Talk:Dipole antenna/Archive 1

Archive 1

impedance at wavelength?

"Impedance is not real but it does becomes real for a length of about .46 λ"; I take it to mean this is because the "Im" curve of the graph shown crosses the x-axis at about .46. But the impedance always has a real part and an imaginary part; the imaginary part is just positive past .46 λ, right? Could somebody with a little more savvy check me on this?

74.193.185.169 17:47, 18 March 2007 (UTC)

An impedance is normally considered real if the imaginary part is zero. An impedance with both a real and imaginary part is normally considered complex. At some length a little under a half-wave, the imaginary part of the dipole's impedance will be zero, so the impedance will be considered real. (The exact length the antenna becomes resonate, and so have zero imaginary part, depends on the diameter to some extent too). So for example, 40 + j 0 Ohms is real, 0 + j 23 Ohms is imaginary and 34 + j 89 Ohms is complex. Drkirkby (talk) 10:04, 30 March 2008 (UTC) Drkirkby (talk) 10:05, 30 March 2008 (UTC)

improvements needed

This page needs VERY desperate attention. This is probably THE most important antenna type, since most antennas are built using arrays of dipoles. Main problem, too much fluff, not enough actual info.

TODO:

• The animated gif needs to go. It should be replaced with a better reprensetation of the contents, maybe xy gain then xz gain. —Preceding unsigned comment added by 192.52.218.43 (talk) 15:21, 30 October 2009 (UTC)
• inventor and first use
  • The dipole is so simple, I would think that it would have had to be invented simulatenously with the radio?
  • Credit is given to Hertz for this one.Kgrr 11:21, 15 January 2006 (UTC)
• (radiation efficiency vs length)
  • What do you mean by "radiation efficiency"? May be the good term is "gain"? Radiation efficiency is almost always near 100%: all power fed to antenna is radiated. Loses in wire resistance are negligible.LPFR 12:18, 27 July 2006 (UTC)
• (frequency tuning)
  • • done Kgrr 12:02, 15 January 2006 (UTC)
• theoretical impedance with infinitely thin (and change due to having thickness)
  • Theoretical resistive part of the impedance can be calculated "easily" with a bit of numerical integration. The reactive part is another history. I put a diagram of measured real and imaginary parts of the impedance of a dipole from $\scriptstyle {0.4\lambda }$ to $\scriptstyle {0.6\lambda }$. Influence of thickness is known (the antenna becomes more capacitive) but I do not think it can be theoretically calculated. There are thing that are far easy to measure than to calculate... antennas for one! LPFR 12:18, 27 July 2006 (UTC)
    • The reactive part of the input impedance CAN be calculated. There are several methods that this can be done - some give a pure analytical result, others need numerical integration and so are more suited for use by computers. See for example Antenna Theory - Analysis and Design by Constantine Balanis, 3rd edition, 2005, page 258 (ISBN 0-471-66782-X). Antennas by Kraus (1988) also has a formula for this, based on the work of Brown and King, which was published in Proc. IRE, volume 22, pages 457-480, (April 1934). Drkirkby (talk) 11:14, 30 March 2008 (UTC)
• theoretical gain over isotropic (2.4 dBi?)
  • done. LPFR 12:18, 27 July 2006 (UTC)
    • done Kgrr 12:15, 15 January 2006 (UTC)
• theoretical radiation pattern
  • A graph would be nice --ssd 09:10, 8 January 2006 (UTC)
    • done Kgrr 11:21, 15 January 2006 (UTC)
    • The diagram of radiation produced by Kgrr has a logarithmic scale. It should be indicated. Diagrams in the sections "short dipole" and "half wave dipole" are in linear scale. LPFR 12:18, 27 July 2006 (UTC)
• that it requires a balanced feed (and reference to balanced and how monopole is unbalanced)
  • • done Kgrr 12:21, 15 January 2006 (UTC)
  • I have added a remark about the correct way of feeding the dipole from coax (using a BALUN)Cadmium 12:21, 13 December 2005 (UTC)
    • yes, ideally but not required; done Kgrr 11:21, 15 January 2006 (UTC)
• change in radiation pattern vs height above ground plane
  • I have posted "Effect of ground" in antenna as the effect of ground is the same for all antennas. LPFR 11:10, 17 August 2006 (UTC)

