3
$\begingroup$

I would like to understand pink (or $1/f$) noise better. However, clearly written resources are difficult to come by, and are usually concerned with its Fourier spectrum, or qualitative descriptions of it as a time-dependent process.

I hope that someone can describe how a pink noise process $P(t)$ depends on its values at earlier times $t' < t$, in the form of a probability density function on $P(t) - P(t')$, in the same way that we can do for Brown noise (i.e. Brownian motion).

Preliminaries

As a 'process', white noise can be modelled as a collection of i.i.d. random variables $W(t)$, where for each $t$ the distribution of $W(t)$ is uniform over some interval $[-a,a]$. Brown or red noise is the time-integral of white noise, in the sense that $$ B(t) = \int_0^t W(s) \,\mathrm d s \;.$$ Note that $B(t)$ is not independent of $B(t')$ for $t' < t$, in that $B_t - B_{t'}$ is normally distributed with mean $0$ and variance proportional to $(t - t')$. Thus we can talk about the expected properties of the trajectory at discrete time intervals in a straight-forward way.

If we take the Fourier transform of a white noise process over a time interval, we obtain a function whose expected modulus-square is constant over all frequencies, albeit with increasing variance for higher frequencies. If we do the same for Brown noise, we obtain a function whose expected modulus-square varies as $1/f^2$ with respect to frequency, again with increasing variance for higher frequencies.

'Defining' pink noise

Intermediate to these is pink noise, which is a process whose Fourier transform has a modulus-square proportional to $1/f$ (so that it is often called $1/f$ noise). For a discrete-time process, where we define $W(t)$ on the integers $t \in [0,T]$ for some $T$, we may mock up a pink noise process by setting $$ \begin{aligned} \hat P(0) &= \hat W(0) \\ \hat P(k) &= \hat W(k) \big/ \sqrt{\lvert k \rvert}, \quad k \in [-T{+}1,T] \setminus \{ 0 \} \end{aligned} $$ and then defining $P$ to be the inverse Fourier transform of $\hat P$: this is what the clearest sources that I can find suggest.

A small amount of numerical experimentation on my part suggests that the resulting function $P(t)$ has a much smaller variance than the original white-noise process $W(t)$. The time-integral $C(t)$ (i.e., a cumulative sum) of $P(t)$ describes a trajectory which is less volatile than the Brown noise process $B(t)$ obtained from the same white-noise function, but otherwise the two time-integrated noise processes follow a similar trajectory. Both of these observations are not very surprising, given that $P$ arises by decreasing the importance of the high-frequency components of $W$.

Question.

There are some qualitative descriptions of pink noise, in terms of how $P(t)$ depends on $P(t')$ for $t' < t$, such as that the autocorrelation of $P(t)$ with $P(t')$ decreases exponentially towards some non-zero value and stays there. However, I would like a quantitative result similar to what we have for Brown noise — specifically: is there a precise description for the probability density function of $P(t) - P(t')$?

(Failing this, is there a better description for how to simulate a pink noise process than the one that I've outlined?)

$\endgroup$
2
  • $\begingroup$ What you call pink noise would be called log-correlated noise in mathematics. See for example arxiv.org/abs/1711.00427 for a non-stationary construction. The construction cannot be made stationary on any space of functions / distributions, but it can be made stationary on the space of Schwartz distributions quotiented by constants. $\endgroup$ Aug 16, 2019 at 15:41
  • $\begingroup$ For my own future reference, I'll note that in the above preprint, Ref. [12] –– B. Duplantier, R. Rhodes, S. Sheffield, and V. Vargas. Log-correlated Gaussian fields: an overview. [arXiv:1407.5605], 2014 –– seems a promising introductory resource. $\endgroup$ Aug 16, 2019 at 15:58

1 Answer 1

4
$\begingroup$

The spectral density of $1/f$ noise is $S(\omega)=c/\omega$ for $\omega_1<\omega<\omega_2$. The corresponding autocorrelation function in the time domain $$P(t)=\lim_{T\rightarrow \infty}\frac{1}{T}\int_0^T x(t')x(t'+t)\,dt'$$ is given by the Wiener-Khinchine theorem, $$P(t)=\frac{c}{2\pi}\int_{\omega_1}^{\omega_2}\frac{\cos\omega t}{\omega}\,dt=\frac{c}{2\pi}\left[C(\omega_1 t)-C(\omega_2 t)\right],$$ with $C(x)$ the cosine integral. The variance of the noise process is $$\mathbb{E}[x(t)^2]=P(0)=\frac{c}{2\pi}\log(\omega_2/\omega_1).$$

$\endgroup$
3
  • $\begingroup$ I'm having some difficulties connecting your post to what I have read. Your remark on spectral density (or the power distribution) is of course standard. But it isn't clear to me why the autocorrelation function should only involve a cosine contribution, or how I should relate the autocorrelation which you denote by $P(t)$ to the expected value (resp. the variance) of the noise process, which in the notation of my post would be $\langle P(t) \rangle$ (resp. $\langle P(t)^2 - \langle P(t)^2 \rangle \rangle$), as $P(t)$ differs from $0$ for finite $t$ (and converges to $0$ as $t \to \infty$). $\endgroup$ Aug 16, 2019 at 16:30
  • $\begingroup$ the Fourier transform of $S(\omega)$ has only a $\cos\omega t$ contribution because $P(t)=P(-t)$ for a stationary process. $\endgroup$ Aug 16, 2019 at 18:04
  • $\begingroup$ This is much clearer, thanks. I must admit that in my numerical experiments (where I am considering a discrete-time variant in which I suppose that I have $\omega_2/\omega_1 = 10^5$), I seem to have a variance a few orders of magnitude smaller than what I would expect from the spectral density, given your analysis. However, it may be that I should more carefully consider my analysis. $\endgroup$ Aug 19, 2019 at 10:45

Your Answer

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

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