How to calculate antenna beam width - Page 1
Forum section: electronics / projects

author: yanir
author_rank: Regular Contributor
timestamp: July 11, 2016, 12:11:53 am
content: Hi Antenna gurus.I'm trying to figure out the beam width for a patch antenna at a specified gain. Let's say the gain of the antenna is 25dbi. If I take an ideal isotropic antenna and take all the energy from the lower hemisphere and direct it equally on the upper hemisphere I'd have a gain of 3dbi and a beam width of 180 degrees right? If I keep doubling my power and narrowing the beam by halving the beam width will I arrive at a decent approximation?Is there a better way do this (without simulation)?Is my line of thinking correct?I question my results because the beam is gets extremely narrow at 25dbi using this method.Thanks!

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 01:19:01 am
content: Your thinking is correct.Quote from: yanir on July 11, 2016, 12:11:53 amTechnically, you don't halve the angle, you halve the surface area and end up with some angle. But for high gain antennas you halve so many times, it does not really matter.And yes, high gain antennas are very directional.

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 01:36:01 am
content: Also, after doing some math, it look like angle of the ideal beam can be calculated as A=arcsin(2/G). This is valid for G >= 2.So in this case A = arcsin(2/316) = 0.0063291 radians = 0.36 degrees.

author: retrolefty
author_rank: Super Contributor
timestamp: July 11, 2016, 01:39:16 am
content: 25 dbi is a lot of gain and it counts both for transmit and receive, care to share the design and frequency?

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 01:42:27 am
content: Quote from: retrolefty on July 11, 2016, 01:39:16 amIt is a typical claim for Chinese 2.4 GHz yagi antennas that are sold all over the place.Just goolge "25 dBi yagi" and you will find a number of places selling them.PS: I have not seen an actual performance measurement, so they very well may be "Chinese" 25 dBi, lust like 1000 W portable CD players

author: retrolefty
author_rank: Super Contributor
timestamp: July 11, 2016, 01:57:24 am
content: Quote from: ataradov on July 11, 2016, 01:42:27 amMost likely a claim not to be belived in most cases. My attention was raised as the OP stated it's a 'patch antenna' and I wasn't aware that they could be made with that much gain. A quick but not through search brought up this nugget:Quote

author: yanir
author_rank: Regular Contributor
timestamp: July 11, 2016, 02:09:27 am
content: QuoteAh interesting. I didn't know that a patch could not have that much directivity, perhaps it's more practical than theoretical?I can't go into much detail on the source antenna, mostly because I don't have the detail. But I'm ultimately trying to estimate the beam width for an antenna with a gain of 25dbi. I usually use patches for this application but it would seem that something else is being used (i'd guess helical).Either way it seems that it's quite narrow.

author: yanir
author_rank: Regular Contributor
timestamp: July 11, 2016, 05:07:18 am
content: Quote from: ataradov on July 11, 2016, 01:36:01 amis G = Gain in dbi? If so where did 316 come from? 0.36 degrees seems too narrow.Are you sure this equation is correct? If i use 25 for G I get 4.58 degrees. Seems possible.

author: jh15
author_rank: Frequent Contributor
timestamp: July 11, 2016, 05:13:55 am
content: Does it have side lobes?

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 05:14:43 am
content: Quote from: yanir on July 11, 2016, 05:07:18 amIt is a linear gain. P(dBi) = 10*log10(P) => P = 10 ^ (P(dBi) / 10) = 10^2.5 = 316.23.Quote from: yanir on July 11, 2016, 05:07:18 amThis equation is derived from some basic geometry.I worked with hemispheres (hence "/2"), since otherwise math get a bit complicated. Let me double check my calculations and I'll type them here after that.

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 05:21:00 am
content: Area of a sphere A1 = 4*pi*R^2, Area of a spot where energy will be concentrated in a directional antenna is A2 = 2*pi*R^2 * sin(alpha), where alpha is a beam width angle. In this case math holds only for alpha = 0..90 degrees (hemisphere).A2 = A1 / G. After some trivial transformations we get sin(alpha) = 2/G, so alpha = arcsin(2/G).Simple check. For G = 2 (or 3 dBi) we get alpha = 90 degrees, or full hemisphere.

author: Howardlong
author_rank: Super Contributor
timestamp: July 11, 2016, 05:43:40 am
content: Quote from: yanir on July 11, 2016, 05:07:18 amI agree, 0.36 is waaay out of the ballpark. Even 4.58 degrees is hopeful, maybe 10 degrees off the top of my head.Also, a single patch on its own is unlikely to have anywhere near 25dBi gain, more like about 7 or 8dBi.

