Why do the orders of hi/low pass filters go in 6 dB increments?

1

A low pass filter - the first order is -6 dB per octave and as the order increases, so the cut is increased by 6 dB i.e. -12 dB, -18 dB, -24 dB, etc.

Why is it 6 dB? Why can't it be 4 dB for instance?

Ryan Ashton

Posted 2015-10-10T09:26:50.677

Reputation: 11

Yeah - fair enough. But I don't think that's the answer. I think there might be a technical/mathematical answer. – Ryan Ashton – 2015-10-10T10:50:21.987

Answers

3

The 6 dB per octave roll-off is simply an approximation of the properties of the first order RC circuit - low pass filter design, i.e. it is not a convention or related to the fact that it is the double of 3 dB.

Actually the roll-off is not exactly 6 dB per octave, it is 20log2 = 6,0205999132796239042... and this 20log2 formula again is a generalization of the tendency of the roll off when you look over multiple octaves (take a look at the roll-off wikipedia article for the mathematical details):

enter image description here

The center frequency or cut-off frequency is more a convenient convention though - at -3 dB the power is reduced to half, which is a good indication that the filter kicked in. With first order filters the center frequency at -3 dB is also easily derived from the component values. There may be other scenarios and filter types where other center frequency conventions are used. Thus the center frequency or cut-off at 3 dB is not strictly tied to the general roll-off properties (the slope). Take a look at this question at the electronics SE for additional details on this topic.

For n-order filters the properties of the first order filter still applies, which is why you see the "6 dB" steps/relations

Other roll-off slopes are thinkable of-course and the actual choice of components may influence how that curve is in reality.

Michael Hansen Buur

Posted 2015-10-10T09:26:50.677

Reputation: 3 612

1"6dB" is totally related to the cut-off attenuation ("3dB"). A half power point (where the cut-off frequency is) is 10log(0.5) = ~-3.0103 dB and not -3dB. Even your graph shows that. – Andy aka – 2015-10-13T18:59:40.017

It may look like that at first hand and yes for first order filters there is definitely a correlation in roll off per octave and cut-off numbers. BUT this is by convention and not strictly derived from the roll off. Take a look here: http://electronics.stackexchange.com/questions/127798/cutoff-frequency-exactly-3-db-of-power

– Michael Hansen Buur – 2015-10-14T08:15:01.763

Yes I know that question - I produced one of the answers! – Andy aka – 2015-10-17T16:41:21.747

1

A simple, natural, 1st order low pass filter (be it mechanical, electrical or any physical type imaginable) reduces the output signal in a natural way. For instance, above the cut-off frequency point, the output amplitude reduces proportional to frequency. An example might be an electrical low pass RC network: -

enter image description here

So, if the cut off frequency is 1 kHz and you put 10kHz in, the output amplitude will have fallen to one-tenth (-20 dB) and, if you put 20kHz in the output will have fallen to one-twentieth (-26.021 dB).

The output amplitude reduces proportional to the frequency.

Relative to 10 kHz, the 20 kHz amplitude is halved. The difference between 10 kHz and 20 kHz is one octave and a halving of the signal is 20*log(2) = approximately -6.021 dB but it's easier to say -6 dB.

Cascading two of these filters produces an attenuation of signal with frequency that is twice the amount of one filter so, a 2nd order filter attenuates at ~12.042 dB/octave. A 3rd order attenuates at ~18.06 dB/octave.

Andy aka

Posted 2015-10-10T09:26:50.677

Reputation: 2 036

-3

Standardization.

You can make your own in whatever increments you feel like. (Down vote?)

EDIT:

I would suggest contacting a company that makes filters if you need the why. (Manufacturing processes, cost, etc Standardization...) The most you'll get out of this forum is a bunch of copy and paste info and audiophile fan boy stuff.

Seriously, it's math. And it's not hard.

I built and installed a passive brick wall hp filter that attenuates nearly everything below 180hz on a guitar. It is sweepable to 220ish before it starts a one octave boost from low mid freqs to around 900hz. It cost all of 5 dollars for parts...

F-N-newb

Posted 2015-10-10T09:26:50.677

Reputation: 1

1Naturally, simple filters are 1st order, 2nd order, 3rd order etc. and these are achieved fairly simply with simple components. Trying to make something that is somewhere between two natural filters is a lot harder and costlier. – Andy aka – 2015-10-17T16:43:01.873

Costlier in a manufacturing environment? You can do whatever you'd like in your own workshop which is what I was getting at, good on the down vote though. Whatever happened to experimentation? – F-N-newb – 2015-10-17T18:56:04.473

It's all simple math, and there is software that let's you experiment with the "expensive parts" before you prototype and build the equipment. Or did the audiophiles just naturally assume that everyone buys equipment? – F-N-newb – 2015-10-17T18:58:21.157

1I design equipment for a living. I'm an electronic analogue engineer and I laugh at the stuff audiophiles come up with. Your answer didn't address the question that's why I downvoted it but I can't speak for others. – Andy aka – 2015-10-17T20:16:48.647