Understanding propagation loss: What does this output mean?

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1. Aug 3, 2015

H Smith 94

Hi there!

I am currently building a simulation to model the propagation of radio waves in seawater in terms of its propagation loss. I have previously discussed the models I've looked at but have settled on a model which depends primarily on the propagation distance $r$, the carrier wave frequency $f$ and the salinity $S$ of the water.

For those interested in the details, the model is:

$$\begin{split} L_\text{total}(r,f,S)\ [\text{dB}] = 20 \log fr - 147.55 &\\- 1.287\times10^{-7} \sqrt{\hat{\epsilon}_\text{r}\left[\sqrt{1+\left[39.974\,C - 6.323 C^{1/2}K + K^2\right] C^{3}\left(\frac{1}{2\pi\epsilon_0\hat{\epsilon}_\text{r}f}\right)^2-1}\right]}\,fr \end{split}$$

where:

$C=\rho_\text{w}S/N_A m_i$, in which $m_i = m_\text{Na}+m_\text{Cl}$;
$\hat{\epsilon}_\text{r}$ is the complex permittivity using the Debye model (discussed here on Math SE.)
The model attempts to model primarily the near-field characteristics of the radio propagation and attempts (if rather poorly) to consider the electrolytic properties of salt-water solutions.

As you can see, this model is represented in decibels, defined in terms of the ITU signal loss:

$$L\ [\text{dB}]=10\log\left|\frac{P_t}{P_r}\right|.$$

My simulation output is as follows, where the vertical is $\mathfrak{R}\{L_\text{total}\}$ (i.e. the real part of the loss):

Parameters:

$1\,\mathrm{m} \le r \le 20\,\mathrm{m}, \Delta r = 1\,\mathrm{m}$
$200\,\mathrm{Hz} \le f \le 500\,\mathrm{kHz}, \Delta f = 100\,\mathrm{Hz}$
$S = 0.0010\,\mathrm{kg}\,\mathrm{kg}^{-1}$

I'm struggling to understand what this output means. If the frequency- and position-dependent signal strength $I(r,f)$ in dB is calculated by

$$I(r,f)\ [\text{dB}] = I(0,f)\ [\text{dB}] + L_\text{total}(r,f,S)\ [\text{dB}]$$​
then how can there be a positive loss? Wouldn't this imply there is a signal gain at these frequencies/distances where $L>0$?

I'm not sure if my interpretation of loss is wrong, my understanding is at fault or whether my simulation model is lacking. Does anyone have any ideas as to how this could be either explained or improved?

Last edited: Aug 3, 2015
2. Aug 3, 2015

Staff: Mentor

The initial log term 20 log fr seems to dominate. Where does it come from?

3. Aug 4, 2015

H Smith 94

I agree! I was shocked to see how little the final signal loss term ended up contributing.

It originates from the free space path loss (FSPL), which is

$$\text{FSPL}\ [\text{dB}] = 20\,\log{f} + 20\,\log{r} - 147.55.$$

Interestingly, the Wikipedia page also reports a similar graph for FSPL alone:

(Sss41 (2014). Free-space path loss. Wikimedia. Free-space path loss plot.)​

4. Aug 4, 2015

sophiecentaur

Agreed. That link makes suggestions about the frequency dependence of radiated power from a given antenna. That, afaics, is making some sort of assumption about the nature of the antenna. (??) For a given transmitted power, the 'free space' loss will just be due to ISL. How can it not be? If the antenna is tuned and matched to the transmitter, the frequency dependence could be a very sharp frequency peak and certainly not inversely dependent on f.

This link discusses that -20log(f) factor but makes it clear how and when it can be useful.

5. Aug 4, 2015

H Smith 94

Thanks for linking this, it looks like a useful resource. From a quick read through it, though, it seems that the calculations are simply the free space loss with an additional antenna gain factor $G$. The $20\log{f}$ factor is simply the frequency-dependency of the free space path loss; it doesn't seem necessarily specific to antennas.

If we use the loss as a negative quantity, sure.

$$\begin{split}I &= I_0 - L \\&\backsim I_0 - 20 \log{fr} = I_0 + \log\left|{(1/fr)^2}\right|.\end{split}$$
So is this the correct way to talk about signal loss? This still doesn't explain the apparent signal gain at particular frequency or distance ranges in the FSPL model.

The antenna is assumed to be lossless at this stage, with no external, additional gain or loss. It is essentially an emission node with no other real characteristics. I want to get the propagation down before considering anything to do with the PHY specifics.

