Common Signal Multiplication Modulation Input Pattern

I’m working on a wavefolding effect, and I want to have the potential for pre- and post- effect for amplitude modulation or ring mod.

I’ve got an input, along with a pre-gain and post-gain. The modulation is optional, but since it is a multiplication with the signal, I need to be careful about nulling out the main signal.

Currently, I’ve been doing something like this, in pseudo-code:

in = \in.ar([0, 0]);
mod = \mod.ar([0, 0]); 
modPreGain = \modPreGain.kr(0);
modPostGain = \modPostGain.kr(0);

modPre = mod * modPreGain; 
modPre = modPre + (modPreGain |==| 0);

modPost = mod * modPostGain;
modPost = modPost + (modPostGain |==| 0);

sig = in *  modPre;

sig = { effect core here } 

sig = sig * modPost;

This works fine at the moment, but I’m wondering if there’s a better way to achieve this type of thing. I’ve thought about Select or ‘.blend’ but if I want control of the gain, then that requires an additional parameter control for pre- and post-.

Thanks,
Boris

Hm, the way you wrote it here, there would be a large discontinuity between e.g. modPreGain = 0.0001 and modPreGain = 0.

Another way to think of it: a dry-wet mix. modPreGain = 0 would correspond to a fully dry signal (no RM), = 1 would be fully wet (100% RM signal), and every mixture in between.

// not modPreGain, but modPreAmt
sig = LinXFade2.ar(sig, sig * mod, modPreAmt * 2 - 1);

Algebraically, this works around to a modulator with an interesting property (which also, when you see it, makes intuitive sense).

// mix = 0 .. 1
linxfade = (b - a) * mix + a

= ((sig * mod) - sig) * mix + sig
= (sig * mod * mix) - (sig * mix) + sig

// here, factoring out sig means
// it's worked back around to ring modulation!
= sig * ((mod * mix) - mix + 1)

// ... i.e., mod has now mul = mix, and add = 1 - mix
// let's plot the final modulator signal then

{
	var mix = Line.ar(0, 1, 0.01);
	SinOsc.ar(1000) * mix + (1 - mix)
}.plot;

ringmod-gain

So the LinXFade2 is the same as scaling the modulator (analogous to gain) while also offsetting it so that the peak is always 1.0 (= no nulling-out).

hjh

1 Like

Would there be a discontinuity? If modPreGain = 0.0001, mod would oscillate between 0.0001 and -0.0001, and then, since modPreGain is not equal to 0, the boolean expression would add 0 to the signal.

eg:

modPreGain = 0.0001;
modPre = mod * modPreGain; // oscillates between +/- 0.0001
modPre = modPre + (modPreGain |==| 0); // since modPreGain != 0, this would be the signal from line above + 0. 

For example:

{
	var sig = SinOsc.ar(1000);
	var modPreGain = Line.ar(0, 1, 0.01);
	var mod = 1;
	var modPre = mod * modPreGain;
	modPre = modPre + (modPreGain |==| 0);
	sig * modPre;
}.plot;
)


Or

(
{
	var sig = SinOsc.ar(1000);
	var modPreGain = Line.ar(0, 1, 0.01);
	var mod = SinOsc.ar(500);
	var modPre = mod * modPreGain;
	modPre = modPre + (modPreGain |==| 0);
	sig * modPre;
}.plot;
)

The thing I lose by using the dry/wet blending is the ability to either overdrive the signal to taste or to use a negative gain value to flip the phase. The phase flipping especially is interesting to me in my current use case… flipping post- but not pre-, for example.

Oh, I see, the discontinuity may be from no AM or ring mod to a tiny bit:

(
{
	var sig = SinOsc.ar(1000);
	var modPreGain = Env([0, 0, 1], [0.005, 0.01]).ar;
	var mod = SinOsc.ar(500);
	var modPre = mod * modPreGain;
	modPre = modPre + (modPreGain |==| 0);
	sig * modPre;
}.plot(0.015);
)

(
{
	var sig = SinOsc.ar(1000);
	var modPreAmt =  Env([0, 0, 1], [0.005, 0.01]).ar;
	var mod = SinOsc.ar(500);
	sig = LinXFade2.ar(sig, sig * mod, modPreAmt * 2 - 1);
}.plot(0.015);
)

Hm, then:

// -1 <= modGain <= 1
sig = sig * ((mod * modGain) + (1 - abs(modGain)))

(
a = {
	var modGain = Line.ar(-1, 1, 0.02);
	var positiveGain = modGain.fold(0, 1);
	// plot modulator only
	var mod = SinOsc.ar(1000);
	[
		// upper: no negative gain allowed
		((mod * positiveGain) + (1 - positiveGain)),
		// lower: with negative gain
		((mod * modGain) + (1 - abs(modGain)))
	]
}.plot(duration: 0.02);
)

hjh

1 Like

Thanks! This is working nicely!