User talk:Catslash

Hello, Catslash, and welcome to Wikipedia! Thank you for your contributions. I hope you like the place and decide to stay. Here are some pages that you might find helpful:

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I hope you enjoy editing here and being a Wikipedian! Please sign your name on talk pages using four tildes (~~~~); this will automatically produce your name and the date. If you need help, check out Wikipedia:Questions, ask me on my talk page, or place {{helpme}} on your talk page and someone will show up shortly to answer your questions. Again, welcome! Timrem 00:24, 5 September 2006 (UTC)[reply]

One gets tired of looking at the red link for your name.  ;-) —Steven G. Johnson 23:45, 7 September 2006 (UTC)[reply]

I don't care to describe myself - but I do want a sandbox (so I hope that creates one here) --catslash (talk) 12:12, 11 June 2009 (UTC)[reply]

Can you upload a larger version of Image:Radiation-patterns-v.png? The text is unreadable and the lines look broken. — Omegatron 08:40, 25 November 2006 (UTC)[reply]

OK; I'll have to re-create it - it'd on my list of things to do --catslash 21:18, 27 November 2006 (UTC)[reply]

What did you create it with? — Omegatron 22:40, 19 January 2007 (UTC)[reply]

I took a couple of screen graps of plots from commercial EM simualtor software, and then tidied them up using the GIMP. Since you mentioned the quality, I've tried printing the plots to a file, and converting them to SVG, but the resulting files are very big - much larger than the image files. Perhaps I should just plot the patterns at double the size, and then shrink them with antialiasing? - or upload the larger plots aand let wikimedia do the shrinking? --catslash 00:57, 21 January 2007 (UTC)[reply]

I assert to have voted for picture 3 --catslash 16:46, 21 February 2007 (UTC)[reply]

Please correct your "diff" link, which should refer to the addres under which you voted. Alvesgaspar 17:32, 21 February 2007 (UTC)[reply]

I further assert that my IP address is 80.176.145.157 --catslash 01:13, 22 February 2007 (UTC)[reply]

Your edit comment here is wrong. Isolators are always made out of 3 port circulators in which one port is terminated. How else do you propose to make an isolator? --Mr. PIM 23:19, 31 August 2007 (UTC)[reply]

Like this [1] and here is a picture of one [2]. And there's another method, using Faraday rotation; here's a picture of one [3], and here's a Wikipedia page which describes an optical frequency version of this sort Optical isolator. --catslash 23:59, 31 August 2007 (UTC)[reply]

And now I've written a page about it: isolator (microwave)! --catslash (talk) 20:43, 25 May 2009 (UTC)[reply]

I have done a draft translation into English which you can find in one of my user pages here. I suggest we keep it there until people feel it is good enough to replace the actual article. Please check it with great care, and leave any comments at the head of the page. Regards, JohnCD (talk) 08:29, 11 February 2008 (UTC)[reply]

• I have followed your advice and dumped my translation into the main article. I had no comments on the translation and some discouraging ones on the maths, which I have put on the article's talk page, in the hope that someone will be inspired or provoked into writing the much more concise article that the comments envisage. JohnCD (talk) 20:49, 19 February 2008 (UTC)[reply]

Just saw your edit summary to this article. The reason is that I had no idea that the "O" in O. Brune stood for Otto Walter Heidrich Oscar when I first wrote the article! Thanks for fixing. SpinningSpark 21:22, 18 June 2009 (UTC)[reply]

Hi Catslash, last month I put Otto Julius Zobel up for FAC. Even though the comments at FAC were addressed, it still failed to pass. This appears to have been due to a shortage of interested editors reviewing the article rather than any identified shortcomings in the article. So this time I am bringing it to the attention of editors I know to have been previously interested in filters or network analysis. I wonder if I could persuade you to take a look at Wikipedia:Featured article candidates/Otto Julius Zobel/archive2. Please don't take this as in any way canvassing you to !vote to promote the article, or even to !vote at all. You will, of course, act as you see fit on the merits of the article. SpinningSpark 16:40, 13 September 2009 (UTC)[reply]

Sorry for late reply; I've been on holiday. I'll take a look at that after deleting mountain of emails, checking watch-list etc. --catslash (talk) 09:57, 20 September 2009 (UTC)[reply]

Thanks for looking. Can always do with more eyes on an article at FA. SpinningSpark 20:29, 20 September 2009 (UTC)[reply]

Good work! Nice to know that someone not only looked at my code, but understood it and made something new with it. The logarithmic mapping is rather ad hoc, but has the nice property that the log of the dielectric constant is just twice the log of the refractive index. And of course it maps one to zero. Seemed like the thing to do at the time. I guess I might like to rescale the values in my diagram by log(4) so that the two diagrams match exactly. 0x30114 (talk) 08:37, 10 November 2009 (UTC)[reply]

Thanks for noticing and commenting! I was wondering if anybody had looked at what I'd done. --catslash (talk) 21:50, 11 November 2009 (UTC)[reply]

You actually read my mind about doing the Maxwell fish-eye, so it was a nice surprise to see it when I logged in again after a longish break. I'd like to do a few diagrams of the other generalized Luneburg lenses-- I've read the paper by Morgan, but I haven't figured out the exact shape of the rays yet. 0x30114 (talk) 11:04, 12 November 2009 (UTC)[reply]

The ray path in the general case is probably non-analytic, or anyway not a shape available in SVG. So you just have to follow the path by integrating

${\displaystyle d\theta ={\frac {h}{r{\sqrt {n^{2}r^{2}-h^{2}}}}}dr}$

numerically (h is the constant for that ray (and n is a function of r of course)). --catslash (talk) 11:36, 12 November 2009 (UTC)[reply]

Yes, I'm aware of that differential equation and that the solution is likely not a simple shape. (By "exact" I meant exact to some degree of numerical precision.) The annoying thing is that one also has to compute n(r) using numerical integration. I was thinking that there might be a way to combine the two steps and make things simpler computationally. Of course, for diagrammatic purposes I could get away with rather crude approximations that are indistinguishable from a more precise result, but it's the principle of the thing, y'know? :^) 0x30114 (talk) 19:13, 21 November 2009 (UTC)[reply]

I made a change to the article. You can use {{Nobots}} (qv) in a similar circumstance, but it should only be a stop gap. If possible I will file a bug report tomorrow. Rich Farmbrough, 23:40, 20 November 2009 (UTC).[reply]

Hi Catslash
Would you like me to ask for you to get rollback? I see that you do a lot of anti-vandalism work. Rollback is much faster than undo, because it requires only one click and adds an automated edit summary. If you would like me to ask for you, just Email me, or leave a message on my talk page. If you want to ask yourself, you can ask here. Thanks! --The High Fin Sperm Whale 18:19, 29 January 2010 (UTC)[reply]

 Done I put in a request for you. --The High Fin Sperm Whale 19:13, 29 January 2010 (UTC)[reply]

