Advanced Signals and Systems - Deterministic Systems

 

19. System Classification.

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.

 

20. Impulse Response, Transfer Function.

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*}

 

21. Linear, Time-Invariant System.

Task

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*}