• fact that monopole over groundplane IS same as dipole in free space
  • Last two points: I think that an entry "ground effect in radiation patterns" is missing. It should settle some issues as "dipole vs. monopole" or "Influence of height of the antenna. In a near future, I will fill this gap. LPFR 12:18, 27 July 2006 (UTC). Done.LPFR 11:10, 17 August 2006 (UTC)
    • The two are not the same. The most obvious difference is that a monopole over a groundplane will radiate only above the groundplane and not below it, whereas a vertically polarised dipole will have a symmetrical pattern above and below the centre. Drkirkby (talk) 10:16, 30 March 2008 (UTC)
• reference to electromagnetic waves
  • At least partially done in "elementary" to "half wave dipole". LPFR 12:18, 27 July 2006 (UTC)

Notes: From Warren L. Stutzman and Gary A. Thiele, "Antenna Theory and Design" and assuming a 100% efficient ideal dipole the theoretical gain would be 1.76dBi ( 10*log(3/2) ).

(Note: this critique was contributed 9 July 2004 by Brandon.irwin. This article has been editted over 20 times since.)

• Possibly the variations of the dipole should be organized and listed together, rather than being spread through the article. --ssd 09:09, 8 January 2006 (UTC)

folded dipole

I don't think it is correct that a folded dipole is a loop. Loops and dipoles are very different. I'll have to double check, but I don't remember seeing the ends of any folded dipoles being connected. --ssd 12:43, 27 December 2005 (UTC)

At a glance, the folded dipole might look like a loop, as the ends of the dipole are joined. However, this is not a loop; it acts like a dipole with an impedance transformer, with the radiation pattern of a dipole, not that of a loop. The best reference for this I could find was W4RNL's folded dipole article. (The Antenna Book mentions the folded dipole, but the information is scattered through chapter 6, with little theory even mentioned.) --ssd 14:40, 28 December 2005 (UTC)

John D Heys (G3BDQ) in his book Practical Wire Antennas (ISBN 0900612878) lists the folded dipole under loops, and he regards the folded dipole as a loop. It is possible by adding a third wire to the folded dipole in increase the radiation resistance yet further. In this book the theroy of the folded dipole is explained. Cadmium 20:49, 29 December 2005 (UTC)

I'll have to look for that text. Does it discuss radiation pattern of a folded dipole? --ssd 00:22, 30 December 2005 (UTC)

the book states on page 37 that the radiation pattern of a folded dipole is the same as that of a normal dipole.Cadmium 10:30, 30 December 2005 (UTC)

That's what I thought. That would mean it isn't a loop, despite appearances. Thanks! --ssd 15:20, 31 December 2005 (UTC)

Folded dipoles and loops behave very differently. A loop is more similar to a quad in its operation and radiation pattern.Kgrr 06:12, 13 January 2006 (UTC)

A loop is to a quad what a dipole is to a yagi. --ssd 06:47, 13 January 2006 (UTC)

Why does the text say a folded dipole is half a wavelength while the image says it's one wavelength? —Preceding unsigned comment added by 99.231.124.188 (talk) 05:00, 13 June 2008 (UTC)

Yeah, the image is incorrect and confusing. I am removing it as it is incorrectly labeled. The total length is 1/2 not each end. Here is the original image