I have granted rollback rights to your account; the reason for this is that after a review of some of your contributions, I believe you can be trusted to use rollback correctly, and for its intended usage of reverting vandalism, and that you will not abuse it by reverting good-faith edits or to revert-war. For information on rollback, see Wikipedia:New admin school/Rollback and Wikipedia:Rollback feature. If you do not want rollback, just let me know, and I'll remove it. Good luck and thanks. Arbitrarily0 ^(talk) 20:49, 29 January 2010 (UTC)[reply]

Thanks The High Fin Sperm Whale and Arbitrarily0. Now I'm all tooled up and itching for some vandals to show their faces. --catslash (talk) 18:42, 30 January 2010 (UTC)[reply]

Thank you for your "awe and admiration" regarding Distributed element filter. It's great to know that my work is appreciated. The article has now been nominated as a Featured article candidate. You might like to leave comments on its nomination page. SpinningSpark 09:08, 3 May 2010 (UTC)[reply]

I don't know if you noticed it, but the FA (assistant) director has deleted all the existing comments and restarted the nomination. That means if you want your comments to count, you will have to make them again. Not sure why this was done, it appears that the page was too long and complicated for them to make a decision. I am not informing everyone that took part (the idea seems to be to attract a new set of reviewers) but I am taking the message you left on my talk page as a request to be kept informed. SpinningSpark 14:25, 13 June 2010 (UTC)[reply]

The mechanical filter article which you have previously edited and/or reviewed has been nominated as a Featured Article. You may give your opinion on whether you think this article should be promoted to Featured Article status by leaving a comment on the nomination page. SpinningSpark 18:48, 17 September 2010 (UTC)[reply]

I have responded to your comments at the Mechanical filter FAC. Would you please revisit, hopefully everything has now been addressed and you can find your way to support. By the way, apologies for the cookie-cutter message above, I thought it best to give everyone the same message. SpinningSpark 14:07, 9 October 2010 (UTC)[reply]

Will do. That I didn't offer an immediate Support does not indicate that I have any reservations, only that I need to review the FA criteria and check the article against them. The article and its subject matter are much more interesting to me than the FA thing, and so my comments (like having more red-blue arrows (which I should have mentioned earlier)), may not always be relevant to attainment of FA status. --catslash (talk) 16:00, 9 October 2010 (UTC)[reply]

That's fine. Support only when you think it is ready, not because I poked you. But it is helpful (both to me and the FA director) if you strike comments that have been adequately dealt with. SpinningSpark 16:48, 9 October 2010 (UTC)[reply]

...handy in building the filter to the have the fractional error in the resonator frequencies << 1/Q (irrespective of what you do with the finished filter...)

I'm replying to you here rather than clutter the FAC with a long post most others won't be interested in. What you say is true if a high-Q filter is being built. However, in just about all applications of mechanical filters the effective Q of the completed filter is way less than the Q of the resonators. If it were only centre-frequency that was being trimmed for, then trimming would not need to be anywhere near as finely done as it actually is. Of far more concern than resonator centre-frequency is the effect off-frequency resonators have on the filter skirt - the steepness relies on the resonators being high-Q and spot on frequency. Think about an LC circuit working into a resistive load. If the components are ideal, the resonator has infinite Q, but that does not mean the frequency accuracy must have zero error, as implied by your statement. The bandwidth of the circuit is determined by the value of the load resistor and 1/BW is the effective Q. So long as the resonator Q >> than the effective Q errors in the resonant frequency will have about the same effect regardless of the value of resonator Q. SpinningSpark 23:31, 10 October 2010 (UTC)[reply]

Yes, I realize that its the loaded Q that matters, and I'm fairly happy with your original text. --catslash (talk) 23:38, 10 October 2010 (UTC)[reply]

I seem to have put the kibosh on this FAC. Sorry, this wasn't my intention. Hope your demotivation has worn off. --catslash (talk) 20:52, 20 October 2010 (UTC)[reply]

I posted an unbalanced version under the "Telegrapher's Equations". It is quite a bit simpler.

I like to call this type of two-port (the kind that has an internal schematic) a white box and the the kind where all you have are the port parameters a black box. —Preceding unsigned comment added by Constant314 (talk • contribs) 12:12, 20 October 2010 (UTC)[reply]

On 2 December 2010, Did you know? was updated with a fact from the article Waveguide flange, which you created or substantially expanded. The fact was ... that a good contact-connection between undamaged plain waveguide flanges can outperform a choke-connection, but a bad one can cause arcing? You are welcome to check how many hits the article got while on the front page (here's how, quick check) and add it to DYKSTATS if it got over 5,000. If you know of another interesting fact from a recently created article, then please suggest it on the Did you know? talk page.

Materialscientist (talk) 06:03, 2 December 2010 (UTC)[reply]

That's a nice little article, congratulations! Thanks for watching over my talk page while I have been away. Anything I can do to help you, please ask. SpinningSpark 17:20, 18 December 2010 (UTC)[reply]

Hi Catslash. Did you verify that those equations you added back are correct? Note that y₃ was recently added by an IP whose previous edit to the article was reverted. In any case I don't think that the article really needs the n=3 cases, especially unsourced. Paul August ☎ 01:12, 27 December 2010 (UTC)[reply]

Hi Paul August, I did think twice about adding these back. Since these expansions are presented as illustrations of the preceding Rayleigh's formulas, I felt that it was sufficient (and better) to add a citation for these general formulae. I did also check that the n=3 cases were free from error. Whether or not going to 3 is superfluous or excessive, I don't know. On the one hand A&S thought that 0 to 2 sufficed, on the other it doesn't take up much space and helps establish the general appearance of these expansions for higher n. If you feel strongly to the contrary, then remove these cases and I will not argue further (please be careful not to revert the naming and citation I added for the preceding general formula). Yes, this IP's previous edit appears to be wrong, but having checked the later edits, I will assume good faith. --catslash (talk) 12:51, 27 December 2010 (UTC)[reply]

It seems that 71.63.60.87's previous edit mistakenly changed the recurrence relation from that applicable to a (cylindrical) Bessel function to that applicable to a spherical Bessel function. --catslash (talk) 14:49, 27 December 2010 (UTC)[reply]

The main reason I think that sticking with n = 0 to 2 is probably best is simply that the ability to link to A&S makes verification trivial, an important consideration for the purposes of maintainability. Paul August ☎ 15:50, 27 December 2010 (UTC)[reply]

Don't be afraid to revert stuff like this, which I have now done. Good faith edits that add something to an article, in general should be cleaned up rather than deleted, but this guy is just tinkering with the formatting and adding tons of errors in the process. Edits which do not leave an edit summary or otherwise explain themselves AND delete or contradict existing references AND use notation that is completely unfamiliar to those skilled in the art, in my opinion deserve to get reverted without too much worrying about a possible good bit buried in there.

