[sdiy] CV inputs with bipolar "amount" knobs

[cheater cheater]

On Sun, Jul 25, 2010 at 02:47, Mattias Rickardsson <mr at analogue.org> wrote:
> Interesting topic.
>
> Here's Roman Sowa's version of the buffer/inverter solution, with a
> shunting resistor at the output giving a "bipolar fake log" response.
> The log ("audio taper") response is good for two reasons: our general
> nonlinear preferred feeling of signal level pots; and the widening of
> the null point since the pot curve is less steep in the middle.
>
>  http://www.sowa.synth.net/modular/inverter.gif

But if you want the curve to be less steep around the center you want
an exponential or even better an nth power response such as x^3. A
logarithmic response makes it more difficult to zero in on the center.
The curve needs to be steeper on the middle so that the function can
cover the whole range while the curve is flatter at the extremes. This
way it lessens the precision at the center while improving the
precision around the extremes. Have you actually tried it? If it
behaves the way you describe, maybe the response isn't actually
logarithmic at all.

> Speaking of null point problems caused by different behaviour of the
> buffer and the inverter - would it be better to use two inverters
> connected in series instead, and send their two outputs to the pot?
>
> Perhaps the limitations of the first inverter such as slew rate and HF
> roll-off would not occur much in the second inverter, removing
> problems with spikes etc passing through in the null point?

No, their frequency responses would accumulate. In the end a simple
model of the situation looks like this: every partial of the signal at
frequency f has amplitude a(f) in the original signal; when put
through the inverter it is raised by d(f) decibels above the original
signal and flipped.

If you only use one inverter and the clean signal and mix between
them, at the zero you get these levels for the frequency f:

The original signal: a(f)
the inverted signal: i(f) = -(a(f)+d(f))

Added:

x * a(f) + (-y) * i(f)
= x * a(f) + (-y) * (a(f) + d(f))
= (x-y)*a(f) - y * d(f)

The problem is that d(f) is not constant across all f. So as much as
you could find an x and y such that some of these zero out, the others
will not.

With two inverters:

The inverted signal: i(f) = -(a(f) + d(f))
The twice inverted signal: j(f) = a(f) + 2d(f)

Added:

x * j(f) + (-y) * i(f)
= x * (a(f) + 2d(f)) + (-y) * (a(f) + d(f))
= (x-y) * a(f) + (2x-y) * d(f)

You're not removing the error, just flipping its sign. When x = y
(i.e. the pot is at the center) then the first formula ends up:

(x-y)*a(f) - y * d(f)
= (x-x) * a(f) - x*d(f)
= 0*a(f) - x*d(f)
= -x*d(f)

whereas the second formula ends up:
(x-y) * a(f) + (2x-y) * d(f)
= (x-x) * a(f) + (2x-x) * d(f)
= 0 * a(f) + x*d(f)
= x*d(f)

so you're just flipping the sign of the error.

Or put simply: it's like an inverter is a low-pass filter.

Cheers,
D.


> To conclude, I'm also curious about what Mark Verbos just asked,
> regarding Buchla's bipolar solution. The 259 has several of them, for
> instance.
>
> /mr
>
> On 20 July 2010 00:08, mark verbos <mverbos at earthlink.net> wrote:
>>
>> On Jul 19, 2010, at 5:50 PM, David G. Dixon wrote:
>>
>>>> If it's a DC thing: as I understand it the only problem is, the null
>>>> point is not exactly in the middle of the pot travel. Crude way to
>>>> take care of this: mount the knob on the pot shaft so that the
>>>> marking on the knob is in middle position at the null point.
>>>
>>> I believe Harry was talking about centre-detent pots, in which case you
>>> can't really choose the centre independently.  However, if you use 1%
>>> resistors and a decent opamp, I can't imagine this should be a problem.  I
>>> mean, after all, if you need absolutely zero CV to your circuit, unplug
>>> the
>>> damn patch cord!  If that's not good enough for you, then a 1k trimmer off
>>> one end of the pot will do the job; which end depends on where the "true
>>> centre" is relative to the detent.
>>
>>
>> A trim pot in parallel with the pot is easier.
>>
>> BTW. is there any reason to have the pot after the opamps rather than
>> sending the signal through a resistor to the wiper and then one side to an
>> inverter and the other side into the summer like Buchla does?
>>
>> Mark
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