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Hermitian Symmetry

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)[reply]

By the way, this is a very, very good article! — Preceding unsigned comment added by Yates (talk • contribs) 21:50, 19 March 2023 (UTC)[reply]

untitled

From sci.electronics.misc:

:Looks like it's STEPHEN Butterworth OBE !

:A Leslie Green (CEng MIEE) replied to a post that I made on the IEE web
:site. Aparently S. Butterworths 1930 article says he worked for a that he
:worked for the  Admiralty Research Laboratory.

:Then "Data Junkie" from the soc.genealogy.britain newsgroup sent me this
:obituary from the Times...

> The Times, Friday, Nov 07, 1958; pg. 15; Issue 54302; col G
> Mr. S. Butterworth
> Category: Obituaries
>
> [Extract]
>
> Mr. Stephen Butterworth, O.B.E., who died recently, joined the Admiralty
> scientific staff in 1921 after having served for a short time in the
> National Physical Laboratory. He retired in 1945.
>
>
> Data Junkie

> The award of an OBE made me look at the London Gazette at
> http://www.gazettes-online.co.uk/archiveSearch.asp?WebType=0&Referer=WW2
> where I found the following in the Birthday Honours for 1942:
>
> ================
> SUPPLEMENT TO THE LONDON GAZETTE, 11 JUNE, 1942
>
> CENTRAL CHANCERY OF THE ORDERS
> OF KNIGHTHOOD.
>
> St. James's Palace, S.W.1.
> 11th June, 1942.
>
> The KING has been graciously pleased, on
> the occasion of the Celebration of His Majesty's
> Birthday, to give orders for the following
> promotions in, and appointments to, the Most
> Excellent Order of the British Empire: -
>
> To be Additional Officers of the Civil Division
> of the said Most Excellent Order: -
>
> Stephen Butterworth, Esq., M.Sc., Principal
> Scientific Officer, Admiralty.
> =================
>
> He is also shown in the Navy List, and the March 1939 edition has him
> as a Principal Scientific Officer at the Admiralty Research Laboratory
> in the Scientific Research and Experiment Department of the Admiralty.

In case we write an article about the man. - Omegatron 12:58, Apr 5, 2004 (UTC)

The article does not address the phase loss associated with the increase in filter order. This is an important point for practical filter usage (and why other filters types are used). The filter exhibits flat (unity) reponse in the pass band region but the trade off is more phase loss versus say a Chebyshev filter of the same order. The phase loss can become important in signal quality when used as a signal filter and can effect system stability when used in a close loop conrtroller.

Images

I am working on finding the general transfer function of the filter, and making a better graph that shows multiple orders like so http://ece.gmu.edu/~gbeale/ece_320/images/sampling_02_01.gif

http://www.freqdev.com/guide/images/fig5.gif

Also we need pole zero plots of the first few orders, like so http://www.ee.bgu.ac.il/~dsp/slides/slides/Butterworth_poles.jpg - Omegatron 15:21, July 19, 2005 (UTC)

First order filters are all the same

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)[reply]

re omegatron's latest change, all first order filters do not have the same response, because it depends on where the pole is. Can we improve the wording or just take out that assertion about all filters being the same? It's not really necessary to state that. I'd improve it myself but I can't think of anything. Pfalstad 05:42, 14 October 2005 (UTC)[reply]

"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)[reply]

Souldn't a pole at w=0 vs. pole at i be considered different shapes? (i.e. simple RC vs. op amp integrator). One is unstable at DC.

 3ch 68.109.164.40 02:28, 13 June 2006 (UTC)[reply]

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)[reply]

But a 1st order filter with only a pole at zero has no cutoff frequency (what is 3 dB down from infinity?) so it -can't- be defined by the cutoff frequency. Also, you can have a pole -and- a zero (even demanding stability). A response like (1-iw/w_1)/(1-iw/w_2) is approx. 1 for w << w_1,w_2 and w_2/w_1 for w >> w_1,w_2. I wouldn't consider that as the same shape as a butterworth either. [What definition of order is used here? pole count? max(number of poles, number of zeros)?] About the strongest statment you can make is that for 1st order a Butterworth is the same as most other common types.

