This article is rated C-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Systems Mid‑importance This article is within the scope of WikiProject Systems, which collaborates on articles related to systems and systems science.SystemsWikipedia:WikiProject SystemsTemplate:WikiProject SystemsSystems Mid This article has been rated as Mid-importance on the project's importance scale. This article is within the field of Control theory. Professional sound production Mid‑importance This article is within the scope of WikiProject Professional sound production, a collaborative effort to improve the coverage of sound recording and reproduction on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.Professional sound productionWikipedia:WikiProject Professional sound productionTemplate:WikiProject Professional sound productionProfessional sound production Mid This article has been rated as Mid-importance on the project's importance scale. Physics: Acoustics Mid‑importance This article is within the scope of WikiProject Physics, a collaborative effort to improve the coverage of Physics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.PhysicsWikipedia:WikiProject PhysicsTemplate:WikiProject Physicsphysics Mid This article has been rated as Mid-importance on the project's importance scale. This article is supported by Acoustics Taskforce.

"It is especially useful in this case because the complex curves that appear in a linear magnitude-frequency plot can often be approximated by straight lines in a Bode plot."

Can't it always be approximated by straight lines? Isn't that how a Bode plot is drawn? It's just that the approximations are not so accurate for certain systems? Specifically, what are the straight line asymptotes for a Chebyshev filter? - Omegatron 16:36, May 17, 2005 (UTC)

I suppose the question is how accurately can it be approximated by straight lines (and how many straight lines you are going to use). RJFJR 16:43, 6 October 2005 (UTC)[reply]

Actually, I think it's more accurate (and correct) if the break frequencies are located at the radial distance from the pole/zero to the imaginary axis... —Preceding unsigned comment added by 99.235.205.120 (talk) 23:02, 18 February 2008 (UTC)[reply]

The comment about RC lowpass filter should ahve an image, but that is a particularly simple case so we probably want a more complicated one too. (Preference to bode-phase plot, I'd say). RJFJR 16:43, 6 October 2005 (UTC)[reply]

Image added. :-) — Omegatron 19:51, 8 October 2005 (UTC)[reply]

The article says: "These asymptotic approximations are useful because they can be drawn by hand following a few simple rules.",

but it doesn't say what they are. I think its talking about the rules:

For a transfer function in the form ${\displaystyle H(s)=A\prod {\frac {(N_{n})^{a_{n}}}{(D_{n})^{b_{n}}}}}$

where s = jω, N is a numerator term, D is a denominator term, and H is the transfer function.

• start with a straight line at H(0)
• at every value of s where N = 0 (a zero), increase the slope of the line by 20*${\displaystyle a_{n}}$ dB per decade.
• at every value of s where D = 0 (a pole), decrease the slope of the line by 20*${\displaystyle b_{n}}$ dB per decade.

For some reason s is equivalent to -ω when graphing. My explanation here isn't very well written, and probably has incorrect explanation. It would be very nice if someone could add this sort of thing onto the page. Fresheneesz 22:16, 4 May 2006 (UTC)[reply]

It's covered in the external links, too. I was going to add it to the article, but haven't gotten around to it. You know how it is... — Omegatron 23:12, 4 May 2006 (UTC)[reply]

Heh, yea, I'm only learning about them, so anything I might add is subject to stupidity. But maybe I'll hazard it anyway. Fresheneesz 01:57, 5 May 2006 (UTC)[reply]

Needs something on complex poles and zeros - I'm not up for it. Fresheneesz 09:37, 14 May 2006 (UTC)[reply]

I have suggest that Bode plotter be Merged to Bode plot. 24.126.199.129 21:51, 22 August 2006 (UTC)[reply]

whadaya think about adding stuff about TFs w/ -s+xn in the numerator or den. or other combinations of that sort? —The preceding unsigned comment was added by 128.46.105.66 (talk) 17:05, 8 December 2006 (UTC).[reply]

