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Im a beginner in signal processing so my question may be obvious.

A white noise has the property to have its autocorrelation function that is equal to

$$\mathbb{E}[f(t+\tau)f(t)]=\sigma^2 \delta(\tau).$$

We say that random signals can be Gaussian white noise, but for me if the signal has a Gaussian distribution it is necesseraly a white noise because as soon as we know the distribution then $\mathbb{E}[f(t+\tau)f(t)]$ is automatically determined.

Am I wrong ?

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When we say the stochastic process $X_{t_1}X_{t_2}\cdots X_{t_i}$ is a Gaussian Process with mean $\mu$ and variance $\sigma^2$ it only means $$X_{t_i}\sim\mathcal{N}(\mu,\sigma^2)$$ So still we know nothing about the relation of the $X_{t_i}$'s with each other. When we add that the process is also white, it means the $X_{t_i}$'s are uncorrelated. That is

$$\mathsf{E}[(X_{t_i}-\mathsf{E}X_{t_i})\mathsf{E}(X_{t_j}-\mathsf{E}X_{t_j})^*]=\begin{cases}\sigma^2,&i=j\\0,&i\neq j\end{cases}$$

Off topic, but notice that the $\tau$ notation is an implicit assumption of the process being stationary.

The process can be Gaussian but colored (rather than white). In such case, the random variables are Gaussian but correlated. For instance, the autocorrelation function in case of stationary Gaussian process you mentioned can be something like this: $$R_{XX}(\tau)=\sigma^2e^{-|\tau|}$$

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  • $\begingroup$ White noise is always assumed to have zero mean, even in the continuous-time case. Absent this property, the power spectral density has an impulse at the origin, and thus the PSD does not have the defining characteristic of white noise: constant value for all frequencies. $\endgroup$ – Dilip Sarwate Nov 24 '16 at 21:56
  • $\begingroup$ Thank you ! The thing that I don't really get in fact is that i a process at time "n" follows a given law of probability I don't see how it could depend of what happened before ? Is it smth like this : I have 10 card with number s from 1 to 10 and 1 chance over 10 to have a given one. Then If I do the experiment 10 times the probability will always be the same. But If I decide to say "if I have the number "x" at a previous experiment, then I put it at the back of my set (in position 10), then there will have a correlation between two experiment ? $\endgroup$ – StarBucK Nov 24 '16 at 21:57
  • $\begingroup$ Yes @DilipSarwate, thank you for adding that. user3183950: I couldn't fully catch your example, but assume a stochastic process $X_{i+1}=aX_{i}+Y$, where $Y$ is another Gaussian random variable, independent of $X_{i}$ and $X_{i+1}$. You can see the process $\{X_{n}\}$ is Gaussian but the samples are correlated. $\endgroup$ – msm Nov 24 '16 at 22:51
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We say that random signals can be Gaussian white noise, but for me if the signal has a Gaussian distribution it is necesseraly a white noise because as soon as we know the distribution then $\mathbb{E}[f(t+\tau)f(t)]$ is automatically determined.

Am I wrong ?

Yes, you are wrong in various ways.

First, by Gaussian noise is meant a random process (not a random function as the OP calls it) -- that is, an infinite collection of random variables, one random variable for each time instant $t$ -- such that if we pick any $n \geq 1$ random variables $X_{t_1}, X_{t_2}, \ldots, X_{t_n}$ from the random process, then these random variables have a $n$-dimensional joint Gaussian distribution. Now, a $n$-dimensional Gaussian distribution is entirely determined by the $n$ means of the random variables and the $n\times n$ covariance matrix of the random variables, but note very carefully that nothing at all has been asserted about the means or the covariance matrix of the $n$ Gaussian random variables. Specifically, nowhere is it written that all the means must be the same or that $X_{t_i}$ and $X_{t_j}$ have the same variance or that $\operatorname{cov}(X_{t_i},X_{t_j}) = \operatorname{cov}(X_{t_i+1},X_{t_j+1})$ etc. All that is meant by Gaussian noise is that the $n$ random variables have a jointly Gaussian distribution (which implies, among other things, that $X(t)$ and $X(s)$ individually are Gaussian random variables as well).

