# Advanced Signals and Systems - Random Signals and Systems

### 22. Spectra of random signals.

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

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

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### Recent Publications

J. Reermann, P. Durdaut, S. Salzer, T. Demming, A. Piorra, E. Quandt, N. Frey, M. Höft, and G. Schmidt: Evaluation of Magnetoelectric Sensor Systems for Cardiological Applications, Measurement (Elsevier), ISSN 0263-2241, https://doi.org/­10.1016/­j.measurement.2017.09.047, 2017

S. Graf, T. Herbig, M. Buck, G. Schmidt: Low-Complexity Pitch Estimation Based on Phase Differences Between Low-Resolution Spectra, Proc. Interspeech, pp. 2316 -2320, 2017

### Contact

Prof. Dr.-Ing. Gerhard Schmidt

E-Mail: gus@tf.uni-kiel.de

Christian-Albrechts-Universität zu Kiel
Faculty of Engineering
Institute for Electrical Engineering and Information Engineering
Digital Signal Processing and System Theory

Kaiserstr. 2
24143 Kiel, Germany

## Recent News

Jens Reermann Defended his Dissertation with Distinction

On Friday, 21st of June, Jens Reermann defended his research on signals processing for magnetoelectric sensor systems very successfully. After 90 minutes of talk and question time he finished his PhD with distinction. Congratulations, Jens, from the entire DSS team.

Jens worked for about three and a half years - as part of the collaborative research center (SFB) 1261 - on all kinds of signal ...