Exercise 4.12Z: White Gaussian Noise

PSD of white noise

A noise signal $n(t)$ is called "white" if it contains all spectral components without preference of any frequencies.

The physical power-spectral density defined only for positive frequencies ⇒ ${\it \Phi}_{n+}(f)$ is constant $($equal $N_0)$ and extends frequency-wise to infinity.
${\it \Phi}_{n+}(f)$ is shown in green in the upper graph. The plus sign in the index is to indicate that the function is valid only for positive values of $f$.
For mathematical description one usually uses the two-sided power-spectral density ${\it \Phi}_{n}(f)$. Here applies for all frequencies from $-\infty$ to $+\infty$ (blue curve in the upper graph):

$${\it \Phi}_n (f) ={N_0}/{2}.$$

The bottom graph shows the two power-spectral densities ${\it \Phi}_{b}(f)$ and ${\it \Phi}_{b+}(f)$ of bandlimited white noise signal $b(t)$. It holds with the one-sided bandwidth $B$:

$${\it \Phi}_b(f)=\left\{ {N_0/2\atop 0}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad |f|\le B \atop {\rm else}}\right.,$$

$${\it \Phi}_{b+}(f)=\left\{ {N_0\atop 0}{\hspace{0.5cm} {\rm f\ddot{u}r}\quad 0 \le f\le B \atop {\rm else}}\right.$$

For computer simulation of noise processes, band-limited noise must always be assumed, since only discrete-time processes can be handled. For this, the Sampling Theorem must be obeyed. This states that the bandwidth $B$ must be set according to the interpolation distance $T_{\rm A}$ of the simulation.

Assume the following numerical values throughout this exercise:

The noise power density – with respect to the resistor $1 \hspace{0.05cm}\rm \Omega$ – is $N_0 = 4 \cdot 10^{-14}\hspace{0.05cm}\rm V^2/Hz$.
The (one-sided) bandwidth of the band-limited white noise is $B = 100 \hspace{0.08cm}\rm MHz$.

Hints:

This exercise belongs to the chapter Power-Spectral Density.
Reference is also made to the chapter Auto-Correlation Function.
The properties of white noise are summarized in the second part of the (German language) learning video The AWGN channel.

Questions

	The ACF $\varphi_n(t)$ has a sinc-shaped progression.
	The ACF $\varphi_n(\tau)$ is a Dirac delta function at $\tau = 0$ with weight $N_0/2$.
	In practice, there is no (exact) white noise.
	Thermal noise can always be approximated as white.
	White noise is always Gaussian distributed.

$\varphi_b(\tau = 0) \ = \ $

$\ \cdot 10^{-6} \ \rm V^2$

$\sigma_b \ = \ $

$\ \rm mV$

$T_{\rm A} \ = \ $

$\ \rm ns$

	The samples are uncorrelated.
	The samples are positively correlated.
	The samples are negatively correlated.

Solution

(1) Correct are the solutions 2, 3, and 4:

The auto-correlation function $\rm (ACF)$ is the Fourier transform of the power-spectral density $\rm (PSD)$. Here:

$${\it \Phi}_n (f) = {N_0}/{2} \hspace{0.3cm} \bullet\!\!-\!\!\!-\!\!\!-\!\!\circ\, \hspace{0.3cm} \varphi_n (\tau)={N_0}/{2} \cdot {\rm \delta} ( \tau).$$

However, there is no "real" white noise in physics, since such a noise would have to have an infinitely large signal power $($the integral over the PSD as well as the ACF value at $\tau = 0$ are both infinitely large$)$.
Thermal noise has a constant PSD up to frequencies of about $\text{6000 GHz}$. Since all (current) transmission systems operate in a much lower frequency range, thermal noise can be said to be "white" to a good approximation.
The statistical property "white" says nothing about the amplitude distribution, which is determined by the probability density function $\rm (PDF)$ alone.
When considering the phase of a bandpass signal as the stochastic variable, it is often modeled as uniformly distributed between $0$ and $2\pi$.
If there are no statistical bindings between the respective phase angles at different times, this random process is also "white".

ACF of band-limited noise

(2) The power-spectral density is a rectangle of width $2B$ and height $N_0/2$.

The inverse Fourier transform yields an sinc–function:

$$\varphi_b(\tau) = N_0 \cdot B \cdot {\rm si} (2 \pi B \tau)\hspace{0.3cm} \Rightarrow \hspace{0.3cm}\varphi_b(\tau = 0) = N_0 \cdot B \hspace{0.15cm}\underline {=4}\cdot 10^{-6} \ \rm V^2.$$

(3) The ACF value at the point $\tau = 0$ gives the power.

The root of this is called the "rms value":

$$\sigma_b = \sqrt{\varphi_b(\tau = 0)} \hspace{0.15cm}\underline {=2 \hspace{0.05cm}\rm V}.$$

(4) The ACF computed in (3) has zeros at equidistant distance from $T_{\rm A}= 1/(2B)\hspace{0.15cm}\underline {=5\hspace{0.05cm} \rm ns}$:

There are no statistical bindings between the two signal values $b(t)$ and $b(t + \nu \cdot T_{\rm A})$,
where $\nu$ can take all integer values.

(5) The correct solution is the suggested solution 2.

The ACF value at $\tau = T_{\rm A} = 1 \hspace{0.05cm}\rm ns$ is

$$\varphi_b(\tau = T_{\rm A}) = {\rm 4 \cdot 10^{-6} \hspace{0.1cm}V^2 \cdot sinc (1/5) \approx 3.742 \cdot 10^{-6} \hspace{0.1cm}V^2} > 0.$$

This result says: Two signal values separated by $T_{\rm A} = 1 \hspace{0.05cm}\rm ns$ are positively correlated:

If $b(t)$ is positive and large, then with high probability $b(t+1 \hspace{0.05cm}\rm ns)$ is also positive and large.
In contrast, there is a negative correlation between $b(t)$ and $b(t+7 \hspace{0.05cm}\rm ns)$. If $b(t)$ is positive, then $b(t+7 \hspace{0.05cm}\rm ns)$ is probably negative.