Talk:Radiation pattern

3D far field plots
Hi - if there's anyone out there. I've created some 3D far field plots (Image:Radiation-patterns-v.png), and intend to put them on this page. In order to have enough text to go with both the existing and new images, I reckon I'll add some words defining cuts (great-circle and maybe conical cuts) through the pattern. Stop me now if you don't agree! --catslash 00:25, 1 August 2006 (UTC)

...but instead I added a section on reciprocity of antenna patterns, as this seemed a major omission. I was going to put this on the reciprocity theorem page, but that article is already long enough, and too nice for me to mess with. But- should this article be called antenna pattern (which currently redirects to this one)? --catslash 22:15, 30 August 2006 (UTC)

Why is the proof so long?
Since you are taking the Lorentz reciprocity theorem as a given, I don't understand why the proof that an antenna works equally well as a transmitter and receiver, for the same radiation pattern, needs to be so long. It seems like you can prove it in a few lines:


 * Lorentz reciprocity, and in particular its special case of Rayleigh-Carson reciprocity, tells you that $$\int \mathbf{J}_1 \cdot \mathbf{E}_2 = \int \mathbf{J}_2 \cdot \mathbf{E}_1$$ for two currents J and the resulting fields E.  In the limit of thin wires, this tells you that if a current in one wire produces a voltage in a second wire, then the same current in the second wire produces the same voltage in the first wire.


 * Suppose an antenna is driven with a current amplitude I, and the resulting field pattern is detected as a voltage amplitude V with a small test antenna someplace else (proportional to the field strength in the direction of the test antenna). Then, by reciprocity, if we drive the test antenna at the same place and orientation with a current amplitude I, then our source antenna will detect a voltage V.  Since the currents and voltages directly give the power transmitted and received (via P=V2/R or P=I2R), it immediately follows that the ratio of transmitted to received power is the same in both cases. Hence, the transmitting and receiving radiation patterns are identical.  Q.E.D.

Note that this proof is equally true in the near and far field, in inhomogeneous materials (e.g. if the antenna is placed inside a waveguide, as an extreme example), and tells you that the polarization sensitivity of the antenna is reciprocal and not just the angular pattern. Hence, it is both simpler and more general than the argument that currently appears in the article. Moreover, the argument currently in the article says that "reciprocity requires that the power transfer is equally effective in each direction," which isn't a very precise statement of the reciprocity theorem.

Am I missing something?

—Steven G. Johnson 01:50, 3 September 2006 (UTC)


 * Yes the proof is very long, but this is because I've put a lot of words it, in an attempt to make it more accessible. I deliberately omitted mention of polarization in order to limit the length, with the intention of adding a section on partial patterns (i.e. w.r.t. to a particular polarization), at a later date.


 * Re. your proposed proof outlined above; most of this relates to the fact that V-I reciprocity implies P-P reciprocity, while the body of the argument is reduced to Hence, the transmitting and receiving radiation patterns are identical (which is too short!). Regarding the V-I reciprocity part, I find three objections; (1) many antennas have a waveguide feeds, no wires being involved, (2) including it would make the existing proof even longer and (3) it's really the province of the reciprocity (electromagnetism) page. It would be good if that page had stated that the S-matrix of a microwave device is symmetrical (provided the device contains no non-reciprocal or non-linear materials) - that would give me S12 = S21 for my pair of antennas (unfortunately, there seems to be no page explaining S-matrices in this context (I may have to write it)). The microwave/RF/antenna material in Wikipedia is somewhat wanting - have you seen antenna (radio)?


 * Re. Generality/environment/near-field; the point is that the radiation pattern is supposed/defined to be a property of the antenna alone, not of the whole system. The environment may affect the power transfer by causing multi-path propagation (e.g. a reflected-off-metal-wall path interferes with the direct path), but this does not affect the antenna's pattern - instead the power transfer now depends on more than one point on the pattern. --catslash 15:21, 3 September 2006 (UTC)


 * The extension to waveguide feeds is a simple matter, because for the problem to be well-defined the waveguide has to be single-mode (at least in some region), as otherwise the radiation pattern will depend upon the excitation conditions of the waveguide. In that case the power is completely determined by the field at a single point, and can also equivalently be driven by a point current at the same point. So, this is a one sentence fix from my perspective.


