# If a MarCO-type CubeSat were in orbit around Bennu, what kind of power would it need to communicate with the Deep Space Network?

This is a follow-up question to: What is the lowest power signal that the DSN can detect?

Essentially, I have determined that the asteroid Bennu (ranging from 25 to 350 million kilometers from Earth) is in view of 2 DSN dishes for 5 hours, followed by another 2 hours, per day.

I have also found that the power received from a copy of the MarCO CubeSat by a 34 meter dishes varies from about -95.7 to -118.6 dBW (these were calculated using the equation and values found in my comment below). The power sent, for reference, is about 35 dBW including transmitter gain, excluding receiver gain and distance losses.

The transponder being used is an Iris v2 (Spec sheet: https://www.jpl.nasa.gov/cubesat/pdf/Brochure_IrisV2.1_201611-URS_Approved_CL16-5469.pdf). According to the spec sheet, the maximum bandwidth available for X-band using the Iris v2 transponder is about 0.05 GHz.

The goal is purely to estimate Bennu's position by performing a differenced doppler with a MarCO CubeSat in orbit around Bennu (https://solarsystem.nasa.gov/basics/chapter13-1/). No extra information, such as sending back images, is necessarily required, though may be useful if possible. Telemetry is not critical, and can be automated on-board. How can I go about finding the ideal SNR, and verifying a successful reading? What rough data rate should be required for an accurate differenced doppler reading? How long should a signal be sent to Earth to get an accurate reading of the doppler shift?

Let me know if I need to specify or clarify anything else. Thank you.

• @uhoh, the power numbers are in dBW. These numbers come from your equation in space.stackexchange.com/a/24343/12102, where Ptx = 3.8W = 5.8 dBW, Gtx = 29.2 dBi, Grx = 68.24 dB (for 34-m dish), and Lfs ranges from 199 to 222 depending on current orbital distances. These are specs for both DSN and MarCO. Those all help to calculate the Prx listed in the question. – SteveMcGroto Mar 15 '19 at 1:02
• Good idea. I will add that information to the original question. – SteveMcGroto Mar 15 '19 at 1:06
• new edit looks really great! I'll need a little time to update my answer accordingly, but I think it's going to be much more interesting now.Thanks! – uhoh Mar 16 '19 at 6:59
• Edit to original post: Added "How long should a signal be sent to Earth to get an accurate reading of the doppler shift?" – SteveMcGroto Mar 19 '19 at 0:46

## 1 Answer

note: based on discussion in comments the question has been revised and more details added, so I will be updating this answer with the day.

I'll address the current title:

What power and signal-to-noise-ratio is required to send a signal from an asteroid to the DSN?

The short answer is that it depends on the required data rate. A 12 bit 1024 x 1024 monochrome image with 10:1 lossless compression (with error correction) would need about 350 bits/sec to be sent in one hour so let's use that as a starting point.

Let's work back from the dish specifications. This is from the really thorough Chapter 2 System Noise Concepts with DSN Applications in Volume 10 of the Descanso book series. (see also When was the idea behind the DESCANSO Book Series first conceived, by whom, and what is it's “mission”?)

Here is all of Volume 10: Low-Noise Systems in the Deep Space Network Let's choose a 34 meter dish at X-band and say the gain is 68 dB and the noise temperature is 33 Kelvin.

For an effective receiver temperature of say 33 Kelvin, the noise equivalent power will be about $$k_B T \times \Delta f$$ where $$k_B$$ is the Boltzmann constant or 4.6E-22 (Joules) times $$\Delta f$$.

Let's do things the normal way and specify a S/N ratio rather than calculate it. With good error correction built into the system it can be pretty low, but let's be very generous and go with +12 dB.

From what I wrote in Am I using Shannon-Hartley Theorem and thermal noise correctly here?

The Shannon-Hartley theorem says that the maximum data rate $$C$$ is given by

$$C = BW \ log_2 \left(1 + \frac{S}{N}\right).$$

Where $$S$$ and $$N$$ are the signal and noise powers within the fully-used bandwidth $$BW$$.

Plug in SN = $$10^{12/10}=15.8$$ and a data rate of 350/sec and the absolute theoretical minimum bandwidth $$BW$$ required is 350/4.1 or 85 Hz. But nobody does this. Lets instead just set the bandwidth equal to the data rate to be a bit conservative. $$\Delta f$$ is 350 Hz.

That gives a system noise of 4.6E-22 (Joules) times $$\Delta f$$ = 1.6E-19 Watts. That's -188 dBW.

Using the chosen S/N of +12 dB, that means your 34 meter antenna must receive -176 dBM.

If instead you'd like 1 MiB/sec, everything scales linearly and you'd need to receive about 3000 times more power or about -141 dBW, for the same specified S/N of +12 dB.

• Thanks for the great answer. To work in reverse, I assume that I can specify a data-rate and find the S/N ratio resulting from that, if I figure out what data-rate I need. That being said, does the S/N ratio not also depend on the thermal noise of the spacecraft components, as well as the cosmic background noise? Is that incorporated somehow in either the table or equation, or is it so minimal such that it is unimportant? – SteveMcGroto Mar 15 '19 at 2:17
• Very good explanation. I did not find the transmitter power needed at the asteroid. – Uwe Mar 15 '19 at 12:46
• @uhoh just an asteroid as an example, for instance the asteroid Bennu (25 to 350 million kilometers) as asked in the question. – Uwe Mar 15 '19 at 16:11
• @uhoh The hardware being used here is an Iris v2 transponder and a MarCo reflectarray antenna. I have worked through Shannon-Hartley with an S and N value, but I worry that my answer for BW is meaningless without determining a good C. Any sources for figuring out what data rate is needed? To perform differenced doppler (solarsystem.nasa.gov/basics/chapter13-1), with NO extra info needing to be sent to Earth. Just sending back the same signal that the DSN sends the satellite for long enough to inspect doppler shift. There is a 2 hour window and a 5 hour window daily for doing this. – SteveMcGroto Mar 16 '19 at 5:32
• @uhoh Your original answer was very helpful for me to get this far, thank you so much. This topic is still new to me so I appreciate it. I will add some more information to the original question right now. – SteveMcGroto Mar 16 '19 at 5:51