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The terms in Equation are all in decibels (dB), which is logarithmically scaled. In Equation, $P_{rx}$ is the received peak power, $P_{tx}$ is the transmitted peak power, $G_{tx}$ is the transmitter gain, $G_{rx}$ is the receiver gain, and $L_{range}$ is the range loss. Additionally, $h_{tx}$ is the transmitter optical efficiency, $h_{atm}$ is the atmospheric transmission efficiency, $h_{rx}$ is the receiver optical efficiency, $T_{point}$ is the transmitter pointing gain, and $R_{point}$ is the receiver pointing loss:

$$P_{rx }= P_{tx }+ G_{tx } + G_{rx} + L_{range} + h_{tx } + h_{atm} + h_{rx } + T_{point} + R_{point}$$

My question is how to compute the following terms:

  1. $h_{tx}$ is the transmitter optical efficiency,
  2. $h_{atm}$ is the atmospheric transmission efficiency,
  3. $h_{rx}$ is the receiver optical efficiency,
  4. $T_{point}$ is the transmitter pointing gain,
  5. $R_{point}$ is the receiver pointing loss.

EDIT 1. This equation was found here:

Barnwell, N., Ritz, T., Parry, S., Clark, M., Serra, P. and Conklin, J.W., 2019. The Miniature Optical Communication Transceiver—A Compact, Power-Efficient Lasercom System for Deep Space Nanosatellites. Aerospace, 6(1), p.2.

It could be finde via scholar google

In p 25, they give only numeric results, Table A3.

enter image description here

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  • $\begingroup$ Can you expand on some definitions of terms? For ex: "atmospheric transmission efficiency". Where can we find a publicly available definition of it? (Wiki, paper, etc...). Same for "transmitter pointing gain". $\endgroup$
    – Ng Ph
    Nov 12, 2021 at 10:12
  • $\begingroup$ The terminology used in the paper you cited (Barnwell,..) is peculiar. At least, I have not been able to find "atmospheric transmission efficiency" anywhere else. From the numerical value (-0.59 dB), it looks like it represents the transmittance of the atmosphere in clear-sky, under certain other assumptions (not given by Barnwell). Also, why there is a "transmitter pointing gain" and a "receiver pointing loss" is above my reasoning capability. Perhaps, we have experts in this field here to clarify all your interrogations. I upvoted BTW. $\endgroup$
    – Ng Ph
    Nov 12, 2021 at 19:31
  • $\begingroup$ If I remember I'm going to add another answer here tomorrow. I've usually just estimated link budget calculations to within a few dB, the way this is parsed they have really added a lot of individual, specific terms, which one would use in a detailed calculation trying to arrive at a pretty accurate number. The problem is that for a lot of these the terms are not at all easy to calculate; there won't be a simple formula for it. For example, the pointing gain/loss will have a complex pattern that comes from detailed measurements, and the atmospheric efficiency depends on humidity and angle. $\endgroup$
    – uhoh
    Nov 16, 2021 at 13:12
  • $\begingroup$ In the mean time you can see some simple formula at this answer to How to calculate data rate of Voyager 1? $\endgroup$
    – uhoh
    Nov 16, 2021 at 13:13
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    $\begingroup$ In a link-budget, I remember that I always encourage my young engineers to make use frequently of the magic "implementation margin" factor. This is a "catch-all" place holder where you lump everything you can't estimate accurately. This is much better than pretending you know, when you can't show how to evaluate, you can't justify the figure that somebody's routine spit out, or even define precisely the meaning when asked. And never give numbers in dB to the second digit after the decimal point for quantities that are not computed analytically. $\endgroup$
    – Ng Ph
    Nov 17, 2021 at 19:06

