Understanding the Friis Transmission Equation

[kevinisfrom]

TL;DR Summary

Limits of Friis Transmission Equation

If we had a transmitter antenna with 40dBi gain, and the same antenna on the receiving end, based on reciprocity, the receiver antenna would be 40dBi too. Assuming a wavelength of 0.1m and transmitted power of 100W over 100m, the received power comes out to be 63W based on this calculator here: https://www.everythingrf.com/rf-calculators/friis-transmission-calculator

Does the Friis equation take into account beam efficiency? That is, the power transmitted in main beam vs the total power transmitted (thus losses from sidelobes)? I doesn't seem like antenna gain tells us how much power was lost from the non-primary beam. For example, if the distance was decreased to 10m, the power received will go to 6kW, which cannot be possible. I guess more broadly, what are the limits properly using the Friis Transmission Equation?

 

[sophiecentaur]

Summary:: Limits of Friis Transmission Equation

Does the Friis equation take into account beam efficiency?

You are right to doubt the high level of received power that's calculated. It doesn't pass the reality check test. Just talking about the actual areas of dish involved implies that much more than 40W will go missing into space.

The formula must assume realistic separation between Tx and Rx. The efficiency of the antenna (normally between 60 and 70%) will only cover the losses involved in the feed vs the dish (you choose between a broad illumination of the dish from the feed and waste a lot over the sides or a narrow illumination with a consequent wider beam width and lower geometric gain). This implies that a bigger dish than the simple calculation would suggest.

The geometry of the model you're working with means that the aperture of the receive antenna would be about 4m which would mean that its area could fall outside the actual 2 degree (or so) transmitter beam width at 100m. The formula would only apply if the transmitter and receiver dishes each lie well within the main beam of the other end. You'd need to re-do the geometry yourself as those are 'mental' calculations of mine but what I say would always have to apply at some stage. (Practical dishes would be bigger than ideal so the effect would be greater)
Those formulae that you find are basically for use with real link planning and the distances are always much greater (or the dish gains are much lower) - because they use standard receiving equipment which expect low powers. Engineers don't want formulae that are bigger than necessary.

 

[kevinisfrom]

I think what you may be describing is aperture efficiency, which for the sake of the question, is assumed to be lossless. I define beam efficiency as the ratio of power in the main beam over total power radiated. https://www.tutorialspoint.com/antenna_theory/antenna_theory_beam_and_polarization.htm

I guess I'm comparing a horn antenna and a dish antenna in my mind. The dish antenna could have higher gain/directivity, but there's also more side lobes thus energy not transmitted in the main beam (and lower beam efficiency). Horn antennas have lower gain, but can have much higher beam efficiency. Assuming clear path, and point to point Tx/Rx, Friis equation would say the dish antenna has more power received, but it's hard to see that to be the case if say 40% of the energy is dissipated in side lobes (despite a very concentrated primary beam), versus say a lower gain but all the energy is contained in the main beam.

This lead me to think whether Friis transmission equation took into account antenna designs that had very high gain but lots of minor side lobe losses (thus main beam total power is less) versus a design with all energy going into primary beam and no side lobes (the beam having less gain, but is 'fatter' as the energy is less densely spread, but all contained in the primary beam)

 

[jasonRF]

Well, when I insert your numbers into
$$
P_{rx} = \frac{P_{tx} G_{tx} G_{rx} \lambda^2 }{(4\pi r)^2}
$$
I get 63 Watts. What do you get?

Usually aperture efficiencies is included in gain. If you don't include efficiencies then it is common to use the term "directivity" instead of gain. Directivity is based solely on the pattern, so includes effects due to sidelobes. In any case, both gain and directivity include the impact of energy lost in sidelobes - just look at the definition.

EDIT: see chapter 16 of
https://www.ece.rutgers.edu/~orfanidi/ewa/
for definitions

EDIT2 (I really should have thought more before posting - sorry!). Sophicentaur is bringing up a good point, though. The gain is actually a directional quantity - so if the receive aperture covers a sufficiently large angular area from the transmit antenna, then you would need to take that into account in your calculation. The naive use of the equation (like I did and the calculator did) will not include that. Also you might not be in the far-field of the dishes (usually the criterion is you need the range to be at least $2D^2/\lambda$ where $D$ is the aperture size).

sorry for my rambling!

Jason

 

[sophiecentaur]

Assuming clear path, and point to point Tx/Rx, Friis equation would say the dish antenna has more power received, but it's hard to see that to be the case if say 40% of the energy is dissipated in side lobes (despite a very concentrated primary beam), versus say a lower gain but all the energy is contained in the main beam.

Here, you seem to be confusing / combining transmit and receive antennae when you talk of energy being dissipated in sidelobes. Sidelobe performance is usually more important in interference rejection, which this thread is not considering. If you take the 3dB points as defining the beam width, there will still be 'most of' the 'wasted' power within the main beam skirts as you have quite a way to go to the first minimum. The total sidelobe power need not be much at all although the rejection of off-beam signals can be significant.

Also you might not be in the far-field of the dishes (usually the criterion is you need the range to be at least 2D2/λ where D is the aperture size).

