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Re: THD


John Woodgate
 

In message <kkdr4u+gvv5@...>, dated Sun, 14 Apr 2013, Echidna <mchambin@...> writes:

Indeed, I found 25*(D/T)^2 by trial.
Running simulations that gave me THD values for various D values. I have uploaded the .asc file of that simulaton.
Thanks.

"enable" is a spelling mistake. I meant "unable".
I just wanted to prevent others being confused.

I did try using Fourier analysis in my manual calculations. Calculation of the Fourrier coefficient is extremly difficult,
Agreed. I will try using Mathcad, but I can't do it immediately.

this is why I tried a workaround. I am puzzled to see such a complex calculation while the THD = 25*( D/T )^2 result is so simple.
It happens sometimes.

This result fits so well with simuulation trials, I doubt it is so by chance.
Maybe.

I notice a small point. Your pulse starts at the zero-crossing. To simulate crossover distortion correctly it should start at time t/2 before the zero-crossing and end at time t/2 after it.
--
OOO - Own Opinions Only. See www.jmwa.demon.co.uk
They took me to a specialist burns unit - and made me learn 'To a haggis'.

John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK

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