Yes, I found 25*(D/T)^2 by trial.
Running simuations that give me THD, I see it fits perfectly with D/T from 0.02 to 0.
Yes, I meant "unable".
Now, I understand better.
The 25*(D/T)^2 is quite normal.
Because I am looking for small D/T the fonction has to behave like.
THD = a*( D/T) + b* (D/T)^2
The only thing that would need a mathematical proof is that there is no D/T term.
I think this is because the THD fonction is and odd fonction of D/T. By odd fonction, I mean f(-x)=-f(x). May be, the proof of this is not too difficult.
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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
