Transcript Chapter 5

Chapter 5
On-Line Computer Control –
The z Transform
Analysis of Discrete-Time Systems
1.
The sampling process
2.
z-transform
3.
Properties of z-transforms
4.
Analysis of open-loop and closed-loop discrete
time systems
5.
Design of discrete-time controllers
y Continuous
signal
Disontinuous y*
signal
y*
y
9
1
3
5
7
9
11
1
t (sec)
3
6
t (sec)
T = 1 sec
y*
(a)
(b)
9
6
3
12
t (sec)
T = 3 sec

y*
(c)
Continuous signal and its discrete-time representation with different sampling rates
y*
Dt
t = nT
(a)
t
Dt
t = nT
y*
t
(b)
Impulse area  nT 
t = nT

12
t
(c)
From the response of a real sampler to the response of an ideal impulse sample
The Sampling Process
Impulse
Sampler
y (t)
y*
y* t   y(0) (t)  y(T)  (t - T)  ....  y(nT)  (t - nT)  ...

  y(nT)  (t - nT)
n 0
1.
2.
At sampling times, strength of impulse is equal to value of input signal.
Between sampling times, it is zero.
Laplacing

y (s)   y(nT) L[ (t - nT)]
*
n 0

or
y (s)   y(nT) e -nTs
*
n 0
The Hold Process :
From Discrete to Continuous Time
discrete
impulses m* (t)

Zero – Order Hold :
m(t)  m* (t)

m (t)
Hold Device
Continuous
output
for nT  t  (n  1)T
Transfer Function :
Response of an impulse input :  (t)
1
 t 
m* (t)
m (t)
t
-Ts

1 e
1
H(s)   1 - e -Ts
s s
s

T
First Order Hold
m(nT) - m[(n - 1)T]
(t - nT)
T
nT  t  (n  1)T n  2,3,4....etc
m(t)  m(nT) 
Response to an impulse input
2
1
T
2T
1
-1
Transfer function:
1  sT  1 - e -sT 


H(S) 
T  s 
2
First Order versus Zero Order Hold
m* (nT)
m (t)
m (t)
0
1T
3T
5T
7T
t
0
1T
3T
5T
(a)
7T
t
(b)
0
1T
3T
5T
7T
t
(c)
Comparison of reconstruction with zero-order and first-order holds, for slowly varying signals.
m* (nT)
m (t)
m (t)
0
2T
4T
6T
8T
10T
t
0
2T
4T
6T
8T
(a)
(b)
0
2T
4T
6T
8T
10T
t
(c)
Comparison of reconstruction with zero-order and first-order holds, for rapidly changing signals.
10T
t
Z-Transforms
Sample
y(t)
yz(t)

y ( s )   y (n )e  nTs
z
n 0
Let z  eTs

y ( s )   y ( n ) z  n
z
n 0

 y( z )



Ζ y (t )  y(z) 
z

 y(n  )z
-n
n 0
Remarks
1.
z-transform depends only on the discrete values y(0), y(‫)ז‬,y(‫)ז‬..etc. If two
continuous functions have the same sampled values , then z-transform
will be the same.
2.
It is assumed that the summation exists and is finit.
3.
We can also view t in the form Z[ y (s) ] = ŷ(z)
Z-Transforms of Basic Functions
1. Unit Step Function
Z[u(t)]  1  1 z 1  1 z 2  1 z 3  ......

1
z

1  z 1
z-1
2. Exponential Function
   e

Ze
-at
 anT
z
n 0
n

  λ n , λ  e -aT z 1
n 0
1
1 λ
1
 -aT 1
1-e z
z

z-e -aT

e -aT z 1  1 for convergence
Z-Transforms of Basic Functions Continued
3. Ramp Function
Z at   S  0  aT(z 1 )  2(aT)z 2  3(aT)z 3  ...
z 1S 
1
aTz  2  2(aT)z 3  
1
(1  z ) S  0  aT(z )  (aT)z
2
aTz 1
 (aT)z  ... 
1  z 1
4. Trigonometric Functions
zsinwT
Zsinwt   2
z  2zcoswT  1
z 2  zcoswT
Zcoswt   2
z  2zcoswT  1
3
Z-Transforms of Basic Functions Continued
5. Translation
Z  f(t - kT)  Z  e
 kTs

f ( s )   f (z)z  k

Z  f(t - kT)   f(nT  kT)z  n
let
n 0
 n - k,assume f( τ)  0 for



f( τ)z 
k
n  k

   f( τ)z - z  k
0

 f (z) z  k
 0
Z-transform for Numerical
Derivative

z f(t - T)  f (z)z 1
z-1 is like a back shift operator
dy ynt  yn 1
 
dt
T
1
 y n  y n 1  1  z
Z

y(z)

