Let's assume we have the following situation:

  1. Transmitter on moon orbit (distance from earth ~400'000km), around 433Mhz.

  2. Transmitter power 1W, antenna gain 10dB, circular polarization. Steered to point to earth.

  3. Receiver is on Earth. Antenna is steerable 25dB, matching circular polarization.

  4. On-off keying modulation.

What data transfer speed is achievable in this configuration?

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    $\begingroup$ Not a lot, with OOK. Why not use a modern modulation? QPSK is pretty basic and common for satellite communications. $\endgroup$ Commented Jan 29, 2017 at 14:22
  • $\begingroup$ @PhilFrost-W8II We are limited by average transmitter power, so transmitter power efficiency is more important where OOK should not be too bad. OOK transmitter also is the simplest and most robust one. $\endgroup$ Commented Jan 29, 2017 at 18:36
  • 1
    $\begingroup$ OOK is a terrible choice under those restraints. It's basically always a terrible choice, unless the most important limitation is the numbers of transistors in your transmitter. Which nowadays is never the case, unless you buy the cheapest china RF remote you can get. Proof is easy: in PSK, bit energy is always average RF power, i.e. errors are evenly likely for all possible symbols. For OOK, bit energy is highly uneven, skewing the bit error probability, because it's impossible to precisely predict received ON-power. It's really the textbook example of a bad choice. $\endgroup$ Commented Jan 29, 2017 at 19:33
  • $\begingroup$ @MarcusMüller but for OOK for same average power bit energy of 1 will be double that of PSK? $\endgroup$ Commented Jan 29, 2017 at 19:57
  • $\begingroup$ @BarsMonster Yes. But you should be transmitting as many 1s as zeros, so that really doesn't help you at all. $\endgroup$ Commented Jan 29, 2017 at 22:49

2 Answers 2


Short answer: Math says max link rate is 2Mb/s if you knew the perfect channel coding. Which is still an unsolved puzzle.

Long answer:

You're calculating a link rate. That is fine, and can be answered using Shannon's Channel Capacity, which gives us the upper limit for bits per second that we can get across a given channel:

$$ C= b\cdot \log_2\left(1+ \frac SN\right)$$

with $b$ being the channel bandwidth, $S$ the signal energy and $N$ the noise power.

This is a upper limit. It's a mathematically proven limit. Over a bandwidth-limited channel with a given SNR, you cannot get more bits per second reliably. There's no technical way around that limit.

Now, you didn't mention $b$ at all. But that doesn't matter, we'll just calculate a "per Hertz of bandwidth" rate, and you can multiply that with the bandwidth of your transceiver.

Now, $S$ is only partly specified – you said what your transmit power $P$ was, and thus, we'll simply apply free space path loss to that (which simply is "since the sphere on which we distribute the energy grows, by how does the field strength go down?"). This incorporates the surface of a sphere, divided by the wavelength:

$$\begin{align} A_\text{free space} &= \left(\frac{4\pi d}\lambda \right)^2\\ &=\left(\frac{4\pi dc_0}f\right)^2\\ &=\frac{16\pi^2 d^2 f^2}{c_0^2}\\ &\approx\frac{160\,\cdot \,4^2\cdot10^{16}\,\text{m}^2\,\cdot\,4.33^2\cdot10^{12} \frac1{\text{s}^2}}{9\cdot 10^{18}\frac{\text{m}^2}{\text{s}^2}}\\ &\approx\frac{48000\cdot10^{16}\,\cdot10^{12}}{9\cdot 10^{18}}\\ &=\frac{48}9 10^{13}\\ &\approx 5 \cdot 10^{13}\\ &\approx 136\,\text{dB} \end{align}$$

So $S\approx \frac{1\,\text W}{5 \cdot 10^{13}} = 2 \cdot10^{-14}\,\text W$.

So, what is your noise power? We don't know, since we don't know your receiver!

Now, let's assume you operate that receiver at room temperature.