2 Answers 2

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First, a general warning. There is widespread agreement on the full link budget equation, but there are many ways to group its components into named pieces, and different authors make different choices. Unfortunately, this means that if you take some of the terms from one source, and some of the terms from another source, there is significant danger that you have just double-counted or omitted something important! To illustrate this, look at the document included in the other answer in this thread. The transmitter gain $G_T = 16/\Theta_T^2$, the receiver gain $G_R = (\pi D_R/\lambda)^2$, and the propagation (free space) loss $L_S = (\lambda/4\pi L)^2$. The thing that looks strangest to me about this setup is the choice of $16$ in the numerator of $G_T$, and $\pi^2$ in the numerator of $G_R$. I usually expect to see it broken up equally, with $4\pi$ in the numerator of each of the gains. Also, note the presence of $(4\pi)^2$ in the denominator of $L_S$. Purely numerical factors like this have a tendency to wander, and sometimes get lost, when someone decides they'll just cancel $16\pi^2$ from both, leaving definitions of $G_T=1/\Theta_T^2$, $G_R=(D_R/\lambda)^2$, and $L_S=(\lambda/L)^2$. Exactly the same answers would come out, but only if you make the substitution in all the right places, and no others! You could even cancel the $\lambda^2$ from $G_R$ and $L_S$, and it would all still work. $G_R=D_R^2$ looks funny to me in a way that $L_S=1/L^2$ doesn't, but all it really means is that you have chosen to define your gain with respect to a reference receiver aperture of "one" in whatever units you happen to be using (kilometers, centimeters, inches, miles, etc.) I would not advise using these formulae, but instead pick a set where all the pieces come from one author. The only thing I actually disagree with this source about is that I like to write gain in a way that does not require uniform illumination of a circular aperture, because many of my antennas are more complicated than that, but even this can be worked around with sufficient footnotes. Also, beware the combination of G and T: $G_T$ means the gain of the transmitter, but $G/T$ means gain (of either the receiver or the transmitter) with respect to the antenna noise temperature, a somewhat confusing concept I will not address here, but perhaps in another post.

The transmitter and receiver efficiencies are more measured than calculated. They could in principle be calculated, but the models to do so are very complicated, and in practice you usually just need to look at the hardware you are using, and see what its numbers are. If you get to choose the hardware, I'm not sure what is considered ideal in optics, but for RF propagation (which has identical equations, just with a different number plugged in for frequency), ideal efficiency is one half, not one, because of impedance matching to minimize power reflection at the antenna. Everyone uses decibels to talk about link budgets, so one half (0.5) is generally written as -3 dB. The table you show uses -3.25 dB for both of these, which equals 0.473, but that's so close to a half that I think it's probably just a fudge factor for implementation error (that is, you were aiming for 0.5 exactly, but you didn't quite achieve the ideal).

Atmospheric efficiency is about distortions and absorptions that occur as the signal goes through it. There is a huge amount of scholarly literature on this point. There are ITU models and NASA studies and other official sources for rain fade, aerosol scattering, tropospheric and ionospheric delay, bending, attenuation, etc. You can use those models if you want, or you can make your own by using something like MODTRAN. This is a lot of work, and has produced an overwhelming number of doctoral theses over the decades. To do this, you also have to pick a specific atmosphere that provides every parameter the models need, which includes things like vertical profiles (tables with one of the two columns being usually pressure rather than height) of temperature, water vapor, and carbon dioxide, trace gas abundance (there's not a lot of methane or ozone or carbon monoxide in the atmosphere, but they have a big impact on the RF and optical radiation behavior, because ordinary nitrogen doesn't have an electric dipole moment; you'd like to have profiles of them as well, but there usually isn't enough data to give good ones), aerosols, and other things.

You need to decide whether you are interested in modeling exactly what one particular atmosphere does to the signal, knowing the actual atmosphere will have changed to something different before you can put all its parameters into your equations; or if you would instead prefer to talk statistically about the sorts of atmospheres one usually encounters at a particular spot on the globe, and calculate the expected impact at various percentiles of badness. Both of these are valid approaches, and both are interesting subjects for study, but please note that neither is a complete answer to the question. You could spend several entire lifetimes trying to model just this one parameter, and many people have chosen to; thanks to them, those of us who just need to make a quick link budget calculation can use their stuff and then move on to something else.

