{t_2} \! |f(t)|^2\ dt</math>
+<math>=\int_0^{2\pi}{|2cos(t)|^2dt}</math>
+<math>=\int_0^{2\pi}{(2(2cos(t)^2-1)+2)dt}</math>
+<math>=\int_0^{2\pi}{2+cos(2t))dt}</math>
+<math>=(2t+sin(2t))|_{t=0}^{t=2\pi}</math>
+<math>=4\pi+0-0-0</math>
+<math>=4\pi</math>
+</font>
+==Power==
+According to formula of Power of a singal,we can get:
+<font size="4">
+<math>P=\frac{1}{2T}\int_{-T}^{T}\!|f(t)|^2\ dt</math>
+<math>=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{|2cos(t)|^2dt}</math>
+<math>=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{(2(2cos(t)^2-1)+2)dt}</math>
-<math>P = \int_0^2 \! |e^(2t)\ dt</math>
+<math>=\frac{1}{4\pi}\int_{-2\pi}^{2\pi}{(2+cos(2t))dt}</math>
+<math>=\frac{1}{4\pi}(2t+sin(2t))|_{t={-2\pi}}^{t=2\pi}</math>
-<math>P = \int_0^2 \! exp(2t) dt</math>      (since exp(t) cannot be negative,take off the absolute value sign)
+<math>=\frac{1}{4\pi}(4\pi+0-(-4\pi)-0)</math>
+<math>=\frac{1}{4\pi}(8\pi)</math>
-<math>P = {1\over 2} ( exp(2t) )\mid_0^{2}</math>
+<math>=2</math>
-<math>P = {1\over 2}(exp(4) - 1)}</math>
 </font>