Ring modulation
Jim Finnis
Can anybody out there enlighten me as to what a ring modulator (as
beloved of old analogue synth makers) actually DOES? I have to knock out
a digital version of one, so I'd be grateful for a mathematical
definition, although I could probably parse an electronic description.
I know this is probably a dumb question, 'cause I feel I really ought to
know the answer, but any help would be vastly appreciated.
Thanks in advance,
Jim
--
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Jim Finnis, | Long sigblocks are the Work of the Anti-Bob!
Clef Digital Systems |
cl...@aber.ac.uk | "Has anybody seen my camel?"
Michael O'Dell
or 4-quadrant, depending on the design.
So the general formula for a 4-quad mixer is
z = x * y
i i i
and this produces outputs (for X=f1 and Y=f2 frequencies) at
f1, f2, f1+f2, and f1-f2.
To get 2-quadrant behaviour, you can add an absolute value
operator to one in put, or even a simple diode function on
one (this assumes signals which are bipolar, -1 to +1 nominally,
so the offsets may be non-zero depending on the DSP representation).
2-quadrant behavior alters the resulting spectrum, of course.
and by using two multipliers, two quadrature phase shifters and
a summer, you can generate real single-sideband (only f1+f2 or
f1-f2 with the f1 and f2 suppressed). This is what the famous
Bode frequency shifter did.
This is all classical frequency mixing and exciter design.
The Radio Amateur's Handbook published by the ARRL has large chapters
covering these areas.
-Mike
Bellcore??? Bellcore isn't allowed to have opinions, and I can't
imagine why they'd have an opinion about this.
Scott Amspoker
>
>Can anybody out there enlighten me as to what a ring modulator (as
>beloved of old analogue synth makers) actually DOES? I have to knock out
>a digital version of one, so I'd be grateful for a mathematical
>definition, although I could probably parse an electronic description.
In a nutshell, a ring modulator inputs two frequencies (f1 and f2) and
ouputs the sum and difference of those frequencies (f1+f2, f1-f2).
--
Scott Amspoker |
Basis International, Albuquerque, NM | Too bad ignorance isn't really
| bliss. Then it could be outlawed.
sc...@bbx.basis.com |
Scott O'Hare
>In a nutshell, a ring modulator inputs two frequencies (f1 and f2) and
>ouputs the sum and difference of those frequencies (f1+f2, f1-f2).
>
...and sounds like bloody hell, too (very useful...)
soh
--
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- Scott O'Hare (s...@gdstech.grumman.com) -
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Georg Mueller
In article <1993Feb10....@aber.ac.uk> cl...@aber.ac.uk (Jim Finnis) writes:
>
>Can anybody out there enlighten me as to what a ring modulator (as
>beloved of old analogue synth makers) actually DOES? I have to knock out
>a digital version of one, so I'd be grateful for a mathematical
>definition, although I could probably parse an electronic description.
In a nutshell, a ring modulator inputs two frequencies (f1 and f2) and
ouputs the sum and difference of those frequencies (f1+f2, f1-f2).
That's way true, but a ring modulator is a time domain device. The
result is in fact as you describe it. The sums and differences are
taken for every frequency pair found in the spectra of both signals.
In the time domain it's simply a four-quadrant multiplication of both
input voltages.
Georg.
Adam MIROWSKI
: That's way true, but a ring modulator is a time domain device. The
: result is in fact as you describe it. The sums and differences are
: taken for every frequency pair found in the spectra of both signals.
: In the time domain it's simply a four-quadrant multiplication of both
: input voltages.
What is a "four-quadrant" mutiplication? You mean you
suppress the continuous component of both signals first?
--
Adam Mirowski, m...@chorus.fr (FRANCE), tel. +33 (1) 30-64-82-00 or 74
Chorus systemes, 6, av.Gustave Eiffel, 78182 Saint-Quentin-en-Yvelines CEDEX
Scott O'Hare
>
> In a nutshell, a ring modulator inputs two frequencies (f1 and f2) and
> ouputs the sum and difference of those frequencies (f1+f2, f1-f2).
>
>That's way true, but a ring modulator is a time domain device. The
>result is in fact as you describe it. The sums and differences are
>taken for every frequency pair found in the spectra of both signals.
>In the time domain it's simply a four-quadrant multiplication of both
>input voltages.
>
what "four-quadrant" multiplication of signals is (complex numbers?). Also
the notion of multiplying every frequency pair of the spectrum boggles my
mind - how are these frequencies determined, and how do they get multiplied?
