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Have we decided on a definition for THD ?

Discussion in 'Electronic Design' started by Adam S, Nov 21, 2006.

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  1. Adam S

    Adam S Guest

    I was writing a little program that needs to calculate Total Harmonic
    Distortion of a waveform. When I go to refresh my memory of what the
    heck THD is, I get conflicting explanations. From what I can find on the
    web there are two variants. One definition expresses THD as ratio of
    _power_ , and other in ratio of _RMS_ volts. Since power is proportional
    to RMS volts squared the two definitions are obviously not equal. Last
    time I checked, x != x^2.

    These web pages offer the power ratio explanation:
    http://en.wikipedia.org/wiki/Total_harmonic_distortion
    http://www.sweetwater.com/expert-center/glossary/t--THD

    While these web pages offer the RMS ratio explanation:
    http://www.vk1od.net/SquareWave/THD.htm
    http://www.tonmeister.ca/main/textbook/node549.html
    http://www.birds-eye.net/definition/acronym.cgi?what+is+THD=Total+Harmonic+Distortion&id=1153665296
    http://www.maxim-ic.com/tools/calculators/index.cfm/calc_id/maxim_sinad

    So which is it guys ?
     
  2. Eeyore

    Eeyore Guest

    When expressed as dB it makes no difference of course.

    In fact, when expressed as dB it's not tricky to see why some 'old chestnuts' like accepting 0.1% THD
    as inaudible are so wrong ( it's only -60dB ).

    I go by the voltage method for the percentage value and all the test gear I know does too.

    Graham
     
  3. Eeyore

    Eeyore Guest

    Which links to....
    http://www.rane.com/note145.html

    It's *voltage*.

    Graham
     
  4. Phil Allison

    Phil Allison Guest

    "Adam S"


    ** Where THD is expressed as a percentage, the ratio is in terms of
    voltages.

    Commercial THD meters do NOT use "true rms" meters, but average responding
    ones calibrated to give the rms value for a sine wave = 1.1111 times the
    average rectified value. The fundamental or test frequency is removed from
    the signal under test and any residual signal measured, including noise, in
    the range up to about 100 kHz.

    The sites you quoted show how to convert a spectrum analysis table into a
    THD figure in -dB or percentage.

    There is no confusion anywhere except YOUR wrong interpretation.




    ........ Phil
     
  5. Adam S

    Adam S Guest

  6. Eeyore

    Eeyore Guest

    I've temporarily edited it.

    Note that power and voltage ratios don't give the same dB value. 1/2 power = -3dB but 1/2 voltage =
    -6dB.

    There's the clue as to why it's wrong to use power ratios for THD..

    Graham
     
  7. Eeyore

    Eeyore Guest

    I've done something temporary.

    Graham
     
  8. Adam S

    Adam S Guest

    power is proportional to RMS voltage squared.
    so to get a ratio of power between two signals, a and , b, you sum the
    square of the RMS values in a, and divide by sum of the square of the
    RMS values in b.

    i.e
    power_ratio = Va1^2 + Va2^2 +... / Vb1^2 + Vb2^2 + ...

    where Va1, Va2,.. is RMS level of each component in signal,a , and
    Vb1,Vb2,... is RMS level of each component in signal b.

    To get ratio of RMS voltages between signals a, and b.

    RMS_ratio = sqrt( Va1^2 + Va2^2 +... ) / sqrt(Vb1^2 + Vb2^2 +...)

    So you see this has is nothing to do about interpretation. They are
    clearly not the same. Put simply , power_ratio = RMS_ratio^2
     
  9. Robert

    Robert Guest

    As long as the Impedances are the same where the two Voltages are being
    measured the two methods are the same.

    Robert
     
  10. Phil Allison

    Phil Allison Guest

    "Adam S"


    ** The confusion is all in your silly interpretation.

    Of course power ratios and voltage ratios produce different numbers, no-one
    even hinted otherwise.

    Where THD is expressed as a percentage, the ratio is in terms of voltages -
    this is a *decades old* convention.

    Where THD is expressed in -dBs, the ratio is that of two powers.

    1% = -40dB

    0.1 % = - 60 dB

    0.01% = -80 dB


    Capice ?




    ......... Phil
     
  11. Phil Allison

    Phil Allison Guest

    "Robert"

    ** The time honoured convention is to ONLY quote THD of a sine wave signal
    as a percentage when the ratio is that of two voltages.

    It must never be quoted as the percentage ratio of two powers - lest you
    be considered as yet another snake oil con artist quoting grossly inflated
    numbers.




    ........ Phil
     
  12. Eeyore

    Eeyore Guest

    What *2 impedances* are you rambling on about ?

    Voltage has nothing to do with impedance.

    Graham
     
  13. joseph2k

    joseph2k Guest

    Neither one. The actual definition _is_ in terms of power (correctly).
    Typical measurement is usually done in terms of voltage. The key idea is,
    that it is not in terms of absolute volts / watts / amperes / VA but in the
    difference in spectral content of harmonics before and after. Then all of
    the ratios are based on the the level of the fundamental at the output for
    the output measurement, and on the level of harmonics relative the
    fundamental at the input. Volts, Amperes, Watts and all that factor out.
     
  14. Phil Allison

    Phil Allison Guest

    "joseph2k"



    ** Will someone out there track this monumental FUCKING MORON down
    to whatever vermin infested shit hole he is lurking in and SHOOT the
    VILE **** in the head ???


    Please >>>





    ......... Phil
     
  15. Eeyore

    Eeyore Guest

    Whose definition ?

    Typical ? ALL THD measurements are done this way.

    No they don't. Percentages never do.

    Graham
     
  16. PeteS

    PeteS Guest


    I think he's referring to the fact that 20log10 V2/V1 is equivalent to
    10log10 P2/P1 provided the resistance (not impedance) that V2 and V1
    are developed across *are equal*.

    It's a pretty straightforward definition.

    Note - the derivation below is simple and if you're offended by
    simplicity, then skip it ;)

    dB = 10log10 P2/P1, so dB = 10log10 (V2^2/R2) / (V1^2/R1)

    doing a little transposition
    dB = 10log10 (V2^2R1)/(V1^2R2). If R1 and R2 are equal, then they
    cancel, leaving

    dB = 10log10 V2^2 / V1^2, -> 10log10 (V2/V1)^2

    As log x^2 = 2 log x, then
    dB = 20log10 V2/V1(provided the resistances were equal).

    Cheers

    PeteS
     

  17. Pathetic loser. Begging for someone to do what you don't have the
    balls to do. No wonder you have no friends. In civilized nations you
    would be charged with truing to hire a hit man, and spend the rest of
    your useless life in prison, till your cell mate got fed you and killed
    you to get some peace and quiet. On the other hand, you wouldn't have
    to walk the streets looking for men, anymore.

    --
    Service to my country? Been there, Done that, and I've got my DD214 to
    prove it.
    Member of DAV #85.

    Michael A. Terrell
    Central Florida
     
  18. Eeyore

    Eeyore Guest

    They have cannibalism in prisons now ?

    Graham
     
  19. Jamie

    Jamie Guest

    Yes, Red meat on fridays..!
     
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