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