 * I'm not convinced that more words makes a proof more understandable; in math we tend to favor proofs that are as short and general as possible. I'm sure my proposed proof could be clarified, of course.  In any case, if you're going to rely on reciprocity, you need to connect with the precise statement of the theorem, not with a vague paraphrase about powers; otherwise, you could easily end up implying things that aren't true, such as that the radiation pattern from the end of a multimode waveguide depends only on the power input.


 * Regarding the influence of the environment, the simple fact is that the radiation pattern is not a property of the antenna alone. If you want to define it as the radiation pattern of the antenna in vacuum, then you should say so.  However, while this is indisputably the most common environment in which the radiation pattern is calculated, it's not universal, especially when dealing with things other than microwave antennas.  For example, one sometimes calculates the radiation pattern of a source in a layered or anisotropic medium.


 * I don't think the reciprocity page should have a proof that is specific to microwave devices; it should remain devoted to general statements about Maxwell's equations. (The S-matrix is more general than microwave devices, and I agree that there should be a page on it and its symmetries.  I don't think it belongs on the main reciprocity page, though...to even define the S-matrix you have to pull in a whole set of new concepts about independent channels/modes that don't belong on the reciprocity page.  Once you've properly defined the S-matrix, the proof of its symmetry from reciprocity follows in a few lines.)


 * —Steven G. Johnson 15:29, 3 September 2006 (UTC)


 * Thanks for responding so quickly!


 * Re. Your extension to waveguide feeds; you are of course entirely correct, though the argument seems a bit circuitous.


 * Re. Number of words; very true; to be helpful, the words must be well chosen; in engineering we tend to favour proofs which are not intellectually challenging. The precise statement of reciprocity I need to connect with does not seem to be available within Wikepedia, which is why I said the reciprocity (electromagnetism) theorem requires that..., having carefully (and long-windedly) defined the powers to make this true. I then expected the interested reader to infer the step detailed in your proposed proof. Re. multimodes; antennas very often have two modes in the feed of course (degenerate ones; horizontal/vertical polarized, or LHC/RHC polarized in a square or circular guide) with some orthogonal[ized] pair being considered as separate antenna ports.


 * Re. inhomogeneous media; this seems to me a bit obscure (though that may be just me). I've come across ground penetrating radar and diathermy of course, but even in these cases, the pattern must be the pattern in the absence of the target? If the relative position of the target was given, then the system would be completely defined (the SAR on the tumour and surrounding tissue, or the echo from the mine would be fixed), so the concept of a radiation pattern would be redundant? Sorry, I'm speaking from ignorance; please enlighten me.


 * Re. S-matrix; then an S-matrix page must be written to plug the gap. I might do this myself. I propose to leave the offending power transfer statement until this is done. --catslash 17:05, 3 September 2006 (UTC)


 * In the case of a multimode antenna with two degenerate modes, the radiation pattern will indeed depend upon which mode is excited. Of course, if they are degenerate by symmetry then the two modes will give radiation patterns related by symmetry so it suffices to specify just one.  However, in such a case a more complicated statement of the result is needed (analogous to the S-matrix case where you have multiple scattering channels)...you can't simply say that "the" receiving and transmitting radiation patterns are the same, because the transmitting pattern isn't unique.


 * I'm not sure why you find the general proof based directly on Lorentz reciprocity to be "circuitous". I find that approach to be the most precise, general, and correct one, although I could go to an effort to make it more clear than the rough draft I posted above.  As opposed to the one currently in the article which I find rather longwinded, with hidden assumptions (like homogeneous media, or the assumption of single-modedness in order to have a unique radiation pattern, or the omission of polarization dependence), and it relies on a rather imprecise statement of the reciprocity theorem (which imprecision is closely related to your hidden assumption of single-modedness).