Recently EngineersExcel.com Excel Bode Plotter was added by User:Nitin Special:Contributions/Nitin.mehta and then removed by User:Dicklyon claiming is was an ad site, about a proprietary product. I've reviewed the site and I can't see anything wrong with it that breaks WP:EL. The Excel spreadsheets can be freely downloaded and used. As an engineer I would find tools like this useful. I propose keeping the link. --Rehnn83 16:58, 8 January 2007 (UTC)

It's OK with me if you decide it's worth keeping. My removal was based on the editor who was doing nothing but linking this site, so presumably a typical conflict-of-interest spam. The proprietary product I refer to is Excel. Dicklyon 18:26, 8 January 2007 (UTC)[reply]

ps. you can see from the site's main page that the editor is associated with it: [1]. So he should not be linking it; he should ask here and see if someone independent deems it worthwhile. Dicklyon 18:28, 8 January 2007 (UTC)[reply]

I can see your point, however I think Excel is a fairly common tool/programme and the worksheet would help with someones understanding of Bode Plots and therefore would benefit the article. For now I'm going to leave the article as it is for now pending further discussion. - Rehnn83 20:27, 8 January 2007 (UTC)[reply]

If a pole (or zero) is negative, its phase contribution should be the opposite, right? Anyone can confirm this? (by the way) we really need to write something about complex poles or zeros) --Arcturus4669 21:56, 5 June 2007 (UTC)[reply]

No, poles and zeros are not signed. Dicklyon 22:04, 5 June 2007 (UTC)[reply]

Sorry, I used the wrong word. I mean "unstable" poles and zeros (that is, having a positive real part leading to a diverging exponential contribute), such as the zero in

${\displaystyle G(t)={\frac {s-3}{s+1}}}$

--Arcturus4669 15:18, 6 June 2007 (UTC)[reply]

Oh, right, you mean right-half-plane poles and zeros (the zeros aren't unstable, but the poles are). Yes, they make the phase go the other way. Dicklyon 17:52, 6 June 2007 (UTC)[reply]

Ok, I added the information. Feel free to change wording to whatever is the usual way to refer to positive-real-part singularities in English (in Italy we generally refer to them as "unstable" zeros and "non minimum phase" systems) --Arcturus4669 13:28, 10 June 2007 (UTC)[reply]

The article is unclear about what are the purposes and applications of Bode plots. Can somebody just add a line about this? Riki 08:34, 18 August 2007 (UTC)[reply]

I have a question please. As the transfer function is complex, and the poles y_n and zeros x_n are real. As I understand the bode plot, is the transfer function as it is on the imaginary axis (s=jw). The question then is, why are poles or zeros on the real axis of the transfer function create corners and phase changes on the imaginary axis, at the same value of frequency as the pole or zero?

is it becasue the form of the transfer function over the complex domain? or is it because at corner frequecies, by definition there is rise or fall of ${\displaystyle {\sqrt {2}}}$ of the transfer function?

Jackovo (talk) 19:26, 27 April 2009 (UTC)[reply]

The poles and zeros can be on the real axis, or anywhere in the complex plain. But in the simple case of real poles and zeros, the answer is as you seem to see; at the corners, you're root-2 further from the pole that you are at DC (at s = 0); near DC, distance doesn' change much with frequency; far above the corner, distance is almost equal to frequency. The tranfer function gain is proportional to that distance (for a zero) or inversely proportional (for a pole). That's about all there is to it. Dicklyon (talk) 23:27, 27 April 2009 (UTC)[reply]

The Bode Plot wiki page says that Bode should be pronunced as "bod'-duh". But I have always heard it pronounced as "bo-dee", and I can't find any other sources on the web that support "bod'-duh". The sources I find are:

• bo-day - - from the UK
• bow-day - [2]
• Boady - [3]
• BO-dee - [4]
• boh dee - [5]
• bodee - [6]
• boh dee - [7]

Shouldn't the Bode Plot wiki page say that the pronunciation is "bo-dee"? I guess it should be pronounced the same way Hendrik Wade Bode pronounced his name - is info available that definitively indicates how his name was pronounced?

