Butterworth Filter Design

Thread Starter

pinkyponky

Joined Nov 28, 2019
152
Hi,

I'm learning butterworth filter designing, and I have gone though the Electronics Tutorial (https://www.electronics-tutorials.ws/filter/filter_8.html). I have a question, please can you clarify my question.

1632860167295.png
1632860251197.png



Where this value '10' is comes from (marked in Red in the above text)?. Is this gain of Amin?.

From the above text, I have illustrated the graph of the gain of max and min.
1632859006932.png
If the above graph is right, then I want to design a circuit Amax=0dB, Amin=-120dB. Therfore, what will be the gain of Amin?.

Please can you answer above two question (marked in Red)?.
 
Last edited:

Audioguru again

Joined Oct 21, 2019
3,632
The graph you show is not real because it does not show the -3dB cutoff frequency that is what a Butterworth filter is designed for.

-20dB is 1/10th the voltage.

The output is -20dB at 4 times the frequency (2 octaves) where the gain is +0.5dB. You need to know the frequency where the gain is at -3dB to see how steep is. The slope a second order Butterworth filter drops at -12dB per octave, a third order filter drops at -18db per octave. The filter is probably a 4th order.
Did you know that -60dB is 1/thousandth and -120dB is one millionth the voltages.
 

Papabravo

Joined Feb 24, 2006
16,938
You may be confusing gain and attenuation. This is an active filter so positive gain would correspond to negative attenuation. However it is common to specify filters in terms of attenuation. What the specifications are telling you is the maximum attenuation at some point in the passband can be no larger than 0.5 dB. Positive attenuation is negative gain. since this is a Butterworth filter and the response is maximally flat, all the values of attenuation for frequencies less than 200 rads/sec will be less than 0.5db. You are given another point where the minimum attenuation is 20 dB. It is allowed to be larger but it cannot be smaller.

Edit: Normally when you do this calculation, you round the value of n up to the next higher integer to guarantee that the attenuation specification is met. In your case you would design a filter of order 3, since ceiling(2.42) = 3

Floor and ceiling functions - Wikipedia

Edit2: You also need to find the corner frequency in order to complete the design.
 
Last edited:

Ian0

Joined Aug 7, 2020
3,459
If you want the response to drop by 120dB in one octave (from 5kHz to 10kHz) then you need a 20th order filter (ten inductors and ten capacitors for a passive filter or ten op-amps 20 resistors and 20 capacitors for an active filter.
It seems only a couple of weeks since we last discussed this.
 

Ian0

Joined Aug 7, 2020
3,459
When one is new to filter design, it is very tempting just to concentrate of the amplitude response, in the belief that a filter than can go from 0dB to -120dB in a single octave must be the best filter ever.
But as you learn about amplitude response, you must also learn about phase response and group delay. Both are more difficult to understand thoroughly than amplitude response.
Then you might understand why filters steeper than 4th order are very rarely encountered: only in very specialised application, but very rarely in audio, and never in power supplies or any system which includes feedback.
 

Thread Starter

pinkyponky

Joined Nov 28, 2019
152
Hi,

I want to pass the frequency of 0.1hz to 5khz, So fc=5khz, and above this frequency should be roll-off with faster/steeper attenuation. For this, I read that chebyshev filter is good choice. But, can you help me how to calculate the Q factor, passive component values (R and C) and order of the filter?.
 

LvW

Joined Jun 13, 2013
1,311
My recommendation: Consult another book or publication about filters. To me, the referenced paper is very unclear:

Examples:
* In the paper, both quantities, Ho as well as ε , are defined as "maximum passband gain Amax".
Now, when you look at the given formula H1=Ho/SQRT(1+ε) you can see that something must be wrong with these definitions.

* Moreover, it sounds not very logical to define in the example that " Amax = 0.5dB at a pass band frequency (ωp) of 200 radian/sec (31.8Hz)". Why we should use the expression Amax for the magnitude at the corner frequency?
Normally, in filter theory the term Amax is the maximum value which can be reached within the passband (for Butterworth Amax=Ao for f=0 Hz)
___________________
I think, the referenced paper is NOT a good tutorial for learning the basics of active filter design. It is really confusing.,
 

Ian0

Joined Aug 7, 2020
3,459
If you are designing a filter to meet an arbitrary specification, then please state as much information as you have. i.e. passband spec, stop band spec etc.

If you are designing a filter for your own purposes then the choice of type and filter and rate of attenuation depends on WHAT you are trying to filter.
What is the signal you wish to keep, and what is the interfering signal above 5kHz that you wish to remove?
How much signal is there between, say, 4kHz and 6kHz?
For instance, if there is a lot of signal between 4kHz and 6kHz, then a high-Q Chebyshev filter may be a bad choice, as there may be some components of your original signal that correspond to the resonance of the filter, and they would be amplified.
If you have a signal with square edges, they will cause ringing at the Chebyshev peak - you may wish to avoid this.
You may wish to preserve the shape of your low-frequency signal - in which case you need good group delay specs.
If there is nothing much between 5kHz and 20kHz in your signal, then you do not need steep roll-off.
 

Audioguru again

Joined Oct 21, 2019
3,632
Why do you want to completely eliminate normal high audio frequencies? Those frequencies are important in music (harmonics) and speech (consonants). We hear to 20kHz.
 
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