 3ch 68.109.164.40 03:19, 17 June 2006 (UTC)[reply]

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)[reply]

""All frequency-normalized filters have the same response?" "All first order filters have the same shape?" something like that?" is saying much something stronger, even alowing frequency normalized to include hipghpass-lowpass mappings.

Also, the def'n I learned was a low pass attenuated high frequencies more than low, so 1/iw and (1-iw/w_1)/(1-iw/w_2), w_2 < w_1 are low pass. You can claim instability disqualifies the first but -not- the second. Order I learned as the highest of the order of the numberator or the denominator but I don't think that is the source of conflict here.

Maybe then you need to state [or link to] your definitions of high pass, low pass, order etc. explicitly?

 3ch Colonel hack 04:24, 17 June 2006 (UTC)[reply]

Butterworths first name- is it important?

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)[reply]

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)[reply]

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)[reply]

Well if its important to you two guys, can we put it somwhere other than the lead para where it detracts from the flow?--Light current 22:39, 4 October 2005 (UTC)[reply]

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

THank you.--Light current 23:28, 4 October 2005 (UTC)[reply]

I see Stephen has found his way back near the top of the article. Whats the point of arguing and agreeing something if its going to get changed back again? Do we really need this at the top of the page -- seriously??--Light current 04:11, 10 January 2006 (UTC)[reply]

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)[reply]

In the chebyshev article the name does not interrupt the flow beacise attention is not drawn to it like it is in this article. If his name was quoted with no comment as 'Stephen Butterworth', I would have no problem.--Light current 04:31, 10 January 2006 (UTC)[reply]

Equations

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}+2s^{2}+2s+1}

From here. Others [1]

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)[reply]

THe equation describing the Butterworth approximation are given in terms of omega (w) on the article page. Substituting s=jw in to this gives you all the above TFs. I think.

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)[reply]

First or second order filters do not have names.

Moved to Wikipedia talk:WikiProject Electronics#1st-_and_2nd-order_filters

Bode Plot

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)[reply]

Aren't we already having a discussion about this elsewhere? — Omegatron 11:53, 10 October 2005 (UTC)[reply]

No I think this is a different one! (similar maybe)--Light current 12:01, 10 October 2005 (UTC)[reply]

I think the other ones about Butterworth in the LPF page. But I could be wrong. I ve forgotten.--Light current 12:05, 10 October 2005 (UTC)[reply]

Yeah youre right! Dammit! Sorry - getting confused.--Light current 12:09, 10 October 2005 (UTC)[reply]

Enormous pic

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

More math

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)[reply]

And these are the "minimum phase" implementations, but other implementations can be had for the same magnitude response, which is what actually defines the filter (if I understand this correctly...) — Omegatron 04:29, 14 October 2005 (UTC)[reply]

I'm strongly in favor of at least substituting exact values for the decimal approximations in the current version, and the complex factored forms would be useful as well.71.193.170.17 23:31, 23 May 2007 (UTC)[reply]

Incorrect Derivation

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)[reply]

Mention of Cauer form

Just one question-- Why are we mentioning Cauer on the Butterworth page?--Light current 04:08, 10 January 2006 (UTC)[reply]

Well I don't know what I'm talking about here, but I believe there is something called Cauer form (unrelated to Cauer filters, just discovered by the same guy) which is a method of realizing an analog filter (converting a set of poles/zeros to components). Pfalstad 04:37, 10 January 2006 (UTC)[reply]

Why does the diagram have a switch in it? Oli Filth 11:52, 23 August 2006 (UTC)[reply]

Its not a switch. The dashes indicate more similar sctions of passive components. --Electron Kid 01:58, 25 August 2006 (UTC)

[reply]

The way you've drawn it looks almost identical to the symbol for a push-to-make switch (see e.g. [[2]], see the diagram near the bottom). I think a better way to represent this would be something like the following:

We should also mention:

The "first" component doesn't have to be a shunt capacitor, it may be a series inductor (in which case all component types are swapped).
This is normalised to $\omega _{c}=1$ , and $R_{s}=R_{o}=1\Omega$ .
Seeing as in normalised form, the equations for capacitor and inductor values is the same, we don't necessarily need both equations.

