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AC waveform question help

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data

    An A.C. voltage, V comprises of a fundamental voltage of 100V rms at a frequency of 120Hz, a 3rd harmonic which is 20% of the fundamental, a 5th harmonic which is 10% of the fundamental and at a phase angle of 1.2 radians lagging.

    i) Write down an expression for the voltage waveform.

    ii) Sketch the waveforms of the harmonic compnents.

    iii) Determin the voltage at 20ms.

    iv) Given an ideal V = 100V rms, what is the percentage error at 20ms


    3. The attempt at a solution


    part i)

    V = Vrms * sqrt2 = 100* 1.414 = 141.4V at 120Hz

    3rd harmonic = 20% of 141.4 = 28.28V at 360Hz

    5th harmonic = 10% of 141.4 = 14.14V at 600Hz


    v = (141.4sin(240∏t)) + (28.3sin(720∏t)) + (14.1sin(1200∏t-1.2))


    does this first part look correct?
     
  2. jcsd
  3. Apr 20, 2012 #2

    gneill

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    Yes, it looks okay.
     
  4. Apr 20, 2012 #3
    Ok thanks. So for the sketch I just need to plot the 3rd harmonic and 5th harmonic. What time period would be appropriate?
     
  5. Apr 20, 2012 #4

    NascentOxygen

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    A couple of cycles (of the fundamental) should do. But at least up to 20ms. (Note: the fundamental can also be referred to as the first harmonic.)
     
  6. Apr 20, 2012 #5

    gneill

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    The fundamental should be plotted too (it's the "first harmonic").

    Choose a time period that will display at least one full cycle of the fundamental.
     
  7. Apr 20, 2012 #6
    should it look something like this?


    harmonics.JPG
     
  8. Apr 20, 2012 #7

    NascentOxygen

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    You're sketch for fifth harmonic should show x5 the frequency!
     
  9. Apr 20, 2012 #8
    i think the ans is right...
     
  10. Apr 24, 2012 #9
    Ok so i have changed the time scale and got this....


    does this look as it should?

    harmonics.JPG
     
  11. Apr 24, 2012 #10

    NascentOxygen

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    That looks more like it. :smile:
     
  12. Apr 24, 2012 #11
    ok great.

    iii) Determin the voltage at 20ms.

    v = (141.4sin(240∏*0.02)) + (28.3sin(720∏*0.02)) + (14.1sin(1200∏*0.02-1.2))

    v = 83.11 + 26.91 - 13.14 = 96.88 v


    does this look correct?
     
  13. Apr 24, 2012 #12

    NascentOxygen

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    It does.

    Might be worth going back and checking to make sure that the problem did not specify a phase angle for the 3rd harmonic.
     
  14. Apr 25, 2012 #13
    Many thanks. I have checked the original question and there is no mention of a phase angle for the 3rd harmonic.
     
  15. Apr 25, 2012 #14
    ok so last part

    iv) Given an ideal V = 100V rms, what is the percentage error at 20ms


    100v - 96.845v = 3.155v

    so would it be 3.155% ?
     
  16. Apr 25, 2012 #15

    NascentOxygen

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    I'm sure you can't mix RMS and instantaneous values to get anything meaningful.
     
  17. Apr 25, 2012 #16
    ok so would i need to work out the RMS at 20ms?

    96.845 / (√2) = 68.5v


    then work this out as a % of the 100v rms?


    100v - 68.5v = 31.5v = 31.5% error?

    am i making any sense?
     
  18. Apr 25, 2012 #17

    NascentOxygen

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    The only calculation that would make sense to me would be based on:

    ideal instantaneous value at that time - actual instantaneous value
     
  19. Apr 26, 2012 #18
    Actual instantaneous value at 20ms v = (141.4sin(240∏*0.02)) + (28.3sin(720∏*0.02)) + (14.1sin(1200∏*0.02-1.2))

    v = 83.11 + 26.91 - 13.14 = 96.88 v


    Ideal instantaneous value at 20ms = 141.42Sin(240∏*0.02) = 83.11v


    error = [(83.11v - 96.845v)/100]*100 = -13.735%
     
  20. Apr 26, 2012 #19
    does the answer above look ok?


    i have ploted a graph to show actual Vs. Ideal

    untitled.JPG
     
  21. Apr 26, 2012 #20

    NascentOxygen

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    I would not have divided by 100 when determining fractional error. Some other data value would seem more appropriate.
     
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