Filter RC calculation

Started by knutolai, March 09, 2021, 04:31:16 AM

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knutolai

Hi

I've been reading on state variable filters and came over Rod Elliot's excellent article on the subject:
https://sound-au.com/articles/state-variable.htm

In figure 2 of his article (pictured below) he uses an interesting pot/resistor configuration to control the center frequency.


The center frequency range is listed as 43 to 285 Hz. Focusing on the U1B integrator I understand why the center frequency is 285 Hz freq when the VR1A wiper is at the top extreme (connects to U1A output): 1/(2*Pi*R6*C1) = 285 Hz

But how do one go about calculating the center frequency when VR1A wiper is at the bottom extreme (connects to R8)? If we rearrange the formula to calculate 'R' the 43 Hz center frequency implies a equivalent resistance of 37k Ohm. I'm failing to see how this product is calculated. Any help?

I was drawn to this pot/resistor config as it seems to offer possibilities for adjusting the linearity of the pot travel. Could a similar pot/resistor config be used for the sallen-key low pass filter topology?

Rob Strand

#1
QuoteBut how do one go about calculating the center frequency when VR1A wiper is at the bottom extreme (connects to R8)? If we rearrange the formula to calculate 'R' the 43 Hz center frequency implies a equivalent resistance of 37k Ohm. I'm failing to see how this product is calculated. Any help?
The important parameter is the current through R6.

When VR1 is at maximum the current going into R6 is I = Vpin1 / R6.

When VR1 is at the minimum the voltage at the wiper of will be lower than when VR1 is on maximum.    So clearly the current going into R6 is less.    That turns out to be the same as making R6 larger.

The thing that complicates the calculations a little bit is R6 loads down the divider formed by VR1A and R8.   There's many way to analyse this circuit but this way is probably the easiest:   Consider a voltage divider with VR1A feeding R8 and R6 in parallel.
R6//R8 = 10k // 5k6 = 3.59k.   Now calculate the voltage division ratio,

  Vmin = Vpin1 * (R6//R8) / ( VR1A + (R6//R8) )
            = Vpin 1 * 3.59 / (20 + 3.59) = 0.152
or,

  Vmin / Vpin1 = 0.152

So one way to look at it is to stop here and say the current into R6 is
I = Vmin / R6 = Vpin1 * 0.152 / R6  = Vpin1 (0.152 / R6)  = Vpin1 / (6.58 * R6)

So the reduction in current going into R6 due to the divider has made R6 effectively look 6.58 times larger than its actual value.

R6_effective = 6.58 * R6  = 6.58 * 5.6k ohm = 36.8 k ohm

So to calculate the frequency use R6_effective in the original formula in place of R6.   It should clear we are just reducing the frequency for the max setting of 285 by  a factor of 6.58, 285 Hz /  6.58  = 43Hz.

Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

jatalahd

#2
Hi,

This type of filter was discussed in thread: https://www.diystompboxes.com/smfforum/index.php?topic=122295.0

I calculated the result and posted that in that thread (see reply #18)

In this circuit you refer to, the formula to get 43 Hz is:

fmin = (R6||R8||VR1A)/(2*pi*VR1A*R6*C1),

where the R6||R8||VR1A is the parallel resistance of R6, R8 and VR1A. This can be calculated as R6||R8||VR1A = 1/(1/R6 + 1/R8 + 1/VR1A).

At least I got an answer of 43.248 Hz from that equation.

Hopefully this helps.
  • SUPPORTER
I have failed to understand.

knutolai

Thank you for two great replies. I look forward to start plotting in excel and reading through the thread you linked!

Any thoughts on the applicability for such a configuration to be used in a sallen-key topology?

Rob Strand

QuoteAny thoughts on the applicability for such a configuration to be used in a sallen-key topology?
It would end-up being a round about way of just varying the resistances directly.

FWIW, it is possible to just vary R6 (and R7) directly in the State-variable filter.   You can see an equivalent idea in the LFO frequency control on the Boss CE2 vs Boss BF2, as well as many sine-wave oscillators.

About  3 months ago there was a thread about simulating a megaphone using filters.  In the later part of that thread I derived a ground referenced Sallen and Key filter.   The reason it was useful in tat case was because the frequency control was being done VCA's.   You might be able to use the filter from that thread but it will take a little bending.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

knutolai

Quote from: Rob Strand on March 09, 2021, 06:21:21 AM
QuoteAny thoughts on the applicability for such a configuration to be used in a sallen-key topology?
It would end-up being a round about way of just varying the resistances directly.

FWIW, it is possible to just vary R6 (and R7) directly in the State-variable filter.
The problem with this approach when varying with a dual linear pot (what I have on hand) is that the filter sweep doesn't feel right. I guess that would be due to octaves being exponential. I'll have to do some plotting but I think with some adjustments this pot-resistor configuration could be the basis for a nicer sounding filter sweep. My idea would be to do something similar to Rod Elliots volume control:


I'll report my findings

Rob Strand

#6
QuoteThe problem with this approach when varying with a dual linear pot (what I have on hand) is that the filter sweep doesn't feel right. I guess that would be due to octaves being exponential. I'll have to do some plotting but I think with some adjustments this pot-resistor configuration could be the basis for a nicer sounding filter sweep. My idea would be to do something similar to Rod Elliots volume control:
Yes,  that's *exactly* what happens.    You need a reverse log taper pot for the series method.   The voltage divider has the advantage you can tweak the taper with resistors.

Interestingly the Boss BF2 LFO pot is a reverse log taper for the same reason.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.

Rob Strand

FYI, notice the 5k6 resistor on the integrator is already providing the loading effect on the pot, without the need to add another resistor to the pot.
Send:     . .- .-. - .... / - --- / --. --- .-. -
According to the water analogy of electricity, transistor leakage is caused by holes.