If you own a GVCO you might be able to help.
So, if I feed a sequence into the volt/octave input that consists of the “C” major scale ie: C D E F G A B C
it will play that back just fine. However if I turn on the VCO’s built in quantiser, set to chromatic, it plays back C sharp D E F G A B C.
I have calibrated the VCO so it’s not that. I’m completely baffled as to why the first C in the scale is sharp.
Is my module faulty or are they all like this?
Quantizing already quantized notes leads to a dark portal of mysteries and small voltage offsets.
Add a small +- voltage offset to your input cv and your half step off will be fixed, or migrate.
Depends on how your quantizer rounds up or down when the input is close to a note.
Quantizing already quantized notes leads to a dark portal of mysteries and small voltage offsets.
Add a small +- voltage offset to your input cv and your half step off will be fixed, or migrate.
Depends on how your quantizer rounds up or down when the input is close to a note.
The notes from the sequencer are not already quantized. Its a digital sequencer so the voltages are just precise.
So the fact that just the first note is getting shifted up a semi tone really makes little sense to me.
But I don't build or design modules so thats not much of a surprise.
I have repeated the experiment but instead of using the built in quantizer I used an external one and the sequence plays as expected.
So it does appear to point to some issue with the internal quantizer to me.
I got bit by this once too. Your quantizer is working correctly.
It's analog, so what does 'precise' really mean?
You need a voltage to play a C2, so you request that precise 2.000 volts from the digital side.
But there's the D/A and wires and jacks and dusty pots and humidity and transistors drifting from heat.
By the time it arrives at the quantizer's A/D converter, maybe it's 2.01 volts, maybe it's 1.99 volts.
Only a 0.02 volt difference, but the quantizer wants to round them to two different semitones.
This behaviour also depends on your specific quantizer's rules for how it rounds to the nearest note.
It's analog, so precise is always relative and non-linear and squishy.
The non-linear part is important too.
What happens at 1.99 vs. 2.01 volts might be different than the response from 2.99 vs. 3.01 volts.
(only the low C went sharp, but not the higher ones)
Think squishy, not digital.
Dieter said it best:
General function and problems of quantizers
When the input CV of the A-156 changes the module converts the incoming voltage into a stepped voltage at the correspondig CV output. For this there the modules uses voltage thresholds in 1/12V steps. If the incoming CV is very close to a threshold value it may happen that this voltage is converted once to voltage 1 and later to voltage 2 (with voltage 2 = voltage 1 +/- one step or +/- 1/12V). The A-156 does not "know" that the voltage comes from the same source. It just converts an incoming non-stepped voltage into a stepped voltage.
Example: think about a sequence with N steps where the voltage of step #3 is close to a threshold. When the sequence is running it may happen that for step #3 two different voltages appear (+/-1 semitone) at different passes. To avoid this flaw the quantizer would have to "know" that it has to convert a sequence with N steps and that after N conversions the same CV as during the last run has to be generated - provided that the voltage is very close to the former value. He would have to memorize the "old" voltages of all N steps and compare them to the "new" voltages. When the difference between an "old" and "new" voltage is below a certain threshold (e.g. less than half a semitone or less than half of 1/12 V, i.e. about 40mV) the old output value is taken. But this job cannot do the A-156 as it does not "know" anything about a sequence structure but simply converts the incoming continuous voltage into a quantized voltage.
To avoid this the quantizer and the sequencer would require a common "supervisor". For example with our Dark Time stand-alone sequencer this would not happen as the quantizers "knows" the values of all controls during the last run because they are stored in an internal memory. When the advance to a new step is triggered the unit compares the new voltage to the stored voltage of the last run. There has to be a significant difference between the two values. Otherwise the same voltage is generated. Without storing the values of the former run this would be not possible. This problem occurs for all quantizers which are not embedded into the sequencer structure because they don't not "know" anything about a sequence but simply convert the incoming non-stepped voltages into stepped voltages.
Even already quantized control voltages should not be used as CV source for the module. In this case similar problems may occur if the voltage steps of the incoming signal are close to the voltage thresholds of the A-156. In this case it may help to add a small offset voltage to the incoming CV signal so that the voltage steps of the incoming signal are no longer close to the voltage thresholds of the A-156. But the general question is: why using a quantizer if the control voltages are already quantized.
--
I really appreciate the vast esxplanation on how a quantizer works and I understand the principal. But I really think the point of what I'm saying is being missed. The Oscillator will play the scale perfectly without the internal quantizer being switched on.
