Advanced Signals and Systems - Stochastic Processes I

 

3. LAPLACE's distribution

Task

Let \(v\) be a random variable with the probability-density function \(f_{v}(V)=\frac{\alpha}{2}e^{-\alpha |V|}\) (density of LAPLACE's distribution).

  1. Determine the probability function \(F_v(V)\).
  2. Determine the mean value and the variance of \(v\).
  3. Determine the characteristic function of LAPLACE's distribution.
  4. Verify the results of (2) with the characteristic function.

Amount and difficulty

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

Solution

  1. In general: \begin{equation*} F_v(V) = \int \limits_{x=-\infty}^V f_v(x) \, dx \end{equation*} For solving the integral a case differentiation for \(V<0\) and \(V\leq0\) is done and the following result can be derived: \begin{equation*} F_v(V) = \frac{1}{2} \cdot e^{-\alpha |V|} + \left[ 1 - e^{-\alpha |V|}\right] \delta_{-1}(V)\end{equation*}
  2. Determining the mean in general: \begin{equation*} \text{E}\left\{ v\right\} = \int \limits_{V=-\infty}^\infty V f_v(V) \, dV = \mu_v \end{equation*} Here we can conclude \begin{equation*} \text{E}\left\{ v\right\} = \int \limits_{V=-\infty}^\infty \underbrace{\underbrace{V}_{\text{odd}} \cdot \underbrace{\frac{\alpha}{2} \cdot e^{-\alpha |V|}}_{\text{even}}}_{\text{odd}} \, dV = 0 \end{equation*} because the integral of an odd function equals to zero.

    Determining the variance in general: \begin{equation*} \sigma_v^2 = \text{E}\{v^2\} - {\text{E}\left\{v\right\}}^2 \end{equation*} Here we can conclude \begin{align*} \sigma_v^2 &= \text{E}\{v^2\} - \underbrace{\text{E}\{v\}^2}_{=0} = \text{E}\{v^2\}\\ &= \int \limits_{-\infty}^\infty V^2 f_v(V) \, dV = \int \limits_{-\infty}^\infty \underbrace{V^2 \cdot \frac{\alpha}{2} e^{-\alpha |V|}}_{\text{even}} \, dV\\ &= \alpha \int \limits_{0}^\infty V^2 e^{-\alpha V} dV \end{align*} After substitution with \(t=\alpha \cdot V\) and applying the gamma function \(\Gamma(x) = \int \limits_0^\infty t^{x-1} e^{-t} dt\) the variance is defined by: \begin{equation*} \sigma_v^2 = \frac{1}{\alpha^2} \cdot \Gamma(3) = \frac{2}{\alpha^2} \end{equation*}
  3. \begin{align*} C(\omega) &= \text{E} \{ e^{j\omega V} \} = \int \limits_{V=-\infty}^\infty e^{j\omega V} f_v(V) \, dV \\ &= \frac{\alpha^2}{\omega^2 + \alpha^2} \end{align*}
  4. Moment theorem: \begin{align*} \text{E} \left\{x^k \right\} &= \left. \frac{1}{j^k} \frac{d^k C(\omega)}{d \omega ^k} \right| _{\omega =0}\\ \Rightarrow \mu_v &= \text{E} \{x \} = \left. \frac{1}{j} \frac{d C(\omega)}{d \omega }\right| _{\omega =0} = \cdots = 0 \\ \Rightarrow \sigma_v^2 &= \text{E} \{x ^2\} = \left. \frac{1}{j^2} \frac{d^2 C(\omega)}{d \omega ^2}\right| _{\omega =0} = \cdots = \frac{2}{\alpha^2} \end{align*}The same results as in (b) can be achieved.

Additional material

Laplace distribution:

Example speech signal:

4. Properties of ACF/CCF.

Task

  1. Show that for the CCF of the complex random signals \(v_1(n)\) and \(v_2(n)\) holds: \begin{equation*}s_{v_1v_2}(\kappa) = s^\ast_{v_2v_1}(-\kappa) \end{equation*} What does this imply for the CCF of real signals \(v_1(n)\) and \(v_2(n)\)?
    How can this relationship be used to determine the ACF of a real signal?
  2. Show that the CAUCHY-SCHWARZ inequality for real random variables holds: \begin{equation*} \left[E\{xy\}\right]^2 \leq E\{x^2\}\cdot E\{y^2\} \end{equation*} What does this imply for the ACF of a real signal?
  3. How can the above results for discrete signals be applied to the CCF and ACF of continuous signals?

Amount and difficulty

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

Solution

  1. Stationarity has to be assumed. According to the lecture, the cross-correlation is defined by \begin{equation*} s_{v_1v_2}(\kappa) = \text{E} \left\{ v_1^*(n) \cdot v_2(n+\kappa) \right\} \end{equation*} Further it can be shown: \begin{align*}s_{v_2v_1}^*(-\kappa) &= \left[ \text{E}\left\{ v_2^*(n) \cdot v_1(n-\kappa) \right\} \right]^*\\ &= \left[ \text{E}\left\{ v_2^*(n+\kappa) \cdot v_1(n) \right\} \right]^*\\ &= \text{E}\left\{ v_2(n+\kappa) \cdot v_1^*(n) \right\}\\ &= s_{v_1v_2}(\kappa)\end{align*} Special case: \(v_1(n), v_2(n) \in \mathbb{R}\) \begin{equation*}s_{v_1v_2}(\kappa) = \text{E} \{ v_1(n) \cdot v_2(n+\kappa)\} = \cdots = s_{v_2v_1}(-\kappa);\end{equation*} Special case ACF: \(v_1(n) = v_2(n) =v(n)\) \begin{align*} v(n) \in \mathbb{C} &\rightarrow s_{vv}(\kappa) = s_{vv}^* (-\kappa)\\ v(n) \in \mathbb{R} &\rightarrow s_{vv}(\kappa) = s_{vv} (-\kappa) \end{align*}
  2. Let \(x,y\) be real random variables and let \(a \in \mathbb{R}\). \begin{align*} \text{E} \{ (ax-y)^2 \} \geq 0 \\ a^2 \, \text{E} \{ x^2 \} +2a\,\text{E} \{ xy \} +\text{E} \{ y^2\}\geq 0 \end{align*} By solving this equation and inserting \(a=\frac{\text{E} \{ xy\}}{\text{E} \{ x^2\}}\) it follows that \begin{equation*}\text{E}^2 \{ xy \} \leq \text{E}\{x^2\} \cdot \text{E}\{y^2\}\end{equation*}
  3. The above results hold true also for continuous signals.

Ergodicity and stationarity.

Additional material

Example 1:

Stationary?

Ergodic?

Example 2:

Stationary?

Ergodic?

Example 3:

Stationary?

Ergodic?

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Recent Publications

P. Durdaut, J. Reermann, S. Zabel, Ch. Kirchhof, E. Quandt, F. Faupel, G. Schmidt, R. Knöchel, and M. Höft: Modeling and Analysis of Noise Sources for Thin-Film Magnetoelectric Sensors Based on the Delta-E Effect, IEEE Transactions on Instrumentation and Measurement, published online, 2017

P. Durdaut, S. Salzer, J. Reermann, V. Röbisch, J. McCord, D. Meyners, E. Quandt, G. Schmidt, R. Knöchel, and M. Höft: Improved Magnetic Frequency Conversion Approach for Magnetoelectric Sensors, IEEE Sensors Letters, published online, 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

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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 ...


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