Also, there was no real need to fetch an admin to do this, as far as Wikipedia content goes, admins have no more authority than any other editor. SpinningSpark 17:24, 21 May 2011 (UTC)[reply]

Thanks. You've changed colour. --catslash (talk) 17:35, 21 May 2011 (UTC)[reply]

I turned down the volume - it was hurting my eyes. SpinningSpark 18:24, 21 May 2011 (UTC)[reply]

Thanks for your comments about the Murder of Julia Martha Thomas article that I wrote! You might be interested to know that I have nominated the article for FA status - please see Wikipedia:Featured article candidates/Murder of Julia Martha Thomas/archive1. If you have any feedback it would be very welcome. Prioryman (talk) 19:42, 22 July 2011 (UTC)[reply]

Dear catslash editor

About your correction to my change, I understand that this issue is confused. I have analyzed previous discussions and I propose now to use the recoomenations of Stephen Meigs (I copy below) Please let me know if you agree

To be perfectly clear, say either "the decibel is a unit whose value is ten times the logarithm to base 10 of the ratio of two power quantities" or say "the decibel is a unit defined as 1/10 the unit whose value is the logarithm to base 10 of the ratio of two power quantities". How about the following as something one could agree on? "The decibel is a unit whose value is ten times the logarithm to base 10 of the ratio of two power quantities. The actual base 10 logarithm of the ratio itself is called a bel. Though less simply defined, the smaller decibel unit has proven to be more popular than the bel unit. Since the value of a measurement using decibels is 10 times that of a measurement using bels, the decibel (as a unit) equals 1/10 bel, whence the "deci" prefix."Stephen A. Meigs (talk) 16:42, 11 August 2010 (UTC)

Regards OscarJuan --OscarJuan (talk) 18:41, 3 December 2011 (UTC)[reply]

Thanks for the addition of the picture to waveguide filter, that's great. I have a request. You don't have to comply if you don't want to, it is only a request. My draft for this article is at User:Spinningspark/Work in progress/Waveguide filter. It would be helpful if you would add any further contributions you may have to this draft page instead of the live page. The reason for this is that a history merge is a lot easier if the live article is not edited in the meantime. I really did not want to create the live page at all for this reason, but my arm was twisted by the GA reviewer during the GAN of waffle-iron filter. In fact, it would be helpful if you added your image to the draft page as well because that puts you in the edit history of the draft as the image provider. The live stub could then be simply deleted when the draft goes live without any loss of contribution history.

The article isn't very advanced at the moment, not much more than a list of types to be covered. If you can think of any major types that are missing, please add them. Thanks, SpinningSpark 18:21, 25 February 2012 (UTC)[reply]

No further contributions from me are planned at present; it was just a drive-by photo-illustration: I saw the empty article, the filter sitting on my desk, and it was an easy photograph as the posts are clearly visible from the outside.

What do you consider to be a waveguide filter? Does this exclude a system of aperture-coupled cavities with coax ports? - only you don't list ceramic resonator (in cavity) filters, or filters with non-nearest neighbour coupling.

Is it OK to create/use a talk page for this user-space page? --catslash (talk) 19:56, 25 February 2012 (UTC)[reply]

I'm not really sure of the scope at this stage. The current list is taken, more or less without any processing, from the Allen Avionics presentation. I do not really agree that some of their list are within scope. My idea would be that any filter that waveguide-couples resonators in a non-TEM mode is a waveguide filter. That would mean that combline and interdigital filters should be taken off the list - which is ok because we have already covered them in distributed element filter.

Non-nearest neighbour coupling. I am familiar with this idea in mechanical filters and covered it at Mechanical filter#Bridging wires. I have sources that describe this method for cavity resonators, but have never seen it used for real. Do you have any examples of a real filter with this design feature? In any case, I would say yes, good idea, that is certainly in scope.

Talk page: of course it is ok to use the talk page. SpinningSpark 20:44, 25 February 2012 (UTC)[reply]

Can you ping me your e-mail again? I seem to have mislaid it - probably on the PC away for repair - and you don't have e-mail enabled on your account. I'll send you the ring coupler papers by return. SpinningSpark 20:37, 10 March 2012 (UTC)[reply]

[4]...I really think you now need to add an entry for "half-wave". SpinningSpark 18:42, 16 March 2012 (UTC)[reply]

Yes, more explanation is needed. Perhaps when introducing modes, it could say there is a standing-wave pattern in the cross-section plane of the guide, even for waves that in the longitudinal direction are purely travelling. Then say that the mode numbers count the number of half-waves or number of extrema of the E-field in this pattern in the width and in the height of the guide. I'd be happy to take a stab at this. --catslash (talk) 23:49, 17 March 2012 (UTC)[reply]

Go ahead, that was the you part of the comment. SpinningSpark 00:09, 18 March 2012 (UTC)[reply]

I have just uploaded two new diagrams of the hybrid ring;

These more accurately reflect the correct relative track widths for 3 dB and 10 dB coupling respectively. If they are any use to you, feel free to use them in the reworked article. Let me know if you need any changes. SpinningSpark 12:43, 24 March 2012 (UTC)[reply]

They look better. The 2134 port numbering, although not the most common, appeals to me because it's convenient for presenting odd-even mode analysis: if 3 is the mirror image of 1 and 4 of 2, then you can write the final 4×4 S-matrix as a 2×2 array of 2×2 matrices, each corner being the sum and the difference of the odd and even mode S-matrices - whereas the more common 1234 numbering gives no obvious pattern.

The rat-race article is one of five or six that I've done the studying for, and just need to apply myself to the writing of. --catslash (talk) 18:17, 24 March 2012 (UTC)[reply]

The 2134 numbering was used simply for consistency with other designs and diagrams in the directional coupler article. Let me know if you want it changed - or I can remove all the annotation altogether and leave it for the prose to explain. There might possibly be a difference in convention between waveguide and planar designers, but I am not at all sure about that point. SpinningSpark 18:35, 24 March 2012 (UTC)[reply]

Grebennikov[5] uses 2134 for planar, PON[6] uses 1324 and various other authors have used other numberings, but 1234 does seem to be the most common. --catslash (talk) 19:05, 24 March 2012 (UTC)[reply]

Thanks to have reverted a part of these advertisings. However all the edits by this user are either advertising of pushing his personal point of view on the history of the subject. I think that all the edits of this user should be reverted. I have done this once, but for not starting an edit war, I will not do it again. See Wikipedia talk:WikiProject Mathematics‎#User:Akritas2's edits in Sturm's theorem and Root-finding algorithm. — D.Lazard (talk) 17:14, 14 April 2012 (UTC)[reply]

Is this a copyvio? The imprint at the bottom of the pages would seem to indicate that it is. It might be ok for use in an educational establishment but it's not ok for Wikipedia. We are not allowed to link to copyvios. Unless you have evidence that it has been released on a free licence... In any case, the "official" journal copy should still be linked in case the link goes dead. SpinningSpark 20:33, 22 August 2012 (UTC)[reply]

Maybe so - if you wish to change the link to ieeexplore, then I have no objection. --catslash (talk) 20:45, 22 August 2012 (UTC)[reply]

Thanks for the catch. I went back put in a less ambitious version of the edit: no discussion of how the fields are actually found, but with an explicit citation of their mathematical forms (from a microwave antenna book). It's surprisingly difficult to find citations for the dipole equations as written in the article; most books keep to an abstract vector notation (n × p, etc), and most only talk in terms of dipole moment, not current (the conversion is straightforward, but it's nice to have a reference that says it explicitly). I retooled the equations a bit to make them closer to what's in the citation; I don't think I introduced any errors this time. Zueignung (talk) 15:56, 13 September 2012 (UTC)[reply]

It's good that you are happy with that reversion; I thought you might be unhappy, as your edit included a lot of formatting changes along with the objected-to text, and I didn't offer much explanation. Although the field expressions for the Hertzian-dipole/current-element are exact solutions of Maxwell's equations, the infinitesimal current-element at the centre is not a very realistic model of a dipole.