So the voltages coming from the sequencer are precise enough to produce the correct pitch. Both the sequencer and the oscillator are digital as well. As I said in my previous post, I have tried putting an external quantizer in the signal path and it has zero impact. All the notes in the scale are played correctly. Only when I turn on the oscillators built in quantizser do I get the problem and its only the first not in the scale that is a semitone sharp. So adding offsets will just put everything else out. I know because I have tried it already. The only explanation I can think of is that that the internal quantizer is in some way faulty or incorrectly programmed.
Can I have your broken oscillator?
Can I have your broken oscillator?
-- noodle_hut
Sure you can.
Where do you live? I will pop it round shortly.
Even already quantized control voltages should not be used as CV source for the module. In this case similar problems may occur if the voltage steps of the incoming signal are close to the voltage thresholds of the A-156.
I can't see how this is relevant in this context. If the module is correctly calibrated a voltage that isn't close to the thresholds should always be correctly quantized. It doesn't matter if it comes from another quantizer, a keyboard or a random S&H.
I don't have any experience with this VCO, but I see that it supports different scales. Do you have the same problem with all scales? Have you tried to give it controlled voltages around the level that should result in a C to see if you can get that C at all on how much it's off?
The calibration mode seems to be quite helpful. As I understand it, it will report the measured voltage. What's it saying about the C that's played as a C#?
Thanks for your input e_v_k.
So I tried a new experiment today and created a 5 step sequence that consists of C D E G A ( pentatonic major ).
Without the quantizer enabled I get that exact sequence played back. If I turn on the internal quantizer and set it to pentatonic major I get D D E G A. So the first note is now a whole tone sharp. If I change to quantizer to chromatic I get C Sharp as the first note followed by all the correct notes.
If I go back to me original sequence and play it back with the quantizer on and set to chromatic I can get the first note to play C by detuning the "tune" knob by about 48 cents but all the other notes that follow are now out of tune.
I'm still sctarching my head on this one.
What I really need is someone else that has a GVCO to try the same experiments and see what results they get.
What's the voltage range you're using? 0-1V or something higher? What about the result for several octaves, is it only the lowest note that's sharp?
Get an analog voltage source (not a digial one) and monitor it with a digital voltmeter, to at least hundreths precision.
Set your Erica to quantize and feed that voltage to it, from 0v to 5v.
Observe the voltage when the oscillator jumps to the next half step.
Is it when the input is just past ~((1volt / 12.0) * note), or just before?
Or does it flip when the to-be-quantized voltage is past half way between?
Round up or round down or mid-nearest. Or maybe some internal rules table or something.
I couldn't find the quantizer's rules in the vco manual, it just says how to turn it on and set it.
Anyway, now read the voltages from your digital sequencer as it runs a chromatic scale.
Are they each a 'precise' 0.0833333333333333333333 volts apart from each other?
Or 0.0830, or 0.0820, or 0.085?
Analog. Squishy.
There is an implied imprecision down in the 0.003333333s, especially when you're adding them together.
And so there is no "C", or 'in tune', just 'sounds in tune', even when it never really can be.
An analog quantizer can only approximate what your intent for 'C' is, and they all can do it differently.
It's a best guess, because your sequencer isn't sending 'A', it's generating ~0.8333333 volts. Give or take.
Now, knowing the vco's quantizing behaviour, you can see why it is doing what it does and when.
I still think your oscillator is working correctly.
Don't quantize already quantized CVs. You'll get stray semitone errors.
I never like it when someone asks for help, and the annoying reply is 'Why are you doing that'?
But, why are you doing that?
"> What's the voltage range you're using? 0-1V or something higher? What about the result for several octaves, is it only the lowest note that's sharp?"
-- e_v_k
The range is 0V to +5V in this experimant. Transposing up tones, semitones or octaves yeilds the same result and just the first note is sharp.
In response to a question you asked earlier that I missed... When the eronious C sharp plays the GVCO tells me its playing C sharp. It doesn't think its playing C but outputting C Sharp.
I did some new experiments with the internal quantizer on this morning and here are the results please take it as read that with the quantizer off all sequences play back as expected as it does if I use an external quantizer.
If I send in a sequence that is C then C one octave up the two notes play back at the same pitch. If I increase the interval between the two Cs by another octave I get a one octave interval, this behaviour is the same when the quantizer is set to chromatic or octaves.
