It is known that the PSD of additive white Gaussian noise (AWGN) is constant and equal to its variance. What about coloured Gaussian noise (CGN)?

For example, given the following PSD of CGN

$$S(f) = \frac 1f $$

Is the spectral density of such noise frequency-dependent? If so, how to get the PDF by some "inverse" autocorrelation function?

  • $\begingroup$ Note that the variance of white Gaussian noise is not defined, although with some hand-waving you could say it's infinite. If the noise PSD is $N_0/2$, then the variance of filtered noise (where the filter's impulse response has energy equal to 1) is equal to $N_0/2$. $\endgroup$
    – MBaz
    Nov 8, 2017 at 17:57

3 Answers 3


Colored Gaussian noise is by definition a wide-sense-stationary (WSS) process; that is, it has constant mean (all the random variables constituting the process have the same mean) and its autocorrelation function $R_X(t_1, t_2) = E[X(t_1)X(t_2)]$ depends only on the difference $t_2-t_1$ of the arguments. It is conventional to use $\tau$ to denote the difference $t_2-t_1$, and abuse notation by writing $R_X(\tau)$ instead of the more prolix $R_X(0, \tau) = R_X(t,t+\tau)$ for the autocorrelation function. The power spectral density (PSD) of the process is then the Fourier transform of $R_X(\tau)$: $$S_X(f) = \int_{-\infty}^\infty R_X(\tau)e^{-j2\pi ft} \,\mathrm dt.$$ The PSD is an even nonnegative function of $f$.

White noise is a zero-mean process for which $R_X(\tau) = K\delta(\tau)$ where $\delta(\cdot)$ is the Dirac delta or impulse and its PSD has constant value $K$ for $-\infty < f < \infty$. Colored noise is a zero-mean process whose PSD is not constant for all $f$. Colored Gaussian noise is a process in which all the random variables are zero-mean correlated (jointly) Gaussian random variables with random variables separated by time $\tau$ having covariance $R_X(\tau)$. Note that the variance of all the random variables is $\sigma^2 = R_X(0)$. The PSD has the connection to the PDF that the PSD determines the variance of the random variables in question via the following corollary to the inverse Fourier transform formula: $$\sigma^2 = R_X(0) = \int_{-\infty}^\infty S_X(f) \,\mathrm df.$$ Note that all the random variables constituting the process have the same (Gaussian) PDF (and so the same mean and same variance) and the variance is not a time-varying function due to the noise being colored.


Assuming $E\{ x\}=0$ $$ E\{|x|^2\}=r_{xx}(0)=\int_{-\infty}^{\infty} S(f) df $$



The PDF is Gaussian, what other PDF are you asking about?

  • $\begingroup$ The PDF which the PSD is not constant $\endgroup$
    – Mateus R.
    Nov 8, 2017 at 20:07
  • 2
    $\begingroup$ I don't understand your comment. PDF= probability density function PSD = power spectral density. power and probability are different things. $\endgroup$
    – user28715
    Nov 8, 2017 at 20:12
  • $\begingroup$ How can I find the PDF of a distribution with given PSD? $\endgroup$
    – Mateus R.
    Nov 9, 2017 at 1:12
  • 2
    $\begingroup$ You can’t but you stated it was a Gaussian process. That means it had a Gaussian pdf. You can get a pdf from the Inverse Fourier transform of its characteristic function. You can get a pdf from its moment generating function. $\endgroup$
    – user28715
    Nov 9, 2017 at 1:20

For example, consider a discrete-time integrator

$$X_{k+1} = X_k + U_k$$

where $X_0 =: x_0$ is the (zero-variance) initial condition and $U_k \sim \mathcal N (0, \sigma^2)$ is the AWGN input.

$$\begin{array}{rl} X_1 &= x_0 + U_0\\ X_2 &= x_0 + U_0 + U_1\\ X_3 &= x_0 + U_0 + U_1 + U_2\\ &\vdots\\ X_n &= x_0 + U_0 + U_1 + U_2 + \cdots + U_{n-1}\end{array}$$


$$\mathbb E (X_n) = x_0 \qquad \qquad \qquad \mbox{Var} (X_n) = n \sigma^2$$

Since the addition of independent Gaussian random variables is still Gaussian, we conclude that

$$X_n \sim \mathcal N (x_0, n\sigma^2)$$

Linear systems preserve "Gaussian-ness". Sometimes, one can do without the frequency domain.

  • 1
    $\begingroup$ What you write is correct but does not address the OP's question at all. $\endgroup$ Nov 8, 2017 at 20:49
  • $\begingroup$ @DilipSarwate Frankly, I have no idea what the OP wants. He seems to be interested in integrators, so I focused on the discrete-time integrator. The variance of the output increases linearly, violating stationarity. $\endgroup$ Nov 8, 2017 at 20:54

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