Folded Dipole Antenna

Hovden Feb 19 2009

Folded dipole is made of wire which is two wavelength long. This mean that folded dipole is one wavelength long from end to end to reach the resonance for 300 ohm impedance. I tested this myself with measurements because I was confused from to many mistakes in literature like Wikipedia. The image is 100 percent correct, the text which says that folded dipole is half wavelength long is incorrect. Put the image back and repair the text. People to much beleive in what is written, next time check it and test it instead of make copy-paste. Joze Kocevar from Ljubljana/Slovenija —Preceding unsigned comment added by 193.189.163.18 (talk) 16:45, 2 December 2010 (UTC)

Hi I thought you were vandalising when you made the update to the article, however I see that it is just a mistake and not deliberately messing it up. The diagram is wrong and normally the total length of wire is one wavelength, with a quarter wavelength protruding in each direction from the feed point. A full wave dipole has a high impedance, and folding it will not change it to a suitable impedance for a normal feedline. Of course you can drive an antenna with double the frequency and get something matching the diagram, but you will have a serious impedance mismatch. Graeme Bartlett (talk) 11:31, 7 December 2010 (UTC)

I got my data with measurements, how did you get it. In that place I would remind you that lie told 100 times became a true. Joze Kocevar from Ljubljana/Slovenija. —Preceding unsigned comment added by 193.189.163.18 (talk) 12:56, 7 December 2010 (UTC)

merge with Half wave radiator

I think Half wave radiator needs to be merged into this article. I think that's really just a huge dipole, right? --ssd 13:11, 27 December 2005 (UTC)

A half-wave radiator is not a dipole. It's actually more like a fullwave antenna. The ground plane acts like a mirror, providing the other half of the dipole. It's electrical impedance and coverage pattern is very different from a vertical dipole.

A quarter-wave vertical antenna is electrically similar to a vertical dipole because the ground plane forms the other half of the radiating elements. However, still, they are different antennas. Kgrr 06:01, 13 January 2006 (UTC)

This statement makes no sense to me, and contradicts what is currently in the article. A quarter wave whip is certainly a different antenna than a half wave radiator, but they are still both dipoles. The quarter wave whip uses a groundplane as the image, as you say, but the half wave radiator is more like an end fed (rather than center fed) dipole. --ssd 06:41, 13 February 2006 (UTC)

I do agree with Kgrr. A quarter-wave antenna is just one-half of a half wave dipole, as is also its impedance (the impedance of a quarter whip is 36+21j ohms). The misunderstanding comes from the fact that a quarter-wave antenna has always a reflective surface (a real or an artificial ground). If this surface does not exist, you do not have a quarter-wave antenna but an end–feed antenna with a bad length. LPFR 12:34, 27 July 2006 (UTC)

Delisted Ga

No references! Being that references are a requirement to reach GA status, the article is being removed from the GA list. If references are properly added, this article should be able to make GA status again. AndyZ 23:27, 6 January 2006 (UTC)

• I would suggest improving the lead and the extremely short sections. AndyZ 04:42, 15 January 2006 (UTC)

baluns and stuff

The balun section is beginning to make me uncomfortable. The original wording in the article is somewhat confused. (Not exactly wrong, but not right either.) I think the details are too much of a tangental issue to all go into the article, but some of this should. Let me try to clarify some things here:

1. Balance and match are different. A mismatched feedline has the wrong impedance. Impedance mismatches cause reflections at the impedance boundary, which adds up and is measured as SWR. (See Antenna (radio)#impedance for a longer explanation.) In a transmitter SWR multiplies normal losses in the feed line and increases voltage spikes and tranceiver heating, nasty stuff. In a receiver, it is still bad; the impedance mismatch will cause ghosts in a video signal, for instance.
2. An unbalanced feed to a balanced antenna causes two separate problems with a related cause.

   common mode current

   Normally, the current in the shield and the current in the center conductor are 180 degrees out of phase, so that their E-M fields cancel, and the power is transmitted through the transmission line. Common mode current happens because when the current traveling in the shield hits the antenna, part of it goes into the shield half of the dipole, and part goes back down the shield. The current traveling back down the shield is in phase (same mode) with the center conductor, so the fields do not cancel, so the feedline radiates, thus becoming part of the antenna, rather than just a feed line.

