# Advanced Signals and Systems - Random Signals and Systems

### 22. Spectra of random signals.

Given the auto-correlation function (ACF) of a discrete random process \begin{equation}s_{vv}(\kappa) = \frac{\Omega_g}{2\pi} \left( \frac{\sin \left( \frac{\Omega_g}{2} \kappa \right)} {\frac{\Omega_g}{2}\kappa} \right)^2 \nonumber \end{equation}

1. Give the mean-square value $$\mu_v^{(2)}$$ and the linear mean value $$|\mu_v|^2$$ as well as the variance $$\sigma_v^2$$ of this process by means of $$s_{vv}(\kappa)$$.
2. In our case the process is not globally uncorrelated: Are there anyway values of $$\kappa$$ for which uncorrelation exists? If so, which are the values?
3. Determine and sketch the power density spectrum (PDS) $$S_{vv}(e^{j \Omega})$$ of the given process. Label your axes.
4. Add to your sketch of part (c) (e.g. with dashed lines) the PDS $$S_{uu}(e^{j \Omega})$$ of a globally, uncorrelated, zero-mean process $$u(n)$$. How do you term such a signal?
5. The signal $$v(n)$$ is filtered (see below) using an LTI system with the frequency response $$H(e^{j \Omega})$$. Determine $$|H(e^{j \Omega})|$$ in the frequency range $$\Omega \in (-\Omega_g, \Omega_g), \Omega_g \leq \pi$$, so that $$y(n)$$ becomes a "'band-limited white"' signal. ## Amount and difficulty

• Working time: approx. xx minutes
• Difficulty: xx

## Solution

1. From the lecture we know

\begin{equation*} s_{vv}(0) = \mu_v^{(2)} = |\mu_v|^2 + \sigma_v^2 \end{equation*}

hence,

\begin{align*} \mu_v^{(2)} &= \frac{\Omega_g}{2\pi}\\ |\mu_v|^2 &= \lim \limits_{\kappa\rightarrow\infty} s_{vv}(\kappa) = 0\\ \sigma_v^2 &= \frac{\Omega_g}{2\pi} \end{align*}

2. A process is regarded as uncorrelated if

\begin{equation*} \Psi_{vv}(\kappa) = s_{vv}(\kappa) -|\mu_v|^2 = 0 \text{ .} \end{equation*}

This holds true in this particular case for

\begin{equation*} \kappa_i = i \cdot \frac{2\pi}{\Omega_g}, \ \ i \in \mathbb{Z} \backslash \{0\} \text{ .} \end{equation*}

3. See solution for (d) for the joint solutions of (c) and (d)!
4. The PDS can be derived by the inverse transform Fourier transform of the auto correlation sequence $$s_{vv}(\kappa)$$ and is

\begin{equation*} S_{vv}(e^{j\Omega}) = \frac{\Omega_g}{2\pi}\cdot \frac{1}{2\pi} \left( R(e^{j\Omega}) * R(e^{j\Omega}) \right)\text{ ,} \end{equation*}

where $$R(e^{j\Omega})$$ corresponds to a rectangular function in the frequency domain. The signal from question (d) is denoted as "white noise".

5. The PDS at the output of the filter is defined by

\begin{equation*} S_{yy}(e^{j\Omega}) = S_{vv}(e^{j\Omega})\cdot |H(e^{j\Omega})|^2 \text{ .} \end{equation*}

To achieve a constant PDS at the output the frequency response becomes equal to

\begin{equation*} |H(e^{j\Omega})| = \frac{1}{\sqrt{S_{vv}(e^{j\Omega})}} \text{ .} \end{equation*}

