Talk:Butterworth filter

message: It states that a general property of Laplace transforms is that H(-j\omega) = \overline{H(j\omega)} (so-called Hermitian symmetry). However, this is only true for the Laplace transform of real signals or systems.  — Preceding unsigned comment added by Yates (talk • contribs) 21:48, 19 March 2023 (UTC)
thread_title: Hermitian Symmetry

message: I though all first order filters were the same. ie they're not designated Butterworth, Bessel, TBT, Chebychev, Gaussian, Cauer etc because there can be no difference between them. So that diagram could be any old RC network. Right?--Light current 02:27, 4 October 2005 (UTC)
replies: text: "All frequency-normalized filters have the same response?"  "All first order filters have the same shape?"  something like that?  it is a good idea to point out that they're the same.  (though this doesn't mean they're irrelevant...)  — Omegatron 06:53, 14 October 2005 (UTC)

text: Yes, of course.  That's why we need to change the wording.  He means that it is not necessary to refer to a first-order filter as Bessel, Butterworth, etc.  A first-order low-pass Butterworth filter looks the same as a Bessel; it's completely defined by the cutoff frequency.  Pfalstad 02:39, 13 June 2006 (UTC)

text: A filter with a pole at zero is not low-pass.  I'm saying a low-pass first order is defined only by the cut-off frequency.  Similarly for high-pass.  "About the strongest statment you can make is that for 1st order a Butterworth is the same as most other common types."  That's what we're saying!  Pfalstad 03:35, 17 June 2006 (UTC)
thread_title:  First order filters are all the same 

message: Do we really need this trivia about his/her first name in the lead para. It has nothing to do with the subject. I removed it once but its found its way back. Can we have some justifiaction for keeping it?--Light current 02:35, 4 October 2005 (UTC)
replies: text: Why not?  It's natural to want to know who this Butterworth guy was.  Unfortunately we don't know enough about him to write a whole article, I guess.  Pfalstad 15:27, 4 October 2005 (UTC)
replies: text: Of course it's important.  Ideally we'd have it in an article about him, but there isn't enough info available to devote an entire article to him.  — Omegatron 15:52, 4 October 2005 (UTC)

text: That whole sentence could be moved lower in the article.  Pfalstad 22:43, 4 October 2005 (UTC)

text: If you look at articles like Fourier series, Bessel function, etc., they all have the name of the inventor in the first sentence.  In this case, we have a little extra parenthetical remark because his name is not known.  This could be put in the article on Stephen Butterworth, if we had one, but the article would be one sentence long, so we don't.  Go look at Chebyshev polynomials, which has the guy's name twice (once in Russian) in the first sentence, if you really want to see something that interrupts the flow.  I didn't move the Stephen sentence (Omegatron did, on Oct 13) and I don't really care if you want it lower in the article, but if it were me, I would put it right at the top, in the first or second sentence, without a separate heading.  Pfalstad 04:27, 10 January 2006 (UTC)
thread_title:  Butterworths first name- is it important? 

message: The actual equations in Laplace domain (not just the magnitude):
1st order:
:H(s)={1\over s+1}
2nd order:
:H(s)={1\over s^2 + \sqrt{2} s + 1}
3rd order:
:H(s)={1\over s^3 + 2 s^2 + 2 s + 1}
From here.  Others 
There's no simple universal equation that works for all of them??  I thought there was.  There is for the magnitude...  — Omegatron 16:41, 8 October 2005 (UTC)
replies: text: Those are just the magnitude half.  These three are the full transfer function, but I don't see the condensed single-equation form.  I'll think about it a bit.  — Omegatron 20:08, 8 October 2005 (UTC)
thread_title:  Equations 

message: How does the magnitude response in the Bode plot differ from, say, a Bessel first order filter>? If it doesnt, why are we crediting Mr Butterworth with it?--Light current 11:07, 10 October 2005 (UTC)
replies: text: Aren't we already having a discussion about this elsewhere?  — Omegatron 11:53, 10 October 2005 (UTC)
thread_title:  Bode Plot 