   distorted radiation pattern

   The current in the center conductor all goes into its dipole half, but the shield current is split. Thus, the center conductor's half of the dipole's pattern will be stronger than the other, and the shield will add a smaller omnidirectional pattern as it radiates. Of course, when you add them, you don't get a circle added to the nice 8 pattern a dipole would normally get, because some of the phases cancel. Instead, you get a distorted cartoid. (There's a nice direction finding antenna that balances these two patterns to cancel completely in one direction and get a single null, but I digress.) Usually the common mode radiation is weak enough to be ignored in the far field, so you just get a distorted 8 pattern. (Imagine the lobes in the nice symmetrical picture on the page drooping a bit.)

At high power, the common mode current will cause your feedline to radiate into all the surrounding equipment, causing all kinds of problems, and maybe even RF burns. At lower power, that can be ignored, and the distortion sometimes is not of great concern either. However, since the coax is now part of the antenna, you've just added a big long inductive piece to the antenna, and the antenna will have a nasty complex impedance with a non-zero imaginary part, which leads to higher SWR, and much higher loss in the feed line. (And if you are low power, you can't afford to lose much power this way.)

Now, here's where the confusion lies:

1. Baluns are typically transformers. If it is only a balun, it is a 1:1 transformer. The balun blocks unwanted common mode current by forcing the current in each half of the transmission line to be the same.
2. Impedance matching can be fixed with a transformer too. For example, a 4:1 balun/transformer can transform a 75 ohm unbalanced coax impedance into a 300 ohm balanced twin lead impedance.

Essentially, one device does two jobs that are not clearly related. (This should be rewritten, and half put in the balun article in a theory section, and half put in the dipole article.) --ssd 08:14, 12 January 2006 (UTC)

Baluns are not essential to the operation of a dipole. Perhaps the balun treatise should be handled under baluns rather than dipole.Kgrr 06:07, 13 January 2006 (UTC)

No argument here. But someone wanted baluns mentioned, and if we're gonna do it, it should be done right, not semi-factually. That's why I put this here, not in the article. Is there too much in the article now? Perhaps not; it is showing the correct way to feed a dipole, at least. --ssd 06:51, 13 January 2006 (UTC)

Another tangentially related item... At low power, the common mode current in the feed line can be lived with by making the feed line a multiple of a quarter wavelength long (adjusted for velocity factor of course). You'll still get radiation from the feed line, but at least the reactance will be zero. Essentially, the feedline becomes part of the antenna, and this tunes it. If it is feeding a vertial dipole anyway, this might actually increase your gain by making it a much larger antenna. RF burns at higher powers are still a danger, though. --ssd 06:51, 13 February 2006 (UTC)

Speed of signal on a wire

I erased the assertion that "the velocity of propagation of electromagnetic waves in wire is slower than that in free space". The propagation of electromagnetic waves in a wire or in air-filled coaxial line, or in a bifilar line is the same as in free space. The reason for the resonance of a dipole at a length a little shorter than half wave is due maybe to a non-sinusoidal current distribution in the dipole (but I am not certain of this). LPFR 14:18, 16 August 2006 (UTC)

Actually, the speed of an electromagnietic wave in a (pair of) wire(s) IS less than that of free space. This is due to the capacitance and inductance between the two wires (the delivery and the return wire) making the transmission line less than ideal. --DrBob127 08:02, 23 November 2006 (UTC)

Speed of signal-propagation-down a wire

• I'm not sure why you erased this reference as the propagation of energy is slower down wire/coaxial cable and anything else except free space-you are correct as far as I know.

