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.

    [​IMG]

    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 :mad:
    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.
  5. Jess_88

    Thread Starter Member

    Apr 29, 2011
    174
    1
    Thats great.
    Thanks :)
     
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