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Principal frequencies of a Waveform

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Identify the principal frequencies in the current waveform

    2. Relevant equations


    3. The attempt at a solution
    From the excel file, I have identified 3 principle frequencies at 52Hz, 246Hz and 351Hz. Is this correct? What does the term principle frequency actually mean?
     

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  2. jcsd
  3. May 11, 2015 #2

    rude man

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    It's "principal" frequency, not "principle".
    Wouldn't you just guess that they're the three peaks on your spectrum drawing?
     
  4. May 13, 2015 #3
    Thanks for your reply. How do I calculate total harmonic distrortion of that data using the attached equation. Since the vertical axis is in I^2, for I1 I have sqrt0.6617 * sqrt2 = 1.15A where 0.6617 is at 0Hz. I dont know what the nth harmonic is. the n(max) i assume is the n of the highest peak which is 3
     

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  5. May 13, 2015 #4

    rude man

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    I can't seem to reopen your original excel file. But you need to determine the principal frequency. Is it the first (largest) peak? Anyway, the principal frequency is not distortion so you need to follow your formula for THD with all peaks above the first. If the spectrum was really I^2 (which sounds suspicious) then you already have I^2 for the two higher peaks. If the spectrum is I though you need to take I^2 in the formula.
     
  6. May 13, 2015 #5

    upload_2015-5-13_19-38-59.png

    This is the waveform, the largest peak is 15.8785 at 52Hz, then there is 8.77 at 246Hz and finally 6.25 at 351Hz. I think that these 3 frequencies are the principal frequencies.
    I think I1^2 is 0.6617.
     

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  7. May 13, 2015 #6
    I dont fully understand how to obtain the variables in the THD formula using my data.
     
  8. May 13, 2015 #7

    rude man

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    I would interpret the waveform data as the principal frequency being 52 Hz and the distortion frequencies as 246 and 351 Hz. So you'd use the THD formula for the latter two frequencies. Notice the summation is from n=2 to however many "spur" frequencies there are, expressed as a fraction of n=1 or I1. I'd say the sum here is over n = 2 to 3.
    The graph does display I^2 so use the y axis numbers directly in the THD formula. So I1 = 16 A^2 etc.
     
  9. May 13, 2015 #8
    1/16*sqrt(8.77+6.25(In))*100% is what I have so far. Am I on the right track? What is an nth harmonic?

    My notes on this question state I1 is the r.m.s. value of the fundamental current, In the r.m.s
    value of the nth harmonic and n(max) is the number of the highest measurable or significant harmonic.
     
  10. May 13, 2015 #9

    rude man

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    No, you are misusing the formula. You need the sum of squares, not the square of the sum. What's "In" doing in your formula?
    {quote]
    What is an nth harmonic?{/quote]
    n = 2 or 3 for you.
    That makes sense. Your fundamental (n=1) is 16 A2 and your harmonics are 8.77A2 and 6.25A2.
     
  11. May 13, 2015 #10
    Ok so; 1/16*(sqrt ((8.77+6.25)/2))*100%. Sum of squares is standard deviation isnt it? n=2 because there are just the 2 harmonics after the fundamental right?
     
  12. May 13, 2015 #11

    rude man

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    This is worse than last time. Why are you dividing by 2? And you're still not doing the square root of the sum of squares.
    Can be but is irrelevant here. You're not doing standard deviations.
    Right.
     
  13. May 14, 2015 #12
    Ok sorry. RMS of 8.77= 8.77/sqrt2 = 6.2 , RMS of 6.25=6.25/sqrt2= 4.419 . Because it is I2 Im gonna assume that numerators may need to be rooted too?

    Going back to the equation; ((1/16*sqrt6.22) + (1/16*sqrt4.4192))*100
     
  14. May 14, 2015 #13

    rude man

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    1/√16 is right but you are still getting the rest wrong.
    There is only one square root to be taken. You have two. Also, sqrt(6.22) obviously = 6.2 etc.
     
  15. May 14, 2015 #14
    I1(16) is also rms(16/√2=11.313) so should that not be 1/11.313*(6.2+4.419)*100%=93.86 which is a lot of distortion. The equation is trickier than it looks.
     
  16. May 14, 2015 #15

    rude man

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    No. I1 = 16 A2 according to your spectral graph. It says "I2" on the y axis but I believe it means "I2".
    Since I12 = 16 A2, I1 = 4 A. Now sum the n=2 and n=3 I2 numbers, then take the square root, then divide by 4.
    I agree though that distortion is very large!
     
  17. May 14, 2015 #16
    1/√16*√(6.2+4.419)*100%= 80.69%.
     
  18. May 14, 2015 #17

    rude man

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    But your n=2 peak is at 8.77 A2 and the n=3 is at 6.25 A2 so why are you again taking rms of those two numbers? You didn't take the rms of 16, and rightly so!
     
  19. May 14, 2015 #18
    Recalculating I come up 96.89%. I thought the question identified In as the rms of those harmonics thats all.
     
  20. May 14, 2015 #19
    Do you know a good step by step method for sketching the original waveform from the 3 principal harmonics?
     
  21. May 14, 2015 #20

    rude man

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    It depends on your signal.
    If there is just a fundamental (n=1) and noise at two harmonics (n=2 and n=3) ("spot noise") then the spectrum is a delta function at each of those two frequencies f2 and f3. Take your example and assume there is no noise at any frequency except at the spot frequencies 246 and 351 Hz. Then your power spectrum (which is what a spectrum of I2 is) should show a delta function at each of those 3 frequencies and nothing inbetween. The tips of the delta functions would be at 16, 8.77 and 6.25 A2. A delta function for you is just a spike at f1, f2 or f3.

    Formally, the power at each spot frequency fn, n=1, 2 or 3, is ∫-∞In2δ(f-fn)df. Then you rss the two noises with the fundamental as you did with your formula to get rss noise.

    If noise is continuous you have to evaluate a different integral. All this is probably a bit advanced for you. I should mention that a trace like your spectrum is what you'll get from a spectrum analyzer even if there are only spot noise frequencies. It will show some noise at all frequencies due to the limitations of the instrument.
     
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