Interestingly (to me, and since you raise the question of the three terms), the 1/r³ terms are exactly the static E-field due to the pair of point-charges produced at the ends of the segment by the current flowing between them. This is necessarily true if the overall expressions are exact, since the 1/r³ terms dominate the other terms in the low-frequency (small-k) limit. For this reason, the 1/r³ terms are sometimes called the quasi-static terms.

Conversely, the 1/r terms dominate in the far-field (or equivalently at high frequency (large k)) and carry energy away and so are sometimes called the radiation terms. The power density is of course the the product of the E and H fields, so this gives you the 1/r²-law for the radiation intensity.

I don't think that the middle 1/r² terms have any very commonly used name. --catslash (talk) 13:06, 14 September 2012 (UTC)[reply]

Have you any idea what the answer is to this? Sorry it is so long. SpinningSpark 21:02, 24 October 2012 (UTC)[reply]

The practical answer is stick to working with voltage wave amplitudes and voltage scattering coefficients, and then there is no confusion - you can convert to dBs when you have finished. Working with power wave amplitudes means dividing by the square root of the characteristic impedance, and then you have to decide what to do when the impedance is complex. The impedance is slightly complex for lines with (non-Heaviside-condition) loss, but nearly imaginary for guides below cutoff (exactly imaginary in the absence of loss). I believe (and will check on this), that there is more than one convention for handling this, but whichever convention you adopt, you will lose some of the nice properties of the s-matrix, interpretations of the amplitudes, coefficients etc.; there are different shortcomings for different choices of convention. I do have some references for this somewhere, and will get back to you when I find them. The S-parameters page follows Kurokawa. --catslash (talk) 22:42, 24 October 2012 (UTC)[reply]

What I'm really struggling with: is there a reflection from a line terminated with a complex conjugated impedance. Using normal Γ says there is, but multiplying the in-phase components of reflected current and voltage can result in reflected power greater than the incident power for some choices of impedances - which can't be right. Or maybe I am just screwing up the maths somehow. Using complex conjugate resolves that issue (reflection = zero) but creates even worse ones. If a line terminated in its characteristic impedance has a reflection (as complex conjugate form says it must) then a piece of line terminated in another piece of line must have a reflection, so there will be finite reflections every infinitesimal along the line and no power would ever reach the load. SpinningSpark 23:47, 24 October 2012 (UTC)[reply]

Power carried by waves doesn't always add, but only when the waves are uncorrelated overall. Consider two superposed waves of the same wavelength and polarization travelling in the same direction in a medium with real Z₀. These waves could interfere constructively everywhere, in which case the total power is twice the sum of the powers in the two waves, or interfere destructively everywhere, in which case the total power is zero. For non-decaying waves travelling in opposite directions (real Z₀) (the usual case for transmission lines), then you can add (or rather subtract) the power in the forward and backward waves - but this is not the case (I think - I'll have to do some algebra to check) for decaying waves in a medium with complex Z₀. --catslash (talk) 00:30, 25 October 2012 (UTC)[reply]

Consider the extreme case of a lossless waveguide below cutoff; Z₀ is pure imaginary (inductive) and for a semi-infinite length no power is transmitted down it (the generator sees an inductance). The E and H fields decay exponentially going along the guide (are evanescent) and are in time-quadrature. Now take a finite length of the guide and put another generator on the other end, but in time quadrature with the first. The E from generator 2 is in phase with the H from generator 1, so power is extracted by generator 2 (say), also the H from 2 is in phase with the E from 1 and (because the polarity of H 2 is the opposite of H 1), it also sucks power into generator 2. Now non-zero power is transmitted down the guide. --catslash (talk) 01:20, 25 October 2012 (UTC)[reply]

Yes, I'm familiar with constructive and destructive interference. I calcualted forward power and reverse travelling power separately (which I think is ok) and took power transmitted as the dot product of the voltage and current phasors. I need to recheck the calculation later though, I may have made a simple arithmetic mistake. SpinningSpark 06:52, 25 October 2012 (UTC)[reply]

Suppose ${\displaystyle \scriptstyle {V_{+}}}$, ${\displaystyle \scriptstyle {I_{+}={\frac {V_{+}}{Z_{0}}}}}$ and ${\displaystyle \scriptstyle {V_{-}}}$, ${\displaystyle \scriptstyle {I_{-}={\frac {V_{-}}{Z_{0}}}}}$ are the voltages and currents associated with forward and backward waves at the same single point in a transmission line (i.e. with ${\displaystyle \scriptstyle {I_{+}}}$ and ${\displaystyle \scriptstyle {I_{-}}}$ defined to have opposite signs when they are in the same direction). Here forward and backward waves is used to mean having the same form as the waves on a semi-infinite line with a generator at one end.

Now

${\displaystyle P_{+}={\frac {1}{2}}Re(V_{+}I_{+}^{*})}$

${\displaystyle P_{-}={\frac {1}{2}}Re(V_{-}I_{-}^{*})}$

${\displaystyle P_{nett}={\frac {1}{2}}Re((V_{+}+V_{-})(I_{+}-I_{-})^{*})}$

and with real ${\displaystyle \scriptstyle {Z_{0}}}$ we have ${\displaystyle \scriptstyle {P_{nett}=P_{+}-P_{-}}}$, but in the general case

${\displaystyle {\begin{aligned}P_{nett}-(P_{+}-P_{-})&={\frac {1}{2}}Re((V_{+}+V_{-})(I_{+}^{*}-I_{-}^{*})-V_{+}I_{+}^{*}+V_{-}I_{-}^{*})\\&={\frac {1}{2}}Re(-V_{+}I_{-}^{*}+V_{-}I_{+}^{*})\\&={\frac {1}{2}}Re\left({\frac {-V_{+}V_{-}^{*}+V_{-}V_{+}^{*}}{Z_{0}^{*}}}\right)\\&={\frac {1}{2}}Re\left({\frac {2iIm(V_{-}V_{+}^{*})}{Z_{0}^{*}}}\right)\\&=-Im\left({\frac {1}{Z_{0}^{*}}}\right)Im(V_{-}V_{+}^{*})\\&=Im\left({\frac {1}{Z_{0}}}\right)Im(V_{-}V_{+}^{*})\\&=Im(Y_{0})Im(V_{-}V_{+}^{*})\end{aligned}}}$

so

${\displaystyle P_{nett}=P_{+}-P_{-}+Im(Y_{0})Im(V_{-}V_{+}^{*})}$

which means that if you define forward and backward waves in this way, then you can't calculate the forward and reverse travelling power separately unless ${\displaystyle \scriptstyle {Im(Y_{0})=0}}$. (can't discuss this further at the moment (at work)). --catslash (talk) 10:31, 25 October 2012 (UTC)[reply]