That means you get -174 dBm/Hz of spectral noise power density. Ok, that means your

$$\frac SN = \frac{-136\,\text{dBW}}{-204\frac{\text{dBW}}{\text{Hz}}\cdot b}= \frac{\text{Hz}}b \cdot 68\,\text{dB}\approx \frac{\text{Hz}}b \cdot 6.3\,\text{MHz}$$

(ignoring Noise Figure For NowTM)

When $\frac SN$ is smaller than 1, the argument of the logarithm becomes roughly 1, and $\log 1=0$, so you can basically get nothing across. So we must postulate:

$$\begin{align} \frac{\text{Hz}}b &< (68-\text{NF}_\text{dB})\,\text{dB} \end{align}$$

Since the logarithm falls slower than $b$ rises, you'll get the best result with a $b_\text{dBHz}\lim (68-\text{NF}_\text{dB})$; assuming a Noise Figure of 5 dB, a 63 dBHz = 2 MHz channel leads to maximized $C$:

$$C_{max} \approx 2\,\text{MHz}\log_2\left(1+3.15\right)\approx 2\log_2(4.15)\frac{\text{Mb}}{\text s} \approx 4\frac{\text{Mb}}{\text s}$$.

You cannot get that with On-Off-keying; you need at least 2 bits per Hz. So the "lowest" modulation that might be able to fulfill this purpose would be QPSK, if zero-roll-off channel filters existed. So you'd probably settle for something like 8-PSK with a whole metric effton of channel coding/error correction on top, to get even close to that. You'll probably be happy if you get a single megabit per second.

  • $\begingroup$ It will be straightforward to have NF of receiver LNA down to ~1dB. But I was under impression that in this band galactic/solar noise will be higher than this intrinsic noise of LNA. What do you think? $\endgroup$ Commented Jan 29, 2017 at 19:42
  • $\begingroup$ Don't know, don't have numbers on that. It really doesn't matter for the formula whether $N$ is dominated by LNA noise or background noise – both are defined by a constant noise power density multiplied by your channel bandwidth (assuming you properly filter) $\endgroup$ Commented Jan 29, 2017 at 19:43
  • 1
    $\begingroup$ Tsys can be much lower... given that you've flown to the moon, you should spash out on a fancy pre-amp. From an Amsat list message by i8cvs, sky temperature of a good antenna should be 20K, and with a 0.5 dB pre-amp, Tsys of 55 K is possible. That should increase the possible data rate by about 5 times. $\endgroup$
    – tomnexus
    Commented Jan 29, 2017 at 20:16
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    $\begingroup$ @phil Yes, but it depends on how much of your beamwidth the warm moon fills. 25 dBi is a 10 or 11 degree beam, and the moon is only 30' diameter. Surface temperature someone says is 140K to 280K depending on phase, and it fills 1/400 of the beam, so even a full moon adds only about 1 K to antenna temperature. Should still be detectable on/off the moon, this is how they found the surface temperature of planets. $\endgroup$
    – tomnexus
    Commented Jan 30, 2017 at 20:22
  • 2
    $\begingroup$ Man, look at this answer. Someone has a lot of time on their hands :D $\endgroup$ Commented Nov 12, 2021 at 17:57

Summary: Theoretical maximum in the neighborhood of 10s of megabits per second. Less than that in practice, perhaps a lot less depending on budget.

Let's start with the Friis transmission equation:

$$ P_{r(\mathrm{dB})} = P_{t(\mathrm{dB})} + G_{t(\mathrm{dB})} + G_{r(\mathrm{dB})} + 147.6 - 20 \log_{10} (rf) $$

Insert the values you've given for antenna gain, transmitter power, frequency and distance, to calculate the signal power at the receiver:

$$ \begin{align} P_{r(\mathrm{dB})} &= 30\:\mathrm{dBm} + 10\:\mathrm{dB} + 25\:\mathrm{dB} + 147.6 - 20 \log_{10} (400,000\:\mathrm{km} \cdot 433\:\mathrm{MHz})\\ &= -132\:\mathrm{dBm} \\ &= 6.31\cdot 10^{-14}\:\mathrm{W} \tag{1} \end{align} $$