### 23. Spectra of random signals.

Let $$\tilde{v}(n)$$ be the sum of a real, stationary random signal $$v(n)$$ and a constant signal $$C$$. The sum signal $$\tilde{v}(n)$$ is input signal to the System $$S$$ below. The probability-density function (pdf) of $$v(n)$$ is given by $$f_v(V) = B \cdot e^{-bV} \cdot \delta_{-1}(V), b>0$$. 1. Determine the constant B and sketch $$f_v(V)$$.
2. Find the mean value $$\mu_v$$ and the variance $$\sigma_v^2$$ of $$v(n)$$.
3. Assume now $$C = -\mu_v$$. Sketch the resulting pdf $$f_{\tilde{v}}(V)$$, and give the mean value $$\mu_{\tilde{v}}$$ and the variance $$\sigma_{\tilde{v}}^2$$.
4. Let the auto-correlation sequence $$s_{\tilde{v}\tilde{v}}(\kappa)$$ be equal to zero for $$\kappa \neq -1,0,1$$ and $$s_{\tilde{v}\tilde{v}}(1)=1$$. Determine the unknown values $$s_{\tilde{v}\tilde{v}}(0)$$ and $$s_{\tilde{v}\tilde{v}}(-1)$$.
5. Find the power-density spectrum $$S_{\tilde{v}\tilde{v}} (e^{j \Omega})$$ and sketch it for $$b=\sqrt{1/2}$$.
6. Determine the transfer function $$H(z)$$ and the magnitude spectrum $$|H (e^{j \Omega})|$$ of the filter ($$a \in \mathbb{R}$$).
7. Choose the constants $$a$$ and $$b$$ such that the output of the filter is a white random signal with variance $$\sigma_y^2=2$$.
8. Find for this case the auto-correlation sequence $$s_{yy}(\kappa)$$ and the power-density spectrum $$S_{yy} (e^{j \Omega})$$ and sketch both.

## Amount and difficulty

• Working time: approx. xx minutes
• Difficulty: xx

## Solution

1. We know that

\begin{equation*}\int \limits_{V=-\infty}^{\infty} f_v(V) dV = 1\text{ ,}\end{equation*}

hence $$B = b$$. 2. We know that

\begin{align*} \mu_v &= \int \limits_{V=-\infty}^{\infty} V \cdot f_v(V) dV = \cdots = \frac{1}{b}\\ \sigma_v^2 &= \int \limits_{V=-\infty}^{\infty} V^2 \cdot f_v(V) dV = \cdots = \frac{1}{b^2} \end{align*}

3. The mean value and the variance of a sum process can be defined by

\begin{align*} \mu_{\tilde{v}} &= \mu_{v} + \mu_c = \cdots = 0 \\ \sigma_{\tilde{v}}^2 &= \sigma_{v}^2 + \sigma_{\tilde{v}}^2 = \cdots = \sigma_{v}^2 \end{align*}

In addition the pdf can be derived by

\begin{equation*} f_{\tilde{v}}(V) = f_v(V) * f_c(V) = B \cdot e^{-b(V+\mu_v)} \cdot \delta_{-1}(V+\mu_v), \ \ b>0 \end{equation*} 4. The auto correlation sequence is an even function, hence,

\begin{equation*} s_{\tilde{v}\tilde{v}}(-1) = s_{\tilde{v}\tilde{v}}(1) = 1 \text{ .} \end{equation*}

From part (c) it follows

\begin{equation*} s_{\tilde{v}\tilde{v}}(0) = \sigma_{\tilde{v}}^2 + \mu_{\tilde{v}}^2 = \frac{1}{b^2} \text{ .} \end{equation*}

5. By transforming the auto correlation sequence into the frequency domain

\begin{equation*} S_{\tilde{v}\tilde{v}}(e^{j\Omega}) \sum \limits_{k=-\infty}^{\infty} s_{\tilde{v}\tilde{v}}(k) =\cdots= \frac{1}{b^2}+2 \cos(\Omega)\end{equation*}

For $$b=\sqrt{1/2}$$ the following sketch can be concluded. 6. \begin{align*} H(z) &= \frac{Y(z)}{\tilde{V}(z)} = \frac{z}{z-a}\\ |H(e^{j\Omega})| &= \frac{1}{|1-a e^{-j\Omega}|} = \cdots = \frac{1}{\sqrt{1 + a^2 - 2a \cos(\Omega)}} \end{align*}
7. The following equation has to hold true:

\begin{equation*} S_{yy}(e^{j\Omega}) = |H(e^{j\Omega})|^2 \cdot S_{\tilde{v}\tilde{v}}(e^{j\Omega}) = \sigma_y^2 \end{equation*}

From the problems before the nagnitude spectrum and power-density spectrum of $${y}(n)$$ is known. Hence,

\begin{equation*} a = -\frac{1}{2} \ \ \ \ \ \text{ and } \ \ \ \ \ b = \frac{1}{\sqrt{5/2}} \text{ .} \end{equation*}

8. The output $${y}(n)$$ is a white random signal. So the auto-correlation sequence and power-density spectrum is given by 