The propagation speed of signals down a coaxial cable, with ordinary (non-magnetic) isolation is $\scriptstyle {v={1 \over {\sqrt {\mu \varepsilon }}}={c \over n}}$ where $\scriptstyle {n={\sqrt {\varepsilon _{r}}}}$ is the refractive index and $\scriptstyle {\varepsilon _{r}}$ is the dielectric constant of the isolation filling the gap between the inner and outer conductors. If the coaxial cable is air-filled, the speed will be the same as the speed of electromagnetic waves in free air (almost the same speed as in vacuum). A single wire in air, as is each branch of a dipole, can be seen as a coaxial cable whose outer conductor is very far. As this "coax" is filled with air, the speed will be the same as in the air, and not 5% smaller as stated the phrase that I erased. -- LPFR 11:42, 27 October 2006 (UTC)

Hi LPFR,

This may be so, however there is also the effect of end capacitance /support insulators that tend to introduce some difference in the physical length of a dipole versus the electrical length. the effect is in the order of from 2% to 5%. --Read-write-services 00:27, 28 October 2006 (UTC)

Merging with "antenna (radio)"

• I was thinking that perhaps this article needs to merge with or, at least link to [[1]]?

Wikipedian self-named Read-write-services is asked to respect wikipedia usages. This point is not an item of "Speed of signal-propagation-down a wire". LPFR 11:57, 27 October 2006 (UTC)

I apologise, not all of us are experts. --Read-write-services 00:29, 28 October 2006 (UTC)

As for the asked merging with antenna, I do not think that it is a good thing. Antenna and dipole articles are too long to be merged. In fact, I plan to split antenna article (see talk:antenna). As for the link to antenna, it is done in the second line. LPFR 11:57, 27 October 2006 (UTC)

Included diagrams are of unreasonable size

The diagrams included in the baluns sections are of a ridiculous dimension, despite being very company file size wise. My browser, literally, had to hit swap for 4 minutes before all the PNG files were loaded into memory and stabilized to a point that I could scroll. It also does not appear to be enough detail in the images to justify the several million pixels used, per diagram. Please resize said images to a more reasonable and readable resolution. 24.80.114.123 19:20, 11 December 2006 (UTC)

Yep, my computer crashed several times while trying to view this page. Sewebster 03:56, 3 February 2007 (UTC)

I switched the balun images to the linkimage template so that the thumbs won't be displayed. Someone needs to resize all of them, reupload, and switch back to the normal image format. I might do it, but I'm not an image upload expert (maintain proper authorship etc etc.) Sewebster 04:09, 3 February 2007 (UTC)

I've taken care of it (hopefully). - mako 05:00, 3 February 2007 (UTC)

Diagram should be in metric (meters), not imperial (feet)

There's a diagram showing a dipole being a length 468/MHz (feet). Given other parts of the article use length in meters, I feel this should too. In any case, the SI unit of length is meters, not feet. So I suggest its changed to l=144/f, where l is in meters and f in MHz.

Also, the diagram says it's public domain, but a robot on Wikipedia says that is no longer acceptable. So I assume the diagram needs removing for that reason too.—Preceding unsigned comment added by Drkirkby (talk • contribs)

I've updated the image, using l=143/f (142.65 is the figure given in the article).--Father Goose (talk) 05:57, 19 July 2008 (UTC)

Dipole reactance needs more information

There is a rather complex derivation of the real part of the input impedance of a half-wave dipole, which shows it's 73.13 Ohms. There is nothing for the reactance. There is a simple graph drawn with GNUPLOT which plots the reactance and resistance vs length, but no idea where the formula comes from. I suspect (know) its only a very rough approximation. For a start, they depend on diameter too.

Ideally we want a formula for both the resistance and reactance of a dipole of arbitrary length and diameter. If nothing else, the exact values for an infinitely thin dipole of half wave.

I think the book by Balanis has the derivations for at least some of this. —Preceding unsigned comment added by Drkirkby (talk • contribs)

Gain of dipole antennas

Many other sources state, that the gain of a full wavelength dipole is about 3.8 dbi and for a 3$\lambda$/2 dipole about 3.3 dbi.

