# Antenna gain reciprocity violation of conservation of energy?

1. Sep 16, 2016

### ande4jo

I expect that others have already asked and answers this question but I could not find it with Google searches. My thought of this apparent antenna reciprocity violation is per below.

Since antenna reciprocity states that an antenna will have same characteristics whether used a transmit antenna or a receive antenna, that means given an antenna with 28db gain (very focesed) used to transmit a radio signal from one such antenna to another antenna that also had 28 db gain used as a receive antenna and separated 10 wavelengths apart, that the receive signal would be larger than original signal by 14db.

Obviously if this could happen we would be getting energy for free. The math for this is very simple in that 2 antennas separated by 10 wavelengths in free space would only attenuate signal by 42 db (any off the shelf calculator on line can verify this and thus I won't show the math unless requested), yet the gain of TX and gain ov Rx antennas combined is equal to 56 db (28 + 28). Thus 56 db of gain less the 42db of loss gives a net result of 14db of free energy if this was actually achievable.

Note I chose the value of 10 wavelengths to keep receive antenna in far field only. Unless I am missing something the most probable violation would be in the antenna reciprocity assumption and obviously not conservation of energy assumption. Any thoughts to help me see what I am doing wrong with this scenario will be appreciated

Last edited by a moderator: Sep 16, 2016
2. Sep 16, 2016

### The Buttered Cat

Antenna reciprocity is not an assumption, and you can't violate that under normal circumstances (meaning the antenna behaves linearly). I don't know much else about your problem, but that is certainly true.

3. Sep 16, 2016

### The Buttered Cat

Can you point me to one of these online calculators you're using to come up with attenuation?

4. Sep 16, 2016

### Merlin3189

10 λ sounds very close. That's where I would think the problem lies.
I would think antenna reciprocity is the most solid notion of all here. It seems to have proofs based on Maxwells equations and on thermodynamics. Pretty solid foundations.
More likely suspects are the Friis equation and the antenna gain.
Since we are considering an ideal situation here, the usual objections, that some Friis assumptions are not achieved in reality, do not apply. So I assume the main problem is in the range >> λ requirement. I don't know how big R needs to be, but 10λ seems incredibly close to me. By the time you put 100λ into the equation, the results are becoming more plausible, but maybe even greater is needed before the calculation becomes reasonably accurate.

As far as antenna gain goes, I first thought of practical errors, but again we are assuming an ideal situation and the antennae are perfectly aligned, polarisation matched and in free space well away from any possible obstructions & reflections. So here I wonder about the concept of antenna gain itself. Not that I doubt that antennae give gain, but is it a simple number that applies under all circumstances?
First I think again that it will only apply when the range is large. How large, I don't know, but 10λ still sounds v.close.
Second, when we are considering very high gains, are there aperture effects? I assume antenna gain works, for the transmitter by narrowing the angle over which most of the power is radiated, increasing the field strength at the receiver. And for the receiver by capturing more energy from the required direction and less from outside. (I find this much harder to grasp than the TX.) But at high gains like 28dB, we are thinking of something like a 10λ diameter dish (at 1GHz, 3m) with a beamwidth of maybe 3o, equivalent to 0.5m width at 10m range.
If we think of two searchlight beams, the larger the aperture, the better, I think, we can focus the beam - more gain. But if the transmitted beam is so narrowly focused that almost all its light falls within the aperture of the receiver, then increasing the aperture of the receiver will no longer increase the amount of light received.
I've no idea whether this is indeed how the concept of fixed gain breaks down, but my instinct is that it might.

Reciprocity aside, there has to be some problem with the calculation, link gain = antenna gains - free space path loss. If we can keep increasing gain of either antenna, we can always overcome the FSPL and end up in profit. Presumably there is an upper limit to the gain actually achievable with an antenna. What other problems it may have, I don't know.

Conservation of energy I don't even think about doubting. Reciprocity I believe. Friis' equation is probably a useful rule of thumb under the right circumstances. Antenna gain is more like politicians percentages. Well, perhaps not that bad, but I wouldn't be surprised if some people measure it in swars .

5. Sep 17, 2016

### Tom.G

"'Directivity' is the ratio of power density in that direction to the power density that would be produced if the power were radiated isotropically."

