Free Space Loss at Low Frequencies: Is It Really Zero?

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The discussion centers on the behavior of free space path loss at low frequencies, particularly when using a calculator that shows nearly zero loss at short distances. Users question how a negative path loss value can occur and whether it indicates gain, especially at frequencies like 1.6 MHz and 738 KHz. It is noted that the calculator's results may not be reliable for short distances due to the breakdown of the inverse square law in near-field conditions, where the radiation pattern of antennas becomes significant. The conversation also touches on the effectiveness of magnetic versus electric field antennas at low frequencies and the implications of antenna size on communication efficiency. Ultimately, the validity of the calculator's output for low frequencies and short distances is debated, with suggestions to consider more comprehensive formulas like the Friis formula for better accuracy.
dnyberg2
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I know that path loss is proportional to the square of the distance between the transmitter and receiver AND is also proportional to the square of the frequency in use but what does it mean when a free space calculator shows a negative number? How is it possible to get zero loss? For instance, this calculator is the one I'm picking on...

http://www.radio-electronics.com/info/propagation/path-loss/free-space-formula-equation.php

If you plug these values in you get nearly zero free space path loss!

Distance: .0305 km (100 feet)
Frequency: 1.7527 MHz
Rx antenna gain: 2 dBi
Tx antenna gain: 5 dBi

How is that possible? No free space path loss? None at all?

If you lower the frequency to 1.6 MHz, the number goes negative!
Is that an indication of gain and if so, how's that possible?
 
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What do you get if you plug in 0dBi of antenna gain at TX and Rx? :smile:
 
about 6dB of path loss but if I lower the freq to 738 KHz it goes back down to zero...

Really??

AM has no path loss? (NOT)
 
The calculator seems okay for more normal numbers (like out at 1km, etc.). It could just be a math problem for that calculator -- there are enough typos in the write-up that I wouldn't necessarily trust the author.

I also doubt what he is saying about the frequency dependence for path loss (decreased antenna aperture), but who knows, maybe that's true for some situations. Others will reply with more info and opinions...
 
That's what I was thinking.

Thanks Berkeman!
 
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dnyberg2 said:
about 6dB of path loss but if I lower the freq to 738 KHz it goes back down to zero...

Really??

AM has no path loss? (NOT)

berkeman said:
The calculator seems okay for more normal numbers (like out at 1km, etc.). It could just be a math problem for that calculator -- there are enough typos in the write-up that I wouldn't necessarily trust the author.

I also doubt what he is saying about the frequency dependence for path loss (decreased antenna aperture), but who knows, maybe that's true for some situations. Others will reply with more info and opinions...

and they are really designed for far field loss calculations

At a few 100 feet at those low frequencies, it is still well inside near field and calculations for path loss don't really workDave
 
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dnyberg2 said:
I know that path loss is proportional to the square of the distance between the transmitter and receiver AND is also proportional to the square of the frequency in use but what does it mean when a free space calculator shows a negative number? How is it possible to get zero loss? For instance, this calculator is the one I'm picking on...

http://www.radio-electronics.com/info/propagation/path-loss/free-space-formula-equation.php

If you plug these values in you get nearly zero free space path loss!

Distance: .0305 km (100 feet)
Frequency: 1.7527 MHz
Rx antenna gain: 2 dBi
Tx antenna gain: 5 dBi

How is that possible? No free space path loss? None at all?

If you lower the frequency to 1.6 MHz, the number goes negative!
Is that an indication of gain and if so, how's that possible?
The two antennas have a certain aperture, so the radiation never starts from a point.
Even an isotropic antenna has a finite size.
That is why the inverse square law breaks down at close distances.
For the case you are quoting, remember the wavelength is large, about 170m, so the antennas will also be large.
To take the idea further, some antennas have a Radiation Near Zone, where, in the case of a dish for instance, the beam remains parallel for some distance before spreading out and following the inverse square law. So again the formula does not work for short distances.
I also want to mention that very close to an antenna we have another region called the Induction Field, or Reactive Near Field, where the currents and voltages on the antenna create strong local fields, but this is not really relevant to your present question.
 
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So let me ask this then, If that is true, is it also true that close in at low frequencies say 200 feet, the problem in no longer in the far field but more near field or H field and not E field? Does that mean in that case that magnetic type antennas would work better than an antenna designed to "radiate" at those frequencies?

Ultimately, does the calculator results, even if not so effective mathematically at those frequencies and short distances, really tell me the truth? Is free space path loss really very low at low frequencies and short distances compared to say a GHz at the same distances?
 
dnyberg2 said:
So let me ask this then, If that is true, is it also true that close in at low frequencies say 200 feet, the problem in no longer in the far field but more near field or H field and not E field? Does that mean in that case that magnetic type antennas would work better than an antenna designed to "radiate" at those frequencies?

Ultimately, does the calculator results, even if not so effective mathematically at those frequencies and short distances, really tell me the truth? Is free space path loss really very low at low frequencies and short distances compared to say a GHz at the same distances?
Q1. The formula you are using breaks down because the radiation comes from an aperture. For a dipole antenna, it is perhaps something like a sphere, so inside this volume the radiation intensity does not increase as we get closer. The formula doe not break down because of the presence of the induction field, which I described.
However, if you want to obtain communication over very short distances, it is true that the induction fields become stronger than the radiation fields if you are closer to the antenna than about a sixth of a wavelength. There is a choice of either E or H induction fields; for a dipole, the H field is strong near the centre and the E field near the ends. For a small loop, the H field is strongest within the loop and the E-field is maximum near the resonating capacitor. As to the best choice, with a loop you need to be within the loop to obtain strong fields, whereas with a dipole the fields extend to about a sixth of a wavelength in the way I have described.
Q2. Path loss between isotropic antennas is low at low frequencies because the isotropic receiving antenna has a correspondingly large aperture and will therefore intercept more energy.
 
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dnyberg2 said:
So let me ask this then, If that is true, is it also true that close in at low frequencies say 200 feet, the problem in no longer in the far field but more near field or H field and not E field? Does that mean in that case that magnetic type antennas would work better than an antenna designed to "radiate" at those frequencies?

Ultimately, does the calculator results, even if not so effective mathematically at those frequencies and short distances, really tell me the truth? Is free space path loss really very low at low frequencies and short distances compared to say a GHz at the same distances?
Further to Q2, I realize you are talking about very short distances and the comparison between LF and GHz frequencies.
An antenna for low frequencies is itself very large, longer than the distance you wish to communicate, so it is not necessarily an attractive idea. It really comes down to the physical size of antennas you can accommodate and also the availability of frequencies and equipment. You might have a look at the Friis formula on Wiki.
 
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