4
$\begingroup$

I want to know what is the difference between white noise and WSS white noise. is there any difference between them or they're equal?
and what about white Gaussian Noise?

$\endgroup$

4 Answers 4

4
$\begingroup$

In most of the engineering literature I'm familiar with, white noise is introduced as an idealized random process $n(t)$ with a flat power spectrum

$$S_N(f)=\frac{N_0}{2}\tag{1}$$

and the corresponding autocorrelation function

$$R_N(\tau)=\frac{N_0}{2}\delta(\tau)\tag{2}$$

The reason for defining white noise in this way is because it closely approximates the properties of thermal noise for frequencies below about $10^{12}$ Hz.

According to above definition, white noise is a WSS random process. Note that $(1)$ and $(2)$ imply that $n(t)$ has a constant mean equal to zero. I would claim that this is the standard definition of white noise in textbooks in the field of signal processing and digital communications.

White noise can also be defined in a less restrictive sense, namely as a process $n(t)$ for which the values $n(t_1)$ and $n(t_2)$ are uncorrelated for all $t_1$ and $t_2\neq t_1$. I.e., the autocovariance function of $n(t)$ has the form

$$C_N(t_1,t_2)=q(t_1)\delta(t_1-t_2),\qquad q(t)\ge 0\tag{3}$$

This definition can be found in Probablity, Random Variables, and Stochastic Processes by Papoulis (p. 295 of the 3rd edition). Eq. $(3)$ implies an autocorrelation function of the form

$$R_N(t_1,t_2)=q(t_1)\delta(t_1-t_2)+\mu_N(t_1)\mu_N(t_2)\tag{4}$$

with $\mu_N(t)=E\{n(t)\}$. Defined in that way, white noise is generally non-stationary and doesn't have a power spectrum in the conventional sense.

The "engineering definition" of white noise given above is obtained from the less restrictive definition by assuming that $q(t)$ is constant and that $\mu_N(t)=0$. Note that if we assume a constant but non-zero $\mu_N(t)$, the process would be WSS but the power spectrum would have a Dirac delta impulse at DC, which wouldn't be a good model for thermal noise.

$\endgroup$
3
  • $\begingroup$ I found the same definition of white noise (same notation!) on page 395 of the 4th edition of Probability, Random Variables, and Stochastic Processes by Papoulis and Pillai and so I wonder if the citation of a book by Proakis above is just a typo. I learned about stochastic processes from the first edition of the book (by Papoulis alone) back in 1968 and don't remember all the bells and whistles of $(3)$ and $(4)$ which I suspect are an addition by Pillai (he took over starting with 3rd edition) trying to be "as general as possible" and failing miserably. $\endgroup$ Commented Jul 20, 2020 at 12:20
  • $\begingroup$ @DilipSarwate: Sure, it's Papoulis, not Proakis. Thanks for pointing out my typo. I was in fact going through several books looking for their definitions of white noise, one of which was the one by Proakis. $\endgroup$
    – Matt L.
    Commented Jul 20, 2020 at 12:53
  • $\begingroup$ @DilipSarwate: By the way, Pillai's name seems to appear only on the 4th edition, not on the third, so Eq. (3) in my answer is very likely not one of Pillai's additions. $\endgroup$
    – Matt L.
    Commented Jul 20, 2020 at 13:05
5
$\begingroup$

White noise is not "WSS by nature" whatever you mean by that phrase but it can be treated as a (zero-mean) WSS process insofar as its effects in linear systems are concerned.

For example, standard linear system theory ways when the input to an LTI system is an ordinary WSS process $\{X(t)\}$ with autocorrelation function $R_X(\tau)$, then the output of the LTI system is a WSS process $\{Y(t)\}$ with autocorrelation function $R_Y(\tau)$ given by $$R_Y = h\star \tilde{h} \star R_X \tag{1}$$ where $h(t)$ is the impulse response of the LTI system and $\tilde{h}(t) = h(-t)$ is the time-reversed impulse response of the LTI system. The power spectral densities are related as $$S_Y(f) = |H(f)|^2S_X(f)\tag{2}$$ where $H(f)$ is the transfer function of the LTI system. If $\{X(t)\}$ is a white noise process with autocorrelation function $K\delta(\tau)$ and we pretend that $(1)$ and $(2)$ are still applicable, we get that $\{Y(t)\}$ is a zero-mean WSS process with autocorrelation function $R_Y = K\cdot h\star \tilde{h}$ and power spectral density $S_Y(f) = K\cdot |H(f)|^2$. Of course, mathematicians would laugh at this calculation but physical experiments using the naturally occurring thermal noise in electrical circuits as a stand-in for a white noise process show that these results are pretty close to reality. As engineers, we seek equations that match the universe as we observe it (physicists seek universes that match their equations while mathematicians don't care) and so we go blithely on our way treating white noise as a WSS process in linear systems and everything works out OK. The troubles start when we start treating white noise as a WSS process in nonlinear systems and the world comes crashing down about our ears and we need to start paying attention to what the math people are saying.

White noise is referred as white Gaussian noise if we pretend or claim or insist that $\{Y(t)\}$ is a Gaussian process which means that not only are all the random variables $Y(t)$ Gaussian random variables, but every finite set $\{Y(t_1), Y(t_2), \cdots, Y(t_n)\}, n \geq 2,$ of random variables is a set of jointly Gaussian random variables. Standard random process theory says that when a Gaussian process is passed through an LTI system, the output is a Gaussian process but this fact does not allow us to reverse-engineer the result and claim that all the $X(t)$'s are also Gaussian random variables.