More eyes are needed on this article. See this edit. SpinningSpark 22:17, 16 November 2012 (UTC)[reply]

Hey Catslash; I'm dropping you this note because you've used the article feedback tool in the last month or so. On Thursday and Friday the tool will be down for a major deployment; it should be up by Saturday, failing anything going wrong, and by Monday if something does :). Thanks, Okeyes (WMF) (talk) 23:08, 13 March 2013 (UTC)[reply]

...is on DYK on the front page right now. SpinningSpark 16:18, 4 May 2013 (UTC)[reply]

This pleases me greatly (and I've chanced to return from holiday (on Loch Katrine), at just the right hour to see it). Once again you have my admiration for the huge amount of work you have undertaken to produce such a high quality article. --catslash (talk) 18:26, 4 May 2013 (UTC)[reply]

Now about getting it to FA...SpinningSpark 19:48, 4 May 2013 (UTC)[reply]

Hey Catslash. I'm contacting you because you're involved in the Article Feedback Tool in some way, either as a previous newsletter recipient or as an active user of the system. As you might have heard, a user recently anonymously disabled the feedback tool on 2,000 pages. We were unable to track or prevent this due to the lack of logging feature in AFT5. We're deeply sorry for this, as we know that quite a few users found the software very useful, and were using it on their articles.

We've now re-released the software, with the addition of a logging feature and restrictions on the ability to disable. Obviously, we're not going to automatically re-enable it on each article—we don't want to create a situation where it was enabled by users who have now moved on, and feedback would sit there unattended—but if you're interested in enabling it for your articles, it's pretty simple to do. Just go to the article you want to enable it on, click the "request feedback" link in the toolbox in the sidebar, and AFT5 will be enabled for that article.

Again, we're very sorry about this issue; hopefully it'll be smooth sailing after this :). If you have any questions, just drop them at the talkpage. Thanks! Okeyes (WMF) 21:55, 1 September 2013 (UTC)[reply]

What does "conductor-backed" mean more specifically? Electron9 (talk) 22:36, 25 September 2013 (UTC)[reply]

Just wrote a description of that. This is worth a read. It might be better to move this discussion to the coplanar waveguide article talk-page. --catslash (talk) 22:48, 25 September 2013 (UTC)[reply]

The previous FA nomination of this article was archived without promotion due to a lack of supporters (no one actually opposed either). As you have previously taken an interest in the article would you please take a look. If you think it is up to scratch you can support at Wikipedia:Featured article candidates/Waveguide filter/archive2. SpinningSpark 17:39, 8 November 2013 (UTC)[reply]

I shall review the FA criteria and then the article nomination. --catslash (talk) 18:03, 8 November 2013 (UTC)[reply]

Are you going to support this article? Just to remind you that you said you would return to the FA "in a couple of weeks" and it has been longer than that. There has been no activity on the page for some time now and it is in danger of being archived again for lack of interest. SpinningSpark 00:31, 20 December 2013 (UTC)[reply]

Apologies. I have left some more comments. Once I am happy that the article meets the comprehensiveness criterion, I will support FA status. --catslash (talk) 01:31, 22 December 2013 (UTC)[reply]

I think your comments are now addressed. If you bold the word support that will make clear to the FA director that you are indeed now supporting the article. SpinningSpark 21:39, 24 December 2013 (UTC)[reply]

Re the discussion on the position of history sections, can I suggest that you make a proposal for the layout of electronics articles at Wikipedia talk:WikiProject Electronics. I could easily do this myself of course, but as it is largely articles written by me that are under discussion and it will probably be me making the changes, I think it would be preferable if this was more than me acting as a "one man show" and imposing my will on the rest of Wikipedia. If we can get something written into the project guidelines it will be a stop to these repeated returns to the same discussion in article after article. Anyone who then disagrees in the future can simply be referred to the Wikiproject and they can try and get it changed there if they wish. SpinningSpark 21:39, 24 December 2013 (UTC)[reply]

You took part in the FAC of Waveguide filter which has now been promoted to a featured article. I have nominated it as a candidate for Today's featured article. If you wish, you can support that nomination here. Regards, SpinningSpark 17:28, 3 January 2014 (UTC)[reply]

Congratulations on getting FA status for this article - what are you planning to work on next? --catslash (talk) 18:55, 3 January 2014 (UTC)[reply]

Well thankyou, but you played more than a small part yourself. Right now, I am going through Category:G13 eligible AfC submissions to see what can be saved. About 200,000 of these pages have been mass deleted. I am able to save between 1 in 10 and 1 in 8 of these with a little work and there are a lot more that could be saved but I pass by for lack of interest. There are still 18,000 articles in the category, so at the rate I am rescuing them there are probably still a couple of thousand worth saving. After that, I have a long to do list but it is not often I work up the enthusiasm to put a major effort into any of them. If you have something in particular you want to collaborate on that would probably be sufficient motivation to get it done. SpinningSpark 19:32, 3 January 2014 (UTC)[reply]

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Dear Catslash, I have been looking at the talk page [7] about solenoidal vector fields. I am interested in the discussion about this (I teach vector calculus and carefully distinguish between vector fields possessing a vector potential and being divergenceless). Some years ago, in that talk page, you mentioned the article about solenoidal vector fields in the Encyclopaedia of Mathematics, vol 9, p. 402. However, I have not been able to arrive to this text using Google. Please, could you tell me or sent to me what is the contents of that article? Thank you for your attention! --Txebixev (talk) 21:03, 23 February 2014 (UTC)[reply]

It seems that this is no longer viewable through Google Books, and I have not got access to it by any other means. Counterexamples in Analysis by Bernard R. Gelbaum, John M. H. Olmsted in Google Books is interesting; they seem to define solenoidal as divergenceless and conclude that solenoidalness does not necessarily imply having a vector potential. Should you find conflicting definitions of solenoidal, then the correct course is to report that some authors define it this way and others that way, and note that the definitions are not equivalent. Sorry not tot be of more help. Good luck sorting out this article. --catslash (talk) 22:46, 23 February 2014 (UTC)[reply]

(talk page stalker) I can see that page on Amazon Look Inside. If it's not available to you (AMI availability is dependent on country and, I think, individual purchasing history) e-mail me and I will copy it back to you. SpinningSpark 00:14, 24 February 2014 (UTC)[reply]

Thank you for your answer Catslash. In my opinion, there is an asymmetry between the well-established conservative-irrotational and the not so well established (?)-divergenceless. Apparently there are 3 ways of saying that the divergence is zero and none to say that there is a vector potential. I will go on looking if there is a reference for this. Spinningspark, I can see only the first pages of the book, but no the ones with the definition of solenoidal vector field. I'd be glad if you could send it to me. --Txebixev (talk) 00:28, 24 February 2014 (UTC)[reply]

