# Advanced Signals and Systems - Z Transform / Pole-zero Plot

### 12. Symmetry relations of Z-transform.

Given the sequence $$v(n)$$ and its Z-transform $$V(z)$$, determine the Z-transform of the following sequences:

1. $$v_1(n) = v(-n)$$
2. $$v_2(n) = v^*(n)$$
3. $$v_3(n) = v(n-k)$$

## Amount and difficulty

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

## Solution

1. $$v_1(n) = v(-n)$$ \begin{equation*} V_1(z) = \sum \limits_{n=-\infty}^\infty v_1(n) z^{-n} = \cdots = \sum \limits_{n=-\infty}^\infty v(n) \left( z^{-1} \right)^{-n} = V(z^{-1}) \end{equation*}

2. $$v_2(n) = v^*(n)$$ \begin{equation*} V_2(z) = \sum \limits_{n=-\infty}^\infty v(n)^* z^{-n} = \cdots = V^*(z^{*}) \end{equation*}

3. $$v_3(n) = v(n-k)$$ \begin{equation*} V_3(z) = \sum \limits_{n=-\infty}^\infty v(n-k)^* z^{-n} = \cdots = z^{-k} \cdot V^*(z^{*}) \end{equation*}

### 13. Z and inverse Z-transform.

Given the sequence $$v(n)$$ and its Z-transform $$V(z)$$, determine the Z-transform of the following sequences:

1. Find the z transform of the sequence $$v(n)$$ $$v(n) = \begin{cases} 1 & ,|n|\leq N\\ 0 & , \text{else} \;\;\;\;\; \text{ .} \nonumber \end{cases} \nonumber$$
2. Given the z transform $$Y(z)$$ of a sequence $$y(n)$$ $$Y(z) = \frac{23z^3 - 34z^2 - 28z + 56}{z^5 - 5z^4 + 6z^3 + 4z^2 - 8z} \text{ ,} \nonumber$$ find $$y(n)$$ using partial fraction expansion.

## Amount and difficulty

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

## Solution

1. The Z-transform can be derived by applying the definition and using a geometric series. \begin{align*} V(z) &= \sum \limits_{n=-\infty}^\infty v(n) z^{-n} = \sum \limits_{n=0}^N z^{n} + \sum \limits_{n=0}^N z^{-n} - 1 = \cdots \\ &= \frac{z^{-N}-z^{N+1}}{1-z} \end{align*}

2. Any ratio of polynomials $$V(z) = \frac{V_1(z)}{V_2(z)}$$ can be expressed by partial fraction expansion \begin{equation*} V(z) = \sum \limits_{\nu=1}^{N_0}\sum \limits_{\mu=1}^{N_\nu} B_{\nu\mu} \frac{1}{(z-z_{\infty\nu})^\mu} \end{equation*} where

• $$N_0=$$ number of distinct pole
• $$N_\nu=$$ order of the $$\nu$$th pole
• $$\nu=$$ index of the pole
• $$\mu=$$ index within a value $$\leq N_\nu$$

But the inverse transform of $$\frac{1}{(z-z_{\infty\nu})^\mu}$$ is not listed in transform tables, therefore, the addend of the partial fraction is extended by $$\frac{1}{z}$$. Hence, the partial fraction is done with $$\frac{Y(z)}{z}$$ where the following poles and coefficients can be found: \begin{align*} z_{\infty 0} &= 0 & z_{\infty 1} &= 0 & z_{\infty 2} &= -1 & z_{\infty 3} &= 2 & z_{\infty 4} &= 2 & z_{\infty 5} &= 2 \\ B_{01} &= 0 & B_{02} &= -7 & B_{11} &= -1 & B_{21} &= 1 & B_{22} &= 4 & B_{23} &= 4 \end{align*} Transforming $$Y(z)$$ by using a transformation table leads to \begin{equation*} y(n) = -7\gamma_0(n-1) + \left[ (-1)^{n+1} +2^{n-1}(n^2+3n+2) \right] \gamma_{-1}(n) \end{equation*}

### 14. Pole-zero plot.

Given is the following pole-zero plot, belonging to the z transform $$X(z)$$ of a causal sequence $$x(n)$$:

In the following it is assumed that $$y(n) = \left( \frac{1}{2} \right)^n \cdot x(n) \nonumber$$

1. Define the poles and the zeros of $$Y(z)$$.
2. Sketch the pole-zero plot of $$Y(z)$$ and the region of convergence (ROC).

## Amount and difficulty

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

## Solution

1. The following z transform can be extracted from the pole-zero plot: \begin{equation*} X(z) = K \cdot \frac{(z+j)(z-j)}{z-\frac{1}{2}} = K \cdot \frac{z^2+1}{z- \frac{1}{2}} \end{equation*} Additionally we do know that \begin{equation*} z_0^n f(n) \ \ \ \circ-\bullet \ \ \ F \left(\frac{z}{z_0}\right) \end{equation*} so it follows \begin{equation*} Y(z) = K \cdot \frac{(2z)^2+1}{2z- \frac{1}{2}} \text{ .} \end{equation*} Hence the poles and zeros are \begin{equation*} z_{\infty 1} = \frac{1}{4} \text{ and } z_{0 1} = \frac{j}{2}, z_{0 2} = - \frac{j}{2} \text{ .} \end{equation*}

2. As we have a causal sequence $$|z|>\frac{1}{4}$$ has to hold true for the region of convergence.