Yes, for off the shelf paraboloids. Using a modified reflector shape could get back some of the inadequacies due to what is really focussing error. A purpose built system could avoid that. But, whatever, the directivity of the beams would need to be tighter than the signal gain would imply. Because of Reciprocity, it would seem that the beams of both dishes would need to cover the whole of the angle subtended by the other dish to minimise 'beam loss'. I think the Friis formula neglects these factors because the Friis formula is for practical link systems and not for maximum RF Power distribution because real dishes are very small for a given path distance. Any normal link assumes a lot of wasted Tx power being sprayed out into space.

 

[kevinisfrom]

Here, you seem to be confusing / combining transmit and receive antennae when you talk of energy being dissipated in sidelobes. Sidelobe performance is usually more important in interference rejection, which this thread is not considering. If you take the 3dB points as defining the beam width, there will still be 'most of' the 'wasted' power within the main beam skirts as you have quite a way to go to the first minimum. The total sidelobe power need not be much at all although the rejection of off-beam signals can be significant.

I guess the way I look at this, we can have say 50dBi gain, but a primary beam width that is infinitesimally small (for argument sake). Thus, the total power transmitted by the primary beam can be infinitesimally small since the beam width (for argument sake) is infinitesimally small. Yet, with the Friis equation, more power is transmitted due to this peak gain value compared to say 30dBi with a null-to-null angle of 10 degrees.

What I am trying to wrap my head around, and maybe I'm completely missing the point here, is that if we just looked at "gain" in the Friis equation, it doenst tells us about the power per m^2. Gain represents max energy density at a single point in space, but I don't think it takes into account the energy density across the area of the primary beam. By only reporting the 'max' value, we miss out on the beam distribution (thus I bring up side lobes) and how much power per area is transmitted in the primary beam.

Please correct me if dBi refers to a density, as that would make things much more clear. Rather, dBi from what I have seen dBi is the max value of the main beam, but does not provide information on the power density, as this could be (as the example above) an infinitesimal beam width (for sake of argument) but it does not seem like Friis takes this into account (that is the integral of the solid angle of the primary beam from null-to-null)

 

[sophiecentaur]

Please correct me if dBi refers to a density, as that would make things much more clear. Rather, dBi from what I have seen dBi is the max value of the main beam, but does not provide information on the power density,

dBi totally refers to Power Density (relates to Field strength). It's a quantity that tells you what the Power Density at the peak of the main beam will be, relative to the same Tx power transmitted from an isotropic radiator. For a given effective receiving 'gathering area', it tells you how much power will get to the receiver. It has to assume that the illumination of the Rx antenna is uniform. The higher the Tx gain (aka angular directivity, aka receiving solid angle ), the less can you assume "i" applies.
You seem to have a problem with the fact that Friis was designed to work for real systems, where that "i" applies; where the Power Density is the same all over the receive antenna effective area. Remember that beam width is between -3dB points and half the power can fall outside that (for a basic antenna like a dish).
Take a 1W laser with a beam that produces a 10mm square spot on a distant target. The most Power you can hope to receive will be 1W, whether you use a 15mm diameter lens or a 1000mm telescope. Friis would disagree. Is he right? Your 1000mm telescope objective will get a useful level of signal from a faint star because the star is Isotropic and the Power Density is the same all over the objective.

 

[kevinisfrom]

Take a 1W laser with a beam that produces a 10mm square spot on a distant target. The most Power you can hope to receive will be 1W, whether you use a 15mm diameter lens or a 1000mm telescope. Friis would disagree. Is he right?

Agreed, this formula would not make sense for a coherent beam.

The higher the Tx gain (aka angular directivity, aka receiving solid angle ), the less can you assume "i" applies.
You seem to have a problem with the fact that Friis was designed to work for real systems, where that "i" applies; where the Power Density is the same all over the receive antenna effective area. Remember that beam width is between -3dB points and half the power can fall outside that (for a basic antenna like a dish).

Interesting, so Friis breaks down the higher the gain. The 'real' systems I'm thinking about is something like:
http://mainland.cctt.org/istf2008/one.asp

"2.5 GHz band over a 1.6 km distance from an antenna dish to a rectenna. The team was able to produce an 84% efficiency with a 30 kW DC output"

The 40dBi-50dBi example in the original post was the typical values for dish antennas. Given the high directivity/gain, and how far away we are from the isotropic case, Friis may not apply in this case. I was looking to see if the values quoted in the above examples could be applied to Friis, but my impression is that Friis specifically applies to comms, rather than a measure of energy transfer for narrow beams. Which answers the question of "more broadly, what are the limits properly using the Friis Transmission Equation?" and the limits are when G-->very large, or A_e-->very large.

This leads me to wonder, is there a more appropriate way to estimate power transfer in very narrow non-coherent beams (for example the antenna dish that sent 30kW over 1.6km)?

 

[sophiecentaur]

I guess the way I look at this, we can have say 50dBi gain, but a primary beam width that is infinitesimally small (for argument sake)

Beam width and Gain are locked together (ignoring practicalities. Your infinitesimally small beam width would correspond to a very high gain.