I am not a total expert on this, but it seems to me the table needs some revision. —Preceding unsigned comment added by 91.1.255.104 (talk) 16:13, 11 September 2008 (UTC)

I've deleted a lot of the table. I don't have any reference for this, but some data from a computer model, based on the method of moments shows 2.15 dBi for 0.5$\lambda$, 3.84 dBi for 1.0$\lambda$, 3.51 dBi for 1.5$\lambda$, 4.03 dBi for 2.0$\lambda$, 4.87 dBi for 2.5$\lambda$, 4.78 dBi for 3.0$\lambda$, 5.87 dBi for 3.5$\lambda$, 5.46 dBi for 4.0 $\lambda$ and 7.36 dBi for 8.0$\lambda$. Note that longer dipoles do not always show higher gain, unlike the table which showed progressively increasing gain for longer dipoles. However, please don't put my figures in, as they are are just quick calculations using MMANA-GAL, which is a program based on the method of moments. I have no reference for them. Note also, for the dipoles which are an integer number of wavelenghts long, the input impedance is very high. It should be possible for find a peer reviewed paper which has some of these gain figures. If so, they can be added. Drkirkby (talk) 11:20, 1 November 2011 (UTC)

Multiple merges of stubs

I've noticed there are handful of stub articles that talk about variations of the dipole antenna. Many of these articles are such bad stubs that they also contain factual errors and misconceptions. I'd like to consider merging all of these into this article, and convert the original into a redirect.

Alternatively, if someone objects to merging any of these into this article, they should rewrite or substantially expand the stub article, and move this article's content into that article, leaving behind a link and a very short summary.

• Inverted vee antenna
• Isotron (seems to be a brand name, rather than an antenna type)
• Whip antenna / Monopole antenna / (Rubber Ducky antenna) (these need some merging, but are prob ok separate from this)

Opinions? --ssd (talk) 15:22, 27 September 2008 (UTC)

Isotron is not an example of a variation of a dipole, but calls itself one. It is coil loaded. It is a brand name, but it is also a unique design, worthy of note here. —Preceding unsigned comment added by 206.245.139.67 (talk) 21:27, 4 February 2009 (UTC)

Calculation of impedance

The calculation of impedance as $20\pi ^{2}(L/\lambda )^{2}$ needs elaboration. Where does the 20 come from? It should be related to the natural quantity $Z_{0}={\sqrt {\mu /\epsilon }}$, the impedance of free space, which is about 377 ohms. The radiation resistance should be something like: $R_{rad}={\pi r^{2} \over I_{0}^{2}Z_{0}}\int _{0}^{\pi }{|E_{\theta }|^{2}\sin \theta d\theta }$ where $|E|$ is the far radiated electric field. Later on, in the discussion of gain, the approximation of $Z_{0}=120\pi$ ohms is used, but a factor of 2 is dropped from the equation....

Similarly for the half-wave dipole, the radiation resistance should be stated in terms of the impedance of free space.

Chris.n.richardson (talk) 12:17, 18 December 2008 (UTC)

Too Technical

Added the template. As a non-scientist, I found that I didn't learn much about Dipole antennas after reading this article. I realize that the subject is inherently technical, but this article - and the talk page - seems to be written by technicians for technicians. Please try to make at least some of the content more digestible to non-techs. Thanks! Iskandar Jamshyd (talk) 23:54, 14 May 2010 (UTC)

"Dipole types" section is redudant

Some of the material in this section needs to be moved up to the "Short dipole" and "Elementary doublet" sections, since they are redundant. -Roger (talk) 05:23, 20 September 2010 (UTC)

Not necessarily half-wave

Article text:

Large constructed half-wavelength dipole towers include the Warsaw radio mast — the only half-wave dipole for longwave ever built — and Blaw-Knox Towers.

