(New page: Given the Signal x(t) = 4sin(2 * pi * 6t), Find the energy and power of the signal from 2 to 6 seconds. [edit] Energy)
 
 
(4 intermediate revisions by the same user not shown)
Line 1: Line 1:
 
Given the Signal x(t) = 4sin(2 * pi * 6t), Find the energy and power of the signal from 2 to 6 seconds.  
 
Given the Signal x(t) = 4sin(2 * pi * 6t), Find the energy and power of the signal from 2 to 6 seconds.  
  
[edit] Energy
+
 
 +
== Energy ==
 +
<math>E=\int_2^6 |x(t)|^2 dt</math>
 +
 
 +
<math>E=\int_2^6 |4sin(12\pi t)|^2 dt</math>
 +
 
 +
 
 +
<math>E=16*\int_2^6 sin(12\pi t)^2 dt</math>
 +
 
 +
 
 +
<math>E=16*\int_2^6 sin(12\pi t)^2 dt</math>
 +
 
 +
<math>E=16*(\dfrac{12 \pi t}{2}-\dfrac{sin(2*(12\pi t))}{4})\mid_2^6</math>
 +
 
 +
<math>E=16*(6\pi t-\dfrac{sin(24\pi t)}{4})\mid_2^6</math>
 +
 
 +
<math>E=96\pi *t-\dfrac{16sin(24\pi *t)}{4}\mid_2^6</math>
 +
 
 +
<math>E=96\pi *6-\dfrac{16sin(24\pi *5)}{4}-(96\pi *2-\dfrac{16sin(24\pi *1)}{4}</math>
 +
 
 +
<math>E=384\pi</math>
 +
 
 +
 
 +
 
 +
== Power ==
 +
<math>P=\dfrac{1}{t2-t1}\int_2^6 |x(t)|^2 dt</math>
 +
 
 +
<math>P=\dfrac{1}{6-2}\int_2^6 |4sin(12\pi t)|^2 dt</math>
 +
 
 +
 
 +
<math>P=\dfrac{16}{4}*\int_2^6 sin(12\pi t)^2 dt</math>
 +
 
 +
 
 +
<math>P=\dfrac{16}{4}*\int_2^6 sin(12\pi t)^2 dt</math>
 +
 
 +
<math>P=\dfrac{16}{4}*(\dfrac{12 \pi t}{2}-\dfrac{sin(2*(12\pi t))}{4})\mid_2^6</math>
 +
 
 +
<math>P=\dfrac{16}{4}*(6\pi t-\dfrac{sin(24\pi t)}{4})\mid_2^6</math>
 +
 
 +
<math>P=\dfrac{96}{4}\pi *t-\dfrac{16sin(24\pi *t)}{16}\mid_2^6</math>
 +
 
 +
<math>P=\dfrac{96}{4}\pi *6-\dfrac{16sin(24\pi *6)}{16}-(\dfrac{96}{4}\pi *2-\dfrac{16sin(24\pi *2)}{16}</math>
 +
 
 +
'''Power =96 *pi'''

Latest revision as of 08:32, 5 September 2008

Given the Signal x(t) = 4sin(2 * pi * 6t), Find the energy and power of the signal from 2 to 6 seconds.


Energy

$ E=\int_2^6 |x(t)|^2 dt $

$ E=\int_2^6 |4sin(12\pi t)|^2 dt $


$ E=16*\int_2^6 sin(12\pi t)^2 dt $


$ E=16*\int_2^6 sin(12\pi t)^2 dt $

$ E=16*(\dfrac{12 \pi t}{2}-\dfrac{sin(2*(12\pi t))}{4})\mid_2^6 $

$ E=16*(6\pi t-\dfrac{sin(24\pi t)}{4})\mid_2^6 $

$ E=96\pi *t-\dfrac{16sin(24\pi *t)}{4}\mid_2^6 $

$ E=96\pi *6-\dfrac{16sin(24\pi *5)}{4}-(96\pi *2-\dfrac{16sin(24\pi *1)}{4} $

$ E=384\pi $


Power

$ P=\dfrac{1}{t2-t1}\int_2^6 |x(t)|^2 dt $

$ P=\dfrac{1}{6-2}\int_2^6 |4sin(12\pi t)|^2 dt $


$ P=\dfrac{16}{4}*\int_2^6 sin(12\pi t)^2 dt $


$ P=\dfrac{16}{4}*\int_2^6 sin(12\pi t)^2 dt $

$ P=\dfrac{16}{4}*(\dfrac{12 \pi t}{2}-\dfrac{sin(2*(12\pi t))}{4})\mid_2^6 $

$ P=\dfrac{16}{4}*(6\pi t-\dfrac{sin(24\pi t)}{4})\mid_2^6 $

$ P=\dfrac{96}{4}\pi *t-\dfrac{16sin(24\pi *t)}{16}\mid_2^6 $

$ P=\dfrac{96}{4}\pi *6-\dfrac{16sin(24\pi *6)}{16}-(\dfrac{96}{4}\pi *2-\dfrac{16sin(24\pi *2)}{16} $

Power =96 *pi

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett