The power spectrum of Poisson noise

In summary: I think you may be confusing a few things here. You originally mentioned Fourier transforming the counts of the sum of signal and noise to obtain the power spectrum. This is different from the Fourier transform of the count rate. Power spectrum is the square of the amplitude of the Fourier transform of the counts. Also, the "a" in your original equation is not the same as the "a" in the second equation you provided. In any case, I'm not sure I can fully answer your question without more context and details. Can you provide more information or specify your equations and variables more clearly?
  • #1
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I thought that if we Fourier transformed the counts of the sum of the signal from the source and the Poisson noise, and obtained the power spectrum, we would get the following,
##P_{j}=P_{j, \text { signal }}+P_{j, \text { noise }}+\text { cross terms }##
but I found the following description.
if it would be true that ##a_j = a_{j,noise} + a_{j,signal}##, if the noise is random and uncorrelated and if many powers are averaged, then Eq is approximately valid.$$ P_j =P_{j, \text { signal }}+P_{j, \text { noise }} $$

I don't understand why the cross term here disappears.
Could you give me some hints?
Thank you.
 
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  • #2
If I understand correctly, then the "cross terms" are the product of your signal and your noise (doesn't matter that it's poissonian, could be any distribution). Since you are given that your noise is uncorrelated with the signal, the cross terms vanish on average. It's in the definition of "correlation".

More concretely, ##P_j = (a_{j,signal} + a_{j,noise})^2 = P_{j,signal} + P_{j,noise} + 2a_{j,noise} a_{j,signal}##. Let's decompose ##a_{j,signal}## into an average value ##\langle a_{j,signal} \rangle## and some uncertain fluctuation ##\delta a_{j,signal}##, so overall ## a_{j,signal} = \langle a_{j,signal} \rangle + \delta a_{j,signal}##. Then the cross term is ##2 a_{j,noise} \langle a_{j,signal} \rangle + 2 a_{j,noise} \delta a_{j,signal}##. The first part averages to zero since ##\langle a_{j,noise}\rangle = 0##. The second part averages to zero because the signal and noise are uncorrelated, i.e. their covariance is zero.
 
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  • #3
Thank you for explaining exactly the part I was wondering about.I would like to ask further questions if you don't mind.

Twigg said:
The first part averages to zero since ##\langle a_{j,noise}\rangle = 0##. The second part averages to zero because the signal and noise are uncorrelated, i.e. their covariance is zero.

I am now thinking about the analysis of data containing Poisson noise, which is the sum of the time variability of a signal and a almost constant value of background noise detected by an X-ray detector.
In these cases, i.e. when the average background noise is not zero, is it possible that the cross term remains?

Or does someone know of any papers that could reference a detailed discussion of this area?

Thank you.
 
  • #4
Sure, you could have an average background that's non-zero, but calling that "noise" is a bit of a misnomer. For clarity, I'd call that a background, as opposed to your signal and your noise (which has average value of zero).

If these were coherent signals (like an AC voltage signal on top of a DC background), then the signal and background would have a non-zero cross term on average. However, for an x-ray detector, your signal and background should be incoherent, meaning they still have no cross term because their phases are uncorrelated. You can think of the cross term for x-rays as interference. Only coherent sources cause interference, and a background signal is rarely coherent (coherent x-rays are very exotic as well).
 
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  • #5
I see, even if there is background, as long as it is uncorrelated with the signal, the cross term disappears.
I rewrote the previous equation in my own way.

##a_{j, \text { signal }}=\left\langle a_{j, \text { signal }}\right\rangle+\delta a_{j, \text { signal }}##, and ##a_{j, \text { bg }}=\left\langle a_{j, \text { bg }}\right\rangle+\delta a_{j, \text { bg }} ##
then cross term is
## 2a_{j,bg} a_{j,signal}=2(\left\langle a_{j, \text { bg }}\right\rangle+\delta a_{j, \text { bg }} )(\left\langle a_{j,signal}\right\rangle+\delta a_{j,signal})##
If we average these over multiple powers, ##\left\langle a_{j, \text { bg }}\right\rangle = 0##. Thus
##\left\langle 2(\left\langle a_{j, \text { bg }}\right\rangle+\delta a_{j, \text { bg }} )(\left\langle a_{j,signal}\right\rangle+\delta a_{j,signal}) \right\rangle ##
##= 2\left\langle \delta a_{j, \text { bg }} \left\langle a_{j, \text { signal }}\right\rangle \right\rangle+2 \left\langle a_{j,noise}\delta a_{j,signal} \right\rangle
\\ + 2\left\langle \delta a_{j, \text { bg }} \delta a_{j,signal} \right\rangle ##

where ##\left\langle \delta a_{j, \text { bg }} \right\rangle = \left\langle \delta a_{j, \text { bg }}\right\rangle =0##, and ##\left\langle \delta a_{j, \text { bg }} \delta a_{j,signal} \right\rangle = \text{Covariance}(\text{signal,bg})=0##
so we can white
## 2a_{j,bg} a_{j,signal}=0##Do I understand it right?
 
  • #6
To re-iterate from my previous post, the fact you are talking about an x-ray detector changes things. What is "a" here? Is it the x-ray field amplitude or the detector voltage?
 
  • #7
It was the Fourier transform of the X-ray count rate.
In other words, it is the one that appears in the definition of the power spectrum, and ##j## is the wave number.
##P_{j}=\left|a_{j}\right|^{2}##
 
  • #8
Ah, ok. Then you can ignore what I said about coherence in my previous reply. Sorry about that.

What you wrote is correct. I might nitpick a little on the subscripts, but the spirit of it is correct. The only thing that seems off is your convention where you let ##\langle a_{j,bg} \rangle = 0##. Normally backgrounds have a non-zero time average, otherwise they'd just be noise. Anyways, it's just semantics. Nice work!
 
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  • #9
No, it's because I didn't explain it well enough. I'm sorry.

About ##\left\langle a_{j, b g}\right\rangle=0##, this was my misunderstand.
So, is there a term that remains in the cross-term?
I mean,
arcTomato said:
If we average these over multiple powers, ##\left\langle a_{j, \text { bg }}\right\rangle = 0##. Thus
##\left\langle 2(\left\langle a_{j, \text { bg }}\right\rangle+\delta a_{j, \text { bg }} )(\left\langle a_{j,signal}\right\rangle+\delta a_{j,signal}) \right\rangle ##
##= 2\left\langle \delta a_{j, \text { bg }} \left\langle a_{j, \text { signal }}\right\rangle \right\rangle+2 \left\langle a_{j,noise}\delta a_{j,signal} \right\rangle
\\ + 2\left\langle \delta a_{j, \text { bg }} \delta a_{j,signal} \right\rangle ##

##2\left\langle a_{j, \mathrm{bg}}\right\rangle \left\langle a_{j, \mathrm{signal}}\right\rangle## does't disapper??I still have some questions, but the part I had been thinking about for a long time has been solved. Thank you very much.
 
  • #10
If this is a Fourier transform, you background is probably highly bandlimited, if not purely DC. You won't see it except near 0, so the cross term should tend to zero for higher frequencies, and also as you increase the integration time.
 
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  • #11
I see.
So you mean that the effect of the cross term is visible at ## j=0## and nearby, but becomes invisible at higher frequencies.

The discussion was very easy to understand. Thank you once again, and sorry for my bad English.
 
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