I also tried another experiment inspired by noodle_hut.
If I create a chromatic sequence by ignoring the voltages coming from the sequencer and just setting them by ear, I can get the GVCO to play back a chromatic scale ( with the internal quantizer on ) but every note is a semitone sharp above what the GVCO its tuned to when the quantizer is off. So in some ways one could argue it is, kinda, working.
But I have access to six other quantizers and not one of them produces the results the internal quantizer does. They all work as expected and what you put in you get out ( unless you put in a rouge voltage and then that gets corrected/adjusted so the resulting note is what was desired )
Get an analog voltage source (not a digial one) and monitor it with a digital voltmeter, to at least hundreths precision.
Set your Erica to quantize and feed that voltage to it, from 0v to 5v.
Observe the voltage when the oscillator jumps to the next half step.
Is it when the input is just past ~((1volt / 12.0) * note), or just before?
Or does it flip when the to-be-quantized voltage is past half way between?
Round up or round down or mid-nearest. Or maybe some internal rules table or something.
I couldn't find the quantizer's rules in the vco manual, it just says how to turn it on and set it.Anyway, now read the voltages from your digital sequencer as it runs a chromatic scale.
Are they each a 'precise' 0.0833333333333333333333 volts apart from each other?
Or 0.0830, or 0.0820, or 0.085?
Analog. Squishy.
There is an implied imprecision down in the 0.003333333s, especially when you're adding them together.
And so there is no "C", or 'in tune', just 'sounds in tune', even when it never really can be.
An analog quantizer can only approximate what your intent for 'C' is, and they all can do it differently.
It's a best guess, because your sequencer isn't sending 'A', it's generating ~0.8333333 volts. Give or take.Now, knowing the vco's quantizing behaviour, you can see why it is doing what it does and when.
I still think your oscillator is working correctly.Don't quantize already quantized CVs. You'll get stray semitone errors.
I never like it when someone asks for help, and the annoying reply is 'Why are you doing that'?
But, why are you doing that?-- noodle_hut
Hello noodle_hut.
I have just measured ( for the last time ) the voltage coming from my sequencer for the first step ( the C ) and its 0.000V
So a precise voltage. I can send you a picture if you do not believe me. Its about as perfect a voltage as you could hope for.
The GVCO plays C when its given this voltage as I would expect it to. If I turn on the intermal quantizer the GVCO produces C sharp. I don't know why you are insisting that this is anything other than wrong.
I don't think we are getting anywhere with this so probably best I stop asking for help now.
Once again, thanks for your input.
On tiny voltages. precision, and irrational (non-terminating) decimal numbers
Zero volts is easy, but what does it give you for an E?
0.416666687 volts? 0.40? 0.4166666666666667?
If you feed the vco a known 0v input, or no input at all, and turn on it's quantize function,
does the oscillator jump from C to C# ? It should according to what you wrote.
If it does, maybe it is broken, or just wrong, or is off calibration.
It's not data. There is no 'C' to be recognized on input.
It doesn't know, and all quantizers can do it differently.
Precision is a compromise that computers and we humans apply when choosing how many digits to represent.
The difference between semitones in 1v/oct is ~0.08333333 volts.
(with more ‘precision’ it’s closer to 0.08333333333333332871)
This is a non-terminating repeating decimal. It has an infinite number of digits,
so you could say it has infinite precision in its exact mathematical form, and internally in our voltages,
but that’s not how we talk about those numbers.
If we’re set up to play a C1 if fed 1.000000000 volts, when does the quantizer jump to C#1?
At 1.00100 volts? Or at the more correct 1.08200 volts?
Will it jump to B if it’s fed 0.999 volts?
It could, and in a very few case, that would be correct.
A quantizer rule that says:
‘only an exact 1.0000000v is a C1, anything below is a B0, anything above 1.0000000v is a C#1’
isn’t going to give you that C1 very often, if ever.
What are possible internal quantizer rules to play a C1?
Here's a jump-when-halfway-to-pitch quantizing design:
A 1v/oct quarter step would be ((1.0/12.0) / 2) = ~0.0417
So, if it is fed a voltage between 0.583v and 1.0417v it outputs 1.000 volt (our C1).
Below 0.583v, it’s a B0, so we round down and output (1.000 - 0.0833) = ~0.917 (our B0).
