# Integration to determine power signal

Discussion in 'Math' started by Jess_88, Aug 6, 2011.

1. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
Hey guys

I'm having a little trouble following a example in my lecture notes.

using
2sin^(2)a = 1 - cos2a
the result is determined to be P = [A^(2)]/2

I'v been trying it... but I still can't work it out
Could someone please show me procedure of integration for this?

Thanks guys

2. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
What I have worked so far is

p = [(A^2)*ω]/2∏ ∫sin^2(wt + θ)dt

let wt + θ = u

(A^2)*ω]/2∏ x 1/2 ∫ 2sin^2(u)du

using 2sin^2(u) = 1-cos(2u)

[(A^2)*ω]/4∏ ∫ [1 - cos(2u)]du

[(A^2)*ω]/4∏ [∫ (1) dt - ∫cos(2u)du

[(A^2)*ω]/4∏ *(2∏/w) - [(A^2)*ω]/4∏ * 1/2∫2*cos(2u)du

A^2/w - A^2*w/8∏ * [sin(2u)]

subbing u = tw + θ

A^2/w - A^2*w/8∏ * [sin(2*tw + 2θ)]

A^2/w - A^2*w/8∏ * [sin(4∏ + 2θ)]

... this is where I am stuck

did I make a mistake???

3. ### Jess_88 Thread Starter Member

Apr 29, 2011
174
1
ok I think I know where I made the mistake... maybe

How dose this look?
A^(2) /2 - [A^(2)/8∏]*[sin(4∏ + 2θ) - sin(2θ)]

can I do this next part???

A^(2) /2 - [A^(2)/8∏]*[sin(4∏) + sin(2θ) - sin(2θ)]

A^(2) /2 - [A^(2)/8∏]*0

= A^(2)/2

4. ### Zazoo Member

Jul 27, 2011
114
43
I didn't work through every step you have there, but you can simplify the problem immensely by noting that at this step:

[(A^2)*ω]/4∏ [∫ (1) dt - ∫cos(2u)du]

you can set the second integral, (∫cos(2u)du), to zero since integrating the sine or cosine over one period (or any integer multiple of T) is zero - the negative and positive halves of the wave cancel each other out. So really all you have to integrate is the first term:

[(A^2)*ω]/4∏ [∫ (1) dt]

Which is much easier and indeed (A^2)/2 as you found.

Mike

Jess_88 likes this.

Apr 29, 2011
174
1
Thats great.
Thanks