message: It's a very nice pic, but a lot of people only have 800x640 screens, you know...  Pfalstad 16:57, 10 October 2005 (UTC)
thread_title:  Enormous pic 

message: n	                                                                            	Factors of Polynomial B_n(s)                               	Magnitude response         	Pole-zero plots
1	(s+1)                                                                       	(s+1)                                                      	1 \over \sqrt{1 + \omega^2}	...
2	(s + e^{j(\pi/4)})(s + e^{-j(\pi/4)})                                       	s^2+\sqrt{2}s+1                                            	1 \over \sqrt{1 + \omega^4}	...
3	(s+1)(s + e^{j(\pi/3)})(s + e^{-j(\pi/3)})                                  	(s+1)(s^2+s+1)                                             	1 \over \sqrt{1 + \omega^6}	...
4	(s + e^{j(\pi/8)})(s + e^{-j(\pi/8)})(s + e^{j(3\pi/8)})(s + e^{-j(3\pi/8)})	(s^2+0.7654s+1)(s^2+1.8478s+1)                             	1 \over \sqrt{1 + \omega^8}	...
5	...                                                                         	(s+1)(s^2+0.6180s+1)(s^2+1.6180s+1)                        	...                        	...
6	...                                                                         	(s^2+0.5176s+1)(s^2+\sqrt{2}s+1)(s^2+1.9318s+1)            	...                        	...
7	...                                                                         	(s+1)(s^2+0.4450s+1)(s^2+1.247s+1)(s^2+1.8022s+1)          	...                        	...
8	...                                                                         	(s^2+0.3986s+1)(s^2+1.111s+1)(s^2+1.6630s+1)(s^2+1.9622s+1)	...                        	...
Just an idea.  (I don't like approximations.) :-)  — Omegatron 04:18, 14 October 2005 (UTC)
thread_title:  More math 

message: In the article, the derivation of the Butterworth coefficients is given as follows:
"The first (2n-1)th derivatives of g are zero for ω = 0. This means that the rate of change of gain with respect to ω is zero i.e., the gain is constant with ω, where n is taken to be the order of the filter."
This is incorrect due to the following:
The premise of the low-pass Butterworth filter is that the passband is maximally flat, having a gain |H(w)| = 1  for w = 0. For a first-order Butterworth filter (or any first-order filter, for that matter), the transfer function is of the form: H(w) = 1/(jw - p0), where p0 is a pole. If n = 1 (first-order), then the value of the pole p0 is required to equal -1 by the constraint that |H(0)| = 1. With any first-order filter, the derivative with respect to w evaluated at w=0 equals zero, regardless of the value of p0. It is only for higher-order Butterworth filters that the requirement that  d^k/dw^k |H(w=0)| = 0  becomes necessary.
Moreover, the formula  (2n-1)  is incorrect due to the fact that requiring (2n-1) derivatives to equal zero for a transfer function of only n poles would create an over-determined system for all  n > 1 . The derivation should read:
 |H(0)| = 1 
 d^k/dw^k |H(w=0)| = 0   for  0 < k < n 
Breebo 21:23, 27 November 2005 (UTC)
thread_title:  Incorrect Derivation 

message: Well, I bit the bullet and did a massive rewrite of the article. The main changes are:
* A new section entitled "Simple Example" which gives the basics for a simple 3rd order filter without a lot of derivation, etc. and relating an actual circuit to an actual gain formula to an actual pole plot.
* Elimination of a number of redundant or unreadable gain plots
* Inclusion of a design section showing how to design a Butterworth filter. (this could be improved, especially the Cauer topology stuff).
* Some derivations, and making it so the meaning and usefulness of the Butterworth polynomials is evident (hopefully)
PAR 15:53, 20 July 2006 (UTC)
thread_title:  Massive rewrite 