There's nothing specific to the Blaw-Knox diamond-cantilever tower that makes it different from any other series-fed tower in terms of this article -- in particular, there's no particular reason why they should be half a wavelength. (The most efficient tower height is about 5/8 wavelength, anyway.) The WLW tower in Mason, Ohio, is 191 degrees; the WSM tower in Brentwood, Tenn., is 192 degrees; the original WFEA tower in Merrimack, N.H., is 175 degrees (although the FCC's records seem to be somewhat confused as to which tower is which). There are plenty of uniform-cross-section AM towers ranging from less than a quarter wavelength to nearly 3/4 wave (WJOI at 243.6 degrees appears to be the current winner), not counting sectionalized towers like Franklins (which as true center-fed, full-wave dipoles are the most appropriate to consider in this article, never mind the fact that there are only two currently extant, both at KFBK). 121a0012 (talk) 05:48, 11 May 2011 (UTC)

After waiting for eight months, I went ahead and deleted the reference to Blaw-Knox. The whole one-sentence "section" should probably go with it, but I'll leave it up to the engineers if they want to rescue it. If anyone does, they should probably be prepared to explain why top-loading works in addition to finding relevant examples of broadcast towers that are exactly half a wavelength tall. 121a0012 (talk) 06:23, 15 January 2012 (UTC)

Removed most of the table on dipole lengths

There was a table showing the gain of dipoles of various lengths, which showed increasing gain up to 8.51 dBi for 8 wavelengths long. I've no idea where someone got these numbers from, but the data is not referenced and does not agree with a computer model. For example, the full wave dipole is shown to have a gain of 2 (3.01 dBi), yet according to a computer model it is 3.84 dBi. For 8 wavelengths the table showed 8.51 dBi, but a computer model shows 7.36 dBi. I've not spent any time on this, and doing so would be a waste of time as it would be considered original research, but unless someone can reference these numbers, they are best removed. I've left the data for a half-wave (2.15 dBi) and very short dipole (1.76 dBi), as they are correct, but I don't believe the others are correct. Drkirkby (talk) 11:01, 1 November 2011 (UTC)

Differential Electric Field

The first electric field equation should be a differential in the electric field should it not? — Preceding unsigned comment added by 24.234.252.52 (talk) 18:36, 26 November 2011 (UTC)

The relevant differential equations are just Maxwell's equations in spherical coordinates. The equations for E and H given in this article are solutions of these. --catslash (talk) 19:43, 26 November 2011 (UTC)

That's not what I'm talking about. The electric field is equated to an incremental current element. It should be an incremental electric field equated to an incremental current element. The actual electric field is the integral of the incremental current elements. This is what is done in the 'short dipole' section where the electric field is derived from the integration of the triangular current distribution. — Preceding unsigned comment added by 70.26.148.210 (talk) 07:35, 7 December 2011 (UTC)

So, strictly it should say

$\delta E_{\theta }={{-iI_{\circ }\sin \theta } \over 2\varepsilon _{\circ }cr}{\delta \ell \over \lambda }e^{i\left(\omega t-kr\right)}$

etc. True, but a quick scan of Google Books shows that it is not conventionally written that way. --catslash (talk) 11:30, 7 December 2011 (UTC)

No, the assumptions is that the antenna is just a very short length, so the entire field is due to that. If you want to integrate it for a larger antenna then convert to an integral form of what is written above. Graeme Bartlett (talk) 12:05, 7 Dec ember 2011 (UTC)

Yes Catslash, that's what I meant. But how can you defend the Google Books. Have you ever seen a macroscopic quantity equated to an infinitesimal? @ Graeme - the infinitesimal current element is part of the dipole which can have a large current variation along it's length. The distant field is due to their integration, regardless of the antenna length. I admit I don't know much about EM radiation, which is why I read the article, but I am mystified by the defense of what appears to be some pretty questionable math.

Archive 1