It is said that the Deep Space Network can detect low-power signals sent from Voyager in the range of 10^-16 W. I can't seem to find any sources that indicate what the minimum power requirement of a received signal is. Also, what would be the minimum required power for the DSN to reliably interpret the signal? Does this figure differ between the 70m and 34m dishes?

  • $\begingroup$ The lowest detectable power depends on the data rate of the signal. But 1 bit per hour is too slow for a real transmission. $\endgroup$
    – Uwe
    Mar 10, 2019 at 10:06
  • $\begingroup$ @Uwe is it, though? Could imagine a space probe flying through the vastness of space doesn't need to update the nuclear power source's operational state every day… $\endgroup$
    – user17550
    Mar 10, 2019 at 10:07
  • $\begingroup$ You can read about things like link budget and SNR here: space.stackexchange.com/a/24343/12102 $\endgroup$
    – uhoh
    Mar 10, 2019 at 11:50
  • 2
    $\begingroup$ @MarcusMüller really slow transmissions are undesirable, there are many spacecraft that need the DSN so you want to transmit your data as quickly as possible. The lowest data rate in use now is the 160 bit/s for Voyager. $\endgroup$
    – Hobbes
    Mar 10, 2019 at 11:54
  • 1
    $\begingroup$ @Hobbes sure, practical systems that exist are more than 0.28 mbit/s, but there's no reason for 160 bit/s being a natural lower boundary – not much going to change on the channel in an hour. $\endgroup$
    – user17550
    Mar 10, 2019 at 12:27

1 Answer 1


There's no minimum power for any receiver if you don't specify the type of transmission and the amount of knowledge about that existing at the receiver!

For example, a GPS receiver has way worse characteristics (for example, a 4-bit ADC) than a digital TV receiver. So, who needs higher power to work?

GPS works well below the noise floor because the data rate received is so low, and the receiver knows what to look for very exactly (mathematically: it uses correlation over a long spreading sequence, meaning that it does an inner product in the number-of-samples-dimensional complex (signal) vector space, which means that the magnitude you get when detecting signal-containing noise is far, far higher than when just doing the same product with a vector that's purely white noise).

TV reception, on the other hand, needs to be high-rate, and thus, the SNR must be way, way better (see: Shannon Capacity C = 1/2 B · log_2(1+SNR) ).

Since the DSN isn't tailored to one specific type of signal but equipped with hardware and software that can be configured/programmed to pick out a desired wavefrom, the sensitivity depends on the transmission type and can't be generally answered.

The only figure of merit you could get would be the system's Noise Figure or the equivalent Noise Temperature, because that's how much worse a signal's SNR gets by going through the system. That would allow you to calculate the theoretical sensitivity for any given transmission. Now, since DSN is extremely distributed, especially among hot and cold objects, that cumulative number can't exist either, sorry.

  • 1
    $\begingroup$ Thanks for the insightful answer, Marcus. For context, and a follow-up question, is that I am trying to figure out if it is possible for a MarCO-type CubeSat to be able to send a signal to Earth from a near earth asteroid (ranging from 0.25*10^8 km to 3.5*10^8 km from Earth). All it needs to do is report its position, so the information it needs to send is very minimal. Using @uhoh's link above, I can calculate that the power received from this theoretical MarCO by a 34-m DSS varies from -95.70588 dB to -118.62844 dB. How can I go about finding the SNR, and verifying a successful reading? $\endgroup$ Mar 10, 2019 at 20:56
  • 1
    $\begingroup$ would be worth posting as separate question. Also: Think about what "position" is, and how a CubeSat would determine its own, and why ground would need to know the position of a satellite that got launched into space with known trajectory. $\endgroup$
    – user17550
    Mar 10, 2019 at 22:11
  • $\begingroup$ @SteveMcGroto I agree, you should consider asking that as a follow-up question. $\endgroup$
    – uhoh
    Mar 14, 2019 at 12:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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