Can you explain how to find omega_c in the article? Do we need to multiply each value of capacitance and inductance by Ts? —Preceding unsigned comment added by 70.55.254.146 (talk) 19:58, 24 September 2007 (UTC)[reply]

Oli Filth 13:44, 25 August 2006 (UTC)[reply]

Explanation of maximum flatness

Perhaps it would be useful to explain in the article why the Butterworth formula causes maximum flatness in the passband. In most textbooks this is done by calculating the McLaurin series and showing that the n first derivatives cancel out at w=0. Does anyone else think it could make the article more didactic?

Barna, 3rd July 2006

Massive rewrite

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)[reply]

S-plane diagram

Why is there a log density plot of the s-plane representation of the filter, instead of the standard pole-zero diagram as would be given in any textbook on the subject? The density plot provides no useful information; the only useful bit is the gain along the $j\omega$ axis, and this is obscured by the axis itself! Oli Filth 12:13, 23 August 2006 (UTC)[reply]

The density plot gives the same information as standard pole-zero diagram and more. They give an idea of how the poles and zeroes influence the gain (along the ω axis), and they give an immediate idea of what a pole or zero IS, for someone who might not know. I could remove the axis lines, does that sound like a good idea? PAR 15:06, 23 August 2006 (UTC)[reply]

The density plot gives no extra information, as H(s) is uniquely determined by the location of the poles. Whilst I agree that it does demonstrate that poles are in some rough sense "maxima" of the function, that's not going to help someone who doesn't already understand the relevance of the s-plot.

As you say, the importance (to a newcomer to the subject) is the connection between pole location and the magnitude (and perhaps phase) along the

j\omega

axis. Personally, I think this would be illustrated better with a "normal" pole-zero plot, along with the geometric interpretation - that the magnitude is the inverse of the product of the Euclidean distances from the poles, and the phase is the sum of the angles. Although this would be better off in another article. Perhaps I will write one... Oli Filth 17:11, 23 August 2006 (UTC)[reply]

Well, sure, H(s) is uniquely determined by the function given in the text, so why have a plot at all? It provides no extra information, but it is helpful for giving an idea of what the function looks like.My only complaint is that it's hard to see the zeros. Pfalstad 17:42, 23 August 2006 (UTC) (Oops, forgot the zeros were at infinity; I was thinking of digital filters. Pfalstad 19:00, 23 August 2006 (UTC))[reply]

My point was that the magnitude of the function for

\sigma \neq 0

is irrelevant; the only important issues are the pole locations, and the magnitude (and phase) along the

j\omega

axis. Which is probably why textbooks draw pole-zero plots - they're the simplest graphical representation that contains all the useful information.

Incidentally, the only zeros here are at infinity, so they're going to be pretty hard to illustrate by any method!! Oli Filth 18:17, 23 August 2006 (UTC)[reply]

Yes, I can't think about why the gain function looks the way it does without thinking about what it looks like in complex frequency space. I am working on similar plots for the Chebyshev filters, and the type II do have zeroes. My work page for modifying the present Chebyshev filter page is at User:PAR/Work8 - check out the gain image for the type II filter. When I get this done, I will replace the present Chebyshev filter with it. PAR 18:20, 23 August 2006 (UTC)[reply]

I guess we'll have to agree to disagree on this one. Whilst I admit these diagrams are kind of cool (especially the Chebyshev Type II) I think that's all they are. They aren't going to help someone new to the idea of s-plots to say "Ah! Now it all makes sense!"

And with no disrepect intended, just because you aren't able to visualise what's going on in pole-zero land without a density plot, I don't see that as a reason to introduce a completely non-standard depiction into an encyclopedic article. People working in control theory, for instance, have to cope with the idea of poles moving around in the s-plane without needing to see density plots.

