Total Harmonic Distortion

[calculus_cuthbert]

Is total harmonic distortion calculation done only on sinusoidal signals or can it be done on triangular waveforms also?

 

[JoannesPaulus]

It has to be a sine wave since it is defined as "the ratio of the sum of the powers of all harmonic components to the power of the fundamental frequency".

 

[LvW]

It has to be a sine wave since it is defined as "the ratio of the sum of the powers of all harmonic components to the power of the fundamental frequency".

Sorry, just the opposite is true.
The THD factor by definition for a sinewave is zero (no harmonics).
Calculation of the THD makes sense only for non-sinusiodal waves (like triangle or everything between sinus and sqarewave).

 

[betwixt]

Also, distortion is a measure of deviation from a perfect waveform. If the perfect waveform is a triangle wave, an exact replica of it will have zero distortion. It should still be possible to quantify differences in terms of harmonics but the math could be horrendous (I'm not volunteering!).

Brian.

 

[LvW]

Also, distortion is a measure of deviation from a perfect waveform. If the perfect waveform is a triangle wave, an exact replica of it will have zero distortion. It should still be possible to quantify differences in terms of harmonics but the math could be horrendous (I'm not volunteering!).
Brian.

Hi Brian,

I agree with you - however, did you ever see a definition of a distortion parameter which defines, for example, the deviation of a triangle wave from its ideal shape?
LvW

 

[FvM]

In my understanding, JoannesPaulus was simply referring to the fact, that the commonly known definition of the term THD
only applies to a sine as (ideal) reference function. For a triangle waveform, you can use a respective definition,
RMS(error)/RMS(ideal triangle). Other than with usual THD, the value couldn't be easily determined from magnitudes of
fourier components. The error term can be interpreted as standard deviation between real and ideal waveform.

 

[betwixt]

That's why I didn't volunteer to do the math

Brian.

 

[LvW]

OK, but I guess the original question from calculus_cathbert was related solely to the THD as it is defined for distorted sinusoidal waves.

 

[JoannesPaulus]

It has to be a sine wave since it is defined as "the ratio of the sum of the powers of all harmonic components to the power of the fundamental frequency".

Sorry, just the opposite is true.
The THD factor by definition for a sinewave is zero (no harmonics).
Calculation of the THD makes sense only for non-sinusiodal waves (like triangle or everything between sinus and sqarewave).

Sorry, I respectfully disagree. As FvM mentioned, the THD defines how far from a pure sinewave is your signal. There are other methods, with no specific names, that are used to measure how far your signal is from, for instance, a triangular wave or a multi-tone signal. So, I believe that if you mention total harmonic distortion you are referring to a sinusoidal signal.

 

[LvW]

..............
Sorry, I respectfully disagree. As FvM mentioned, the THD defines how far from a pure sinewave is your signal. There are other methods, with no specific names, that are used to measure how far your signal is from, for instance, a triangular wave or a multi-tone signal. So, I believe that if you mention total harmonic distortion you are referring to a sinusoidal signal.

Sorry, JoannesPaulus, I respectfully must confess that there was a misunderstanding between us. You are, of course, completely right that THD is "referred to a sinusoidal signal". No doubt about this.
But remember the original question: Is total harmonic distortion calculation done only on sinusoidal signals or can it be done on triangular waveforms also?

For my understanding, this sounds as if Calculus_Cuthbert thought, the THD principle could be applied either on sinusoidal or (!!!) for example on triangle waveforms. And - trying to be 100% correct - my first answer was based on the assumption that a signal which is called "sinusoidal" is really a clear sinewave with a THD=0% (knowing that this, of course, is an unrealistic case) and that for all other waveforms which deviate from this "ideal sinus" the concept of THD applies. OK?
Regards
LvW

 

[JoannesPaulus]

No harm done. I always enjoy your answers!