You need to e-mail me first so I get your address. Attachments cannot be sent through the wikimail feature. Did you try putting "solenoidal" in the search box? SpinningSpark 00:33, 24 February 2014 (UTC)[reply]

Thanks.So on p 402 is an article by A. B. Ivanov, who says For a continuously-differentiable vector field to be solenoidal it is necessary and sufficient that div a = 0 at all points of V.... This is followed by Editorial comments: ... The condition div a = 0 is necessary for a vector field to be solenoidal. It is sufficient on, for example, convex domains. The general additional condition is that the second homology of the domain vanishes. This can easily be seen from the de Rham cohomology theory. There are examples of vector fields on 3-space with one point removed which have vanishing divergence, but are not solenoidal. --catslash (talk) 00:42, 24 February 2014 (UTC)[reply]

Dear Catslash,

I am currently trying to reproduce the figure you created and uploaded the to Mie Scattering entry. Would you be able to share with me the code you used to generate this figure? If not, might you be able to point me to the equation you used to do this modeling? Any help will be greatly appreciated. -ke0m (talk) 00:37, 14 July 2015 (UTC)[reply]

Unfortunately the code I used isn't in the form of a simple stand-alone function, but rather several libraries of mostly irrelevant junk cobbled together. I will take a look at it in the next few days to see how easy it would be to reduce this to a single simple function. It's even more unfortunate that the Mie scattering article does not give any formulae relevant to actual Mie scattering - it really does need fixing. --catslash (talk) 02:03, 15 July 2015 (UTC)[reply]

Thank you very much for your help. If you are not able to reduce it down to a simple stand-alone function, pointing me to the library would also be greatly appreciated. Hopefully in the future, I can contribute to the article. --ke0m (talk) 22:37, 20 July 2015 (UTC)[reply]

I found the code that I used to create the graph, and it should be possible to get something usable out of it. It's a bit messy because complex-number arithmetic is needed, and I used c (c89), which doesn't have any provision for complex numbers (though I believe that there are (incompatible) complex arithmetic libraries in c99 and c++) - I should probably have stuck to FORTRAN IV. It wasn't clear to me how serious you were in your enquiry - whether it merited a detailed response - but as you have asked again, I will see if I can do something to help (hopefully before too much longer). --catslash (talk) 23:37, 20 July 2015 (UTC)[reply]

Probably I consulted PW Barber and SC Hill, though my code isn't directly based on it. --catslash (talk) 00:08, 21 July 2015 (UTC)[reply]

Thank you again for your help and the reference. --ke0m (talk) 05:46, 21 July 2015 (UTC)[reply]

It transpires that because it's a very special case ((1) RCS, so want only far field (2) monostatic RCS, so only want scattering for minus incident direction (3) perfectly reflecting sphere), then the result can be reduced to a relatively simple expression, namely

${\displaystyle {\frac {\sigma }{\pi R^{2}}}={\frac {1}{(KR)^{2}}}\left|\sum _{n=1}^{\infty }\left({\frac {{j}_{n}(KR)}{{h^{\mathrm {(2)} }}_{n}(KR)}}-{\frac {{j}_{n-1}(KR)-{\frac {n}{KR}}{j}_{n}(KR)}{{h^{\mathrm {(2)} }}_{n-1}(KR)-{\frac {n}{KR}}{h^{\mathrm {(2)} }}_{n}(KR)}}\right)(2n+1)(-1)^{n}\right|^{2}}$

(I reckon), where ${\displaystyle \scriptstyle {\sigma }}$ is the RCS, ${\displaystyle \scriptstyle {R}}$ is the radius of the sphere, ${\displaystyle \scriptstyle {K}}$ is the wavenumber (radians per metre,${\displaystyle \scriptstyle {\frac {2\pi }{\lambda }}}$), ${\displaystyle \scriptstyle {{j}_{n}}}$ is the spherical Bessel function of the first kind and ${\displaystyle \scriptstyle {{h^{\mathrm {(2)} }}_{n}}}$ is the spherical Bessel function of the fourth kind (spherical Hankel function of the second kind).

If you would like, I can email you a self-contained (including the Bessel function evaluation) c code snippet that returns the value of the above expression. You can email me your email address using the Email this user link on the side of this page - or better (for anonymity), you could tick Enable email from other users on your user preference page. --catslash (talk) 01:29, 23 July 2015 (UTC)[reply]

I have ticked the Enable email from other users on my preference page so that you may send me the code. If that is not working, I can also send you my email. Once again, your help is greatly appreciated. --ke0m (talk) 23:00, 23 July 2015 (UTC)[reply]

On second thoughts forget email as the code is small enough to post here:

/*----------------------------------------------------------------------------*/
#include <math.h>
/*----------------------------------------------------------------------------*/
void
j_spheric(double j[], int nmax, double z)
  /*determine jn(z) for n = 0 -> nmax (incl.) using recursion formula*/
{
  /*j[n] and y[n] satisfy f[n-1](z) + f[n+1](z) = (2n+1) f[n](z) / z
  /*[Abramowitz & Stegun: 10.1.19] - for j[n] this recurrence should be
  /*stable for _decreasing_ n and for y[n] either way (probably)
  /*so use J. C. P. Miller's algorithm for j[n]
  */
  int    n, nstart;
  double modz;
  double jnp2, jnp1, jn;
  double dead_small, scale_factor;


  /*arbitrary but small number*/
  dead_small = 1e-145;

  if (z == 0.){
    /*this would make the recurrence relations indeterminate*/
    n = nmax;
    while (n) j[n--] = 0.;
    j[0] = 1.;
  }
  else if (nmax < 1){
    /*going to assume nmax >= 1 in general case*/
    if (nmax == 0) j[0] = sin(z) / z;
  }
  else{
    /*decide what n to start recurring from - formula from [Barber & Hill]*/
    modz = fabs(z);
    nstart = (int)(modz + 4.05 * pow(modz, 1./3.) + 1.);
    if (nstart < nmax) nstart = nmax;
    nstart += (int)(sqrt(101. + modz) + 1.);

    /*starting from arbitrary (but small) values recur down to j[nmax - 1]*/
    n = nstart;
    jnp2 = 0.;
    jnp1 = dead_small;
    while (n-- > nmax - 1){
      /*this n is (n - 1) of the above quoted formula*/
      jn = (double)(2 * n + 3) * (jnp1 / z) - jnp2;
      /*next n*/
      jnp2 = jnp1;
      jnp1 = jn;
      if (fabs(jnp1) > 1.){
        /*prevent arithmetic overflow*/
        jnp1 *= dead_small;
        jnp2 *= dead_small;
      }
    }
    /*should now have correct _relative_ values of j[nmax](z) and j[nmax-1](z)*/
    /*j[nmax    ] = jnp2;*/
    /*j[nmax - 1] = jnp1;*/