 * Whether inhomogeneity matters depends upon whether the inhomogeneity is in the near or far field. If it is in the far field (e.g. for the radar target), then the radiation pattern of the source can be defined nearly independently of the inhomogeneity.  If an inhomogeneity is in the near field, then it cannot be separated so cleanly.  For example, if you sit your antenna on the ground it will have a different radiation pattern than if you hold it up in the air...it's not just a matter of taking the original radiation pattern and reflecting/refracting it, because there can be non-negligible feedback directly on the source.  (As a more extreme case, if you surround your antenna by a photonic crystal with a complete bandgap at the operating frequency, then the antenna will radiate no power at all!) —Steven G. Johnson 19:34, 3 September 2006 (UTC)


 * PS. Although I know people in the microwave community (e.g. the great, late Hermann Haus at MIT), my own work is in nanophotonics, primarily at shorter wavelengths. So, you'll have to excuse me if my terminology is slightly different than that in the antenna/microwave community.  In nanophotonics, the components almost always interact in the near field and surrounded by inhomogeneities, which is perhaps why I find an emphasis on far-field interactions in homogeneous media to be overly restrictive.  —Steven G. Johnson 19:38, 3 September 2006 (UTC)


 * Interesting - I hope you're not bored of this discussion yet.


 * Re. Channels; I take it you don't think the article should say that if there's more than one socket on the back of an antenna, then it might matter which one you connect to? That's a well-known property of electrical gadgets. But there's a better reason why we only need to mention 1-port (1-channel) antennas; any N-port device is also a M-port device with for all M such that 0 &le; M &le; N; you just have to leave N - M of the ports in a fixed condition (matched, shorted etc.). This is analogous to currying mathematical functions. So although the assumption of a single channel is 'hidden', it's not unwarranted.


 * Re. Circuitous; Sorry that wasn't a fair description. I meant do we need to go from the reciprocity theorem stated in terms of spatial integrals, down to a result for a single point, then (via our knowledge of the possible modes), to the amplitude of the signal (which is an integral over the feed cross-section). I suspect there must be a more direct way, because this method breaks down once we consider more than a handful of modes (because there will be no suitable point).


 * Re. Objects in the near field; I'm happy to have such objects (a ground plane perhaps), but then the radiation pattern is the radiation pattern of the antenna plus objects. I'm also happy to have a periodic background medium. I'm not happy about defining a near-field radiation pattern because (1) it's an oxymoron, (2) it's pointless - if you place a receiver there it will affect the transmitter and (3) I'v.e never heard of such a thing (which is the relevant point for Wikepedia) --catslash 00:52, 4 September 2006 (UTC)


 * OK - I stand corrected. From the IEEE Standard Dictionary of Electrical and Electronics Terms:


 * radiation pattern (1) (fiber optics) Relative power distribution as a function of position or angle [my italics]. Notes:. Near-field radiation pattern describes the radiant emittance (W × m^-2) as a function of position in the plane of the exit face of an optical fiber. Far field radiation pattern describes the irradiance as a function of angle in the far-field region...)


 * ...and of course reciprocity between a given point on the exit face and a given fibre mode applies immediately.


 * This near-field 'radiation pattern' is fundamentally different to the far-field pattern, because it's a function of position not of angle. It's going to require a re-organization of the page. Damn! --catslash 15:08, 4 September 2006 (UTC)


 * OK - I've got a stack of books defining radiation pattern, near-field pattern and the strange term near-field radiation pattern, so I'll fix this (unless somebody beats me to it), before moving back to the question of precision regarding the reciprocity theorem. --catslash 20:36, 4 September 2006 (UTC)

You wrote: ''Re. Circuitous; Sorry that wasn't a fair description. I meant do we need to go from the reciprocity theorem stated in terms of spatial integrals, down to a result for a single point, then (via our knowledge of the possible modes), to the amplitude of the signal (which is an integral over the feed cross-section). I suspect there must be a more direct way, because this method breaks down once we consider more than a handful of modes (because there will be no suitable point).''

Yes, I do think you need to start from the reciprocity theorem stated precisely (i.e. as integrals of the fields/currents). The essential fact is this: you have to define what you mean by the "power" in a "channel", and in particular relate it to the fields. For a single mode, the power can be defined in terms of the field at any single point, so this is easy. For multiple modes/channels, you have to be able to decompose them somehow, which you normally do by orthogonality&mdash;that is, instead of using the field at a single point, you use an integral of the field against a mode pattern, or a current with the same pattern as the mode, so that you only couple to a single mode.

I would say that the theorem about the "radiation pattern" being reciprocal implicitly assumes a single operating mode in order for the "radiation pattern" to be uniquely defined. (There can be other modes, of course, but you can only use a single superposition of them in this simple version of the theorem.) As soon as you get to the more general multi-mode case you are really talking about the S-matrix, which should go into a separate article. My inclination would be to prove only the single-mode case in this article using the simple single-point argument, and refer to the (still to-be-written) S-matrix article for the multi-mode situation. (Or you can just prove the general S-matrix case on its page, and refer to reciprocity here as a special case. However, the single-mode case follows so simply from Lorentz reciprocity that I think it is probably worthwhile to prove it separately for pedagogical reasons.)

