I am a newbie and trying to understand the operation of this White Noise Coloration Filter - I have contacted the original author but at time of writing have bot received a response.

The circuit is as follows (I have retained nomenclature from original post for clarity)

EDIT: Error in original diagram, RV1 should be 10K not 100R enter image description here

I can't figure out how the bottom half of this is a high pass filter (unless the resistor to ground is R5 and RV1). I can sort of see how the top half is a low pass filter but cant understand how/why the potentiometer and the first 4k7 resistors function.

The OP says

R4/C4/R5 and R6/C5 form low- and high-pass passive filters respectively. The pot pans between them by proportionally grounding one, or the other, or somewhere in between

I've tried to look at this by splitting it into two halves but not sure if that is valid and have made limited progress in my comprehension.

Any pointers on how to go about decomposing this to understand how it works would be gratefully appreciated

  • \$\begingroup\$ Filters imply states and resistors have none. Therefore the active elements are the capacitors here, and if you look at the bottom section there is a series capacitor (and at the top a parallel one). \$\endgroup\$ May 25, 2021 at 15:48
  • \$\begingroup\$ @aconcernedcitizen I wouldn't state it like that. a) "active" elements have a special meaning in electronics (and a capacitor is none) and b) a capacitor without a resistor is not a filter, at all. \$\endgroup\$ May 25, 2021 at 16:11
  • \$\begingroup\$ @MarcusMüller I was thinking in terms of actively participating to filtering due to their reactance. It does sound a bit strange, I agree. However, a capacitor does filter even if only by itself. If you meant in terms of a needed current path, then an OTA+C wouldn't need one, and it would be an integrator. I'm not talking about real-life considerations (current generator limited resistance, parasitics, imperfections, etc). Here it makes sense if you consider that the input and the output have an implicit load. \$\endgroup\$ May 25, 2021 at 16:37
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    \$\begingroup\$ I suspect the pot should be more like 10K than 100R for proper operation. \$\endgroup\$ May 25, 2021 at 16:56
  • 1
    \$\begingroup\$ yes a 100 ohm pot becomes a -40 dB pad \$\endgroup\$ May 25, 2021 at 17:22

2 Answers 2


Try to analyze the equivalent circuit when the RV1 potentiometer "slide" is at the max position (left side and the right side).

Case one:


simulate this circuit – Schematic created using CircuitLab

Case two


simulate this circuit


R4/C4/R5 and R6/C5 form low- and high-pass passive filters respectively.

Yes, if they existed in isolation. However, they are tied together through R5-C5, so this is a strong simplification – they can't really be modeled as individual filters, sadly:

The pot pans between them by proportionally

No. Let's assume RV1 is somewhere in the middle, i.e. both "halves" are ca. 50Ω. RV1 strongly loads R4 (which is orders of magnitudes larger), thereby very strongly changing the frequency-dependent behaviour of R4/C4.

Even worse for C5: we can basically assume that for all but extreme settings of RV1, the part of current that flows through RV1 instead of C5 dominates the current through R6, thereby very much changing the high-pass behaviour.

This isn't trivial to decompose by hand (you can do it; start by splitting RV1 into two resistors Ra, Rb to ground, then consider one of them to be in parallel to C4, Z_1 = C4||Ra; then do a delta-Y transform on R4,R5,Z_1, then simplify further... it doesn't become pretty). I'd put it into SPICE.

What you can do is analyze the "edge" cases of RV1 being 0 between one end and ground. But that's kind of the "boring" case, because you just ground half of your circuit.


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