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Power percentage, square wave, Fourier series

  1. Mar 31, 2013 #1
    1. The problem statement, all variables and given/known data

    What is the percentage of power (out of the total power) contained up to the third harmonic (power in DC component, a1 , a-1 , a2 , a-2 , a3 , a-3 ) of the square waveform shown above? (the duty cycle = D = τ/T0= 0.5)

    2. Relevant equations


    3. The attempt at a solution

    Hey all,
    The following question refers to the attached diagram. I thought the question was simple enough due to the amplitude being 1, and the value of D being 0.5.

    The simplified formula I calculated for the answer is 2/(k*pi) for odd harmonics (as even values of k result in a value of 0)

    -However when adding 2/pi and 2/(3*pi) i get 0.8488 and hence 84.9%

    However the multiple choice answers are 96.7, 95, 72.5 and 73.3%

    I am not sure whether this is a simple mistake or a large misunderstanding of the theory.

    Any help would be much appreciated!
     

    Attached Files:

    Last edited by a moderator: Apr 1, 2013
  2. jcsd
  3. Apr 1, 2013 #2

    berkeman

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    The waveform looks like it also has a DC component...?
     
  4. Apr 1, 2013 #3
    Sorry only just starting out this topic, would the DC component be 0.5? How does this component chang emy calculations?
     
  5. Apr 1, 2013 #4

    berkeman

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  6. Apr 1, 2013 #5

    berkeman

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    BTW, that article was the first hit on the list of my Google search for square wave harmonics amplitude. :smile:
     
  7. Apr 1, 2013 #6
    Yes I have actually looked at that page, but it just isn't clicking, here is my understanding.
    -The zero harmonic is the DC component and hence 0.5
    -Even harmonics have a value of 0
    -Odd harmonics have a value of 2/(pi*n)
    -Total amplitude is 1
    -So percentage power should be ((2/pi) + 2/(3*pi))/1 = 0.8488
     
  8. Apr 1, 2013 #7

    berkeman

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    That page is listing voltage component values. How is the power related to the voltage?
     
  9. Apr 1, 2013 #8
    P = (V^2)/R
    Hence power percentage would be 0.8488^2.!
    Hopefully I have finally gotten that one right! Thanks for your persistence with me
     
  10. Apr 1, 2013 #9

    berkeman

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  11. Apr 7, 2013 #10
    i think its
    (2/pi)^2 + (2/3pi)^2
    divided by
    (2/pi)^2 + (2/3pi)^2 +(2/5pi)^2 + (2/7pi)^2

    which gives you 95%
     
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