6. Aug 4, 2015

sophiecentaur

But, if you go along with this and extend it to light, frequencies no power would reach more than a few metres!!! That factor is only useful as a rough guide to show how the directivity / gain is likely to vary for a directive antenna of given physical size (like a microwave dish). Rather than an antenna, take a small aperture, much less than a wavelength so that its pattern is pretty well a hemisphere, off axis, illuminated from behind (RF or light) There will be a power flux at 100m and 1/4 of that flux at 200m. Frequency has nothing to do with it. It's called the Inverse Square Law and applies to loads of physical quantities.
You seem fixed on the idea that it is relevant to your problem because you read about it in the context of a different problem. Consequently, you have spotted that the answers are difficult to accept.
Loss, in dB will be a negative quantity because the word "loss" implies reduction. However, in the calculation of total path loss, there may be 'positive dB' factors, such as the focussing of a dish antenna or the Ionosphere. These factors are a sort of 'negative loss'.
Not if the 1/f2 factor is included.
Consider my first sentence in this post. Think about it. Does it not make sense?

7. Aug 4, 2015

H Smith 94

Okay so the FSPL is dependent upon the antenna properties. Thank you for bringing this to my attention.

8. Aug 4, 2015

marcusl

Free space propagation loss L is defined for isotropic (unity gain) antennas, making it frequency (or wavelength) dependent $$L=\left(\frac{4\pi r}{\lambda}\right)^2$$ The Wikipedia article explains it well. In terms of power and gain at the transmitter and receiver, $$L=\frac{P_t G_t G_r}{P_r}$$ The product P_t G_t is called effective isotropic radiated power, E_t. Since the gain of an antenna of aperture area A is $G=\frac{4\pi A}{\lambda^2}$, the received power is $$P_r=\frac{4\pi AE}{\lambda^2 L}=\frac{AE_t}{4\pi r^2}$$ This makes sense: received power is proportional to receive aperture area and transmit EIRP, and is inversely proportional to the spherical surface area at the receiver. It is independent of frequency (and light still propagates).

Last edited: Aug 4, 2015
9. Aug 4, 2015

sophiecentaur

If you are saying this:
Then you would need to define what you mean by ' Free space loss'. i.e. what ratio does it represent? It can't have relevance for light wavelengths if it represents the flux density at a distance, relative to total radiated power - nW for kW even at a distance of 1m is not what we find. The wiki article simply states it and I can't see any justification there. - except where it makes the point that near field signals do not drop off according to the ISL (which is fair enough) and that may have some bearing on it. But how many transmitter - receiver paths are there in which the two antennae are a big proportion of the separation distance? And, if we are talking 'near field' then there can be no simple formula. Talk of an isotropic radiator (a figment, in actuality) implies a very small radiator but then it also mentions aperture (???). It's altogether very confusing.
If you have a transmitter, you can define its output power. You can calculate the feeder loss and antenna gain in a particular direction (this will be frequency dependent but it will hardly be proportional to the frequency. (Antennae - even optical ones- are designed to give maximum gain at their design frequency.) That will tell you the power flux per unit area at a distance r. That formula seems to assume a particular quality for every antenna.
So can you explain what the actual message in that formula is and where it comes in a link budget?

10. Aug 5, 2015

H Smith 94

It may be worth noting that I was so quick to accept any loss due to frequency is because I know how much the attenuation of radio waves through seawater is dependent on the carrier frequency.
That's a very nice derivation, thank you! How could I incorporate this to also include properties of the propagation medium? I am working with an electrically lossy medium which has drastically different properties than free space.

Problems with the $f$-dependence also come in when we start considering the loss in decibels because we find that $P_t/P_r = r^2c^2/A_tA_rf^2$, and thus we are left with a decibel loss of $$\begin{split}L_\text{dB} &= 10\log{P_t/P_r} = 10\log\left|{r^2c^2/A_tA_rf^2}\right| \\&= 10\left(2\log\left|{\frac{rc}{f}}\right| - \log{A_tA_r}\right)\end{split}.$$
I suppose it represents the ratio of distance to wavelength, which explains why both distances between antennas and antenna aperture sizes are often quoted in number of wavelengths.
Most sources I link to or models I've considered have been using primarily near field and LF to VLF radio. Although it may be worth considering more general models for a lot of applications, I've focused mainly on those cases.

11. Aug 5, 2015

sophiecentaur

Which really means that you don't know (??) haha. Loss - particularly when quoted in dB - is a ratio of Powers. Any proper definition should be in terms of two values of power, (which values, taken where?) not distances. It seems to me that the frequency dependent bit is a generalised 'frig factor' that makes general assumptions about the nature of an antenna with a given physical aperture(?) If it's an omni and matched with a lossless network, then all the power into the feed point should emerge into space (the radiation resistance), whatever frequency is used. The Wiki article doesn't define it but perhaps there is something in one of their references.
BTW, the formulae for propagation losses are much less problematical (at first sight, at least).