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 05:50:04 am
content: Well, I've shown my math, show yours if you have better version.It is possible that 25 dBi is not actually achievable in real life, so your real life beams are way wider. But if you were to get 25 dBi, the beam would be very narrow.

author: Howardlong
author_rank: Super Contributor
timestamp: July 11, 2016, 05:50:11 am
content: This is the rule of thumb calculation for reasonably directional antennas:Beamwidth (degrees) ~= sqrt ( 32400 / ( 10 ^ ( GdB / 10 ) ) )

author: Howardlong
author_rank: Super Contributor
timestamp: July 11, 2016, 05:59:00 am
content: Quote from: ataradov on July 11, 2016, 05:50:04 amI also have the added benefit of being able to rely on something more than just equations ;-) somewhere in my career I designed antennas for S band satellite groundstations. Just sayin'I think your maths is assuming only two dimensions, i.e. the beam is squashed like a doughnut in a press, spreading out uniformly in a plane, rather than being squeezed out of a toothpaste tube in one direction.

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 05:59:11 am
content: Where does 32400 come from? And what is "reasonably directional"?Actually in my case you get half of the full cone angle, so total beam width would be A = 2*arcsin(2/316) = 0.73 degrees.

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 06:02:26 am
content: Quote from: Howardlong on July 11, 2016, 05:59:00 amI always respect the real life experience, but then I want to see why math is wrong, especially if math predicts much worse results. Usually things are the other way around.Quote from: Howardlong on July 11, 2016, 05:59:00 amI don't understand .The math is for ideal conical section of the sphere.

author: Howardlong
author_rank: Super Contributor
timestamp: July 11, 2016, 06:03:23 am
content: Quote from: ataradov on July 11, 2016, 05:59:11 amReaonably directional is so thatsin x ~= xfor small angles.Fudge factor is quite a common one, but is here,https://en.m.wikipedia.org/wiki/Directivity

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 07:01:41 am
content: Ok, it looks like I'm wrong somewhere. It is tool late today to think clearly.One thing to note so far. That formula produces solid angle, which is not the same as an angle of a cone approximating the beam. Although they are reasonably close for 25 dBi gain.The logic behind Wikipedia article is basically the same as mine, so it is very possible that I made some stupid mistake. I'll double check this tomorrow.

author: Howardlong
author_rank: Super Contributor
timestamp: July 11, 2016, 07:18:00 am
content: Quote from: ataradov on July 11, 2016, 07:01:41 amIf you take the square root of your result (in radians) you're not a million miles off, hence my analogy to squashed doughnut/squeezed toothpaste.If it's any consolation, for me, it was a bit early.

author: ataradov
author_rank: Super Contributor
timestamp: July 11, 2016, 07:24:09 am
content: Ok, I found a mistake. My handwriting has failed me and I mistook R for AAfter correcting this mistake final formula for the cone angle is theta = acos(1 - 2 / D). Or for D = 316, we get 0.1125 radian or 6.5 degrees.This matches Wikipedia results, and they do essentially what I did - calculate ratio of surface areas of the full sphere and a directional spherical segment.PS: Again, this is half of the angle, so full cone opening angle will be 13 degrees.

author: Howardlong
author_rank: Super Contributor
timestamp: July 11, 2016, 07:48:56 am
content: If the OP wants to get more gain from a patch antenna, there are two common methods.(1) Create a patch array, feeding each element in phase with an appropriate power splitter/phasing harness. The downside of this, in a passive configuration, is that beyond a certain number of patches the losses on the power splitter can become significant. This is especially true of arrays fabricated on FR4 which is becomes quite lossy at higher frequencies. In short, it's the law of diminishing returns.(2) Use a passive reflector, typically this would be a dish. The problem with using patches with dishes is that it's not always easy to efficiently illuminate the dish without some overspill, which leads to noise in your receiver from the warm earth if the antenna's space-pointing. Choosing the right dish f/D is also important for this. This is why horn antennas or their derivatives are more commonly used as feeds. Having said that, I've had a lot of practical success with patch feeds on dishes.