"Gain includes antenna losses; thus gain is the field intensity produced in the given direction by a fixed input power to the antenna. Gain is related to Directivity by efficiency" Gain = Directivity x Efficiency

from: "Reference Data for Engineers", seventh edition, pg 32-3, Howard W. Sams & Co. Indianapolis, Indiana, USA

6. Sep 17, 2016

### Baluncore

Conservation of energy says that no new energy will be created by the passive antennas.
High gain antennas do not amplify the power. They simply send it in a narrower directional beam.
Two 28dBi dishes facing each other will be reasonably efficient at transferring energy because little energy will be lost sideways.
You cannot win, or even break even. The efficiency cannot be 100%.

7. Sep 17, 2016

### ande4jo

I agree that a passive antenna cannot create energy and also agree that on the transmit end focusing the energy is what is meant by gain in the antenna (due to shaping the signal in the near field). My thought is on the receive side whereby it is not able to focus the signal that is in the far field and thus the idea of gain on the receive side seems flawed. I wonder if gain on the receive side is just gain relative to the other azimuth. Thus I think a directional receive antenna actually attenuates signal in all other azimuth except for the main beam and the delta of the attenuation to the 0 db gain of main beam gives a false assumption that we are getting gain on the receive antenna.

8. Sep 17, 2016

### Baluncore

Not so.
A bigger dish has a higher gain because it captures more energy from more of the wavefront.
Think of the gain as being a function of the effective aperture area.

dB gain relative to what. You must specify dBi or dBd.
Antenna gain is meaningless in the near field. It is a far field parameter.

9. Sep 17, 2016

### tech99

According to "Antennas" by J D Kraus, page 570, a uniformly illuminated circular aperture with a gain of 28dBi has a diameter of 10 lambda.
We can take the limit of the Radiation Near Field to be approximately equal to the Rayleigh Distance, that is D^2 / 2 lambda = 50 wavelengths. This is much more than the 10 you are assuming.
So the reason for your anomaly is that you are within the Radiation Near Field. In such circumstances, there tends to be no spreading loss involved, the antennas being coupled by a parallel beam. It is possible to have zero overall loss at certain spacings but not less than zero.
.

10. Sep 17, 2016

### tech99

It is interesting to notice that with reflector antennas, which are unidirectional, the path loss tends to zero at close spacings because all energy is trapped. But in the case of antennas such as dipoles, which are not unidirectional, this is not the case. The RX antenna can now re-radiate energy away from the TX antenna, and we see a minimum path loss of 3dB rather than zero. Even when the two dipoles are in contact, the path loss is 3dB, half the energy of the system being re-radiated.

11. Sep 17, 2016

### jim hardy

I thought db of antenna gain was relative to a simple dipole.
i must've missed something.
It's a two way street,
simple dipole loses 42 db each way, your 28db superantenna loses 42-28 = 14 each way.

If all the energy absorbed by the receiving antenna were re-radiated there'd be another 42-28 = 14 db loss on the return trip.
84 - 56 = 28 db loss for the round trip ?

What did i miss ?

12. Sep 17, 2016

### Baluncore

The fundamental reference is the isotropic radiation of energy. If gain is referenced to isotropic then it is specified in dB isotropic = dBi. The gain of an antenna can also be specified relative to a dipole in which case dBd is specified. The gain of a dipole is about +2.15 dBi.
Therefore dBi = dBd + 2.15

13. Sep 18, 2016

### tech99

Further to the Baluncore comment, your calculation is correct for dipoles, but the antennas having 28dBi gain are too close for the formula for spreading loss to apply.

14. Sep 18, 2016

### Merlin3189

So presumably that means, the higher the gain of antennae, the further apart they must be for the spreading loss to apply?

15. Sep 18, 2016

### Baluncore

True. Close antennas couple to form a single EM structure, a bit like a transmission line or waveguide with a coupler at each end. The coupled pair of elements will have terminal characteristics dependent on their spacing and wavelength.

Yes. For a dipole the far field is considered to begin beyond about 60 wavelengths. If the antenna is more complex, with a reflector or array of elements then the far field begins beyond 60 times that linear dimension. The divergence of 1 in 60 represents an angle that subtends just less than 1 degree.

16. Sep 18, 2016

### tech99

I am not sure that for engineering purposes in the case of a dipole I would agree with 60 wavelengths - the requirement for 1 degree subtended angle seems a bit arbitrary. I have made tests in which the Radiation Near Field seemed to extend to about lambda/3. It is interesting, however, that the Power Flux Density (equatorial plane) remains rather constant when closer than this.

17. Sep 18, 2016

### Baluncore

Arbitrary, yes. The world of RDF and RADAR during WW2 was reliable only to about one degree under the best conditions. That was the realistic limit resulting from real world things like multiple paths and variations in atmospheric diffraction.

