# Advanced Signals and Systems - Deterministic Systems

### 19. System Classification.

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.

### 20. Impulse Response, Transfer Function.

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*}

### 21. Linear, Time-Invariant System.

When the input to a LTI system is

\begin{equation} v(n) = 5 \cdot \gamma_{-1}(n) \nonumber \end{equation}

the output is

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

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*}