Above 1.0417, we round up to ~1.083333v (our C#1).
We spread this rule across a 10v range and we’ve got a workable chromatic scale quantizer.
But the start and end points of those threshold windows are arbitrary and implementation specific.
Maybe it’s a hard-coded table. Maybe it has to accommodate a slow or meh resolution A/D converter.
(Remember we’re dealing in differences of less than 1/12 of a volt). They all can (and will) do it differently.
If your feed device’s idea of start-end windows are different than your quantizer's, and they probably are,
you’ll get those semitone jump errors that you’re hearing.
--
noodle_hut,
Please stop now. I know how a quantizer works. I know you are trying to be helpful but you keep dismissing what I say and just plough on with the same repeated info. Just stop before you respond and think for one second about what I am telling you.
I feed into the GVCO 0.000V and it produces a C. Its internal tuner tells me its a C and my Peterson Strobe tells me its a C.
I turn on the internal Quantizer and the pitch changes to C Sharp. This is shown by the GVCO's inbuilt tuner and my Peterson.
Just forget everything else for now. There voltage is 0.oooV so its not near the cusp of where the quantizer wants to round up to the next semitone. We are not changing the voltage entering the GVCO at all. Just turning on its quantizer.
Are you telling me that this is a quantizer that is working as it should?
Anyway, as I said earlier. Lets draw a close to this now as we are not really getting anywhere other than arguing back and forth and I don't want to continue.
I thank you again for your time though.
On tiny voltages. precision, and irrational (non-terminating) decimal numbers
Zero volts is easy, but what does it give you for an E?
0.416666687 volts? 0.40? 0.4166666666666667?
If you feed the vco a known 0v input, or no input at all, and turn on it's quantize function,
does the oscillator jump from C to C# ? It should according to what you wrote.
If it does, maybe it is broken, or just wrong, or is off calibration.
It's not data. There is no 'C' to be recognized on input.
It doesn't know, and all quantizers can do it differently.
Precision is a compromise that computers and we humans apply when choosing how many digits to represent.
The difference between semitones in 1v/oct is ~0.08333333 volts.
(with more ‘precision’ it’s closer to 0.08333333333333332871)
This is a non-terminating repeating decimal. It has an infinite number of digits,
so you could say it has infinite precision in its exact mathematical form, and internally in our voltages,
but that’s not how we talk about those numbers.
If we’re set up to play a C1 if fed 1.000000000 volts, when does the quantizer jump to C#1?
At 1.00100 volts? Or at the more correct 1.08200 volts?
Will it jump to B if it’s fed 0.999 volts?
It could, and in a very few case, that would be correct.
A quantizer rule that says:
‘only an exact 1.0000000v is a C1, anything below is a B0, anything above 1.0000000v is a C#1’
isn’t going to give you that C1 very often, if ever.
What are possible internal quantizer rules to play a C1?
Here's a jump-when-halfway-to-pitch quantizing design:
A 1v/oct quarter step would be ((1.0/12.0) / 2) = ~0.0417
So, if it is fed a voltage between 0.583v and 1.0417v it outputs 1.000 volt (our C1).
Below 0.583v, it’s a B0, so we round down and output (1.000 - 0.0833) = ~0.917 (our B0).
Above 1.0417, we round up to ~1.083333v (our C#1).
We spread this rule across a 10v range and we’ve got a workable chromatic scale quantizer.
But the start and end points of those threshold windows are arbitrary and implementation specific.
Maybe it’s a hard-coded table. Maybe it has to accommodate a slow or meh resolution A/D converter.
(Remember we’re dealing in differences of less than 1/12 of a volt). They all can (and will) do it differently.
If your feed device’s idea of start-end windows are different than your quantizer's, and they probably are,
you’ll get those semitone jump errors that you’re hearing.
Like I said, quantizing already quantized notes leads to a dark portal of mysteries and small voltage offsets.
--
noodle_hut,
Stop.pretty please. and for the very last time....I'm not quantizing an already quantized note.
It would maybe be possible to examine this further, but it's not going to change the fact that your module is behaving strangley. It would indeed be interesting to hear from someone with the same module.
It would maybe be possible to examine this further, but it's not going to change the fact that your module is behaving strangley. It would indeed be interesting to hear from someone with the same module.
-- e_v_k
I'm guessing its not a very common or popular module ? No one with one has chipped in at all. I have sent a video to Erica Synths but not heard anything back on it so far.