    /*continue recurring down - renormalize to prevent arithmetic overflow*/
    j[nmax    ] = dead_small;
    j[nmax - 1] = j[nmax] * (jnp1 / jnp2);
    n++;
    while (n--){
      /*this n is (n - 1) of the above quoted formula*/
      j[n] = (double)(2 * n + 3) * (j[n + 1] / z) - j[n + 2];
      /*next n*/
    }

    /*scale all j[n](z) to get j[0](z) = sin(z) / z*/
    scale_factor = (fabs(j[0]) > fabs(j[1]))?
                     sin(z) / (z * j[0]):        /*already checked z != 0.*/
                     ((sin(z) / z - cos(z)) / (z * j[1])); /*and nmax >= 1*/
    n = nmax + 1;
    while (n--) j[n] *= scale_factor;
  }
}
/*----------------------------------------------------------------------------*/
void
y_spheric(double y[], int nmax, double z)
  /*determine yn(z) for n = 0 -> nmax (incl.) using recursion formula*/
{
  /*j[n] and y[n] satisfy f[n-1](z) + f[n+1](z) = (2n+1) f[n](z) / z
  /*[Abramowitz & Stegun: 10.1.19] - for j[n] this recurrence should be
  /*stable for _decreasing_ n and for y[n] either way (probably)
  */
  int    n;


  /*first two n - infinite for z == 0*/
  if (nmax >= 0) y[0] = -cos(z) / z;
  if (nmax >= 1) y[1] = (y[0] - sin(z)) / z;
  n = 1;
  while (n++ < nmax){
    /*this n is (n + 1) of the above quoted formula*/
    y[n] = (double)(2 * n - 1) * (y[n - 1] / z) - y[n - 2];
    /*next n*/
  }
}
/*----------------------------------------------------------------------------*/
double
metal_sphere_relative_rcs(double kr)
{
  /*returns  monostatic_RCS / projected_area*/
  /*kr = circumference / wavelength = wave_number * radius = radians_in_radius*/
  /*ok for  0.1 < kr < 100  but gives garbage much outside that range         */
  double  num;
  double  den_r, den_i, mod_den_2;
  double  term_r, term_i;
  double  sum_r, sum_i;
  int     nmax, n, etc;
  double j[121], y[121];


  nmax = 120;

  /*get spherical bessel functions for all n*/
  j_spheric(j, nmax, kr);
  y_spheric(y, nmax, kr);

  /*sum series ove n*/
  sum_r = 0.;
  sum_i = 0.;
  n = 0;
  while (n++ < nmax && y[n] > -1e20){
    /*term = j[n] / h2[n]*/
    den_r =  j[n];
    den_i = -y[n];
    num   =  den_r;

    mod_den_2 = den_r * den_r + den_i * den_i;
    term_r =  den_r * num / mod_den_2;
    term_i = -den_i * num / mod_den_2;

    /*term -= (j[n-1] - n * j[n] / kr) / (h2[n-1] - n * h2[n] / kr)*/
    den_r =   j[n-1] - n * j[n] / kr;
    den_i = -(y[n-1] - n * y[n] / kr);
    num   = den_r;

    mod_den_2 = den_r * den_r + den_i * den_i;
    term_r -=  den_r * num / mod_den_2;
    term_i -= -den_i * num / mod_den_2;
     
    /*sum += term * (2 * n + 1)(-1)^n */
    etc = (2 * n + 1) * (1 - 2 * (n & 1));
    sum_r += etc * term_r;
    sum_i += etc * term_i;
    /*next n*/
  }

  return (sum_r * sum_r + sum_i * sum_i) / (kr * kr);
}
/*----------------------------------------------------------------------------*/

The graph in the Mie scattering article is a log-log plot - it looks very different on linear axes.

The original plot was produced for the testing of various full-wave electromagnetic solvers (numerical software) against an array of analytically-tractable problems. For a sufficiently large work-space and sufficiently fine mesh, all the solvers converged on the Mie solution (though with the computer resources available, none was able to go as far up in frequency as kr = 100). So I have a high degree of confidence in the correctness of the original plot.

Hope it is of use to you. --catslash (talk) 18:35, 24 July 2015 (UTC)[reply]

Catslash, my apologies for the delayed response. Thank you very much for your help. This code is exactly what I needed. --ke0m (talk) 18:55, 31 July 2015 (UTC)[reply]

My implementation above of J. C. P. Miller's algorithm for the spherical Bessel function j_n(z) was flawed. The result was poorly determined near non-zero-integer multiples of π. That was very stupid of me - I have now amended it. catslash (talk) 02:10, 13 March 2016 (UTC)[reply]

Hi,
You appear to be eligible to vote in the current Arbitration Committee election. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to enact binding solutions for disputes between editors, primarily related to serious behavioural issues that the community has been unable to resolve. This includes the ability to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate, you are welcome to review the candidates' statements and submit your choices on the voting page. For the Election committee, MediaWiki message delivery (talk) 16:08, 23 November 2015 (UTC)[reply]

Just to say thanks, for your illustration of the double-pendulum (Double-compound-pendulum.gif). I think the double pendulum is a neat example for the development of "chaos" in systems and your illustration of it demonstrates this as well as being seriously entertaining! Much better than a Newton's cradle. Thanks again. LookingGlass (talk) 13:07, 15 February 2016 (UTC)[reply]

It's very pleasing that you liked it. I would still like to address the question of whether the motion is chaotic for all energies or just for some (but I'm becoming worse than ever at getting around to things). --catslash (talk) 02:15, 13 March 2016 (UTC)[reply]

Hello, Catslash. Voting in the 2016 Arbitration Committee elections is open from Monday, 00:00, 21 November through Sunday, 23:59, 4 December to all unblocked users who have registered an account before Wednesday, 00:00, 28 October 2016 and have made at least 150 mainspace edits before Sunday, 00:00, 1 November 2016.

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Do you have an understanding of hybrid modes? I understand the definition of longitudinal section modes, LSE and LSM. But what about HE and EH modes? Are these the same thing (synonymous), or an alternative way of characterising hybrid modes, or a different kind of hybrid mode altogether? Can you point me to a link online where I can read up on this? SpinningSpark 15:06, 8 February 2017 (UTC)[reply]

Coincidentally, I am just looking at modes in dielectric and inhomogeneous guides at work - but sorry, I am not (as yet) familiar with the mode nomenclature (also I will be away for a few days, and so unable to discuss this further). This (8th page, numbered 306), seems to imply that HE and EH modes are distinct from LSE and LSM modes. --catslash (talk) 16:16, 8 February 2017 (UTC)[reply]

On the other hand, this says that EH and HE are synonymous with LSE and LSM. The definitions given for EH and HE are very vague however, so it may not be a reliable source. --catslash (talk) 17:44, 8 February 2017 (UTC)[reply]

Just a guess, but perhaps LSE and LSM are special instances of HE and EH, specific to case where the interface between different media is planar. --catslash (talk) 17:57, 8 February 2017 (UTC)[reply]

You could be right. It is certainly true that LSE and LSM come up in connection with planar formats in inhomogeneous media. It is also the case that LSE/LSM modes can be decomposed into a linear superposition of a TE and TM mode, decidedly not planar modes. So I guess another question is are there hybrid modes that can't be so decomposed? And are there linear superpositions of TE and TM that are not LSE or LSM? A straightforward definition of HE/EH is what I'm looking for, but not finding it anywhere. SpinningSpark 21:32, 8 February 2017 (UTC)[reply]

Hi. Thank you for your recent edits. Wikipedia appreciates your help. We noticed though that when you edited Coplanar waveguide, you added links pointing to the disambiguation pages Quadrature and Ferrite. Such links are almost always unintended, since a disambiguation page is merely a list of "Did you mean..." article titles. Read the FAQ • Join us at the DPL WikiProject.

