Part of Respawned Fluff's answer to this recent question on headphones got me thinking about low-pass filters:

It seems they actually invert the transfer function of the dummy head/ears via software because they say right before that that "Theoretically, this graph should be a flat line at 0dB."... but I'm not entirely sure what they do... because after that they say "A “natural sounding” headphone should be slightly higher in the bass (about 3 or 4 dB) between 40Hz and 500Hz." and "Headphones also need to be rolled-off in the highs to compensate for the drivers being so close to the ear; a gently sloping flat line from 1kHz to about 8-10dB down at 20kHz is about right." Which doesn't quite compile for me in relation to their previous statement about inverting/removing the HRTF.

This is talking about headphones, not circuits, but it made me wonder if it's possible to create such a transfer function with an analog circuit. First-order filters have a slope of -20 dB/decade. Is there anything weaker? I suppose the transfer function would be something like this:

$$H(s) = \frac 1 {1 + \sqrt{s / \omega_c}}$$

  • 1
    \$\begingroup\$ In RF, you could use the skin effect to get a \$1/\sqrt{f}\$ characteristic. I imagine at lower frequencies you could use a combination of discrete elements to form both poles and zeros to approximate that characteristic over some band. Or find a material (like in a ferrite bead, for example) with a convenient frequency dependence. \$\endgroup\$
    – The Photon
    Commented Nov 12, 2015 at 23:47
  • 2
    \$\begingroup\$ First order filters have a maximum slope of 20dB/decade. Now consider one with an R in series with the C (as in lag-lead compensation) - its ultimate slope is ... 0 again, and it may never reach 20dB/decade. A network of several of these at different frequencies can deliver a slope as shallow as you want. \$\endgroup\$
    – user16324
    Commented Nov 13, 2015 at 0:10
  • \$\begingroup\$ I disagree that a headphone needs the bass boosted and the treble reduced. Use the "cut" parts of a Baxandall treble tone control. Replace its pot with two resistors. \$\endgroup\$
    – Audioguru
    Commented Aug 12, 2019 at 2:34

3 Answers 3


Yes it is but it's more complex because you have to use breakpoints formed by a multiple array of resistors and capacitors:

enter image description here

The above is a piecemeal 3dB per octave (10 dB per decade) filter. It was designed to convert white noise to pink noise. See this link.

Here's another white to pink noise filter using an op-amp with a few more breakpoints:

enter image description here

You could convert it to 2 dB per octave or 4 dB per octave but the accuracy comes from the number of breakpoints and therefore the number of RCR stages.

Note that pink noise rolls off at 3 dB per octave and here's the final "circuit" and graph:

enter image description here

enter image description here

  • \$\begingroup\$ hey Andy, thanks for the circuits. could you get a better photo of the first passive circuit or tell us what the resistor and capacitor values are. the numbers are pretty badly smudged. \$\endgroup\$ Commented Nov 13, 2015 at 0:55
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    \$\begingroup\$ @robertbristow-johnson quite smudged indeed, but kind of readable: top-left R: 6.3k, middle 3 Rs: 3k, 1k, 300, middle 4 Cs: 1uF, 270nF, 2x 47nF. the rightmost cap seems to be 33nF. \$\endgroup\$
    – Kroltan
    Commented Nov 13, 2015 at 3:10
  • \$\begingroup\$ @robertbristow-johnson this was the site I took the picture from: decodesystems.com/pink-noise.html - the left resistor I reckon is 6k8 but it is the general principle that is important here. If you look at the 3rd circuit - it has values that are all 10x the first circuit. \$\endgroup\$
    – Andy aka
    Commented Nov 13, 2015 at 10:22

...it made me wonder if it's possible to create such a transfer function with an analog circuit. First-order filters have a slope of -20 dB/decade. Is there anything weaker?

The answer is yes. The keyword to this is fractional order filters and there is some literature on this topic, although not much. These filters are based on fractional order elements, which are usually approximated with conventional, lumped-element circuits. Optimization techniques or Padé approximations may provide a close-enough implementation. Wikipedia has an article on Fractional-order systems which may be a starting point to learn about these.


I would use a series of five high-shelf filters spread evenly at octave spacings, each one contributing a cut of -2dB. The first would have a corner at 1kHz, the next at 2kHz, the next at 4kHz, the next at 8kHz, and the last at 16kHz. That would meet your spec pretty nicely, giving you a -2dB cut from 1-2kHz, a -4dB cut from 2-4kHz, a -6dB cut from 4-8kHz, a -8dB cut from 8-16kHz, and a -10dB cut above 16kHz.

(If you really HAD to have an even more gradual roll-off than that, you could even use 10 high-shelfs, at half-octave spacings and each contributing a cut of -1dB, but I really think that would be serious overkill.)


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