—Steven G. Johnson 21:13, 4 September 2006 (UTC)


 * OK - but bear with me a moment longer - the result


 * $$\oint_S (\mathbf{E}_1 \times \mathbf{H}_2) \cdot \mathbf{dA} = \oint_S (\mathbf{E}_2 \times \mathbf{H}_1) \cdot \mathbf{dA} $$


 * appears to be within spitting distance of the result I want, provided that the surface A can be disjoint. That is if A can be a pair of bubbles, one enclosing the transmitter (and so cutting the feed to its antenna), and one enclosing the receiver (and similarly cutting its feed). Then if E1,H1 and E2,H2 are the fields for transmission in opposite directions (with a particular mode excited in the feed of whichever antenna is transmitting). Then... but I don't know the rest. --catslash 22:00, 4 September 2006 (UTC)


 * The surface can be disconnected, but then both integrals are over both surfaces. I'm not sure this is going where you want.   Why don't you first state the theorem that you would like to prove? —Steven G. Johnson 22:23, 4 September 2006 (UTC)

Near field radiation pattern
I'm going to replace the article lead (?), to emphasize the distinction between the far-field pattern and the near-field pattern, and to reflect the fact that radiation pattern can mean either (I'll also make this distinction in the reciprocity section). Here's what I'm snipping out..


 * A radiation pattern refers to the spatial variation of the electromagnetic fields, the field intensity, and/or the irradiance (power density), produced in the far-field from a localized source. Most commonly, the radiation pattern is described for a source in a homogeneous medium (typically vacuum or air).


 * For example, it can refer to the radiated field of a radio or microwave antenna, the power distribution radiated from the end of an optical fiber, the pattern of light from a light-emitting diode or laser, and similarly for many other devices. The radiation pattern generally depends on the precise source geometry and the wavelength of the radiation.


 * The radiation pattern is often represented graphically by the power distribution as a function of angle, in the plane of maximum radiation or in the E-plane and H-plane for linearly polarized sources.



--catslash 13:25, 7 September 2006 (UTC)


 * The new version looks good to me. —Steven G. Johnson 23:44, 7 September 2006 (UTC)

Thanks for the vote of approval - you'd better write a page on integrated optics now! There's still a lot to do on this page of course. --catslash 14:46, 8 September 2006 (UTC)

Missing logic?
I don't get the logic in the end of the proof: "Analysis of a particular antenna (such as a Hertzian dipole), shows that this constant is, where λ is the free-space wavelength. Hence, for any antenna". It seems to me like a generalization to the general from the particular that is not obvious. —Preceding unsigned comment added by 80.216.130.104 (talk) 10:11, 7 March 2009 (UTC)


 * It's intended to be in the immediately preceding sentence: Furthermore, the constant of proportionality is the same irrespective of the nature of the antenna, and so must be the same for all antennas. --catslash (talk) 15:00, 7 March 2009 (UTC)

Pattern Plotting Concept
One question I always had on antenna gain plotting is the difference between rotating the antenna (with signal source attached) and recording the signal strength from a fixed location (same as walking around a fixed antenna with a receiver, taking the path of a constant radius circle and recording received signal strength), and walking around with a receiver and following the path of constant received amplitude. The latter is what you would see if the transmitted energy was visible, the true shape of the field in space. The former collects data about signal strength and therefore antenna gain at different angles, but if you plot it on linear or log scale, you don't get the real world spacial distribution of the energy. If you superimposed an antenna pattern on a map, and assumed that people (with a receiver of a certain threshold) could receive the signal if they were inside the pattern, it would be the latter pattern only that would work, not the standard log gain plot. In ideal space, the "walk around" iso-amplitude pattern would follow 1/r^2 rule after the antenna gain were applied in each direction. Another issue here is if, in the walk around process, you had an antenna null that caused you to travel into the near field area of the antenna, some strange things would happen. You would have to use enough power into the antenna so that your receiver would work everywhere outside the near field while walking the constant signal strength path.