Agreed, this formula would not make sense for a coherent beam.

A beam of microwaves is assumed to be coherent (CW) for the purpose of calculations. A finite signalling bandwidth will affect the details of the beam pattern and the matching / combining networks but not the basic Gain; the fractional bandwidth of most microwave systems is not large and the antennae are simple.

This leads me to wonder, is there a more appropriate way to estimate power transfer in very narrow non-coherent beams (for example the antenna dish that sent 30kW over 1.6km)?

Yes, from what I said about around half the total power spilling outside the 3dB beam width, it's pretty clear that more details would be needed than just the nominal gain. A system to achieve 84% efficiency would definitely need to be treated 'specially'. Let's face it, if 84% were easily achievable then I bet there would be such power links all over the place. Frankly I would worry about health and safety as much as about efficiency.
Friis is for Engineers using established tech for common link arrangements. It's so simple that it obviously make big assumptions which Engineers can ignore in the wider scheme of things. If you want to design a Power Link then that's another ball park, I think. Good luck though.

PS I wonder if that 84% refers to Mains Power in at one end and Maind Power out at the other.

 

[kevinisfrom]

Yes, from what I said about around half the total power spilling outside the 3dB beam width, it's pretty clear that more details would be needed than just the nominal gain.

Is this generally true for all directional transmitters? Even if you have 100% of the beam energy in the main lobe (perfect sidelobe suppression), the 3dB beam width (assuming beam is symmetrical) will contain at most ~50% of the energy? Is there a way to calculate the smallest 3dB beam width possible as a function of wavelength (under ideal antenna conditions)?

PS I wonder if that 84% refers to Mains Power in at one end and Maind Power out at the other.

Digging into this, the 84% was for microwave energy at the receiver to power out. It's basically the efficiency of the rectenna... Apparently the RF power at the transmitter was 450kW(!)

 

[sophiecentaur]

the 3dB beam width (assuming beam is symmetrical) will contain at most ~50% of the energy?

This is just down to how the beam width is defined i.e. between 3dB points. The specific pattern can improve on the 50% loss if an antenna array is fed appropriately. ( It's the same Maths as for filter design.) At the expense of gain, you can get a beam with a very flat top and 3dB points further apart so the 50% problem doesn't arise. I guess that could be used to provide a flat illumination of a receiving array. The approach of considering flux density and receive area rather than simple gain could optimise link performance - but it would involve larger aperture antennae.
Actually, Broadcast antennae use this sort of approach (both vertically and horizontally ) to maximise the area that's fed with the broadcast signal but with minimal loss outside the main beam pattern. But broadcasting involves big angles - nothing like the same as for a system to direct power to a single receiving array.

 

[tech99]

The Friis formula is based on the path loss between isotropic antennas in free space. If we use directive antennas, to apply the formula they must be in free space and this means they must be further apart than their Radiation Near Field (sometimes referred to as the Rayleigh Distance). The Radiation Near Field extends to a distance of approxiumately Diameter^2/2 Lambda. In cases where received power comes out greater than transmitted power, it is caused because the antennas are too close and lie within each other's Radiation Near Field. The reason is that within this region, the beam tends to be parallel and not diverging with the expected beamwidth.
Regarding antenna efficiency, the gain of a dish is ((pi D/lambda)^2) x efficiency factor. This efficiency factor may be 50 to 65% and is mainly caused because the illumination across the aperture of a dish is naturally tapered due to the varying distance from the feed to the centre and to the edge of the reflector.

 

[kevinisfrom]

Beam width and Gain are locked together (ignoring practicalities. Your infinitesimally small beam width would correspond to a very high gain.

Is this true? Is there a formulaic relationship? If gain is a power density, makes sense beam width would be a factor as you'd need to integrate over the solid angle to get m^2. I've never see this correlation before, but would be interesting to understand further.

 

[sophiecentaur]

Is this true?

In principle it has to be. The gain value tells you the equivalent Power of a transmitter with an isotropic radiator. As I have already said, that assumes the receive antenna is receiving a plane wave that's uniform in amplitude over the whole of its (effective) aperture.

I've never see this correlation before,

How come? It's the whole basis of the Gain concept. After all, there is no 'Amplifier Gain' involved so how else could it work?

 

[sophiecentaur]

This efficiency factor may be 50 to 65% and is mainly caused because the illumination across the aperture of a dish is naturally tapered due to the varying distance from the feed to the centre and to the edge of the reflector.

Yes. This is all about Engineering. The actual power delivered to a receiver or gathered from a transmitter always has to be viewed in the context of OTHER users and signals. You have to avoid power getting radiated into the service areas of other transmitters and to reject unwanted signals from them. You sacrifice gain for a desirable overall pattern. Radiotelescopes also need to taper the feed illumination strongly to avoid receiving radiation from a 'hot' Earth, under the scope. Signal to Noise Ratio and radiated interference are always important factors - especially in RF Power Transmission systems where the spilled Power could be a damned nuisance. This will apply even where a particular channel is reserved but where such high power levels are involved; you are a real 'noisy neighbour when you are operating a 30+dB relative to other users. There is no such thing as 'free spectrum space'.