Standard pole-zero sketches are completely intuitive, IMHO - as you move along the

j\omega

axis, if you get closer to poles, the gain goes up, and if you get closer to zeros, the gain goes down. They're really only useful as indicative guides of the general trends of the function ("does the gain go up or down?") and to illustrate the geometric relationships between the various poles and zeros ("the poles are equi-spaced on the unit circle", "the pole is to the left of the zero", "this pole is dangerously close to the jw-axis").

Incidentally, if we are sticking with the density plots, we need to be consistent. The Butterworth plot only contains poles for H(s), whilst the Chebyshev and Elliptic ones show the poles for |H(s)|^2. Oli Filth 19:07, 23 August 2006 (UTC)[reply]

I agree on the consistency point and I will work to fix that. PAR 19:21, 23 August 2006 (UTC)[reply]

I think they do help.. I would have appreciated these diagrams when learning the subject. I'm sure people working in control theory can get along without them, but they are not the target audience for an introductory article on Butterworth filters. The fact that we're leaving out the standard form of the diagrams might be a problem, though. Pfalstad 20:31, 23 August 2006 (UTC)[reply]

Yes, I think that's more the issue I have - the lack of standard diagrams. If these were present, I'd have no issue with the density plots being there too. Oli Filth 23:00, 23 August 2006 (UTC)[reply]

Do you have any links to a suitable standard diagram?PAR 01:20, 24 August 2006 (UTC)[reply]

I don't know specifically where one can find them on the web, but I'm happy to create some. Oli Filth 13:54, 25 August 2006 (UTC)[reply]

How about a digital implementation section?

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)[reply]

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)[reply]

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)[reply]

sallen key topology

$C_{2}(R_{1}+R_{2})=2\cos \left({\frac {2k+n-1}{2n}}\right)$

I'm not sure what's happening here, where do the k and n come from?
It seems like that should be $C_{2}(R_{1}+R_{2})={\sqrt {2}}$

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)[reply]

A simple example

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)[reply]

Unclear sentence

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)[reply]

Obviously "only ... but also ..." should be "not only ... but also ...". WP:SOFIXIT. Dicklyon (talk) 00:16, 15 August 2009 (UTC)[reply]

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)[reply]

too much confusion

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)[reply]

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)[reply]

Questions and comments

1. Both Butterworth and Chebyshev filters can have a passive implementation with the Cauer network. Schematically, they look just about identical. I look at the equations for the inductors and the capacitors for both filters and they look different so I assume that the only difference between the two schematics would be the values assigned to the passive components. Am I saying it right?

2. I read the section about the Sallen–Key topology. In the last equation of the same section, if C₁, C₂, R₁ and R₂ are chosen, I assume that the choice of the "two undefined components" would fall on k and n. I also assume that n is the number of poles and that k is the number of stages. Correct me if I'm wrong. Also, if I cascaded two identical Sallen–Key topology blocks (like in the image), would that make a total of 4 poles? If that is the case, does it mean that the Sallen–Key topology can only offer even-order filters?

3. The digital implementation of the filter should be expanded to include a circuit schematic. Some equations could also be added to complement the analog part discussed above.

ICE77 (talk) 20:35, 11 July 2011 (UTC)[reply]

4. The "Cauer topology" section shows two equations to determine the values of capacitors and inductors. What is the relationship with frequency? It's not shown anywhere. I don't see how a filter can be designed unless there is a reference to a specific frequency.

ICE77 (talk) 18:16, 18 July 2011 (UTC)[reply]

1. Yes correct, and for many other classes of filter including the Cauer filter after who the topology is named, although Cauer was not actually the first to use this filter topology.

4. The section is following the widespread convention of normalising the filter design to ω_c=1. The results have to be scaled to the actual frequency required. See prototype filter.

2. I am not sure what you mean by "fall on k and n". You are free to choose the value of two components. The other two are then completley determined by k and n. If that is what you meant then that is correct. Note, however, that the values are normalised as above in point 4 so scaling for frequency will effectively determine a third component. The fourth component is not usually chosen randomly either - most often the designer will specify an input impedance for the stage based on the requirements of the previous stage.

