2
$\begingroup$

When performing a link budget for satellite communications, a key component is defining an acceptable signal-to-noise-ratio in order to obtain the necessary bit-rate. How does one determine what the needed SNR is?

Additionally, what type of SNR do deep space satellites, such as Voyager, require?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
6
$\begingroup$

There's no such thing as a globally applicable "good" SNR. To make a earthbound comparison:

While your good old analog TV needs maybe an SNR of 40 dB to be somewhat enjoyable, GPS reception on very similar frequencies can work with signal well below the noise floor, so let's say -5 dB.

What SNR you'll need depends on how fast you want to transport data. And as there's very different mission needs for communication, there's very different needs for SNR.

Voyager sends low-rate. If I read this very related question correctly, Voyager's signals have a power of -152.6 dBm (that's 10⁻¹⁵ milliwatt) at earth.

Depending on the communications mode, different bandwidths are used, but documentation suggests a 12 dBHz bandwidth, leading at earth room temperature to a noise receiver power of -174 dBm/Hz + 12 dBHz + NF. We can build good amplifiers these days, so NF=0 dB is a reasonable approximation.

We end up at -162 dBm noise, and -153 dBm signal – a solid SNR of 9 dB.

For the BPSK modulation used, together with the high degree of channel coding applied, that's plenty.

You could, assuming sufficient synchronization and stability, work with way, way less – rule of thumb says BPSK works down to ca 4 dB, I think.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ Thanks, Marcus. A couple additional questions: How can GPS work below the noise floor? Also, is there a source for the 4 dB SNR rule of thumb for BPSK? Thanks! $\endgroup$
    – SteveMcGroto
    Mar 19, 2019 at 17:01
  • 1
    $\begingroup$ re: GPS If I had a reference, it wouldn't be a rule of thumb. It's, however relatively easy to derive: The bit error probability in BPSK is the symbol error probability, is the probability that a zero-mean normal random variable with variance $\sigma$ is larger than 1. Set that to an acceptable bit error rate, e.g. $10^{-2}$, and you get $P(X\sim\mathcal N(0,\sigma)>1) \overset!=p_{B} = 10^{-2}$, solve for $\sigma$ through $\Phi$ table lookup $\endgroup$
    – user17550
    Mar 19, 2019 at 17:19
  • $\begingroup$ Correct me if I am wrong, but is the result of that derivation this plot (montana.edu/aolson/ee447/EB%20and%20NO.pdf on page 3)? Jumping ahead to that plot, if we use a BER of 10^-2, we do get an Eb/N of about 4dB. From there, the same source suggests that SNR = (Eb/No)*(fb/B). So to figure out an acceptable SNR, you must first determine a data rate fb. But using Shannon-Hartley theorem fb = B*log2(1+SNR), you need an SNR value in order to find the acceptable data rate fb...am I missing something obvious here? What is the general approach here? $\endgroup$
    – SteveMcGroto
    Mar 19, 2019 at 18:55

Your Answer

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

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