message: Sure would be nice if this article included a section on digital implementation, like in the Chebyshev Filters article.  I would want to know how to go from the analog filter, once designed, to the digital filter coefficients and the algorithm that uses those coefficients.  I'd do it, but it's not my competency (by any stretch!).
:The bilinear transform procedure is essentially the same for any analogue prototype.  IMO, the Chebyshev article does not need this; as the general procedure should be in the article on the bilinear transform. Oli Filth 00:46, 22 September 2006 (UTC)
replies: text: This strikes me too. Most of the filters are explained in math terms, and are accompanied by hardware implementations, but never code examples. Personally I would understand a code example (C, Basic, Perl, whatever), but not the math stuff. Why favor hardware over software? JoaCHIP (talk) 15:35, 15 February 2009 (UTC)
replies: text:  Because the code example would be identical for any digital filter implementation; all that changes are the values of the coefficients.  Oli Filth(talk|contribs) 22:53, 15 February 2009 (UTC)
thread_title:  How about a digital implementation section? 

message: C_2(R_1+R_2)=2\cos\left(\frac{2k+n-1}{2n}\right)
I'm not sure what's happening here, where do the k and n come from?
It seems like that should be C_2(R_1+R_2)=\sqrt2
unsigned edit by user:71.193.170.17 23:27, 23 May 2007
:The k and n are the kth pole of an nth order filter.  Yes, it would be \sqrt{2} for a second order filter which would only require one Sallen-Key stage, but for higher orders several stages are needed in cascade and the coefficients will be different.  Also, you have not got the expression quite right, it is;
::C_2(R_1+R_2)=-2\cos\left(\frac{2k+n-1}{2n} \pi \right)
:The minus sign is missing out of the article as well. SpinningSpark 00:29, 16 February 2009 (UTC)
thread_title:  sallen key topology 

message: Under this title it says the log10(gain) is plotted. It is just the plain gain that it plotted though, isn't it?  —Preceding unsigned comment added by 86.150.1.188 (talk) 20:17, 17 January 2008 (UTC)
thread_title:  A simple example 

message: The article says: "There are a number of different filter topologies available to implement a linear analogue filter. These circuits differ only in the values of the components, but not in their connections."  I don't understand that sentence as the different topologies have different values and different connections.  Billgordon1099 (talk) 23:10, 14 August 2009 (UTC)
replies: text: Obviously "only ... but also ..." should be "not only ... but also ...".  WP:SOFIXIT. Dicklyon (talk) 00:16, 15 August 2009 (UTC)
replies: text: Actually, I think the intended meaning was probably that a given topology differs only in component values when implementing different classes of filter.  I think the second sentence should be struck altogether as it is not very relevant to an article specifically on Butterworth. SpinningSpark 10:32, 15 August 2009 (UTC)
thread_title:  Unclear sentence 

message: I was hoping that this article would explain better how to pick the values of inductors, capacitors, and resistors to get the desired passband, roll-off, and impedance. ie: there is a nice butterworth filter provided in LTspice educational examples, that has five inductors and five capacitors of various values. but reading this article does not seem to enlighten the subject. -- Waveguy (talk) 21:57, 23 February 2010 (UTC)
replies: text: There are many ways to implement an electronic filter. Each has its own topology. The way it is now, if you know you want a Butterworth filter, use the transfer function provided in this article, then go to the article on that particular topology in order to implement that transfer function. This is explained under the "Filter design" section, but not very clearly, I agree. PAR (talk) 22:56, 23 February 2010 (UTC)
thread_title: too much confusion

