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It seems to be silly, but is it possible to deduce mathematically that a white noise is necessarily zero mean and uncorrelated?

I have seen some people defining a white process as a zero-mean and constant-variance process with uncorrelated samples. However, the corollary that I saw on books is that, if $x(n)$ is a discrete-time white noise, then $R_x (\tau) = \frac{N_0}{2} \delta (\tau)$ and $S_x (f) = \frac{N_0}{2}$. Nothing more. It is called "white" because of the analogy with the white light that has all frequencies. I can infer the following statement from it:

  1. Since $$ \begin{align} R_x(\tau) = E[x(n) x(n+\tau)] = 0\text{ for }\tau \neq 0 \label{1} \tag{1} \end{align} $$ we have that $x(n)$ and $x(n+\tau)$ are orthogonal.

To be an uncorrelated sequence, the covariance shall be $$ \begin{align} C_x(\tau)=R_x(\tau)-m_x ^2 =0\text{ for }\tau \neq 0. \label{2} \tag{2} \end{align} $$

If I guarantee that it is true, then $$ \begin{align} m_x=E[x(n)] = 0. \label{3} \tag{3} \end{align} $$

How to prove that?

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  • $\begingroup$ @V.V.T corrected. $\endgroup$ Commented Jun 6, 2022 at 1:59
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    $\begingroup$ Have you seen this question? Do the answers answer your question? $\endgroup$
    – Matt L.
    Commented Jun 6, 2022 at 10:37
  • $\begingroup$ @MattL. No, I haven't. But this question doesn't answer my question directly. I just found the answer the posted here :) $\endgroup$ Commented Jun 6, 2022 at 19:07
  • $\begingroup$ For a discrete-time white noise process, the delta in the autocorrelation function is generally expressed as $\delta[n]$, the Kronecker delta, and not as $\delta(t)$ which denotes the Dirac delta. $\endgroup$ Commented Jun 6, 2022 at 22:45
  • $\begingroup$ @DilipSarwate depends on which reference you are basing. For Oppenheim, you are right, Kronecker delta must be denoted as $\delta[n]$, whereas $\delta(t)$ indicates the Dirac delta. However, for many other authors (Proakis, Diniz,... etc), both discrete- and continuous-time are denoted in parentheses. But again, there is nothing wrong with that, it is just a matter of notation. $\endgroup$ Commented Jun 6, 2022 at 22:52

2 Answers 2

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A white noise is a random signal having a constant power spectral density (PSD): $$S_x(f) = \frac{N_0}{2}.$$ From that, you can deduce everything else.

The autocorrelation and the PSD are linked through the Wiener-Khinchin theorem:
  • The PSD is the Fourier transform of the autocorrelation.
  • The autocorrelation is the inverse Fourier transform of the PSD.
The inverse Fourier transform of a constant is a Dirac, therefore the autocorrelation of a white noise is: $$R_{xx}(\tau) = \mathcal{F}^{-1}(S_x(f)) = \mathcal{F}^{-1}\left(\frac{N_0}{2}\right) = \frac{N_0}{2} \delta(\tau).$$

This means that the autocorrelation is zero for any delay different than zero. Then, think of the meaning of the autocorrelation. The autocorrelation is a measure of similarity of a signal with itself in the future, or a measure of how the future is related to the present. An autocorrelation of zero means that there is no similarity, hence the value in the future $x(t+\tau)$ is uncorrelated with the present value $x(t)$.

