Calc. amplitude harmonics

The rule of thumb is that you use a Fourier transform.

No. Triangle waves have all the even harmonics IIRC, whereas square waves have all the odd ones (with amplitudes s.t. like

3rd: 1/3, 5th: 1/5... or was it 3rd: 1/9, 5th: 1/25, ...?) Anyhow you get the idea. For a certain waveform, you apply the Fourier transform and it gives you the frequency and phase of all the harmonics - that's what it's for.

"Silvia Marks"

** No.

For a triangle wave see:

For a square wave , the relative harmonic amplitudes are in inverse proportion to the (odd ) harmonic number.

The *actual* amplitude of the fundamental however is 1.137 ( 0.5 + 2/pi) times that of the square wave.

........ Phil

The rules of thumb that I use are derived from Fourier Analysis. You need to work out what frequencies are present, and then what their levels are.

1) The harmonics get less the higher you go, in proportion to "frequency to the power of n", where n is 1 for discontinuities in the voltage (like a square wave suddenly changing voltage), n is 2 for discontinuities in the voltage slope (like a triange wave, where the voltage doesn't change suddenly but the rate of change does), etc. up to a pure sine wave where there are no harmonics, n is "infinite" because the sine wave has a smooth waveform however many times you take the slope, the slope of the slope, etc.

So a square wave's harmonics are at the relative levels of

1/3, 1/5, 1/7, ... (no even harmonics, as explained below)

but a triangle wave has

1/(3^2), 1/(5^2), 1/(7^2)... = 1/9, 1/25, 1/49, ...

This kind of rule works better and better for higher frequencies. It works excellently for all frequencies with the simple examples here!

I've not said anything about the fundamental here. A rule of thumb eludes me at the moment...

2) You more often get odd harmonics, but the even harmonics are absent when the wave is the same backwards as upsidedown (I think the proper term is skew-symmetric).

This is a square wave, but you can't tell me whether I made it upsidedown or backwards:

| |-----| |-----| |-----| |-----| | | | | | | | | | | | | | | | | -> | | | | | | | | | | | | | | | | | | | | | | | | | |-----| |-----| | | |-----| |-----|

so it has no even harmonics. Same with a nice symmetrical triangle wave.

This one is not quite so square, having a duty cycle not 50%:

| |-----| |-----| |---| |---| | | | | | | | | | | | | | | | | -> | | | | | | | | | | | | | | | | | | | | | | | | | |---| |---| | | |-----| |-----|

and you can tell I turned it upsidedown, so it has even harmonics. Same with a saw-tooth wave.

So your example has a triangle wave at 1000Hz. I assume that it is symmetrical. OK? The frequencies present are the odd harmonics:

1000, 3000, 5000, 7000, ... You started with 100V of triangle wave. That has approx 80V of sine wave at 1000Hz, so 80V / 9 = 9V at 3000Hz 80V / 25 = 3V at 5000Hz etc

OK?!

Jean

You recall wrong.

This is right.

No, those are the harmonic amplitudes in a triangular wave.

It should be easy enough to remember, if you first recall that an triangular wave is the integral of a square wave -which means that the harmonic frequencies are the same, but the amplitudes are reduced in proportion to the harmonic number.

It can be useful to think of a square wave as the integral of two alternating Dirac spikes of equal area but opposite polarity - the Fourier transform of a Dirac spike has all the harmonics - odd and even

- at equal amplitude, and for two alternating spikes of equal but oppisute amplitude, all the even harmonics cancel while the odd harmonics sum.

-- Bill Sloman, Nijmegen

"jean.easter"

** 80 % = 8 / pi squared ?

........ Phil

Quite. I would have put Intergral(x from 0 to pi/2)((2x/pi)Sin x)dx / Intergral(x from 0 to pi/2)(Sin x)^2dx = 8/pi^2 but that is not a rule of thumb and is the Fourier Analysis that we are trying to simplify. We could recommend visualising a sine wave and a triangle wave of "similar power" (or some such vague term) and claiming that the sine wave would peak at 80% of the triangle wave peak, but I suspect that we could only do that kind of thing ourselves with conviction because we have experience of Fourier Analysis.

It's that elusive rule of thumb again. Any ideas, anybody?

For square waves, the amplitudes of the harmonics are

(f0 / f),

where f0 is the fundamental and f is the frequency of the harmonic.

For triangle waves, it's (f0 / f) ^ 2

Mark

p.s. this applies to the odd harmonics only; the even ones have zero amplitude.

for a tutorial

hundreds of Fourier calculators in both JavaScript and PostScript appear at

--
Many thanks,

Don Lancaster                          voice phone: (928)428-4073
Synergetics   3860 West First Street   Box 809 Thatcher, AZ 85552
rss: http://www.tinaja.com/whtnu.xml   email: don@tinaja.com

Please visit my GURU\'s LAIR web site at http://www.tinaja.com

The rule of thumb I remember is that square waves have odd harmonics that roll off at 6 dB/oct and tri waves have odd harmonics that roll off at 12 dB/oct

It looks like we agree then.

Mark