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John Fields
 
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Default Distorsion percentage, power or voltage?

On 16 Jan 2004 14:33:31 -0800, (Svante)
wrote:

John Fields wrote in message
On 16 Jan 2004 06:40:38 -0800,
(Svante)
wrote:

Harmonic distorsion is expressed as the ratio between the distorsion
components and the fundamental. What surprises me is that it is the
VOLTAGES that are compared (in the electrical case) not the POWERS. So
if we have a second harmonic 40 dB down, the second harmonic
distorsion is 1 %, not 0.01 %.
(In this case the voltage of the harmonic is 1% of the fundamental,
and its power is 0.01% of the fundamental)

What is the reason for this convention? I'd think that power would be
more logical.


---
It is _much_ easier to notch out the fundamental(s) and measure the
voltage of the remaining component(s) than it is to measure their power.

Then, knowing the voltage of the offending component(s) appearing across
the load and the impedance of the load at the various frequencies at
which distortion rears its ugly head, their contribution to the power
being dissipated in/by the load can be easily calculated.


Sure, but adding the partials a total amount of distortion would be
much easier if the percentages were power percentages. 2% 2nd, 1% 3rd
and 0.5% 4th harmonic would simply add up to 3.5%. Being voltage
percentages one would has to do:
sqrt(0.02^2+0.01^2+0.005^2)=2.3%. OK, nowadays, the computer would do
this for us, so maybe it doesn't matter much.
Certainly I would not argue that such a well-established standard
should be changed, and I would probably have no success at all trying
to do so, I just think it seems a bit akward and worth reflecting
over.


---
When measuring _total_ harmonic distortion, the contribution of each of
the individual partials is immaterial in that what's being determined is
the contribution to distortion that _all_ of the harmonics due to the
fundamental's presence contribute.


Furthermore, even if the contributions of the individual partials were
to be measured, their voltages would each be measured using a tuned
voltmeter and then the process of determining their contribution
determined mathematically. As a matter of fact, in order to measure the
power directly, the normal load would have to be disconnected and a
bolometer with precisely the same impedance as the load substituted for
the load. Expensive and more than just a _bit_ awkward.

--
John Fields