For the zero mean, we can look at the autocorrelation of a signal that is not zero mean. Let's decompose a signal as $x(t) = x_\mathrm{DC} + x_\mathrm{AC}(t)$, where $x_\mathrm{DC}$ corresponds to the DC part of $x(t)$, i.e. it is a constant value corresponding to the mean of $x(t)$, and $x_\mathrm{AC}(t)$ corresponds to the AC part of $x(t)$, i.e. it contains only the fluctuations of $x(t)$ and has zero mean. Then, the autocorrelation of $x(t)$ is $$ \begin{align} R_{xx}(\tau) &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} x(t) x(t+\tau) \mathrm{d}t \\ &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} (x_\mathrm{DC} + x_\mathrm{AC}(t)) (x_\mathrm{DC} + x_\mathrm{AC}(t+\tau)) \mathrm{d}t \\ &= \lim_{T \to \infty} \frac{1}{T} \int_{-T/2}^{T/2} x_\mathrm{DC}^2 + x_\mathrm{DC} x_\mathrm{AC}(t) + x_\mathrm{DC} x_\mathrm{AC}(t+\tau) + x_\mathrm{AC}(t) x_\mathrm{AC}(t+\tau) \mathrm{d}t \\ &= \lim_{T \to \infty} \left( \frac{1}{T} \int_{-T/2}^{T/2} x_\mathrm{DC}^2 \mathrm{d}t + x_\mathrm{DC} \int_{-T/2}^{T/2} x_\mathrm{AC}(t) \mathrm{d}t + x_\mathrm{DC} \int_{-T/2}^{T/2} x_\mathrm{AC}(t+\tau) \mathrm{d}t + \int_{-T/2}^{T/2} x_\mathrm{AC}(t) x_\mathrm{AC}(t+\tau) \mathrm{d}t \right). \end{align} $$ Since $x_\mathrm{AC}(t)$ has zero mean, we have $$ \lim_{T \to \infty} \int_{-T/2}^{T/2} x_\mathrm{AC}(t) = 0. $$ Therefore, the autocorrelation is $$ \begin{align} R_{xx}(\tau) &= \lim_{T \to \infty} \left( \frac{1}{T} \int_{-T/2}^{T/2} x_\mathrm{DC}^2 \mathrm{d}t + \int_{-T/2}^{T/2} x_\mathrm{AC}(t) x_\mathrm{AC}(t+\tau) \mathrm{d}t \right) \\ &= x_\mathrm{DC}^2 + \lim_{T \to \infty} \int_{-T/2}^{T/2} x_\mathrm{AC}(t) x_\mathrm{AC}(t+\tau) \mathrm{d}t. \end{align} $$ In order that the autocorrelation of $x(t)$ is zero for all $\tau \neq 0$, it requires that the autocorrelation of $x_\mathrm{AC}(t)$ is $-x_\mathrm{DC}^2$, i.e. a negative constant. Having a negative constant is not possible, because it means that for any $\tau_1$ and $\tau_2$ different than 0, you have $x_\mathrm{AC}(t+\tau_1) = - x_\mathrm{AC}(t+\tau_2)$, which is not possible (it can be the case for some delays, but not all).
Therefore, if a white noise autocorrelation is zero everywhere except for a zero delay, it necessarily means that the white noise has zero mean.

The zero mean property can also be deduced directly from the PSD. If the mean was nonzero, then there would be a Dirac at 0 Hz in the PSD, which is not the case. Indeed, it is not because the Fourier transform or the PSD has a nonzero value at 0 Hz that it means there is a DC component. For instance, an infinite sequence of rectangular pulses of value +1 or −1 randomly has zero mean, but its PSD is shaped as a sinc² with a nonzero value at 0 Hz. DC component means Dirac at 0 Hz, and vice-versa.

To conclude, when we see sentence as "$x(t)$ is a zero mean additive white Gaussian noise", the zero mean is redundant because it is already implied by the white noise.

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I've found a straightforward prove for my question on this book, in PDF's page 538, or book's page 523 :)

One can observe that "$R_x (\tau)$ approaches of the square of the mean of $x(n)$ as $\tau \rightarrow \infty$". Mathematically, $$ \begin{align} \lim_{\tau \rightarrow \infty} R_x (\tau) = E[x(n)]^2 = m_x^2 \end{align} $$ then $$ \begin{align} m_x = 0. \label{4} \tag{4} \end{align} $$

With it and my question's introduction in mind, it becomes very easy. Substituting the Equation $(4)$ into $(2)$, we have that $$ \begin{align} C_x (\tau) = 0, \text{ for } \tau \neq 0 \label{5} \tag{5} \end{align} $$

Therefore a white noise is always zero mean and uncorrelated.

edit: here, we are considering a WSS (wide-sense stationary) white noise, since a white noise is not necessarily WSS. However, such statement is commonly assumed.

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    $\begingroup$ I am glad that you have found an answer that is acceptable to you, but be aware that there are WSS processes (not necessarily noise processes) for which $\lim_{\tau\to\infty} R_x(\tau)$ does not exist in the sense that $R_X(\tau)$ is not converging to a constant. An example is a WSS process $\{\cos(\omega_0 t+\Theta)\}$ where $\Theta\sim \mathcal U[0,2\pi)$ whose autocorrelation function is $R_X(\tau) = \frac 12\cos(\omega_0\tau)$ which doesn't approach the limit $0$ as $\tau\to\infty$. $\endgroup$ Commented Oct 19, 2022 at 21:49

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