Advanced Signals and Systems - State Space Description

27. State-space description.

Let a system be described by the following state equations:

\begin{align} \textbf{x} (n+1) &= \textbf{A} \cdot \textbf{x}(n) + \textbf{B}\cdot \textbf{v}(n) \nonumber \\ \textbf{y} (n) &= \textbf{C} \cdot \textbf{x}(n) + \textbf{D}\cdot \textbf{v}(n) \nonumber \end{align}

where

\begin{align} \textbf{A}=\begin{pmatrix} -4 & -6\\ 3 & 5 \end{pmatrix}, \textbf{ B}=\begin{pmatrix} 1 & 4\\ -1 & -2 \end{pmatrix}, \textbf{ C}=\begin{pmatrix} 3 & 3\\ 2 & 2 \end{pmatrix}, \textbf{ D}=\textbf{ 0} \nonumber \end{align}

1. Draw the signal-flow graph of the system.
2. Determine the transfer matrix $$\textbf{ H}(z)$$.
3. What is the impulse response matrix $$\textbf{ H}_0(n)$$.

Amount and difficulty

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

Solution

1. The signal-flow graph of the system is given in the figure below.

2. The transfer matrix $${\bf H}(z)$$ is defined by

\begin{align*} {\bf H}(z) &= {\bf C}\left[ z{\bf I} - {\bf A}\right]^{-1}{\bf B}+{\bf D}\\ &= \begin{bmatrix} 3 & 3\\ 2 & 2 \end{bmatrix} \left[ \begin{bmatrix} z & 0\\ 0 & z \end{bmatrix} - \begin{bmatrix} -4 & -6\\ 3 & 5 \end{bmatrix} \right] ^{-1} \cdot \begin{bmatrix} 1 & 4\\ -1 & -2 \end{bmatrix} \\ &= \cdots \\ &= \begin{bmatrix} 0 & \frac{6}{z+1}\\ 0 & \frac{4}{z+1} \end{bmatrix} \end{align*}

3. The impulse response matrix $${\bf H}_0(n)$$ can be derived by taking the inverse Z transform of the transfer matrix.

\begin{align*} {\bf H}_0(n) &= \mathcal{Z}^{-1} \left\{ {\bf H}(z) \right\}\\ &= \cdots \\ &= \begin{bmatrix} 0 & 6 \cdot (-1)^{n-1} \cdot \gamma_{-1}(n-1) \\ 0 & 4 \cdot (-1)^{n-1} \cdot \gamma_{-1}(n-1) \end{bmatrix} \end{align*}

28. State-space description.

Consider the following signal-flow graph:

1. Assign (a necessary number of) system states $$x_i(n), i= 1,\cdots N$$, to the signal-flow graph above (more than one correct answer is possible) and determine the matrices of the state-space description (for your choice of system states) below.

\begin{align} \textbf{ x} (n+1) &= \textbf{ A} \cdot \textbf{ x}(n) + \textbf{ B}\cdot \textbf{ v}(n) \nonumber \\ \textbf{ y} (n) &= \textbf{ C} \cdot \textbf{ x}(n) + \textbf{ D}\cdot \textbf{ v}(n) \nonumber \end{align}

2. Determine the transfer function $$\mathrm{H}(z)$$ of the system.

Hint: For the above system, $$\mathrm{H}(z)$$ can be determined without complicated calculations like matrix inversions by simply evaluating the difference equation in time or frequency domain.

3. Determine the poles, zeros and the frequency response of the system for $$a_0 = 1/2$$ and $$a_1=1$$. Express the poles and zeros in two ways, by real and imaginary part and by magnitude and phase.

Assume the following transfer function $$\mathrm{H}(z)$$ for the subparts (d) and (e):

$$H(z) = \frac{1}{2} \cdot \left( 1+ \frac{z+3/2}{z^2+z+1/2} \right) \nonumber \text{ .}\label{eq:}$$

1. Determine the impulse response $$h_0(n)$$ of the transfer function $$\mathrm{H}(z)$$.
2. Show that the impulse response $$h_0(n)$$ is real-valued.

Amount and difficulty

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

Solution

1. The following system states have been assigned to the signal-flow graph:

It follows the state space description:

\begin{align*} {\bf x} (n+1) &= \begin{bmatrix} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 \\ a_1 & 1 & -a_1 & -a_0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} x_1(n) \\ x_2(n) \\x_3(n) \\ x_4(n)\end{bmatrix} + \begin{bmatrix} 1 \\ 0 \\ a_0\\ 0 \end{bmatrix} \cdot v(n) \nonumber \\ {\bf y} (n) &= \begin{bmatrix} a_1 & 1 & -a_1 & -a_0 \end{bmatrix} \cdot {\bf x}(n) + \begin{bmatrix} a_0 \end{bmatrix} \cdot v(n) \nonumber \end{align*}

2. The transfer function can be found by means of the difference equation which is transformed with a z transform.

\begin{align*} Y(z) &= a_0 V(z) + a_1V(z) z^{-1} + V(z) z^{-2} - a_1Y(z) z^{-1} - a_0Y(z) z^{-2} \\ &\Rightarrow H(z) = \frac{Y(z)}{V(z)} = \frac{a_0+a_1z^{-1}+z^{-2}}{1+a_1z^{-1}+a_0z^{-2}} \end{align*}

3. The poles are found by solving

\begin{align*} & z_{\infty}^2+a_1z_{\infty}+a_0 = 0 \\ \Rightarrow z_{\infty,1,2} &= - \frac{a_1}{2} \pm \sqrt{\left( \frac{a_1}{2} \right)^2 - a_0} = \cdots \\ &= - \frac{1}{2} \pm j \frac{1}{2} = \frac{1}{\sqrt{2}} \, e^{\pm j \frac{3}{4} \pi} \end{align*}

and the zeros by

\begin{align*} & z_{0}^2+\frac{a_1}{a_0}z_{0}+\frac{1}{a_0} = 0 \\ \Rightarrow z_{0,1,2} &= - \frac{a_1}{2a_0} \pm \sqrt{\left( \frac{a_1}{2a_0} \right)^2 - \frac{1}{a_0}} = \cdots \\ &= - 1 \pm j = \sqrt{2} \, e^{\pm j \frac{3}{4} \pi} \end{align*}

4. The transfer function $${H}(z)$$ can be separated in two transfer functions, where

\begin{align*} H(z) = \frac{1}{2} \left( H_1(z) + H_2(z) \right) \text{ .} \end{align*}

Transforming these transfer functions separately into the time domain, by means of a partial fraction expansion, leads to the following impulse response:

\begin{align*} H(z) = \frac{1}{2} \left( \gamma_0 (n) + \left(A \cdot z_{\infty,1}^{n-1} + B \cdot z_{\infty,2}^{n-1} \right)\gamma_{-1} (n-1) \right) \text{ } \end{align*}

Where $$A=1/2-j$$, $$B=1/2+j$$, $$z_{\infty,1}=1+j\,1/2$$ and $$z_{\infty,2}=1-j\,1/2$$.

5. All poles and zeros are appearing in complex conjugated pairs and, hence, the system is a real-valued system.

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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 ...