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Hello, Catslash. Voting in the 2017 Arbitration Committee elections is now open until 23.59 on Sunday, 10 December. All users who registered an account before Saturday, 28 October 2017, made at least 150 mainspace edits before Wednesday, 1 November 2017 and are not currently blocked are eligible to vote. Users with alternate accounts may only vote once.

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Would you care to look over Wikipedia:Peer review/Planar transmission line/archive1? SpinningSpark 22:10, 4 June 2018 (UTC)[reply]

I shall take a fresh look at this article. Since you particularly requested non-technical appraisal, I have thus far been ignoring the review.

It's good to see that the Elite could spare you for long enough to create the Electric bath (electrotherapy), Distributed element circuit, Warren P. Mason and Clydesdale Motor Truck Company articles. --catslash (talk) 01:14, 5 June 2018 (UTC)[reply]

Hello, Catslash. Voting in the 2018 Arbitration Committee elections is now open until 23.59 on Sunday, 3 December. All users who registered an account before Sunday, 28 October 2018, made at least 150 mainspace edits before Thursday, 1 November 2018 and are not currently blocked are eligible to vote. Users with alternate accounts may only vote once.

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Planar transmission line has been nominated as a Featured Article. Comments can be left on the nomination page. I am sending this message to everyone involved in previously reviewing the article. For those that are not familiar with the FA process, articles only become featured if multiple editors say they support its promotion, so your participation is important. Thanks, SpinningSpark 14:46, 1 December 2018 (UTC)[reply]

Thanks for all your help in getting through FA in the past. I have a new one at Wikipedia:Featured article candidates/Distributed element circuit/archive1 that needs reviewing. I've added a section on tapers per your suggestion at the Peer Review. There is also a new section on fractal circuits, which seem to have become a big thing recently ("recently" for me is anything that happened after 1990). SpinningSpark 18:07, 8 August 2019 (UTC)[reply]

To create your personal sandbox (as to forge a personal user page - or anything else) this link should work: User:Catslash/sandbox.

My sandbox is instead at User talk:Catslash/Sandbox. A sandbox at User:Catslash/sandbox may (I'm not certain) require the existence of User:Catslash, which I do wish. catslash (talk) 01:51, 9 November 2019 (UTC)[reply]

No, it doesn't require the existence of the parent page. Such pages are explicitly exempt from speedy deletion per WP:G8. SpinningSpark 10:52, 9 November 2019 (UTC)[reply]

Thank you for answering that. catslash (talk) 18:04, 9 November 2019 (UTC)[reply]

But I came here to ask you to take a look at Wikipedia:Reference_desk/Science#Getting_to_the_bottom_line, TIA, אילן שמעוני (talk) 10:35, 6 November 2019 (UTC)[reply]

It seems that user:Amble has answered your question to your satisfaction. Specifying both the E and B fields at some point in time determines the subsequent evolution of the EM field.

This is analogous to specifying both the initial positions and velocities of bodies in a lossless mechanical system. The velocities determine the rate of change of the positions, and the positions determine the rate of change of the velocities (by determining the inter-body forces and hence the accelerations).

For the diagram that you linked to, it means that both waves must travel from left to right.

Your original question asked about the conservation of energy. As mentioned by other respondents, this is assured in all circumstances by Poynting's theorem. Nevertheless, if one cannot say where the energy goes in various different cases, one cannot really claim an understanding, so it is worthwhile pursuing the question. catslash (talk) 01:51, 9 November 2019 (UTC)[reply]

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Hello! Voting in the 2021 Arbitration Committee elections is now open until 23:59 (UTC) on Monday, 6 December 2021. All eligible users are allowed to vote. Users with alternate accounts may only vote once. The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail. If you wish to participate in the 2021 election, please review the candidates and submit your choices on the voting page. If you no longer wish to receive these messages, you may add {{NoACEMM}} to your user talk page. MediaWiki message delivery (talk) 00:17, 23 November 2021 (UTC)[reply]

Hello! Voting in the 2022 Arbitration Committee elections is now open until 23:59 (UTC) on Monday, 12 December 2022. All eligible users are allowed to vote. Users with alternate accounts may only vote once.

The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail.

If you wish to participate in the 2022 election, please review the candidates and submit your choices on the voting page. If you no longer wish to receive these messages, you may add {{NoACEMM}} to your user talk page. MediaWiki message delivery (talk) 00:35, 29 November 2022 (UTC)[reply]

I hatted it because the OP is a banned user. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:22, 14 May 2023 (UTC)[reply]

That is a valid reason for refusing to engage with the enquirer. Is there an alternative hat or other device that could be used to make that clear? catslash (talk) 18:27, 14 May 2023 (UTC)[reply]

I could have simply deleted the whole thing. Banned users are not allowed to edit. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:29, 14 May 2023 (UTC)[reply]

Hello! Voting in the 2023 Arbitration Committee elections is now open until 23:59 (UTC) on Monday, 11 December 2023. All eligible users are allowed to vote. Users with alternate accounts may only vote once.

The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail.

If you wish to participate in the 2023 election, please review the candidates and submit your choices on the voting page. If you no longer wish to receive these messages, you may add {{NoACEMM}} to your user talk page. MediaWiki message delivery (talk) 00:30, 28 November 2023 (UTC)[reply]

Heh. Thanks for the chuckle. GA-RT-22 (talk) 00:26, 18 January 2024 (UTC)[reply]

Hello! Voting in the 2024 Arbitration Committee elections is now open until 23:59 (UTC) on Monday, 2 December 2024. All eligible users are allowed to vote. Users with alternate accounts may only vote once.

The Arbitration Committee is the panel of editors responsible for conducting the Wikipedia arbitration process. It has the authority to impose binding solutions to disputes between editors, primarily for serious conduct disputes the community has been unable to resolve. This includes the authority to impose site bans, topic bans, editing restrictions, and other measures needed to maintain our editing environment. The arbitration policy describes the Committee's roles and responsibilities in greater detail.

If you wish to participate in the 2024 election, please review the candidates and submit your choices on the voting page. If you no longer wish to receive these messages, you may add {{NoACEMM}} to your user talk page. MediaWiki message delivery (talk) 00:16, 19 November 2024 (UTC)[reply]