z-transformation
$$z = e^{j\omega T}$$
When we sample a signal, be may get poles repeated up and down the imaginary axis. The z-transformation solves this by wrapping the imaginary axis around the unit circle of the z-plane.
$$ X(z) = \mathcal{Z}{x(t)} = \sum_{n=0}^{\infty} x(n)z^{-n} $$ $X(z)$ converges (is stable) if $|x| < 1$.
Zero points in the center only alter the phase, not the amplitude.
Transfer Function
$$H(z) = \frac{Y(z)}{X(z)}$$ $X(z)$: Input sequence $Y(z)$: Output sequence
We cannot create a z-plane without a sample rate.
Lookup Tables
Standard Transfer Functions
1. Order
$$H(z) = \frac{Y(z)}{X(z)} = \frac{a_{0} + a_{1}z^{-1}}{1+b_{1}z^{-1}} = \frac{a_{0}z + a_{1}}{z + b_{1}}$$
2. Order
$$H(z) = \frac{Y(z)}{X(z)} = \frac{a_{0} + a_{1}z^{-1} + a_{2}z^{-2}}{1+b_{1}z^{-1}+b_{2}z^{-2}} = \frac{a_{0}z^2 + a_{1}z + a_{2}}{z^{2} + b_{1}z + b_{2}}$$
Poles and Stability
Poles within the unit circle -> stable system Poles outside the unit circle -> unstable system
Relation to Laplace Transformation
$$ \begin{cases} \mathcal{L}{x(t)} = \sum_{n=0}^{\infty} x(n)e^{-snT} \s s = \sigma + j\omega \ \mathcal{Z}{x(t)} = \sum_{n=0}^{\infty} x(n)z^{-n} \end{cases} \s\Rightarrow\s s = e^{sT} = e^{\frac{\sigma}{f_{s}}} \cdot e^{2\pi} $$
The Real Axis
The real axis in the $z$-domain is mapped to the z-plane’s real axis from $0$ to $\infty$. The negative part is mapped to the range $[0, 1]$.
Inverse z-transform
This is done with a table lookup. See tables. See slides.
If the expression if not found in the table try partial fractions.
Make sure that the system is stable before converting back
$$ >Y(z) = \frac{z}{(z-0.5)(z-0.25)} >$$
$$ >\frac{Y(z)}{z} = \frac{k_{1}}{z-0.5} + \frac{k_{2}}{z-0.25} >$$ Now we find $k_{1}$: let $z=p_{1} = 0.5$ (pole one). $$\frac{Y(z)}{z} \cdot (z-0.5) = k_{1} + k_{2}\frac{z-0.5}{z-0.25}$$ $$ >\frac{\frac{z}{\cancel{(z-0.5)}(z-0.25)} \cdot \cancel{(z-0.5)}}{z} = \frac{\frac{z}{z-0.25}}{z} = k_{1} >$$ Substitute $z = 0.5$ $$ >k_{1} = \frac{\frac{0.5}{0.5-0.25}}{0.5} = 4 >$$
Impulse Response (note)
For the Unit Sample
The impulse response $h(n)$ for a unit sample sequence can be found by taking the inverse z-transformation of the transfer function.
$$h(n) = \mathcal{Z}^{-1}{H(z)}$$
Stabilitet
See slides.
Stabilt system
Et system er stabilt hvis dets impulsrespons $h(n)$ går mod nul når n går med uendelig. $$\lim_{n\to\infty}\ |h(n)| = 0$$ Her ligger **alle poler inden for enhedscirklen** i z-domænet. $$|p_{i}| < 1, \s \forall i \in \set{1,2,3,\dots,N}$$
Marginalt stabilt system
Et system er marginalt stabilt hvis dets impulsrespons $h(n)$ går mod konstant værdi forskellig fra nul eller oscillerer med konstant amplitude og frekvens når $n$ går mod uendelig.
Mindst én pol skal ligge på enhedcirklen og resten indenfor.
Ustabilt system
Et system er ustabilt hvis dets impulsrespons $h(n)$ vokser ubegrænset når $n$ går mod $\infty$. $$\lim_{n\to\infty}\ |h(n)| = \infty$$ Hvis bare **én pol ligger udenfor enhedscirklen** er systemet ustabilt. $$|p_{i}| > 1, \s \exists i \in \set{1,2,3,\dots, N}$$
Frekvensrespons
See slides. See also grafisk bestemmelse.