# Advanced Signals and Systems - Deterministic Systems

## Task

Let a discrete system be described by $$y(n) = a \cdot v(n-1)+b \cdot v^2(n)$$. Is the system linear, shift-invariant, causal, dynamic, stable?

## Amount and difficulty

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

## Solution

A one-dimensional, discrete system $$S$$ is given and is described by

\begin{equation*} S\{v(n)\}=y(n)=a\cdot v(n-1)+b\cdot v^2(n) \text{ .} \end{equation*}

• Linearity:

The system is linear if the following relation is fulfilled:

\begin{align*} \alpha_1 \: y_1(n)+\alpha_2 \: y_2(n) &= \alpha_1 \: S\{v_1(n)\}+\alpha_2 \: S\{v_2(n)\}\\ &= S\{\alpha_1 \: v_1(n)+\alpha_2 \: v_2(n)\} \end{align*}

$$\longrightarrow$$ The given system is not linear

• Shift-invariance:

The system is shift-invariant if

\begin{equation*} y(n-k) = S\{v(n-k)\} \end{equation*}

holds true.

$$\longrightarrow$$ The given system is shift-invariant.

• Causality:

The system is causal if

\begin{equation*} y(n_0) = S\{v(n)\} |_{n\leq n_0} \end{equation*}

holds true.

$$\longrightarrow$$ The given system is causal.

• Dynamic:

A system is dynamic if $$y(n_0)$$ does not only depend on $$v(n_0)$$, but also on $$v(n\neq n_0)$$.

$$\longrightarrow$$ The given system is dynamic.

• Stability:

A system is said to be BIBO-stable if

\begin{align*} |v(n)| \leq M_1 < \infty, \ \ \ \forall \ n \\ |y(n)| \leq M_2 < \infty, \ \ \ \forall \ n \end{align*}

$$\longrightarrow$$ The given system is BIBO-stable.

## Task

Determine the impulse response, step response, and transfer function of the following systems:

1. $$y(n) = \sum \limits_{k=-\infty}^{n}v(k)$$
2. $$y(n) = \frac{1}{m} \sum \limits_{k=0}^{m-1}v(n-k)$$

## Amount and difficulty

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

## Solution

The general affiliations between the different functions are defined in chapter VI of the lecture.

• Transfer function:

The transfer function can be found by applying a z transform and the following equation.

\begin{equation*} H(z) = \frac{Y(z)}{V(z)} = \frac{z}{z-1} \end{equation*}

• Impulse response:

\begin{equation*} H(z) \ \ \ \bullet-\circ \ \ \ h_0(n) = \gamma_{-1} (n) \end{equation*}

• Step response:

\begin{equation*} h_{-1}(n) = \sum \limits_{k=-\infty}^{n} h_0(k) = \begin{cases} 0 , &n<0\\ n+1, &n\geq 0 \end{cases}\end{equation*}

• Impulse response:

\begin{equation*} h_0(n) = \frac{1}{m} \sum \limits_{k=0}^{m-1} \gamma_{0} (n) = \frac{1}{m} \left( \gamma_{-1}(n) - \gamma_{-1}(n-m) \right)\end{equation*}

• Step response:

\begin{equation*} h_{-1}(n) = \sum \limits_{k=-\infty}^{n} h_0(k) = \begin{cases} 0 , &n<0\\ \frac{n+1}{m}, & 0\leq n < m-1\\ 1 , & n\geq m-1\end{cases}\end{equation*}

• Transfer function:

\begin{align*} h_0(n) \ \ \ \bullet-\circ \ \ \ H(z) &= \frac{1}{m} \left[\frac{z}{z-1}-\frac{z}{z-1} z^{-m}\right] \\ &= \frac{1}{m} \left[ \frac{z-z^{-(m-1)}}{z-1} \right] \end{align*}

## Task

When the input to a LTI system is

$$v(n) = 5 \cdot \gamma_{-1}(n) \nonumber$$

the output is

$$y(n) = (2\cdot 0.5^{n} + 3 \cdot (-0.75)^{n}) \cdot \gamma_{-1}(n) \text{ .} \nonumber$$

1. Find the transfer function $$H(z)$$ of the system. Plot the zeros and the poles of $$H(z)$$ and indicate the region of convergence. Is the system stable?
2. Find the impulse response $$h_0(n)$$ of the system for all values of $$n$$.
3. Define the difference equation that characterizes the system.

## Amount and difficulty

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

## Solution

1. The transfer function is defined by

\begin{equation*}H(z) = \frac{Y(z)}{V(z)} \end{equation*}

Transforming the input and output sequences to the z domain and inserting this to the given equation yield to the desired transfer function

\begin{equation*}H(z) = \frac{z(z-1)}{(z-0,5)(z+0,75)} \ \ \ |z|>0,75 \text{.} \end{equation*}

The Figure shows the zeros and poles of the system. As the poles lie within the unit circle the system is stable.

2. Using the partial fraction expantion for $$\frac{H(z)}{z}$$ and transforming this back to time domain the desired impulse response can befound.

\begin{equation*}h_0(n) = -0,4\cdot (0,5)^n\cdot \gamma_{-1}(n) + 1,4 \cdot (-0,75)^n \cdot \gamma_{-1}(n)\text{.}\end{equation*}

3. The difference equation can be found by rearranging the transfer function and applying an inverse z transform.

\begin{align*} \frac{Y(z)}{V(z)} &= \frac{1-z^{-1}}{1+\frac{1}{4}z^{-1}-\frac{3}{8}z^{-2}} \\ \rightarrow &y(n) = v(n) -v(n-1)-\frac{1}{4}y(n-1)+\frac{3}{8}y(n-2) \end{align*}

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### Recent Publications

T. O. Wisch, T. Kaak, A. Namenas, G. Schmidt: Spracherkennung in stark gestörten Unterwasserumgebungen, Proc. DAGA 2018

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

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New PhDs in the DSS Team

Since January this year we have two new PhD students in the team: Elke Warmerdam and Finn Spitz.

Elke is from Amsterdam and she works in the neurology department in the university hospital in the group of Prof. Maetzler. Her research topic is movement analysis of patients with neurologic disorders. Elke cooperates with us in signal processing related aspects of her research. Elke plays ...