In article , Peter John Lawton
writes:

Avery Fineman wrote:

In article , Peter John Lawton
writes:

The above will hold true at any fundamental frequency provided the
rise and fall times are equal and each equal to 0.02 times the
repetition period. Those numbers will change given faster or slower
rise/fall times. All db calculated as 20 x Log (voltage). Width is
determined at the baseline, not the 50% amplitude point.

Len Anderson
retired (from regular hours) electronic engineer person

I wonder what happens to these numbers as the rise/fall time tends to
zero?

The harmonic content will increase...but also show dips depending
on the percentage width relative to the period. I could present those
(takes only minutes to run the program and transcribe the results)
but that is academic only. The rise and fall times will NOT be zero
due to the repetition frequency being high (repetition time short).

Consider that a 3 MHz waveform has a period of 333 1/3 nSec and
that Paul is using a TTL family inverter to make the square wave.
Even with a Schmitt trigger inverter the t_r and t_f are going to be
finite, possibly 15 nSec with a fast device (and some capacitive
loading or semi-resonant whatever to mess with on- and off-times).

15 nSec is 4.5% of the repetition period, quite finite...more than I
showed on the small table given previously.

I'm sure someone out there wants to argue minutae on numbers but
what is being discussed is a squarish waveform with a repetition
frequency in the low HF range. Periods are valued in nanoSeconds
and the on/off times of squaring devices are ALSO in nanoSeconds.
There's just NOT going to be any sort of "zero" on/off times with
practical logic devices used by hobbyists.

I just wondered from a theoretical point of view what the program would
say about the harmonic content as you decreased the values you put into
it for t_r and t_f.

The harmonic values will change, approaching that of an ideal
square wave. That's a truism. With zero rise and fall it IS the
same as an ideal square wave.

There's NO accurate little formula, saying, or myth that will
predict any particular harmonic value. That's the reason for using
nice, very quick number-crunching computer programs.

What is not intuitive to me (and to others) is that harmonic energy
of a rectangular waveform drops drastically by the 5th harmonic
and is certainly lower than "obvious" numbers bandied about.

This is connected with my question. I am pondering why the energy
available for higher harmonics is less than for the fundamental and also
how your program works out this energy.

My program was developed while at RCA Corporation, specifically in
the time period of winter 1973-1974 using the core of three ideal
waveforms: rectangular, rising triangle, falling triangle. They relate
to a singular waveform using a time-delay formula multiplier so that
the rising triangle butts up to (in time) to the start of the rectangular
waveform and the falling triangle starts at the end of the rectangular
waveform. Entry is rise-time (the rising triangle), fall-time (the falling
triangle), and 50% amplitude pulse width which is the rectangular
waveform length and the length of the rising and falling triangles
adjusted for their inputted times. [draw it out to see it better]
Each basic waveform generates its own Fourier coefficient set. All
sets are simply added algebraically. Mathematically okay to do that.

A quick form of proof of that is to use a simple frequency-to-time
transform that works at each specified point in time along the
repetition period of the waveform. The original was a time-to-
frequency transform, mathematically different than the opposite.
If a reconstruction of the frequency-to-time results in the original
entry specifications, then it is called accurate enough. I didn't
derive the reconstruction transform since it was already in a book.
Neither did I derive any of the basic ideal waveforms which were
already in the ITT Blue Bible. The delay multiplier used to set
rise, fall, and 50% width was another book value, simplified to
faster calculation simplicity because the original was a math
problem thing with more terms than needed.

As to WHY of the energy distribution, that's up to any person who
has the textbook formulas and math smarts to fool around with.
I can't sum that up in one message. I doubt anyone can. I do
know this: Using the formulas and the program, then setting up a
test with careful adjustments of a pulse generator and using a
well-calibrated spectrum analyzer, the numbers agree within the
tolerances of the analyzer calibration. To me, and lots of others,
that is all the proof needed. Beyond that, its too much time and
nobody paying me to do this...

Its like pushing the baby on the swing in the park, you only need to
give it the occasional push or pull in the right direction. A 5f
resonator gets has to go for 2.5 cycles in between refuelling from a
square-wave (1:1) of frequency f.

Use any analogue you want. I don't agree with the above, but
feel free and I not going further on that...


What do you mean by intuition here?
My intuition suggests to me that as the rise and fall times get shorter,
the energy available for the harmonics approaches that for the
fundamental. In other words, as a square wave approaches perfection it

For any ideal rectangular shape, the harmonic energies have a
(SinX / X) locus. That's explained in textbooks also. Harmonics
of a repetitive waveform Fourier transform will NEVER have more
energy than the fundamental. That's also basic book stuff.

If the rise and/or fall times are finite, the harmonics will drop their
energy levels compared to the zero rise and fall time ideals. As the
rise and fall times get longer and longer the harmonic energy gets
less and less. By the time one gets to a sinusoid waveshape,
there are NO harmonics in any Fourier transform, its all
fundamental frequency (1 / repetition-period).

Recess.

Len Anderson
retired (from regular hours) electronic engineer person