Principles of Automatic Control

This is where I keep the notes related to the principles of automatic control.

S-Z transform table (Table of s- and z-transforms)

	$f(t)$	$F(s)$	$f(kT)$ or $f(k)$	$F(z)$	name
1	$\delta (t)$	$1$	$\delta(k)$	1	unit impulse
2	$\delta (t-t_d)$	$e^{-t_ds}$	$\delta (k-n)$	$z^{-n}$	delayed unit impulse
3	$u(t) = 1$	$1 \over s$	$u(k) = 1$	${z} \over {z-1}$	unit step
4	$t$	$1 \over {s^2}$	$kT$	${zT} \over {(z-1)^2}$	ramp
5	$t^2$	$2\over{s^3}$	$(kT)^2$	${z(z+1)T^2} \over {(z-1)^3}$	parabola
6			$a^k$	$z \over {z-a}$	geometric sequence
7	$e^{-at}$	${1} \over {s+a}$	$e^{-akT}$	${z} \over {z-e^{-aT}}$	natural exponential
8	$1-e^{-at}$	${1} \over {s(s+a)}$	$1-e^{-akT}$	${(1-e^{-aT})z} \over {(z-1)(z-e^{-aT})}$	unit step exponential
9	$te^{-at}$	${1} \over {(s+a)^2}$	$kTe^{-akT}$	${Tze^{-aT}} \over {(z-e^{-aT})^2}$
10			$ka^{k-1}$	${z} \over {(z-a)^2}$
11			$a^k-b^k$	${(a-b)z} \over {(z-a)(z-b)}$
12	$sin \omega t$	${\omega} \over {s^2+\omega^2}$	$sin \omega kT$	${z sin \omega T}\over{ z^2 - 2zcos\omega T + 1}$	sine
13	$cos \omega t$	${s} \over {s^2+\omega^2}$	$cos \omega kT$	${z(z - cos \omega T)}\over{ z^2 - 2zcos\omega T + 1}$	cosine
14	$e^{-at}sin \omega t$	${s} \over {(s+a)^2+\omega^2}$	$e^{-akT}sin \omega kT$	${ze^{-aT} sin \omega T}\over{ z^2 - 2ze^{-aT}cos\omega T + e^{-2aT}}$	natural underdamped
15	${df}\over{dt}$	$sF(s) - f(0)$	(empty)	(empty)	first derivative ($1^{st}$ derivative)
16	${d^2f}\over{dt^2}$	$s^2F(s) - sf(0) - \frac{df(0)}{dt}$	(empty)	(empty)	second derivative ($2^{nd}$ derivative)
17			$f(k-n)$	$z^{-n}F(z)$	backward shift of n
18	$f(t-t_d)$	$e^{t_ds}F(s)$			time delay
19	(empty)	(empty)	$f(k+1)$	$zF(s) - f(0)$	forward shift of one
20	(empty)	(empty)	$f(k+2)$	$sF(s) - f(0)$	forward shift of two
21	$f(0)$	$\lim\limits_{s\rightarrow\infin}sF(s)$	$f(0)$	$\lim\limits_{z\rightarrow\infin}F(z)$	initial value
22	$f(\infin)$	$\lim\limits_{s\rightarrow 0}sF(s)$	$f(\infin)$	$\lim\limits_{z\rightarrow 1}(z-1)F(z)$	final value

Where:

• For $t<0$, the function $f(t)=0$
• For $k<0$, the function $f(kT) = f(k) = 0$
• $k = 0, 1, 2,\dots$

The Z-transform

1. Difference equation method

2. Power series method

3. Partial fraction method

4. Residue method

The inverse Z-transform

1. Difference equation method

Example: given $m(k) = \begin{cases} 1, &k \text{ even} \\ 0, &k \text{ odd}\end{cases}$ , when $k = 0, 1, \dots , 4$,

find the value of the function $y(k) = m(k) - m(k-1) - y(k - 1)$.

Solution:

$$y(k) = m(k) - m(k-1) - y(k - 1) \Rightarrow Y(z) = M(z) - z^{-1}M(z) - z^{-1}Y(z)$$

$$\Rightarrow \frac{Y(z)}{M(z)} = \frac{1 - z^{-1}}{1 + z^{-1}} = \frac{z-1}{z+1}$$

$\because m(k) = \begin{cases} 1, &k \text{ even} \\ 0, &k \text{ odd}\end{cases}$

$$\Rightarrow M(z) = 1 + 0 + z^{-2} + 0 + z^{-4} + \dots = \frac{1}{1 - z^{-2}} = \frac{z^2}{z^2 - 1}$$

$$\therefore Y(z)= \frac{Y(z)}{M(z)} \times M(z) = \frac{z-1}{z+1} \times \frac{z^2}{z^2 - 1} = \frac{(z-1)z^2}{(z-1)(z+1)^2} = \frac{z^2}{(z+1)^2}$$

$\because Z[y(k)] = Y(z) \Rightarrow y(k) = Z^{-1}[Y(z)]$

$$\therefore y(k) = Z^{-1}[\frac{z^2}{(z+1)^2}]$$

2. Power series method

3. Partial fraction method

4. Residue method

Important

This article was translated by Hermes Agent; the original text can be found at here. The translation is for reference only. Translation time: 2026-11-05