2
$\begingroup$

I am designing a telecommunications device capable of redundant communication from Earth to Saturn's moon Enceladus which ranges from 8 to 11 AU from Earth.

I have decided on an X-band high-gain antenna for the communication on a Entry-Descent-Landing vehicle, however, I am having troubles finding equations on how to determine the diameter of the antenna, power usage, transmission rate, and how to determine the strength of the signal being sent back to Earth.

I was hoping that there would be a book on where I can find all these specs or even a NASA mission PDR where I can digest more information on thereof. Any help will be greatly appreciated.

$\endgroup$
4
  • $\begingroup$ The question is too big to be answered shortly. A similar question space.stackexchange.com/questions/50415/… $\endgroup$ – A. Rumlin Apr 8 at 18:35
  • $\begingroup$ In Russian you can download the book "Radio systems of interplanetary spacecraft", where you can find theoretical and practical data on the systems of long-range space communication of interplanetary stations of the Soviet Union.sovams.narod.ru/Telemetry/amsradio.html $\endgroup$ – A. Rumlin Apr 8 at 18:46
  • $\begingroup$ I've added an answer to your question, let me know if it is helpful or you have questions how to use it. Welcome to Stack Exchange! $\endgroup$ – uhoh Apr 9 at 0:22
  • 1
    $\begingroup$ @A.Rumlin well I didn't write a book, but hopefully the answer will be at least somewhat helpful. $\endgroup$ – uhoh Apr 9 at 0:22
3
$\begingroup$

So we are looking to specify an

X-band high-gain antenna for the communication on a Entry-Descent-Landing vehicle

at circa 10 AU, and want to

determine the diameter of the antenna, power usage, transmission rate, and how to determine the strength of the signal being sent back to Earth.

This is what is called a link budget calculation.

As @A.Rumlin points out it's a big task, but you can make some estimates fairly easily by seeing this answer to How hard is it to receive direct signals from vehicles on the surface of Mars, and has anyone other than the DSN done so? But I'll add an even shorter answer here based on your application.

From here:

Link Budget

From this answer which is from this answer:

$$ P_{RX} = P_{TX} + G_{TX} - L_{FS} + G_{RX} $$

  • $P_{RX}$: received power by spacecraft
  • $P_{TX}$: transmitted power by wristwatch
  • $G_{TX}$: Gain of wristwatch's transmitting antenna (compared to isotropic)
  • $L_{FS}$: Free space Loss, what we usually call $1/r^2$
  • $G_{RX}$: Gain of spacecraft's receiving antenna (compared to isotropic)

$$G \sim 20 \times \log_{10}\left( \frac{\pi d}{\lambda} \right)$$

$$L_{FS} = 20 \times \log_{10}\left( 4 \pi \frac{R}{\lambda} \right).$$

I'll work an example, then you can put in your own numbers.

Let's assume you are using X-band at 7.5 GHz and your spacecraft as a 1.5 meter dish. The wavelength $\lambda = c/f$ is 0.04 meters so your spacecraft's dish has a gain (relative to isotropic) of about 41 dBi.

Let's say your receive antenna is one of the three big 70 meter Deep Space Network dishes, then $G_{RX}$ is 75 dBi

The "free space path loss" $L_{FS}$ (basically $1/r^2$) at 10 AU is about 293 dB. Yikes!

If your spacecraft's transmitted power $P_{TX}$ is 5 Watts or about 7 dBW, then we have

$$ P_{RX} = 7 + 41 - 293 + 75 = \text{-170 dBW}$$

That's $10^{-17}$ watts.

If your receiver is cooled to "say 20 Kelvin, the noise equivalent power will be about $k_B T \times \Delta f$ where $k_B$ is the Boltzmann constant. For 10 kHz that's -189 dBW or about $1.3 \times 10^{-19}$ watts, much lower than the received signal strength. For more on that see this answer for example.

This means that you could trim down the size of the transmit or receive antenna, or increase the bandwidth.

You also have to perform the same calculation for the uplink, but this one is usually easier because all the terms are the same except the transmit power and the receiver noise temperature.

$\endgroup$
5
  • 1
    $\begingroup$ what does a negative value of P_rx indicate? $\endgroup$ – Jose De La Pena Apr 15 at 22:40
  • 1
    $\begingroup$ @JoseDeLaPena these are dB units which are logarithmic. For -170 dBW we convert to real watts using $$10^{-170/10} = 1 \times 10^{-17} \ \text{ watts}$$ So the minus sign is in the exponent, which means the number is positive but very small. See Decibel watt. $\endgroup$ – uhoh Apr 15 at 23:32
  • $\begingroup$ Also, for P_tx, is that 5 Watts the power needed to be supplied by the batteries? Or is there another way to find out the power needed to be supplied to antennas? $\endgroup$ – Jose De La Pena Apr 16 at 0:17
  • $\begingroup$ @JoseDeLaPena "I'll work an example, then you can put in your own numbers." I just chose 5W as an arbitrary starting point. You'll have to decide the data rate that you need, whether you want to receive with a 70 m DSN dish or a 34 m dish, and how you want to design the power system of your spacecraft. You can choose solar or [RTG}(en.wikipedia.org/wiki/Radioisotope_thermoelectric_generator) for your main power source, and if you need more power you can use them to charge batteries, then discharge at a higher rate during high power data transmission. $\endgroup$ – uhoh Apr 16 at 13:19
  • 1
    $\begingroup$ @JoseDeLaPena I recommend that you ask a new question (or two!) about powering your spacecraft next. By posting a new question it gives an chance for many people to post new answers. $\endgroup$ – uhoh Apr 16 at 13:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.