Pointing loss is all about the antenna beam pattern, and where in each other's pattern the transmitter and receiver are located. In traditional propagation theory, antennas are not amplifiers. Amplifiers will be in the circuit somewhere (possibly multiple places), but the antenna as such, in the link budget equation, only redistributes power. It does not change the total power emitted, it just changes how much goes in or comes from each direction. The gain numbers reported are usually measured in what we call "dBi", meaning "decibels with respect to isotropic", where an isotropic antenna is a purely theoretical object which if it did exist would transmit power equally in all directions. There is an important numerical ambiguity embedded in the definition of gain, because some authors think the antenna pattern (the gain or loss as a function of direction) should integrate to one, and others think it should integrate to $4\pi$, because that's the area of the unit sphere. If you measure your antenna only with respect to another antenna (even if it's only an idea), however, you can specify its gain uniquely in dBi, because the factor $4\pi$ will be present or absent in both, and thus cancel either way.

Antenna pattern can be a very complicated thing to plot or calculate, because antennas themselves can get quite complicated. The one I'm going to discuss is important both theoretically and practically, because it is relatively simple and widely used: a parabolic reflector with circular symmetry about its axis, a.k.a. the standard dish you see on houses with satellite TV, or NASA TDRSS, or many other operational satcom systems. This antenna's pattern looks like:

Airy disk Bessel 20 log_10 (2 J_1(x)/x)

The vertical axis is decibels. Note that the highest thing on it is 0 dB, and everything else is negative. This is because the image here is true for all circular parabolic reflectors, and is meant to be used with an antenna specified as in the table you gave, with only its peak gain (83 to 126, above) given. The function plotted is $20 \log_{10} (2 J_1(x) / x) $, where $J_1$ is the Bessel function of the first kind with order one, $10\log_{10}$ converts to decibels, 20 because power is amplitude squared, and I will explain $x$ soon. The derivation of why this should be so is described on wikipedia under Airy disk . It is closely related to the ways a circular drum head oscillates when struck (which are Bessel functions, not sinusoids). The picture for the antenna as a whole is symmetric under rotation about the peak, which is on the left axis in this plot.

To use this thing, you need to know what's packed into the definition of $x=ka\sin\theta$. $a$ is the radius of the aperture (a fancy word for "hole" that is always used in this situation), which in this case is the radius (half the diameter) of the reflecting dish (RF name) or mirror (optical name for same thing). $\theta$ is the angle in radians off the boresight (central axis) of the antenna at which the other antenna is located in this antenna's pattern. $k=2\pi/\lambda$ is called the wavenumber of the radiated signal, $\lambda=c/f$ is its wavelength, $f$ is its frequency, and $c$ is its speed (normally light, but other things, such as sound, can be expressed similarly). One important thing to note here is that $x$ is proportional to $a/\lambda$, which measures the size of the dish in units of the wavelength of the radiation you are emitting or receiving. This means that using the same antenna with a higher frequency (and thus lower wavelength) will make the antenna act more directionally restricted than using the same antenna with a lower frequency (higher wavelength). To figure out what physical angle of pointing off the axis corresponds to any particular place on the plot, you have to plug in $x$, $k$, and $a$, and solve for $\theta$. Note this means for certain combinations of radius and wavelength, the range of $x$ which is physically accessible may be much less, or much more, than I have shown here.