12. Aug 5, 2015

H Smith 94

Well no, that's why I presented it as a supposition -- to encourage debate.

That makes sense. Could one not use marcusl's definition of received power and simply divide the numerical transmitted power (i.e. looking only at the power input and assuming a lossless antenna) by it? Then you'd have the ratio $$L = \frac{P_t}{P_r} = P_t \frac{4\pi r^2}{A_\text{eff}E_t}$$ as a dimensionless expression for loss (which could, of course, be extended to dB.)

Yes that makes sense in an ideal scenario but non-free-space losses do depend on frequency, especially in lossy transmission media. I agree that the FSPL in itself should physically not depend on the frequency.

I've tried looking at their citations but the FSPL Wiki article is not well referenced, unfortunately.

What are the formulae for propagation loss? I can only find general path loss.

Last edited: Aug 5, 2015
13. Aug 5, 2015

sophiecentaur

ISL just deals with the power flux per unit aperture - anything else has to be bolted on afterwards (antenna gain etc.)
I think the FSPL thing was probably lifted from some other source that had a specific application in which the frequency dependence was part of a calculation. There is a wavelength dependent relationship between field strength (V/m) and Power radiated because a long resonant dipole will have more volts across it, for a given field strength (i.e. more metres). Conversely, you need to put more Power into a long dipole than a short one for the same field strength. The radiation resistance for all resonant dipole is the same at their resonant frequency. So perhaps the calculation was to do with field strength and not power??

14. Aug 5, 2015

H Smith 94

Ah, I see. So ISL is the primary source of free-space loss?

It's certainly starting to seem that way after this conversation! Perhaps the Wikipedia page could do with some updating to reflect this.

15. Aug 5, 2015

marcusl

I did--see the equation for L in post #8. It is the loss between two isotropic (unity gain) antennas, or the proportionality quantity that relates received power to EIRP x G_r.
There's no reason that antenna size must be comparable to the separation, the formula applies to unity gain antennas, i.e., small. Whether we like it or not, this quantity is used by some practicing professionals (I've seen it in NASA link budgets and in those produced by aerospace contractors) and is formally defined in standards documents such as IEEE Std 145-1993, 149-2013 and ITU-R P.525-2. In a link budget it always ends up being used with other quantities like transmit power and gain and receive gain.

I have to run, will return with more later.

Last edited: Aug 5, 2015
16. Aug 5, 2015

sophiecentaur

This is my problem. For the transmitting antenna, it will radiate all the power it is fed with and the power flux will be uniform over a sphere. Of course it needs to be matched but that is not as much of a theoretical problem as actually making an isotropic antenna. . If it is a theoretically unity gain antenna then each m2 on a sphere will get that proportion of the power. This is independent of wavelength. Your L value must be to do only with the receiving antenna and its effective aperture to intercept the power flux. Is there something fundamental which says that an isotropic antenna must have an equivalent aperture that depends just on the wavelength? If this is true then a dish antenna, with a fairly well defined cross section would need a truly massive (area) gain to make up for the F (1/λ2 function.
You say you have seen link budget calculations that include the F factor. Well, I have seen link budgets that (I am pretty sure) did not. Here's the thing: if they both get the same answer, where is this factor put in 'by implication'? I think it could be to do with the fact that the receiving area of a dish is the same for all frequencies (at least over a practical range). In fact, perhaps that's it. A dish with a tighter beam can still only get its area's worth of the power from a distant transmitter. So the gain figure, calculated in terms of beam shape, will be higher for a high frequency but the notional increase in input power must be offset by the F factor, keeping the Power received as just a function of the aperture.
I think this may have resolved my problem. Than you for insisting politely enough to make me think harder.
Nevertheless, I have a problem with the terminology which seemed to include this under the heading of "free space loss" when the loss is very much to do with the antenna and not the space between. Wouldn't 'aperture loss' - or even a negative 'wavelength correction factor' be more appropriate?

17. Aug 5, 2015

H Smith 94

The Wikipedia entry also notes the same thing:

18. Aug 5, 2015

sophiecentaur

I think your comment about antenna area is spot on. In radar, in fact, a commonly used figure of merit is $(EIRP)_t A_r = P_tA_rG_t$ which goes directly to your comment that the receive aperture area specifies sensitivity. The FSPL appears in the transmission equation when gain is used in place of area. As an example, the power received by a radar in terms of receive gain is $$P_r=\frac{P_tG_tG_r \sigma F^4}{4\pi R^4} \left(\frac{\lambda}{4\pi}\right)^2$$ which is inversely proportional to the FSPL. (F here is the factor that accounts for other propagation losses, such as attenuation and multipath. No relation to your F factor.)