MOTM

If you feed a sine into both inputs of a ring modulator, you get a sine an
octave higher, not a rectified sine. It outputs the sum and difference
signals, so the sum of "1" and "1" is "2 (doubled frequency)," and the
difference is zero. This only works for sines, of course, because any more
complex signal gets immediately non-linear.

I'm not sure where you're getting the rectification from. You can do some
rectification with a few diodes, but that's usually used for AC power. For
audio applications, there're ways of rigging an op-amp to do that, I think.

I wonder if you could... have your signal split up and fed into two VCAs,
one inverted phase, and also use the signal as the control signal for the
VCAs... naw, that's crazy... ;) 

-----Original Message-----
From: Thomas Hudson [mailto:thudson@...] 
Sent: Wednesday, 05 December, 2001 7:33 PM
To: MOTM
Subject: [motm] My failed experiment


I had this bright idea for creating an envelope follower
from the current offering of modules, and while it does
work somewhat, it is not quite "there," and I'm trying to
figure out why.

Here's what I did. I took the output from my guitar preamp
and ran it into both sides of the ring modulator. This
should create a sort of rectified signal. I then took the
output and ran it into the lag processor, which should
give independent attack and release times for the rectified
signal. I set UP (attack) to min and DOWN (release) to about
5. I ran this output to the FM input of a filter, and my
guitar preamp out to the input of the filter. The filter
does follow the input amplitude somewhat.

I guess I'm wondering if my assumption about the ring mod
doing rectification is correct. I know it is true for a sin
wave. Perhaps if I used the 700 to switch between +/- 5
on the zero crossing of the guitar signal, and used this
for the Y-input of the ring mod?

Tomy

Tkacs, Ken schrieb:
> 
> If you feed a sine into both inputs of a ring modulator, 
> you get a sine an 
> octave higher, not a rectified sine. It outputs the sum 
> and difference 
> signals, so the sum of "1" and "1" is "2 (doubled 
> frequency)," and the 
> difference is zero. 

> I'm not sure where you're getting the rectification from. 

Very true. However, "zero" means "zero Hz" (and not "zero voltage").
That is, the difference frequency becomes a DC voltage!
(DC = "AC @ 0Hz" with amplitude = DC voltage). This
adds to the double frequency sine wave, so you get a sine
of 2f that is *raised* above 0V - and here's where the similarity
to "rectification" comes in.

With the old formula (from memory - I hope it's right, and it's
easier with cosinus to write down):

cos(x) * cos(x) = 1/2 + 1/2 * cos(2x) >= 0.

JH.

On Thursday, December 6, 2001, at 06:23 AM, Tkacs, Ken wrote:

>
> If you feed a sine into both inputs of a ring modulator, you get a sine 
> an
> octave higher, not a rectified sine. It outputs the sum and difference
> signals, so the sum of "1" and "1" is "2 (doubled frequency)," and the
> difference is zero. This only works for sines, of course, because any 
> more
> complex signal gets immediately non-linear.
>
> I'm not sure where you're getting the rectification from. You can do 
> some
> rectification with a few diodes, but that's usually used for AC power. 
> For
> audio applications, there're ways of rigging an op-amp to do that, I 
> think.

But isn't a ring mod a four quadrant multiplier? If the input voltage 
at x
and y is both -1, won't the result be 1? So if I had a guitar signal 
into X,
and square wave swinging from +1/-1 at every zero crossing of the guitar
signal, wouldn't that result in rectification? If I do it with a triangle
and square:


X
+1    /\           /\
     /   \        /    \
  0 +------+-----+-------+-------------------------------
            \   /
-1          \/

Y
+1 -------       --------

  0 -------------------------------------------------

-1        -------


Degrees:
0:   0 * 0 = 0
45:  .5 * 1 = 1
90:  1 * 1 = 1
...
225: -.5 * -1 = .5
270: -1 * -1 = 1

Isn't that rectifying the triangle?

Tomy

I think you're thinking "voltage" instead of "frequency." A ring modulator
IS a four-quadrant multiplier, kind of a "through-zero" VCA, but you have to
think in terms of frequency. You get the sum of the two input frequencies
and the difference of them.

So if you put a 100 Hz sine into x & y, you get 200 and 0.

If you put a 100 Hz sine into one and 200 Hz into the other, you get out a
300 Hz waveform and a 100 Hz waveform (the difference).

With non-sinus waveforms, this addition/subtraction is happening with every
harmonic in each input waveform.

Nothing is being rectified, here, unless as JH pointed out, you drive one of
the output frequencies to zero, which is effectively DC, and that would
"lift" the waveform above the ground-crossing.

I'm really not sure what you're doing with the voltage calculations, there.

-----Original Message-----
From: Thomas Hudson [mailto:thudson@...] 
Sent: Thursday, 06 December, 2001 1:24 PM
To: Tkacs, Ken
Cc: 'motm@yahoogroups.com'
Subject: Re: FW: [motm] My failed experiment


On Thursday, December 6, 2001, at 06:23 AM, Tkacs, Ken wrote:

>
> If you feed a sine into both inputs of a ring modulator, you get a sine 
> an
> octave higher, not a rectified sine. It outputs the sum and difference
> signals, so the sum of "1" and "1" is "2 (doubled frequency)," and the
> difference is zero. This only works for sines, of course, because any 
> more
> complex signal gets immediately non-linear.
>
> I'm not sure where you're getting the rectification from. You can do 
> some
> rectification with a few diodes, but that's usually used for AC power. 
> For
> audio applications, there're ways of rigging an op-amp to do that, I 
> think.

But isn't a ring mod a four quadrant multiplier? If the input voltage 
at x
and y is both -1, won't the result be 1? So if I had a guitar signal 
into X,
and square wave swinging from +1/-1 at every zero crossing of the guitar
signal, wouldn't that result in rectification? If I do it with a triangle
and square:


X
+1    /\           /\
     /   \        /    \
  0 +------+-----+-------+-------------------------------
            \   /
-1          \/

Y
+1 -------       --------

  0 -------------------------------------------------

-1        -------


Degrees:
0:   0 * 0 = 0
45:  .5 * 1 = 1
90:  1 * 1 = 1
...
225: -.5 * -1 = .5
270: -1 * -1 = 1

Isn't that rectifying the triangle?

Tomy

On Thursday, December 6, 2001, at 11:20 AM, Tkacs, Ken wrote:

>
> I think you're thinking "voltage" instead of "frequency." A ring 
> modulator
> IS a four-quadrant multiplier, kind of a "through-zero" VCA, but you 
> have to
> think in terms of frequency. You get the sum of the two input 
> frequencies
> and the difference of them.

I understand that with frequency you get the sum and difference. But we 
wouldn't
call it multiplier unless somewhere multiplication was going on. If I put
DC 1.5 volts into X and DC 2.0 volts into Y, won't I get DC 3.0 volts on
the output? And if I put two negative DC voltages into X and Y, won't the
output be positive?

Okay, I just went and played around with the O-scope. Obviously my 
assumptions
were half wrong. Putting DC -1.5 and -2 into a ring modulator results in
an output of +.5. A triangle wave into both X and Y results in a waveform
resembling a upside down full rectified sin. So this means the response
is exponential? Is there such a creature as a linear response four 
quardrant
multiplier.

Tomy