# Calc. amplitude harmonics

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I am looking for a rule of thumb calculation for determing the
amplitude of harmonic frequencies relative to the fundamental.

For example, if the fundamental is a 1000Hz triangle wave, and we
assign it an amplitude of 100V, what would the amplitudes be for 10,
100, 2,000 and 4,000Hz?

Does the same rule apply for odd harmonics with a squarewave?

Silvia Marks

Re: 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.

Re: Calc. amplitude harmonics

Clifford Heath wrote:

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

Re: Calc. amplitude harmonics

"Silvia Marks"

**  No.

For a triangle wave see:

http://en.wikipedia.org/wiki/Triangle_wave

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

Re: Calc. amplitude harmonics

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

Re: Calc. amplitude harmonics

"jean.easter"

**  80 %   =   8 / pi squared  ?

........   Phil

Re: Calc. amplitude harmonics

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?

Re: Calc. amplitude harmonics

Silvia Marks wrote:

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.

Re: Calc. amplitude harmonics

http://www.tinaja.com/glib/muse90.pdf for a tutorial

hundreds of Fourier calculators in both JavaScript and PostScript appear
at http://www.tinaja.com/magsn01.asp

--
Many thanks,

Don Lancaster                          voice phone: (928)428-4073
We've slightly trimmed the long signature. Click to see the full one.
Re: Calc. amplitude harmonics

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

Re: Calc. amplitude harmonics

It looks like we agree then.

Mark