Now think back to what I said earlier: the sole function of an antenna or mirror or other fancier setup is to redistribute how much radiated power goes in which direction. The bigger the antenna is, the more directional it makes the signal. It only makes the signal stronger in directions near where it is pointing, because to do that it must reduce the power that goes in other directions, in a precisely zero-sum way. It can only put more power in the middle by taking power away from the sides. Therefore, the larger the antenna, the smaller the region of highest gain. This means that a bigger antenna only gives you a stronger signal if you know exactly where to point it! If you point two antennas in slightly the wrong direction, by the same angle, then the larger antenna will suffer more from the same mismatch. That is "pointing error" in dB, sometimes known as "pattern loss". If you don't know what the right direction is, or you can't point right at it for some reason (some other part of the spacecraft is in the way, or has its own pointing requirement, or maybe your attitude control isn't quite right, or you have too much jitter, or many other possibilities), then the larger antenna gives you less signal in the desired direction, because it took away more from that direction in order to make a higher central peak (the "main lobe" or "main beam" of the antenna pattern).

If you have an antenna that can't point --- for example, the antenna on your cubesat is fixed with respect to the body, and you need attitude control to operate the solar panels while the antenna is just carried along for the ride, or the antenna is fixed always to nadir --- then it should have the least gain possible, because then it can work all day long, instead of just the one minute per pass where its antenna just manages to point close enough to the ground station antenna.

On a map, one typically draws a contour at -3 dB or -10 dB (and minuses in this case are usually spoken, in English at least, as "3 dB down" and "10 dB down", and calls that the "beam spot" or "footprint" or similar term. Note that you have a sequence of hills, which are separated by very narrow, very deep valleys. The hills are called "side lobes" of the antenna pattern (sometimes "interference fringes" in optics), and the valleys are called "nulls". The reason people don't usually draw, say, the contour at 20 dB down, is that the peak of the first side lobe (in this one example antenna pattern, not all of them) is at -17.57 dB, so the contour is a spot, surrounded by a blank space, surrounded by a ring (the first side lobe, which wraps all the way around), surrounded by more blank space.

Other antennas have other characteristic shapes. If you keep the same kind of design parabolic mirror with sensor at focus, but stretch it out much longer in one direction than the other, you have made a "sectoral" antenna, which has a wider beam in the direction in which the antenna is narrower, and a narrower beam in the direction in which the antenna is wider. Ratios of 5 or 10 are fairly common, as are found in the giant spinning air traffic control radars long used at airports. Their physical extent is horizontally wide and vertically narrow, so their beam pattern is horizontally narrow and vertically wide. This is used to catch aircraft at many different altitudes and distances (spreading vertically) while looking only at a small range of compass bearing (tightly grouped horizontally). The number of different shapes used, and consequently the number of patterns to learn, was originally rather small, until someone invented phased array antennas (for radar) and adaptive optics (for light), in which you use a few hundred to a few thousand tiny little pieces, and arrange them in a huge number of ways that can change very quickly in a precisely controlled way.

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  • $\begingroup$ This is the type of answer that illustrates the difference between an engineer and a scientist. An engineer plays with numbers until "it works" then "moves on to other problems". A scientist wants to explain why it works. $\endgroup$
    – Ng Ph
    Nov 17, 2021 at 18:47
  • $\begingroup$ @NgPh by academic training, I am a theoretical particle physicist. As such, I know all too well that what scientists do all day is solve equations we know are wrong but are useful anyway because they model real behavior well enough to enhance understanding. I have worked with experts in atmospheric chemistry and radiative transfer physics and meteorology and all the many other disciplines necessary to build and operate functional weather satellites, and appreciate how hard they all worked. There simply is not enough time to know it all, much less explain more than a tiny fraction. $\endgroup$
    – Ryan C
    Nov 17, 2021 at 22:20
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    $\begingroup$ Don't get me wrong. It was not a criticism. I am a retired engineer and sometime am amused at how scientists struggle with engineers' practice. You did a good job explaining this. Barnwell's paper is clearly written for engineers. The reproaches I have on this paper are: (i) the confusing terminology (efficiency, gain,loss, attenuation, ...) and (ii) somewhat a lack of modesty (honesty ?). When you can't justify precisely, it's ok to say "this is my margin". BTW, Barnwell has a 3 dB "implementation margin" too, on top of the numbers given in Annex. $\endgroup$
    – Ng Ph
    Nov 18, 2021 at 10:13
  • $\begingroup$ @NgPh In that case, all is well. :) My biggest hurdles in explaining link budgets are always the statistical nature of signal detection (even at higher SNR than "closes" your link, there is still nonzero BER) and the controversy over whether human error --- the probability that a programmer made a mistake in the software, or that someone installed the hardware wrong, or that the operator pressed the wrong button --- counted as something that could be changed, or just endured (like weather), and where (if anywhere) we were allowed to carry a margin for that. $\endgroup$
    – Ryan C
    Nov 18, 2021 at 15:39
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    $\begingroup$ We don't consider human errors, roundings, using wrong units, etc... in link-budgets. But we do take into account seriously the statistical nature of the prediction, especially when weather is a significant factor. Link-budgets for TV broadcasting for ex., have the concept of worst-month. There are ITU models to compute the rain attenuation not exceeded for X% of the worst-month, for a given frequency and a given category of climate (temperate, tropical, ...). $\endgroup$
    – Ng Ph
    Nov 18, 2021 at 18:51
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I have found this publication, maybe it will be helpful. Free-space optical links for space communication networks which is also a chapter from the book “Springer Handbook of Optical Networks” (pp. 1057-1103) https://doi.org/10.1007/978-3-030-16250-4