Second, white noise is a mythical process that is unobservable in all its glory in nature (probably just as well since it is infinitely powerful and would lead to an immediate solution to the energy crisis). We poor mortals can only observe white noise through some kind of device that necessarily limits what we can observe -- kind of like watching a solar eclipse through special glasses-- and thus what we observe is a pale imitation of the real thing. Well, it has been observed that if an observation device is modeled as a linear filter with transfer function $H(f)$, then (with an open circuit at the filter input), the filter output is a random process with power spectral density $\sigma^2|H(f)|^2$. This is consistent with an assumption that the input to the filter is a white noise process with autocorrelation function $\sigma^2\delta(t)$ where $\delta(t)$ is a Dirac delta or impulse and power spectral density $S(f) = \sigma^2, -\infty < f < \infty$ if we simply plug in $\sigma^2$ for the input power spectral density in the power spectral density equation $$S_{\text{output}}(f) = |H(f)|^2 S_{\text{input}}(f).$$ Never mind that many mathematicians will cringe at the cavalier treatment where we are ignoring that the above formula implicitly assumes that the input process is a finite power process (which white noise is definitely not); we are engineers and we don't care. We are used to treating impulses like ordinary functions.

Finally, turning to discrete-time random processes, remember that one cannot sample the mythical beast called white noise -- it does not exist in nature -- and the sampler is necessarily a device that observes the random process for a very short but nonzero time $\varepsilon$, and thus the sample $X[n]$ is actually something proportional to $\int_{nT-\varepsilon/2}^{nT+\varepsilon/2}X_t \mathrm dt$ which has variance $\sigma^2\varepsilon$ if $\{X_t\}$ is a white noise process. So, a discrete-time white noise process is a collection of zero-mean independent random variables $X[n]$. Yes, many DSP texts say that uncorrelatedness suffices, and in the case of white Gaussian noise it does because Gaussianity brings in the jointly Gaussian property with which uncorrelatedness implies independence; but for arbitrary white noises, it is best to insist on independence and not on just zero correlation.

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Yes, that's wrong. Here is a counterexample. We can form a periodic signal (Fig. 1) with a Gaussian distribution:

$$f(t) = \Phi^{-1}\big(\operatorname{frac}(t)\big),\tag{1}$$

where $\Phi^{-1}$ is the inverse cumulative distribution function of the standard $(\mu = 0,\,$ $\sigma^2 = 1)$ normal (Gaussian) distribution, and $\operatorname{frac}(t),$ with $0 \le \operatorname{frac}(t) < 1,$ gives the fractional part of the real number $t.$ The function $\Phi^{-1}$ is also called the probit function. It can be used to convert numbers coming from the standard uniform distribution into the standard normal distribution (see Inverse transform sampling).

enter image description here
Figure 1. The periodic signal $f(t).$ Each period equals the probit function. At integer $t$ the function approaches $\pm\infty.$

Because the signal is periodic with period 1, then:

$$f(t + 1) = f(t)\quad\text{for all }t\in\mathbb{R}.\tag{2}$$

It follows that:

$$\mathbb{E}[f(t+\tau+1)f(t)]=\mathbb{E}[f(t+\tau)f(t)].\tag{3}$$

If the signal would be white, then, plugging in from the question the equation $\mathbb{E}[f(t+\tau)f(t)]=\sigma^2 \delta(\tau)$ for the autocorrelation function of white noise, we would get:

$$\sigma^2\delta(\tau + 1)=\sigma^2\delta(\tau)\tag{4}$$

and with non-zero $\sigma^2,$ for $\tau = 0:$

$$\delta(1) = \delta(0),\tag{5}$$

which is not true according to the definition of the Dirac delta function. The conclusion is that the signal $f(t)$ cannot be white.

Periodic signals are composed of harmonic frequencies only.

The plot was generated in Octave by:

x = ([-200:299]+0.5)/100;
y = (sqrt(2)*erfinv(2*(x-floor(x))-1));
plot(x, y);
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Gaussian is for the distribution, but white for the correlation. When we say the sample $z_1,\,z_2\,\cdots,\,z_n$ are white Gaussian random variables, it means that each sample follows a Gaussian distribution with certain mean and variance, but the samples are uncorrelated and thus independent. The samples could be Gaussian be not white (it's called colored), in which case the samples are correlated.

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