—Preceding unsigned comment added by 216.112.109.138 (talk) 22:51, 13 August 2009 (UTC)[reply]

You are right. His colleagues pronounced his name as "Boh-dee" and this is the prevalent pronunciation. His family preferred the traditional dutch pronunciation of "boh'-dah". Please see also Hendrik Wade Bode. Dr.K. logos 00:17, 14 August 2009 (UTC)[reply]

The edit (this diff) that said "boady" cited a source that gave three pronunciations (this pdf); that's why I reverted it. If we're going to change the article and cite a source, we need to at least be a little more true to what it says. Dicklyon (talk) 00:40, 14 August 2009 (UTC)[reply]

Historian Valkenburg in his IEEE "In memoriam" paper specifically mentions that he wanted to disambiguate the matter once and for all and provided verbatim the pronunciation that I just inserted into the article. I am sure you can verify this paper by going to any university Library. Dr.K. logos 01:19, 14 August 2009 (UTC)[reply]

In the same paper Van Valkenburg mentions that his family preferred the Dutch version "Boh-dah" but that his colleagues at Bell Labs called him "Boh-dee" and that Bode did not object. Dr.K. logos 02:57, 14 August 2009 (UTC)[reply]

"...His colleagues at Bell Labs called him "'Boh-dee'" -- but that does not explain how he, himself, pronounced the name. If the family pronounced it per the Dutch "Boh-du, then that would be correct despite his non-objection to the mispronunciation by his colleagues.

The integral couldn't be split that way, if the latter part didn't vanish, because the rule of splitting integrals does only hold true for finite integrals. Therefore one can skip directly to the limit. The condition h has to fulfill to lead to a well-defined limes is vanishing faster than ${\displaystyle {\frac {1}{t}}}$ for large t, because u is bounded and the term therefore of the form ${\displaystyle {\frac {c}{t^{\alpha }}}}$ for ${\displaystyle \alpha >1}$, which is an Integral known to converge. — Preceding unsigned comment added by 92.73.143.160 (talk) 22:05, 8 December 2015 (UTC)[reply]

This last sentence in the lede doesn't make much sense to me. How does the second part follow from the first? Should probably also have some kind of source. Should be removed? ediss 12:37, 28 September 2017 (UTC)

I'm looking at the first bode plot example and I was thinking that it would help readers understand the solutions if they had a better idea of what's happening between each step. Is that something that would be appropriate to do here? Does the wikipedia community have any rules of thumbs on this kind of edit?--IanVG (talk) 05:03, 14 December 2020 (UTC)[reply]

Hello. I understand what you are asking, but Wikipedia articles are not supposed to be how-to manuals or text-books. But the line between a text-book and just a good comprehensive article is fuzzy and crossed all the time. There is some discussion at WP:NOTTEXTBOOK. On the other hand, if you want to improve the article, you are welcome to try. Constant314 (talk) 06:11, 14 December 2020 (UTC)[reply]

I apologize for the double post here, but I think the second example is a little wordy and confusing. What exactly is the transfer function here? Could something like (s-z)/(s-p) do the trick here? I am having trouble connecting the images and plots to the concept of the Bode plot. Anyone with more knowledge than I should take some initiative here.--IanVG (talk) 05:08, 14 December 2020 (UTC)[reply]

In control engineering the term which is currently labeled as A_OL (open-loop gain), is usually called forward gain, while the gain of the complete loop beta*A_OL is called loop-gain (or open-loop-gain i.e., describing the complete loop before it is closed). In the context of negative-feedback amplifiers these terms are commonly used interchangeably, but it would be better to change the wording of the first sentence to indicate that it is an example, or to adopt the control engineering definitions throughout (which is way more work). Any chance that an edit in that direction will be accepted? SplashChoke (talk) 13:20, 24 June 2024 (UTC)[reply]

• Agree. Proceed under WP:BEBOLD. Consider changing the name of A_FB also.

Constant314 (talk) 13:26, 24 June 2024 (UTC)[reply]