Estimating THD of a system 4

[2sb18]

Hi again,

If I have one buffer with a THD of .3% feeding another buffer with a THD of .6%, can I multiply the two numbers together to get a good estimate of the total THD?

ie 1.003*1.006=1.009 (.9% THD)

Thanks,
Steve.

 

[Skogsgurra]

No. You should square the numbers, add them and take square root. Like this: sqrt(.003^2 + .006^2), which would be something like 0,006708204. As you see, the dominant distortion figure determines overall distortion.



Gunnar Englund

http://www.gke.org

 

[elfgriper]

skogsgurra,
Just a question, How does your formula work if you input a signal into an Ideal .3% thd amp would you not have a circuit with < .3 thd? Then that signal into a .6 Ideal amp which would add another .6 thd. This would seem to be true if the circuits were independent. If not independent ( feedback from device #2 back to device #1) then it would seem that the thd of the last device would be the determining factor. Also this would not take into consideration any thermal noise and such from other circuit components.
-elf

 

[Skogsgurra]

elfgriper,

Not so sure what you mean. Sqrt(0.0^2 + 0.3^2) = 0.3 So the 0.3 distortion is conserved. Commutative law of multiplication implies that the order of amplifiers has no effect of outcome of the 0.3 and 0.6 THD experiment. Do you have experimental evidence saying otherwise? The sqrt(sum-of-squares) has always given me reasonable accurate results.

Regarding noise: That is a completely different question that was not covered in my answer. But, if there are several independent (stochastic) noise sources, they shall be added in the same way.

Gunnar Englund

http://www.gke.org

 

[elfgriper]

Hi Skogsgurra,
Yes your right about the question of noise, just adding to the conversation. Yes, I have no experimental evidence that would indicate your wrong on your sqrt of the sum of the squares formula. I think it just seemed logical to me that the THD's would be additive. I think that I was relating to amplication where if stage one is A=10 and Stage 2 is a=10 then the overall gain would be 100 not 14.1.
Thanks -elf

 

[Skogsgurra]

This may be one of those cases where self-evident isn't so self-evident. I may have been using the sqrt(sum-of-squares) without really knowing if it is right or wrong. Need to do some deep digging here...

Gunnar Englund

http://www.gke.org

 

[sreid]

Wouldn't the general answer be no since distortion implies non-linearities and superposition doesn't apply to non-linear systems?

 

[Skogsgurra]

Thanks sreid, you did half the digging for me!

Gunnar Englund

http://www.gke.org

 

[logbook]

THD is a tricky subject. Suppose both amplifiers produce only third harmonic distortion. If the phase of the harmonic distortion produced by one amplifier is opposite to that produced by the other amplifier, the resulting distortion will be less than the worst amplifier’s distortion. In the worst possible case, two amplifiers with equal but opposite distortions could give a non distorted signal!

 

[Skogsgurra]

I would call that the BEST possible case ;-)

Perhaps one could say that the sqrt(sum-of-squares) gives an approximation to the likely outcome in a typical situation?

I really have to dig deeper...

Gunnar Englund

http://www.gke.org

 

[2sb18]

Thanks for all your help guys,

All I need is just a rough estimate. I'm going to be taking a design into a lab to have some tests done, I just need a reference point so I know how far off from the ideal I am.

Thanks,
Steve.