[entropicgravity]

For sure they do not work the way the "Path Loss Equation" would have you believe they do. The path loss equation violates conservation of energy ie the frequency or wavelength term depending on how it's structured cannot be in the equation. And the receiving antenna does not have any 'gain' other than physically getting bigger or smaller, though the transmitting antenna can have gain depending on shape and size. That is, the transmitting antenna and the receiving antenna work very differently. Yes, end to end the path loss equation gives the right answer but in between it's scientifically illiterate.

[nestes]

Yes and no. I emphatically agree that the way the Path Loss Equation (Friis) is taught is misleading. I much prefer the way you interpret it, with the transmit antenna represented with gain and the receiving antenna having only an effective receive area. It's much more intuitive because I can visualize a spherical shell of power radiating outward.

That said, a receive antenna does absolutely have "gain", which is evident by the antenna receiving a stronger or weaker signal depending on its orientation with respect to the transmit antenna. The key is this: for an arbitrary antenna, the (transmit, if you like) gain has a one-to-one relationship to the "effective receive area" at a given frequency, so talking about area and gain are equivalent, if not intuitive. We usually assume for point-to-point links that the antennas are oriented at each other, and in such cases (for good aperture antennas), you are absolutely right that the physical area and effective area are approximately equal. For ideal wire antennas, however, the physical area of the antenna is 0, but the effective area is nonzero (because of magic).

Now, I disagree that the path loss equation violates conservation of energy. The link to the effective area and gain depends on the wavelength. When I increase the frequency of operation but I keep the gain of the antennas constant, the areas decrease, so my receive antenna is physically smaller and the power goes down. Not breaking physics. A lot of people will say "path loss gets worse as you go up in frequency", and this is extremely misleading if not "scientifically illiterate" as you pointed out. Sure, there are molecular absorption bands from oxygen/water that literally dissipate power in the atmosphere, but generally speaking, the path loss didn't get worse, your receive antenna just got smaller.

Now wait a minute, what if I just made my receive antenna larger? Well, you can do that! The problem is that because gain and area are linked, efficiently receiving power in a given LARGE area (with respect to the wavelength) implies high gain. High gain implies a very narrow beam (more like a laser pointer than a normal dipole spilling energy everywhere). So it becomes really important that I "point" my receive antenna perfectly at the transmitter. Satellite dishes are really big, and they absolutely have to be pointed accurately at the satellite.

[fourier54]

How can an equation that does not represent a balance of energy violate energy conservation?

With path loss equation I assume you refer to Friis equation which is just the ratio of power received at an antenna to power given to the transmitter. It is correct and does not violate conservation of energy since it says nothing about the power not received at the receiver

[nestes]

What they're saying is that the geometrical interpretation of an outwardly expanding spherical shell of power shouldn't depend on frequency. In this respect they are correct and they have a good intuition for the problem.

Now here's the catch: If the receive area were not changing as a function of frequency when the receive antenna gain is kept constant (it does), this would break physics (it doesn't). However, the effective area of an antenna with fixed gain varies as 1/lambda^2. In effect the geometric interpretation is still correct, but the variation of antenna area with gain resolves the seeming paradox and saves physics.

[fourier54]

> the geometrical interpretation of an outwardly expanding spherical shell of power shouldn't depend on frequency

I think nobody says that is does. I believe the problem is to call Friis transmission equation "Free-space loss". Actually the Friis formula is composed of 3 terms: the receiving and transmitting antennas gain and the actual free space loss which has the 1/R^2 dependency (which actually isn't a "loss" in energy balance terms, since it's not lost energy, just energy not received at a certain point, so we could argue about that term too...)

[nestes]

Yep! Fully agreed with all your points, I was just trying to get at the original poster's line of thinking.

[hilbert42]

"And the receiving antenna does not have any 'gain' other than physically getting bigger or smaller..."

Well, it depends on one's definition of gain! If you were to say to the designers of the ELT (the Extremely Large Telescope) that it had no gain over isotropic then they'd fall about laughing (remember, its method of operation also relies on collecting and concentrating incoming EM radiation as do RF antennae). An antenna's effective gathering aperture and directivity for both RX and TX is just about everything, and the coupling efficiency from the antenna to the feeder and RX/detector, and vice versa for the TX just about covers the rest.

"...though the transmitting antenna can have gain depending on shape and size."

Uh? How? What's the difference? Physics says the law of reciprocity applies, a good transmitting antenna also makes just as good a receiving antenna. The only proviso being that a transmitting antenna has to be designed to withstand high RF power levels (even then, this only applies to TX power levels where I²R losses can cause enough heating to damage the antenna and feed lines, similarly, high power TX levels can lead to very high voltages which can arc over; TX antennae are designed to handle this.)

I used to work with microwave transmitters and receivers and my microwave dishes and other types of antennae were directly interchangeable—in fact, they were identical.

Re the Path Loss Equation, it works in the practical sense and is used everywhere. Fighting over technicalities here is akin to arguing the difference between laws of motion under Newton and when they're subject to the rules of Einstein's Relativity. It's damn obvious when one's applicable and the other is not.

[bigbillheck]

> The path loss equation violates conservation of energy ie the frequency or wavelength term depending on how it's structured cannot be in the equation.

Why is that?

[entropicgravity]

It's because energy created by the transmitter must degrade as one over R squared in the far field. The frequency (or wavelength, have your pick) has nothing to do with the energy transmitted because energy must be conserved. Putting in the frequency term then violates conservation of energy between the antennas. Then, at the receiving antenna the error of conservation of energy is then patched up by assigning a bogus 'gain' at the receiver. The transmitter and receiver are asymmetric but the path loss equation pretends that they are because that's easier for most people to understand and it works out 'end to end'.

[nestes]

Absolutely I agree that the geometry of the problem dictates 1/R^2 dependence, regardless of frequency. The gain, which I agree is a misleading way to think about the area, is related to the area of receive through the frequency terms. If you don't like that form of the path loss equation, I understand (I don't either!), but physics is not broken.

Where the "bogus" gain really shines, though: I can take my original receive antenna, operate it as a transmitter (so gain is now relevant), receive with my original transmit antenna (where I now care about area) and get the exact same result in terms of loss!

[bigbillheck]

The formula on wiki has a distance squared term in the denominator tho?

[entropicgravity]

Yes it does but it's the lambda symbol that's the impostor. An analogy of how it really works is to think of the transmitter as a water hose that is spraying water into a bucket, which is the receiver.

[nestes]

Short answer: it doesn't, though I understand why it's misleading. Read my response above.

[sobriquet9]

Transmitting and receiving antennas work the same way. Flip the sign of time in Maxwell’s equations, and radio waves will run perfectly backwards.