8.1.3 Link budget

The link budget is the key method to determine the overall performance of a lasercom system under a set of operating conditions. The basic link budget is given by equation (8.15) and relates the received power $P_R$ to the transmitted power $P_T$, the transmission and reception gain $G_T$ and $G_R$, the losses of the transmitter $L_T$ and the receiver $L_R$, the atmospheric losses $L_A$, the pointing losses $L_P$ and the free-space losses $L_S$.

$$P_R = P_T G_T L_T L_P L_S L_A L_R G_R \tag{8.15}$$

The most significant parameters in the link budget can be easily quantified, which allows making a quick preliminary analysis of the link. The transmitted gain $G_T$ and received gain $G_R$ can be calculated with equations (8.16) and (8.17), where $\Theta_T$ is the full transmitting divergence angle in radians, $D_R$ is the telescope aperture diameter, and $\lambda$ is the wavelength. The pointing loss $L_P$ is defined by the equation (8.18), where $\Delta_{\Theta}$ is the pointing accuracy. The free-space loss is given by equation (8.19), where $L$ is the distance between terminals.

$$G_T = \frac{16}{\Theta_T^2} \tag{8.16}$$

$$G_R = \left( \frac{\pi D_R}{\lambda} \right)^2 \tag{8.17}$$

$$L_P = \exp \left(-2 \left( \frac{\Delta_{\Theta}}{\Theta_T} \right)^2 \right) \tag{8.18}$$

$$ L_S = \left( \frac{\lambda}{4 \pi L} \right)^2 \tag{8.19}$$

Table 8.1 and Fig. 8.13 show an example of a basic link-budget calculation for the LEO-to-ground SOTA mission carried out by NICT (Japan)11. The conditions of this link budget are as follows: the telescope’s elevation is 30° for a link distance of 1,107 km between the ~600-km SOTA orbit and the NICT’s OGS in Koganei (Tokyo, Japan) during the pass on December 9th, 2015; the operating wavelength is 1549 nm; the receiver’s aperture is 1 m; the optical signal is coupled into multi-mode optical fiber; and the transmitter, receiver and pointing losses are based on experimental measurements. As a reference...

11A. Carrasco-Casado, H. Takenaka, D. Kolev, Y. Munemasa, H. Kunimori, K. Suzuki, T. Fuse, T. KuboOka, M. Akioka, Y. Koyama and M. ​Toyoshima, "LEO-to-ground optical communications using SOTA (Small Optical TrAnsponder) – Payload verification results and experiments on space quantum communications," Acta Astronautica, vol. 139, pp. 377-384, 2017.

(original screenshot)

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