message: I think the example is wrong.  I did the math out and I think that L_3 needs to be set to 2 henrys in order for H(s) to equal 1/(1+2s+2s^2+s^3) 68.230.151.20 (talk) 13:13, 4 April 2014 (UTC)
replies: text: Your calculation is wrong.  My calculation gets the same answer as the article.  They are also the same component values as shown in the table in Matthaei et al. page 107. SpinningSpark 14:27, 4 April 2014 (UTC)
replies: text: Okay, I believe you, but would you mind looking at my work so I can understand what I did wrong? http://68.230.151.20:4050/butterworth.jpg 68.230.151.20 (talk) 20:37, 4 April 2014 (UTC)
replies: text: You are going wrong on the C2/(L1+C2) term. C2 is in parallel with (L3 and R4 in series) and the equivalent impedance of that whole block must be used instead of C2.  Also, your notation is odd, you seem to be using L1, C2 etc to mean the impedance of L1, C2 rather than the normal meaning of their inductance/capacitance.  However, you have correctly substituted in the correct impedance for these elements. SpinningSpark 20:56, 4 April 2014 (UTC)
replies: text: Yeah I guess I was supposed to use Z.  I got the calculation to work, thank you.  Glad I posted here before changing things.  68.230.151.20 (talk) 01:18, 5 April 2014 (UTC)
thread_title: Example circuit

message: hi folks, i wanted to give the article, which is totally cryptic to non-mathematicians, an example in the intro which shows the usage in a practical context. simplification of an animation curve in my example uses the filter in a high class CGI software. key issue: it more or less keeps the curve flow intact. butterworth‘s own graph might be historically interesting, but it‘s as cryptic to the general public as the whole lemma.
reminds me a bit of a discussion we had in the german wikipedia about the angular momentum, german: Drehmoment, from drehen = to rotate. the intro of that article was so focussed on quantum mechanics that the word „drehen“ was not mentioned at all.
so, if you consider my example with the computer animation curve wrong, pls go ahead and give a handy application of the filter to the interested public in the first paragraph. and don‘t collaterally revert my rectification of an exotic image format. Maximilian (talk) 20:13, 1 February 2019 (UTC)
replies: text: I consider that example wrong, not just unconventional.  I think they just adopted the name Butterworth for some reason.  For that application, you need a zero-phase interpolating filter.  The Butteworth is neither linear phase nor interpolating, so it's hard to see why they call it that. Dicklyon (talk) 21:52, 1 February 2019 (UTC)

text: As for the "general public", I don't understand how they could be interested if they don't know what a frequency response it. Dicklyon (talk) 01:29, 2 February 2019 (UTC)

text: It says it "removes noise from data without affecting the FCurve’s minimum or maximum values".  This is not something a Butterworth filter would do.  The illustration suggested it was an interpolating filter; also not something Butterworth would do.  And symmetric left–right; again unlike Butterworth.  No idea why they're calling it Butterworth. Dicklyon (talk) 04:40, 3 February 2019 (UTC)
thread_title:  practical example revert 

message: :G(\omega) = {\frac{1} \sqrt{1+{\varepsilon^{2}(\frac{\omega}{\omega_c})}^{2n}}},
"\varepsilon is the maximum passband gain, and ωc  is the cutoff frequency. If ωc= 1, then \varepsilon=1, and the amplitude response of this type of filter in the passband is 1/ ≈ 0.707, which is half power or −3 dB."
This can't be right. 
At  \omega = 0 the gain of the filter is 1 not {\varepsilon}.
And the gain of the filter at \omega = \omega_c is 
\frac{1} {\sqrt{ 1 + \varepsilon^{2} }} which is \frac{1} {\sqrt{2}} if \varepsilon = 1. This is −3 dB. If \varepsilon \ne 1 it is not −3 dB. — Preceding unsigned comment added by Bodger Boffin (talk • contribs) 19 jun 2019 16:41 (UTC)
:I reverted it.  Hopefully we'll hear here from the editor who changed it that way. Dicklyon (talk) 19:37, 19 June 2019 (UTC)
thread_title:  Gain 

message: 'Worsens' seems like a strange choice of word. After all, an 'ideal' low-pass filter would have a sinc impulse response, which rings forever. (This does make it 'acausal', which makes it impossible to implement when your filter output is a function of time (but not if, e.g., it's a function of position).) Bruce Mardle (talk) 13:28, 6 June 2020 (UTC)
thread_title:  "ringing in step response, which worsens with increasing order"