2. You are correct that a 4-pole filter can be built by cascading two Sallen-Key stages. However, a new calculation of component values is required - the coefficients are not the same in a higher order filter as they were in a second order filter. Odd order filters are possible. The usual trick is tack on a RC circuit at the end to provide the odd pole.

3. A digital implementation does not have a circuit schematic as such. The algorithm is implemented entirely in software. The only real circuitry involved is the A to D converter which got the signal into the digital domain in the first place. Spinning Spark 17:16, 5 August 2011 (UTC)[reply]

SpinningSpark, thanks a lot for the comments. You must know a good deal about filters.

2. "fall on k and n" simply means that k and n have to be chosen. Would the RC circuit follow the 2-pole Sallen-Key block to make a 3-pole filter? That's how I picture it.

3. Interesting: I didn't know that digital filters where entirely implemented with software. I assume those filters go into FPGAs or something like that.

4. I didn't know about normalization and frequency scaling.

ICE77 (talk) 17:58, 11 August 2011 (UTC)[reply]

The one-pole section can go before or after the two-pole section so long as it is buffered. Putting it after means you can use the same op-amp to buffer it. It is possible to make multi-pole Sallen-Key filters using just one op-amp, although they tend not to work very well above three poles. See here for instance, and here. Spinning Spark 17:45, 15 August 2011 (UTC)[reply]

SpinningSpark, the second link you provided points to a book I find rather interesting. I think I'll try go get a copy and take a look at it. Thanks again for the pointers.

ICE77 (talk) 05:00, 30 August 2011 (UTC)[reply]

Example circuit

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)[reply]

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. Spinning Spark 14:27, 4 April 2014 (UTC)[reply]

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)[reply]

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. Spinning Spark 20:56, 4 April 2014 (UTC)[reply]

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)[reply]

practical example revert

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)[reply]

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)[reply]

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)[reply]

could anyone have a look at the (brief) documentation page: ➡️ Maximilian (talk) 08:45, 2 February 2019 (UTC)[reply]

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)[reply]

Gain

G(\omega )={\frac {1}{\sqrt {1+{\varepsilon ^{2}({\frac {\omega }{\omega _{c}}})}^{2n}}}},

" $\varepsilon$ is the maximum passband gain, and ω_c is the cutoff frequency. If ω_c= 1, then $\varepsilon$ =1, and the amplitude response of this type of filter in the passband is 1/√2 ≈ 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 $\varepsilon \neq 1$ 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)[reply]

"ringing in step response, which worsens with increasing order"

'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)[reply]

Is there a typo in the "transfer function" section ?

Based on the following equation:

s_{k}=\omega _{c}e^{\frac {j(2k+n-1)\pi }{2n}}\qquad k=1,2,3,\ldots ,n.

it seems that the transfer function

H(s)=G_{0}\prod _{k=1}^{n}{\frac {\omega _{c}}{s-s_{k}\omega _{c}}}

should actually read as

H(s)=G_{0}\prod _{k=1}^{n}{\frac {\omega _{c}}{s-s_{k}}}

.

Am I right? If so, someone should please fix this. Ffellay (talk) 01:37, 11 February 2023 (UTC)[reply]

After implementation and test of the transfer function polynomials written in complex form for a frequency domain study, I can confirm that it seems I am right. Therefore, I edited the main article with the minor correction proposed above (just for you to know). Ffellay (talk) 00:49, 17 February 2023 (UTC)[reply]

I also proposed to write the expression of the transfer function in terms of the complex poles more explicitly, like this:

H(s)=G_{0}\prod _{k=1}^{n}{\frac {\omega _{c}}{s-s_{k}}}=G_{0}\prod _{k=1}^{n}{\frac {\omega _{c}}{s-\omega _{c}e^{\frac {j(2k+n-1)\pi }{2n}}}}

.

This is sacrificing concision for some more clarity and, hopefully, this will help to avoid confusion (that already led to frequent editions of this equation in the past). Ffellay (talk) 21:28, 22 February 2023 (UTC)[reply]