Now knowing the signal power, we can use the Shannon-Hartley theorem to determine the maximum theoretical channel capacity $C$ in the presence of noise power $N$, signal power $S$, and channel bandwidth $\Delta f$:

$$ \begin{align} C &= \Delta f \log_2 \left ( 1 + {S / N} \right) \\ \tag{2} \end{align} $$

We've already calculated the signal power, but what's the noise power? Let's assume there's no interference or ambient RF noise, and that your receiver introduces no noise of its own. The only noise then is Johnson-Nyquist noise, which is a function temperature $T$ (in kelvin), and the bandwidth of the channel $\Delta f$, and Boltzmann's constant ($k_B \approx 1.38 \cdot 10^{-23}$)

$$ P_\mathrm{noise} = k_B \: T \: \Delta f \tag{3}$$

It's also the case for OOK (with appropriate pulse shaping) that the bandwidth of the signal is equal to the bitrate. That is:

$$ \Delta f = C \tag{4} $$

Combining theoretical channel capacity (2), thermal noise (3), and the bandwidth efficiency of OOK (4), we get:

$$ C = C \: \log_2 \left( 1 + {S \over k_B \: T \: C} \right) $$

Solving for C:

$$ \begin{align} 1 &= \log_2 \left( 1 + {S \over k_B \: T \: C} \right) \\ \ 2 &= 1 + {S \over k_B \: T \: C} \\ \ C &= {S \over k_B \: T} \\ \end{align} $$

Let's approximate Boltzmann's constant and add the received signal power calculated in equation 1:

$$ \begin{align} C &= {6.31\cdot 10^{-14} \over 1.38 \cdot 10^{-23} \: T } \\ \\ &= {4.57\cdot 10^{9} \over T } \end{align}$$

In sunlight, the moon can be about 400K, and you'd then need to integrate the moon's temperature (and the 4K or so for the night sky) over your antenna's pattern to get the noise temperature. The comments in Marcus Müller's answer have a little discussion.

That will give you an upper bound on the possible channel capacity. It's in the neighborhood of tens of megabits per second.

I'd also note that you should be able to do better with QPSK: it has the same bit error rate for a given energy-per-bit as BPSK, while being able to fit the same bit rate within half the bandwidth of OOK or BPSK. A narrower channel bandwidth means less noise, and thus a higher SNR and channel capacity.

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    $\begingroup$ ah, $(4)$ is a clever step if OP actually is restricted to OOK! but:$\frac{10^{-a}}{10^{-b}}=10^{b-a} \ne 10^{-a-b}$ $\endgroup$ Commented Jan 29, 2017 at 20:40
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    $\begingroup$ @MarcusMüller Thanks for the catch! I knew it was too horrible to be true :) Now it almost looks too good to be true, but I guess the noise temperature of systems mere mortals like me can build is orders of magnitude higher than what's possible with a space agency budget. $\endgroup$ Commented Jan 30, 2017 at 1:53
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    $\begingroup$ @PhilFrost-W8II Valiant effort but the bandwidth of OOK is not equal to the capacity. I'm not sure how you came to that conclusion. $\endgroup$
    – KillaKem
    Commented Jan 30, 2017 at 18:24
  • 2
    $\begingroup$ @KillaKem Consider OOK transmitting alternate 1 and 0: that's equivalent to multiplying the carrier by a raised sine wave. Say that sine wave is at 0.5 Hz, and for each cycle you get 2 bits (a 1 and a 0). That's 1 symbol per second, and the multiplication with the carrier will give mixing products at -0.5Hz and +0.5Hz from the carrier frequency: a bandwidth of 1Hz. $\endgroup$ Commented Jan 30, 2017 at 18:36
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    $\begingroup$ @PhilFrost-W8II What you are saying is not accurate, more especially the last line. I will point you to a question that YOU yourself seem to have answered about bandwidth and let you convince yourself that indeed in a OOK modulation scheme the Bandwidth is infinite because the signal has sharp discontinuities and also has the ability to be sparse in time. In this case, like all other modulation schemes, the bandwidth is determined by how wide the output TX filter bandwidth is. $\endgroup$
    – KillaKem
    Commented Jan 31, 2017 at 9:19

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