I've seen in literature that loop performance is inversely proportional to loop SNR. The higher the loop SNR, the lower the variance, the better the performance. My question is how do we calculate the loop SNR of a Costas or PLL? Is there a direct formula for converting from "conventional SNR" to loop SNR?


The definition of loop SNR I'm looking for can be found in Appendix C of Michael Rice's "Digital Communication: A discrete-time approach". An excerpt with the definition is shown below:

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    $\begingroup$ "I've seen in literature". What literature? Did you see it once, or do you see it all over? If you have a link, please edit your question with the link. If it's a printed book, please quote the text where this is said. In particular, because different authors choose different terminology, point to (or quote) the author's or authors' definition of "loop SNR". $\endgroup$
    – TimWescott
    Commented Oct 20, 2021 at 15:02
  • $\begingroup$ @TimWescott I have edited the question and included the reference $\endgroup$ Commented Oct 22, 2021 at 12:11

1 Answer 1


$N_0$ is the noise spectral density at your PLL input. $B_n$ is your PLL bandwidth. $P_{in}$ is the power of the signal being received. If you work out the units, all the Watts and Hz and whatnot will cancel, and you're left with a pure ratio. So as long as you know those three numbers -- you have your answer.

This works because a PLL (or Costas loop) works like a low-pass filter to the input phase noise, and it only lets noise through in the bandwidth $B_n$. $\frac{N_0}{P_{in}}$ is the input phase noise power per $\mathrm {Hz}$, so when you multiply that by the filter (PLL) bandwidth, you're left with a signal to noise ratio.

  • $\begingroup$ So it seems like loop SNR is just another way of expressing PLL error variance. One question about the derivation. How will the derivation look like when we assume a purely digital system where the loop bandwidth is expressed in radians per samples instead of Hz? $\endgroup$ Commented Oct 29, 2021 at 9:38
  • $\begingroup$ There's a lot of twists and kinks in the answer to that. It makes it worthy of a new question. $\endgroup$
    – TimWescott
    Commented Oct 29, 2021 at 14:55
  • $\begingroup$ Done. The new question can be found at dsp.stackexchange.com/questions/78915/… $\endgroup$ Commented Nov 1, 2021 at 10:57

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