Tim Wilson
s...@gdstech.grumman.com (Scott O'Hare) writes:
In article <GEORG.93F...@dali.nlp.physik.th-darmstadt.de>
ge...@dali.nlp.physik.th-darmstadt.de (Georg Mueller) writes:
>
> In a nutshell, a ring modulator inputs two frequencies (f1 and f2) and
> ouputs the sum and difference of those frequencies (f1+f2, f1-f2).
>
>That's way true, but a ring modulator is a time domain device. The
>result is in fact as you describe it. The sums and differences are
>taken for every frequency pair found in the spectra of both signals.
>In the time domain it's simply a four-quadrant multiplication of both
>input voltages.
>
This is very interesting - could you elaborate just a bit? I'm not sure
what "four-quadrant" multiplication of signals is (complex numbers?). Also
the notion of multiplying every frequency pair of the spectrum boggles my
mind - how are these frequencies determined, and how do they get multiplied?
(Hi, Scott.)
"Four-quadrant" multiplication is jargon: the signs of the signals
are preserved. A "two-quadrant" multiplier is an amplitude modulator
where one signal is supposed to determine the absolute value of the
other, not the local value.
Bottom line: a ring modulator is a multiplier. It's frequency
shifting effect has more to do with the signals being multiplied than
with the ring modulator. The frequencies are determined by your input
signals. The signals get multiplied, and the frequencies add and subtract.
Let's say you've got two sinusoidal signals:
x1(t) = cos(w1 t)
x2(t) = cos(w2 t)
When you multiply x1 by x2, you get
x(t) = x1(t) x2(t)
= cos(w1 t) cos(w2 t)
And when you do the trig identities you find
x(t) = (1/2) cos((w1 - w2) t) + (1/2) cos((w1 + w2) t)
For example, I multiply a 200 Hz sine wave by a 110 Hz sine wave; I
end up with sine waves at 90 and 310 Hz, which aren't harmonically
related.
Imagine having one periodic signal represented as a Fourier
sum with harmonic amplitudes and phases A(n) and phi(n), etc. (I'm
not gonna try to do the math notation in ascii.)
x1(t) = Sum A(n) cos(n w1 t + phi(n))
x2(t) = cos(w2 t)
The product has partials at (n w1 - w2) and (n w1 + w2):
x(t) = (1/2) Sum A(n) cos((n w1 - w2) t + phi(n))
+ (1/2) Sum A(n) cos((n w1 + w2) t + phi(n))
The entire spectrum of x1 has been shifted up and down by w2. If
x1(t) had frequency 200 Hz, it has harmonics at 200, 400, 600, etc.
If x2(t) has frequency 110 Hz, the product signal has partials at
90, 290, 490, etc. and at 310, 510, 710, etc. Again, the signal is
inharmonic.
If x2(t) is an arbitrary periodic signal with some harmonic
series representation, too, then one gets the spectrum of x1 shifted
up and down by each of the frequencies in the spectrum of x2: w2,
2 w2, 3 w2, etc. Using the same numbers as above, one gets a spectrum
with partials at 90, 290, 490,..., and 310, 510, 710,... (from 200,
400, 600,... and 110) and 20, 220, 420,..., and 380, 580, 780,...
(from 200, 400, 600,..., and 220) and -130, 70, 270,..., and 530, 730,
930,... (from 200, 400, 600,... and 330), etc. Regarding spectral
energy, the -130 acts the same as 130 Hz. So just considering three
terms from each series, we've got partials at 20, 70, 90, 130, 220,
270, 290, 310, 380, 420, 490, 510, 530, 580, 710, 730, 780, 930.
Yow! Wadda mess. Incredibly rich. Lotsa fun.
But, to digitally implement a ring modulator, all you need to do is
multiply the two signals one sample at a time.
--
Tim Wilson
Internet: t...@ear-ache.mit.edu
UUCP: mit-eddie!mit-athena!tim
Bad Colorado. No business.
Rick Jansen
LM1496, MC1496, Signetics S5596. Another name for these
chips is 'double-balanced modulator'. I guess the data sheets
will also tell you something about how they work.
The 1496 chips only need a few resistors for biasing etc.
And they're not expensive either. Elektor published a
ringmodulator for the Formant synthesizer. (Hello Georg :)
-Rick.
--
ri...@sara.nl
She's a module and she's looking good