Last edited: Sep 18, 2016
18. Sep 18, 2016

### Aaron Crowl

It has to be the case that link budget approximations are only true where (antenna gain)dB << ABS[(attenuation loss)dB]. If that were not true then you would get the violations of conservaton that the op noticed. That may be a more convenient way to describe this problem than wavelengths.

19. Sep 18, 2016

### Baluncore

Antenna gain is only defined in the far field. To get a higher antenna gain requires a larger aperture, and so it will be further to the far field. If the aperture is doubled to get 3dB more gain, then the distance to the far field is also doubled, and the path loss rises by 6 dB due to the inverse square law.

20. Sep 18, 2016

### Aaron Crowl

I agree with everything you say. I'm just offering a different perspective. We could say that a caveat of the Friis equation is that the antenna gain must be much less than the absolute value of attenuation loss. I think it could be proven at length.

I looked in my old antenna textbook and the only definition I could find for farfield is in terms of wavelength. It must be the case that antenna gain extends the boundary of the far field as you say. Maybe these textbooks could use some updating.

Edit: I'm surprised that there isn't much info on this. This problem must be encountered again and again as engineers analyze link budgets for Wifi, wireless tolls, RFID, Bluetooth devices, home electronics, CCTV, etc. It probably didn't come up much in the past when radio was generally a long distance proposition.

Last edited: Sep 18, 2016
21. Sep 18, 2016

### Baluncore

Likewise, I don't disagree with you, but we need to stop trying to define the exact position of a blurry region where the field approximation formulas gradually change their form. We only calculate an RF energy budget if in our experience link reliability might be questionable. Reality sets in and we must get the job done.

Link budgets are never needed for near field signal levels. For a 3 GHz dipole antenna the wavelength is 10 cm, so in free space the far field begins at about 6 metres. Nearby scatterers make the path analysis much more complex. Where devices and operators move about the environment is dynamic. Things like diversity reception, frequency hopping, circular or crossed polarisation and error correction all work to overcome the deep nulls that may be encountered in the near field.

The OP's hypothetical paradox only arises if the far field antenna gain is misapplied in the near field where it is not defined.

22. Sep 18, 2016

### Merlin3189

Just for amusement, I looked up designs for Yagi ants to see how long a high gain antenna would need to be. None of the design sites I looked at went above 20dBd. The online calculators simply ignored my settings or said they were impossible. The best I found was a 70cm yagi with 31 elements and overall length about 7.5m (equivalent to 10λ) for 19.9dBd (estimated). To get more gain it was suggested you could stack these at just over 2m spacing. So to get 28dBi you'd end up with an array at least 2m x 2m x 7.5m, or 8m x 0.4m x 7.5m. If the driven elements were 10λ apart, the antennae would almost completely overlap. If they faced each other not quite touching, the driven elements would be about 20λ apart. I think by any standards this can't be "far field".

Of course you could stack more shorter Yagis to get more air between the Tx and Rx, but it would still be an imposing sight.

23. Sep 19, 2016

### davenn

something us ham radio guys like to get into
not mine, but definitely impressive 16 x 8 element yagis on 2m ( 144MHz) for moon bounce
approx. 22dBi gain

Dave

Last edited: Sep 19, 2016
24. Sep 19, 2016

### tech99

A Yagi array is an end fire antenna and I think the original question was implying the use of a dish or aperture antenna, which is essentially a broadside array.
The Radiation Near FIeld is simply an array effect. It marks the boundary between Fresnel conditions close in (where the rays arriving at an observer are not parallel), and Fraunhofer conditions far out (where the rays are parallel). For an end-fire array of point sources seen on-axis, the rays arriving at an observer are parallel at all distances, so there is no Fresnel Region. However, there is a sort of Radiation Near Field, because at close distances the distance to each source, and hence the spreading loss, now differs. So gain measurements made on an end fire array such as a Yagi will vary with distance, although it does not posses the property of creating a parallel beam near to the antenna as does a broadside array.

25. Sep 19, 2016

### tech99

I agree that the boundary between Fresnel and Fraunhofer conditions (Near and Far Radiation Fields) is very blurry. For example, if we take the the Rayleigh distance as our boundary (the maximum distance at which a lens may be focused), this is D^2/2 lambda. On the other hand if we take a criterion often used for determining the length of a test range it is 2 D^2 / lambda. For the case of a path using dipoles, I have found experimentally that the ordinary path loss formula will give good results at spacings as close as half a wavelength.