Rational Z-Transform and Transfer Function

Sources

1. Rational Z-Transform and Transfer Function (Lecture Slides & Annotations)
2. Lecture Notes

Example 1 - Poles and Zeros Part 1

Problem 1

Determine the pole-zero plot for the signal

$$x[n]=a^{n}u[n]$$

$$\begin{array}{rl}X(z)& =\frac{1}{1-a{z}^{-1}}\cdot \frac{z}{z}\end{array}$$ $$\boxed{\begin{array}{rl}X(z)& =\frac{z}{z-a}\\ \text{zero: }{z}_1& =0\\ \text{pole: }z-a& =0\\ z& =a\text{, }{P}_1=a\end{array}}$$

Example 2 - Poles and Zeros Part 2

Problem 2

Determine the pole-zero plot for the signal

$$x[n]=\{\begin{array}{ll}{a}^n& 0\le n\le M-1\\ 0& \text{elsewhere}\end{array}$$

$$\begin{array}{rl}X(z)& =\sum _{n=0}^{M-1}{a}^n{z}^{-n}\\ & =\sum _{n=1}^N(a{z}^{-1}{)}^n\\ & =\frac{1-(a{z}^{-1}{)}^M}{1-a{z}^{-1}}\\ & =\frac{1-(a{z}^{-1}{)}^M}{1-a{z}^{-1}}\cdot \frac{z^M}{z^M}\\ & =\frac{z^M-a^M}{z^M-a{z}^{M-1}}=\frac{(z-a)(z-z_1)(z-z_2)\dots (z-z_{M-1})}{(z-a)z^{M-1}}\\ X(z)& =\frac{(z-z_1)(z-z_2)\dots (z-z_{M-1})}{z^{M-1}}\end{array}$$

Without pole-zero cancellation

$$\begin{array}{rlr}{z}^M& =a^M& \text{has M roots}\\ z_k& =a{e}^{j\frac{2\pi k}{M},}& k=0,1,2\dots ,M-1\end{array}$$ $$\begin{array}{rl}\text{if }M& =8\\ z_k& =a{e}^{j\frac{2\pi k}{M},}\\ & =a{e}^{j\frac{2\pi k}{8},}\\ z_k& =a{e}^{j\frac{\pi }{4}k}\end{array}$$

Therefore, each poles and zeros are $45^{\circ }$ or $\frac{\pi }{4}$ away from other poles and zeros

Tip 2. Summation of $A^n$

$$\sum _{n=1}^N{A}^n=\frac{1-A^{N+1}}{1-A}\text{, }|A|<1$$

Example 4 - Transfer Function

Problem 4

Determine the system function & unit sample response based on the difference equation:

$$y[n]=\frac{1}{2}y[n-1]+2x[n]$$

Taking the z-transform of both sides

$$\begin{array}{rl}Y(z)& =\frac{1}{2}{z}^{-1}Y(z)+2X(z)\\ \left(1-\frac{1}{2}{z}^{-1}\right)Y(z)& =2X(z)\\ H(z)& =\frac{Y(z)}{X(z)}\\ & =\frac{2}{1-\frac{1}{2}{z}^{-1}}\end{array}$$

Transfer function:

$$\boxed{H(z)=\frac{2}{1-\frac{1}{2}{z}^{-1}}}$$

Take the inverse z-transform of H(z)

$$\boxed{h[n]=2{\left(\frac{1}{2}\right)}^{n}u[n]}$$

Example 5 - Causal LTI System

Problem 5

If input:

$$x[n]={\left(\frac{1}{2}\right)}^{n}u[n]-\frac{1}{4}{\left(\frac{1}{2}\right)}^{n-1}u[n-1]$$

then output:

$$y[n]={\left(\frac{1}{3}\right)}^{n}u[n]$$

a.) Does $H(z)$ satisfy the foregoing conditions?

b.) What is the difference equation that describe this system?

c.) Is the system stable?

a.) Transfer Function

$$H(z)=\frac{Y(z)}{X(z)}$$ $$\begin{array}{rl}\text{From }y[n]& ={\left(\frac{1}{3}\right)}^{n}u[n],\\ Y(z)& =\frac{1}{1-\frac{1}{3}{z}^{-1}}\end{array}$$ $$\begin{array}{rl}\text{From }x[n]& ={\left(\frac{1}{2}\right)}^{n}u[n]-\frac{1}{4}{\left(\frac{1}{2}\right)}^{n-1}u[n-1]\\ X(z)& =Z\left\{{\left(\frac{1}{2}\right)}^{n}u[n]\right\}-\frac{1}{4}{z}^{-1}Z\left\{{\left(\frac{1}{2}\right)}^{n}u[n]\right\}\\ & =\left(1-\frac{1}{4}{z}^{-1}\right)Z\left\{{\left(\frac{1}{2}\right)}^{n}u[n]\right\}\\ X(z)& =\frac{1-\frac{1}{4}{z}^{-1}}{1-\frac{1}{2}{z}^{-1}}\end{array}$$ $$H(z)=\frac{\frac{1}{1-\frac{1}{3}{z}^{-1}}}{\frac{1-\frac{1}{4}{z}^{-1}}{1-\frac{1}{2}{z}^{-1}}}$$ $$\boxed{H(z)=\frac{1-\frac{1}{2}{z}^{-1}}{\left(1-\frac{1}{3}{z}^{-1}\right)\left(1-\frac{1}{4}{z}^{-1}\right)}\text{, }ROC:|z|>\frac{1}{3}}$$

b.) Difference Equation

$$H(z)=\frac{Y(z)}{X(z)}=\frac{1-\frac{1}{2}{z}^{-1}}{1-\frac{7}{12}{z}^{-1}+\frac{1}{12}{z}^{-2}}$$ $$\left(1-\frac{7}{12}{z}^{-1}+\frac{1}{12}{z}^{-2}\right)Y(z)=\left(1-\frac{1}{2}{z}^{-1}\right)X(z)$$ $$z^{-1}\left\{Y(z)-\frac{7}{12}{z}^{-1}Y(z)+\frac{1}{12}{z}^{-2}Y(z)=X(z)-\frac{1}{2}{z}^{-1}X(z)\right\}{z}^{-1}$$ $$y[n]-\frac{7}{12}y[n-1]+\frac{1}{12}y[n-2]=x[n]-\frac{1}{2}x[n-1]$$ $$\boxed{y[n]=\frac{7}{12}y[n-1]-\frac{1}{12}y[n-2]+x[n]-\frac{1}{2}x[n-1]}$$

c.) Is the system causal and stable

ROC: $|z|>\frac{1}{3}$

The system is causal and stable