Re: Have we decided on a definition for THD ?
- From: Adam S <not.valid@nosuchaddress>
- Date: Tue, 21 Nov 2006 17:27:56 +1100
Phil Allison wrote:
"Adam S"
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.
** 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.
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
.
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