2 identical signals generated by the same PLL/VCO should have the same phase noise profile and their respective phase noises should be correlated.

Suppose one of them is delayed and I start sampling both simultaneously , if I now compare the phase noise of their respective samples taken at a certain time instance would the phase noise still be correlated? and if not how long would the delay have to be in order for their phase noise to be uncorrelated (if decorrelation is involved)?

My guess is that since phase noise has a trapezoidal shape and has a low pass filter frequency response, it is roughly the reciprocal of the loop BW that determines the length of the delay that would cause phase noise to become uncorrelated , the wider the loop BW i.e. the shorter the delay.

Does my reasoning make sense?

  • $\begingroup$ Hi! We've done this for a number of your previous questions, so I think it's time to do this on your own: Please add empty lines to logically structure your question. Like this, ist's just a wall of text and unnecessarily hard to read and hence, you're making it unnecessarily hard to answer. $\endgroup$ Nov 14, 2018 at 22:45
  • $\begingroup$ Thank you very much! This is much nicer to read :) $\endgroup$ Nov 14, 2018 at 23:23

1 Answer 1


No, it has nothing to do with loop BW (as you stated that you are observing the final output, so it is the same signal at all offset frequencies; as a very simplified explanation the loop bw indicates the demarcation in phase noise components between the VCO and the reference frequency).

If the output of the PLL/VCO is the same, meaning we split it coherently into two outputs and observe both outputs with no delay between the two, then they will be completely correlated at all offset frequencies. However, if there is a delay between the observations of the two signals, the components of the phase noise at very small frequency offsets from the carrier will still be correlated, while offsets much further away from the carrier will be decorrelated. As the delay between the two observations increases, the "frequency corner" where we start to decorrelate will go down in frequency as given by the inverse of the delay. The corner frequency is the inverse of the delay (in radian frequency for delay in seconds), and the decorrelation for the phase noise component goes down as $1/\sqrt{f}$, where $f$ is the offset frequency from the carrier.

To determine the amount of correlation assuming the two waveforms have the same scaling (as long as there is no amplitude difference, which would apply if we simply delayed one of them as a delay does not change the amplitude), we can simply subtract the two and measure the resulting output power of the waveform (averaged detector). The amount of cancellation indicates the degree of correlation; if the waveforms are completely correlated at all frequencies we would see complete cancellation, and if they were completely uncorrelated at all frequencies we would see a summation in power (+3 dB). In the continuous time domain such a delay and subtract operation has a high pass response, as suggested by the equivalent filter of that process shown in the graphic below, where the frequency shown represents offset frequency from the carrier:

equivalent filter

The frequency response to a noise process of the delay and subtraction operation, is given by:

$$H(\omega)= {1 - e^{-j \omega \tau}}$$

This shows the decorrelation intuitively, as components fully correlated would subtract, creating a cancellation, while the components fully uncorrelated, being random noise, would add in power (thus +3 dB higher). The frequency response given above is the "comb response" that would actually be achieve from a delay and subtract; the in the "high pass" region the response peaks periodically at +6 dB, and due to the periodic nulls, half the power is removed resulting in the net +3 dB gain. In terms of phase noise specifically this is clear when you consider a "slice" of phase noise at a a 3 Hz offset from the carrier (for example, as shown in the graphic below), represents a noise term that is sinsusoidally varying at a 3 Hz rate (at that given power level within 1 Hz of BW at that offset as given by the typical dBc/Hz scale of the vertical axis in phase noise spectrum plots). In comparison, consider a phase noise component from an offset of 0.03 Hz; that represents phase varying much more slowly (it takes 33.3 seconds to go through one cycle); surely you can see how even with a 1 second delay would still be quite correlated to this term, with a significant cancellation ratio (from such a subtraction).

LO phase noise components

The graphic below is specific to a slide I have on explaining such considerations with regard to carrier tracking loops, however directly applies the the explanation above, showing the cancellation of the lower frequency phase noise offsets.


Relationship to Cycle-to-Cycle Jitter It is worth noting at this point how this explanation unifies the relationship between phase noise and cycle-to-cycle jitter; as the measurement of cycle-to-cycle jitter involves subtracting the rising edge of a clock signal from the previous rising edge; which is the block diagram in the first figure above where $\tau$ is the time duration of one clock cycle. Thus if you multiply the frequency response of that filter with the phase noise spectrum (to filter it), and then integrate to get the total rms phase noise, the rms jitter is found by converting phase to time based on the clock frequency used.

  • $\begingroup$ o.k. so basically what you are saying is that the inverse of the delay determines the frequency at which correlation will start fading, right? The reason I mentioned loop BW as a rough is because phase noise underneath the loop BW is about 2/3 of the total phase noise, so again roughly almost the whole phase noise is there. Imagine a PLL with an extremely wide loop BW , its phase noise would look line AWGN and there will be no correlation even if the delay is very short, right?.It is just a way of figuring out things when compared with the case of AWGN $\endgroup$ Nov 17, 2018 at 19:27
  • $\begingroup$ Yes to inverse of delay is the corner freq. The one point I am not confident in is the 1/sqrt(f) roll-off (or is it 1/f). I want to simulate as some point to confirm that but as for the rest of it is from my own experience with phase noise and delay vs decorrelation. Loop BW is internal to the source of the final output. At the point for this discussion it doesn't matter how the source is created, and all has to do with the spectral density at various frequency offsets. Bringing loop BW in only causes confusion as it is not a factor. WIth an extremely wide loop BW it would not look like AWGN; $\endgroup$ Nov 17, 2018 at 19:41
  • $\begingroup$ As you get closer and closer to the carrier you go from white FM, to white PM to flicker to random walk where the noise power spectral density is increasing at a higher and higher rate. (It becomes a non-stationary process to longer you observe-- -and the longer you observe implies the closer to the carrier you are able to see---- perhaps a good way to picture that is to know that we can never observe at 0 offset from a carrier just like we cannot observe a DC signal: to see a DC signal you would have to observe forever, so the time duration of observation is itself a high pass process) $\endgroup$ Nov 17, 2018 at 19:44
  • $\begingroup$ For a certain fixed delay, I repeat for a certain fixed delay I believe that loop BW will be a factor. I may be looking at it from the "half empty glass"point of view $\endgroup$ Nov 17, 2018 at 19:52
  • $\begingroup$ If you change the loop BW, it will change the shape of the phase noise certainly. But the shape of the phase noise has no effect in determining if a particular frequency component of the phase noise is correlated or not to itself after a fixed delay. Phase noise is just signal power density due to phase fluctuations at various frequency offsets. For any given offset regardless of signal its correlation to itself after a delay is simply determined by how much delay is introduced. We can increase and decrease that power level (which is what the loop BW change will do), but that will not impact $\endgroup$ Nov 17, 2018 at 19:56

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