$\endgroup$
5
  • $\begingroup$ @MattL. Thanks for careful reading and deletion of misplaced words. $\endgroup$ Commented Jul 19, 2020 at 21:24
  • $\begingroup$ Our discussion under my answer aside, what is the definition of "white" that the math people use if it doesn't require WSS – the technomathematicians I've asked aren't aware of that! (references also welcome) $\endgroup$ Commented Jul 20, 2020 at 12:00
  • 2
    $\begingroup$ @MarcusMüller Eight years ago, I asked the very same question What is meant by a continuous-time white noise process? over on math.SE, and you can read the answers there (don't forget to read all the comments too!) to get a feel for what mathematicians think about the matter. $\endgroup$ Commented Jul 20, 2020 at 14:54
  • $\begingroup$ @DilipSarwate what is different about that behavior of white noise in a non-linear system that wouldn’t occur in a “truly” WSS process? (In a way that makes the definition of WSS come crashing down)? Can you clarify that in your answer? $\endgroup$ Commented Jun 6, 2022 at 16:18
  • $\begingroup$ @DanBoschen Consider the random process $\{(X(t))^2\colon -\infty < t < \infty\}$ where $\{X(t)\colon -\infty < t < \infty\}$ is white Gaussian noise. What pdf do you ascribe to $(X(t))^2$? Bear in mind that it is not necessarily the case that the individual $X(t)$'s are all Gaussian random variables, or if you wish to insist that they are indeed Gaussian random variables, then their common pdf is the limiting value of a zero-mean Gaussian pdf as the variance increases without bound towards $\infty$. What is their pdf? What is the pdf of $(X(t))^2$? $\endgroup$ Commented Jun 6, 2022 at 20:11
2
$\begingroup$

I will interpret "Is white noise WSS by nature" as "Is white noise necessarily WSS" since it is unclear what "by nature" means. The answer is: no.

Papoulis has stated that a process $v(t)$ is a white noise if its values $v(t_i)$ and $v(t_j)$ are uncorrelated for every $t_i\neq t_j$. That is,

$$C_v(t_i, t_j) = 0, \text{ for } t_i \neq t_j \tag{1}$$

On the other hand, if $v(t)$ were necessarily WSS, then its first- and second-order statistics of would be constants. That is, its mean $m_v(t_i) = m_v$ would be constant, and its covariance function would only depend on the time difference, $\tau = t_i - t_j$, i.e.,

$$C_v(t_i, t_j) = C_v(t_i-t_j) = C_v(\tau)$$

In the way that Papoulis wrote the Equation (1), we cannot infer that the covariance depends only on the difference $t_i - t_j$. Note that even though $C_v(t_i, t_j) = 0$ for $\tau = t_i - t_j \neq 0$, its covariance may vary when $t_1 = t_2$, being thus variant to time shift.

That is exactly what Papoulis means by "nontrivial white-noise process":

$$C_v(t_i, t_j) = q(t_i) \delta(t_i - t_j).$$

It is important to mention, however, that the white noise is usually defined along with the the WSS assumption.

$\endgroup$
1
$\begingroup$

Disclaimer: this might very well be wrong. Still pondering it, but Dilip Sarwate has convincing points.

When you say "white" you assume it's WSS to begin with. For non-WSS processes, "white" isn't defined, since no only lag-dependent autocorrelation can be found. (And a process is white, exactly if its autocorrelation takes the form of a delta dirac impulse.)

So, yes, any process that is called "white" is inherently WSS.

"Gaussian white noise" is white noise whose amplitude is Gaussian-distributed. Amplitude distribution has nothing to do with whiteness or stationarity: a non-stationary process can still be Gaussian distributed at any point in time.

$\endgroup$
17
  • 1
    $\begingroup$ I don't agree with much of this answer, $\endgroup$ Commented Jul 19, 2020 at 15:12
  • 1
    $\begingroup$ @DilipSarwate I must admit although I find my math to check out, I kind of trust you on autocorrelation. $\endgroup$ Commented Jul 19, 2020 at 18:16
  • 2
    $\begingroup$ "I think we are on the same page" Absolutely not. Your statement "WSS process is that the joint distributions of the RV at (all sets of) different times is shift-invariant (your "…for all choices of $t_1,t_2,\ldots,t_n$" is the definition of strict stationarity, not wide-sense-stationarity. $\endgroup$ Commented Jul 19, 2020 at 20:09
  • 1
    $\begingroup$ You might want to add in the constant mean requirement to your revised definition of WSS processes to make it more complete, and then ponder how the boundedness of $R(\tau)$ reconciles with the $K\delta(t)$ of white noise when you insist that white noise is a WSS process. Or is $R_X(0)$ not required to be bounded (or defined), only $R_X(\tau)$ for $\tau \neq 0$? $\endgroup$ Commented Jul 19, 2020 at 20:21
  • 1
    $\begingroup$ Marcus: What you said above is where statistics and DSP differ greatly. In statistics, the autocorrelation function is independent of whether it's a non-zero mean because the mean is subtracted out ( in the formula ) anyway. It sounds like this is not the case in DSP so I'm staying out of this discussion. But it's interesting to read nonetheless. $\endgroup$
    – mark leeds
    Commented Jul 20, 2020 at 3:57

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.