Square pulse, Duty cycle and odd/even harmonics.

[Antoniog2]

To KL7AJ or any one else who could explain…
I was looking for information about relation between a pulse signal and production of odd/even harmonics, when I read this tread #16118 on 11-18-2008

Mark / space ratio - does it matter?
KL7AJ => “However, when you have UNSYMMETRICAL square waves, … … you start getting additive and subtractive components....products of your fundamental square wave frequency PLUS what you would have if your shortened pulses WERE symmetrical! For example, if you have a 1KHz square wave with a 25% duty cycle, you have not odd harmonics of 1KHz, but you have odd harmonics of 4 KHZ, PLUS all the sums and differences possible!”

As I understand-it, a 25% duty cycle (250uS_On, 750uS_Off) will give with this pulse, a symmetrical signal of 250uS_On, 250uS_Off, or 2kHz and all its odds harmonics (2kHz, 6kHz, 10kHz, 14kHz…), “plus all the sums and differences possible”. (4kHz, 8kHz, 12kHz, 16kHz, …)
All this is, in this case, the even harmonics of 1kHz, with no odd ones. Is it right?

When I use a simulator as this little applet
http://www.falstad.com/fourier/
with a 25%_On duty cycle, (move arrow under the bottom of the square, in way to change Duty cycle) I find only harmonics of this order : 1, 2, 3,…,5 ,6 ,7 ,…,9, 10, 11, … with all 4x harmonics missing.

Do you have a simple way to explain me what I’m missing?
No argument intended!
I want only to understand how to quickly find odd and/or even harmonics out of asymmetrical pulses, out of simple fractions of duty cycle, as 50/50 (easy, only odd), 33/66, 25/75, 20/80…

Your way to consider a pulse as a part of a 1/2 symmetrical wave, giving only odd harmonics is very appealing, but not easy to follow for me now.

Thank you.
Antonio

[t_n_k]

If n is the notional harmonic and D is the duty then if nxD is an integer the nth
harmonic value is zero.

If D=0.25 then the 4th, 8th, 12th, etc... harmonics are zero.

If D=0.1 then the 10th, 20th, 30th etc are zero.

and so on.....

[Antoniog2]

Thanks, tnk.
This part of my question is clear now.