Textbook:
Phillps & Nagle
Digital Control System Analysis & Design
References:
1.
B.C. Kuo : Digital Control System
2.
Ogata : Discrete – time Control System
3.
Katz : Digital Control Using Microprocessors(松崗)
4.
Haykin : Adaptive Filter
Analog
Digital
Discrete-time
Continuous
Discrete
Sampled-data
(Deterministic)
(Probabilistic)
Brief introduction
Digital Computer – 1946 – (Vacuum Tube)
Chapter 2 Discrete time systems & the Z-Transform
dy
yx
dt
dy yk  yk 1

dt
t
y  yk 1
RC k
 yk  xk
t
t
1
yk 
yk 1  RC xk
t
t
1
1
RC
RC
 a0 xk  b1 yk 1
RC
R
y
x
Let a0  0.1 , b1  0.9 , y ( 1)  0
xk
a0

yk

b0
D
yk 1
discrete-time approximation
continuous-time
-
y=1-e
t
RC
Amortization
y (k ) --- the money owed after the kth payment
u (k ) --- the amount of the kth payment
r --- interest rate
u (k )  P , k  1, 2,....N
y (0) is the initial debt
assume
y (0)  d
moreover u (1)  P
y (1)  (1  r ) y (0)  u (1)
or
y (1)  (1  r )d  P
y(2)  (1  r ) y(1)  P  (1  r) (1  r)d  P  P  (1  r) 2 d  (1  r) P  P
……………
(1  k ) k  1
y (k )  (1  r ) d  P
r
k
kN
y( N )  0
(1  k ) N  1
 (1  r ) d  P
r
N
N
r
(1

r
)
∴ P
d
N
(1  r )  1
If
d  100000
r  0.01 (1%)
N  10
P  10570
,k  1, 2,...., N
Cadzon , Discrete Time system
Model of Rabbit Papulation
y (k )  y (k  1)  y (k  2)
y (k ) --- No of pairs of rabbit present at the end of Kth month
 e (t )

A/D
Digital
Computer
D/A
m(t )
sensor
e (t )
e(kt )
X (k 1)T 
(k  1)T
kT
t
(k  1)T
Plant
t
m(t )  K P e(t )  K I  e(t )dt
0
x(kT )  x  (k  1)T   Te(kT )
PI controller
x is the approximation of the integral of e(t)
t  kT
m(kT )  K P e(kT )  K I x(kT )
 K P e(kT )  K I x  (k  1)t   K I Te(kT )
 ( K P  K I T )e(kT )  K I x  (k  1)T 
Difference Equations V.S. Differential Equations
x(k )  bn e(k )  bn 1e(k  1)  ...  b0e(k  n)
 an 1 x(k  1)  ...  a0 x( k  n)
d n e(t )
y (t )   k
 ...   0 e(t )
n
dt
d n e(t )
dy (t )
 n

...


1
dt n
dt
Digital compensator design: two approaches
•
Design an analog compensator, then convert it into a digital one
•
Design digital compensator with exact methods
Control Engineers need to
•
Determine T, the sampling period;
•
Determine n, the order of the difference equation;
•
Determine ai and bi .
The design of digital compensator also have the following problems
•
Computer wordlength
•
Round off
e* (t )  e(0) (t )  e(T ) (t  T )  e(2T ) (t  2T )  ...
E * ( S )  e(0)  e(T )e  ST  e(2T )e 2 ST  ...
Let Z  e ST
E ( Z )  Z  e(kT )   e(0)  e(1) Z 1  e(2) Z 2  ...

  e( k ) Z  k
k 0
e (t )
e (t )
t
e* (t )
Fundamentals of z transform
• Will work on the board
2.5 Solution of Difference Equation
EX2.8
m(k )  e(k )  e(k  1)  m(k  1), k  0
1, k even
e( k )  
 0, k odd
k  0 , m(0)  e(0)  e( 1)  m( 1)
k  1 , m(1)  e(1)  e(0)  m(0)
• A MATLAB program segment like the
following
• Can do the job.
m(k )  an 1m(k  1)  ...  a0 m(k  n)
 bn e(k )  bn 1e(k  1)  ...  b0e(k  n)
1
1
 M ( Z )  an 1Z M ( Z )  ...  a0 Z M ( Z )
1
1
 bn E ( Z )  bn 1Z E ( Z )  ...  b0 Z E ( Z )
bn  bn 1Z 1  ...  bn Z  n
E (Z )
 M (Z ) 
n
1
1  an 1Z  ...  an Z
• By long division, will work on the board.
What do initial conditions here mean?