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.
"enable" is a spelling mistake. I meant "unable".
I did try using Fourier analysis in my manual calculations. Calculation of the Fourrier coefficient is extremly difficult,
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. This result fits so well with simuulation trials, I doubt it is so by chance.
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--- In LTspice@..., John Woodgate <jmw@...> wrote:
In message <kkckve+t28o@...>, dated Sat, 13 Apr 2013, Echidna
<mchambin@...> writes:
Using LTspice simulations I find THD = 25*( D/T )^2 In this simulation
I used a 1000 HZ SINE combined with a PULSE with values of D like 0.1u
0.2u 0.5u 1u 2u 5u 10u 20u 50u These simulations gave me a THD that
perfectly fits with 25*( D/T )^2
I think you mean you found 25*(D/T)^2 by trial. This can be misleading.
I once found a power function that matches half a sine wave very
accurately, but outside the range 0 to pi it doesn't.
Here is the issue.
I was enable to prove this result with maths
'unable', I think.
(instead of simulations ).
I started computing the RMS value of the error signal (the dents at the
crossovers) With sin omega*t = omega*t ( valid for t << T ) I don' t
get the 25*( D/T )^2 result, I do get a fonction of D/T, so far so
good, but I don't get the right exponant. It seems my approach is wrong.
The function you do get might match 25*(D/T)^2 over a range of D/T.
Is there a signal theory / maths guru who can give the proof that for
such a signal S(t) ( with D << T ) we have THD = 25*( D/T )^2
Did you try using Fourier analysis in you manual calculations?
--
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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
