I have a signal $R(t)$ that is DC with shot noise. It has units of events per second.

I am trying to understand this signal in a continuous-time picture.

In time-domain, I can model it as a series of delta peaks. Let $R_0$ be the expected value (again, in units of events per second). If I average over any time interval $t_\mathrm{int}$, the variance is $R_0/t_\mathrm{int}$, by the definition of a Poissonian process.

In frequency domain, I expect the spectrum to be flat, because the Poissonian noise is uncorrelated (if we ignore the DC term...). What is the PSD of this noise?

My gut thinking is this:

If $1/t_\mathrm{int}$ is the equivalent noise bandwidth of the 'filter' from above, then the variance I calculated is simply the integral of the PSD over the bandwidth: $$ R_0/t_\mathrm{int} = \int_{-1/t_\mathrm{int}}^{{1/t_\mathrm{int}}}S \mathrm d\! f. $$

Solving for $S$, which I assume to be frequency-independent, I find $$ S = \frac{R_0}{2}. $$

Is that anywhere close to correct?

I am particularly skeptical if the averaging in time domain can be translated into frequency domain brick wall filtering like this.


1 Answer 1


I found a satisfying answer here: https://123.physics.ucdavis.edu/shot_files/ShotNoise.pdf

Let me reproduce the important steps, with the notation from the question.

As mentioned, the signal is a series of Dirac pulses: $$R(t) = \sum_k\delta(t-t_k),$$ with the added constraint that, on average, there are $R_0$ events per unit time.

To compute the power spectrum, we first compute the auto correlation function $A(t')$. (I am calling it $A$ because I unfortunately already used the symbol $R$.) $$A(t') \\= \lim_{T\rightarrow\infty}\frac1T\int_{-T/2}^{T/2}R(t)R(t+t')\mathrm dt\\ = \lim_{T\rightarrow\infty} \sum_k\sum_{k'}\int_{-T/2}^{T/2}\delta(t-t_k)\delta(t-t_{k'}+t')\mathrm dt\\ = \lim_{T\rightarrow\infty} \frac1T \sum_k\sum_{k'} \delta(t_k-t_{k'}+t')\\ \sim R_0\delta(t').$$

Admittedly the last step is somewhat steep. More details in the link above.

By the Wiener–Khinchin theorem, I can then compute the PSD: $$S(f) = R_0.$$

Note: the reference above includes an extra factor 2, which is because the author considers the PSD for positive frequencies only.

  • $\begingroup$ My answer was wrong because I computed the PSD of the expected signal, not the expected PSD of a signal. In my calculation I ended up counting interference between pulses on different realizations of the noise. Sorry for that. $\endgroup$
    – Bob
    Commented Feb 15, 2022 at 12:01
  • $\begingroup$ I very much appreciate your help regardless! Looks like we both learned something :) $\endgroup$
    – polwel
    Commented Feb 15, 2022 at 17:05
  • $\begingroup$ Yes definitely, what I did is ok for a set of orthogonal signals, but not for a generic set of signals. $\endgroup$
    – Bob
    Commented Feb 16, 2022 at 12:18

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