I wish this Wikipedia article addressed these not so obvious issues with measuring and plotting radiation patterns. I personally don't feel qualified to do this. 192.80.95.243 (talk) 16:33, 28 December 2010 (UTC)


 * Yeah, to find the far-field pattern you want to stay out of the near field and transition regions, which can extend as much as 10 wavelengths from the antenna, which (as you say) might be difficult with your second method. I'm pretty sure that at antenna testing ranges they use your first method, recording the signal strength at a constant distance.  The radial coordinate of radiation plots is signal strength (usually dBi), not range.     However (and here's what I think you were asking) on a decibel radiation plot, the radial coordinate IS proportional to the range contour that would give a constant signal strength!  The logarithmic nature of decibels removes the inverse square factor.  So a standard dB radiation pattern is also a pattern of the relative range you would get.  Just rescale the decibels to miles. -- Chetvorno TALK 17:50, 28 December 2010 (UTC)


 * An amplitude plot, (one where the radial scale is proportional to the field strength (amplitude) at a fixed (sufficiently large) distance), is also a plot of the distance travelled before the signal strength drops to some given constant level. This is because the field strength drops off like 1/r (the power is proportional to the square of the field strength, and so drop off like 1/r2). If you double the received field strength in a particular direction, then you have to stand twice as far away to bring it back to the same level. Amplitude plots are often used with low-gain antennas (like dipoles) and dB-scale plots used with high-gain antennas (like dishes). The radiated power, gain or directive gain plotted on a dB scale is also a plot of the logarithm of the distance travelled before the received power drops to a given level. All this assumes the distances involved take you out of the near-field of the antenna (if the largest dimension of the antenna is D, then further away than D2/λ and further away than (say) λ/2, whichever is the greater). I agree that distance reached is a nice intuitive way of understanding the plots - however I believe it's non-standard, and so would count as original research. --catslash (talk) 19:39, 28 December 2010 (UTC)

Why include Proof of reciprocity at all ?
Someone remarked the proof of this is long, but I don't see why such a proof is needed. That's a completely separate topic. You wont find the proof of the reciprocity theorem in any antenna book in a section devoted to the radiation pattern.

It just strikes me someone knows a proof of it, so they thought they would write it in the section on radiation pattern. IMHO, the reciprocity theorm should just be reffered to, and a link to it on another Wikipedia page. It is unneeded in a section on radiation pattern. Sure, it is helpful to know about it for antenna analysis, but so are dB too, and its useful to know what a transmitter is. Is there a need to explain all them?

IMHO, the whole section should be removed. Drkirkby (talk) 23:15, 15 April 2012 (UTC)
 * Yes the proof I wrote is rather long. Yes it's true that you won't find the reciprocity theorem in the radiation-pattern-section of an antenna book. But that's not the theorem that is described here. The theorem that is described here is: the receiving and radiation patterns of an antenna are identical. You will surely find that in any serious description of radiation patterns, together with a version of the proof given here. A shorter version of the proof would be: If a transmitter is not appreciably affected by the presence of a receiver, then we can write the transmitter-to-receiver coupling as the product of two independent factors: (1) the directive gain of the transmitting antenna towards the receiver and (2) the directive (receiving) gain of the receiving antenna in the direction of the transmitter. If these two factors are independent and if the reciprocity theorem holds between the transmitter terminals and the receiver terminals, then the receiving pattern of each antenna must be the same as its radiation pattern. --catslash (talk) 00:50, 16 April 2012 (UTC)


 * Hubregt pp75-80 not only gives a similar proof of what is referred to as antenna reciprocity (sec. 4.4), but in passing pauses to offer a proof of the Lorentz reciprocity theorem (sec. 4.4.1).
 * Silver gives a similar proof of what it refers to as Reciprocity between the transmitting and receiving patterns of an antenna
 * Bakshi, Bakshi & Bakshi offer a similar proof, but refers to it as equality of directional patterns.
 * Likewise Kumar
 * Blahut proves the antenna reciprocity theorem.
 * This is standard content. Perhaps if the sections were renamed e.g. reciprocity -> antenna reciprocity it would avoid confusion? --catslash (talk) 21:26, 16 April 2012 (UTC)

Directivity Plot
The figure showing a rectangular radiation plot is mislabeled as showing directivity in dBi. It is actually showing a normalized power pattern in dB relative to peak, because a 0 dBi antenna can only be an isotropic radiator.