T
 T



Properties of z-Transforms
1. Linearity


Z a f a f  a f (z) a f (z)


1
1

2

2
1
2
1
2
2. Final Value Theorem
lim y(t)  lim (1-z )yˆ (z)
t 
z 1
-1



Proof: (1-z )fˆ y(z)(1 z )  y(nT ) z  n   y(nT ) z  n   y(nT ) z  n 1
n0
n0
n0
-1

-1





lim   y( 0 )-y( 0 )z -1  y(T )-y(T)z-1 z1 y( 2T)-y( 2T)z -1 z 2.... y(nT)
n

 


 



As z1
lim (1 z1 )yˆ (z) lim y nT  lim y(t)
z 1
n   t 
Numerical Integration in z-transform
t nT
y(t)   f t dt
0
nT
y nT  y (n-1 )T 
 f(t)dt
(n-1 )T








Using Trapezoidal Rule
f(nT )  f  n-1T 

T
 y (n 1 )T  
2


yˆ (z)  yˆ (z) z 1  T  fˆ(z) z 1 fˆ(z)
2 

or solving
yˆ z   T 1 z fˆ z 
2 1 z
1
1
Inversion of z-transforms
1.Partial fraction expansion
cn
Q( z 1 )
c1
c2
yˆ ( z ) 



...

P( z 1 ) 1 ( z 1 ) 2 ( z 1 )
n ( z 1 )
λ1, λ2,… λn are low-order polynomials in z-1 compute c1,c2,…cn.
Invert each part separately, we able
z
z 1
yˆ ( z )  2

z  4 z  3 1  4 z 1  3 z  2
z 1
c1
c2



(1  z 1 )(1  3 z 1 ) 1  z 1 1  3 z 1
z 1
c1 
1  3 z 1

z 1 1
1
2
z 1
1
c1 

1
1  z 1 z 1 3 2
1 2
12
 yˆ ( z ) 

1  z 1 1  3 z 1
From Tables of z-transforms
y(nT) = -1/2 + 1/2 e11n
y:0,1,4,13,0,…
z 1
yˆ ( z ) 
1  4 z 1  3z  2
2.Inversion by Long-Division
1-4z-1+3z-2
y(0) = 0
y(T) = 1
y(2T) = 4
y(3T) = 13
1z-1+4z-2+13z-3
z-1
z-1-4z-2+3z-3
4z-2+3z-3
4z-2-16z-3+12z-4
13z-3-12z-4
z-transforms of various functions
Function
in time domain
unit impluse
Lalpace transform
1
unit step
1/s
ramp: f(t) = at
a/s2
f(t) = tn
n!/sn+1
f(t) = e-at
1/s+a
f(t) =te-at
1/(s+a)2
z-transform
1
1
1  z 1
aTz 1
(1  z 1 ) 2
n
1
(1)
lim
a n 1  e  aT z 1
a 0
1
1  e  aT z 1
e  aT z 1
(1  e  aT z 1 ) 2
n
z-transforms of various functions
Function
in time domain
f(t) = sinωt
f(t) = cosωt
f(t) = 1-e-at
f(t) =
e-at
sinωt
f(t) = e-at cosωt
Lalpace transform

s2   2
s
s2   2
a
s(s  a)

(s  a)2   2
sa
(s  a)2   2
z-transform
z 1 sin T
1  2 z 1 cos T  z  2
1  z  cos T
1  2 z 1 cos T  z  2
(1  e  aT ) z 
(1  z 1 )(1  e  aT z  )
z  e  aT sin T
1  2 z 1e  aT cos T  e  2 aT z  2
1  z 1e  aT cos T
1  2 z 1e  aT cos T  e  2 aT z  2
Discrete-Time Response of systems
In computer control:
measurements are taken periodically and
control actions implemented periodically,
This results in a discrete input/discrete
output dynamic system.
cn
en
Discrete System
Example of Discrete Systems
Let
dc
 ken  c
dt
a discrete time approximation is



T
cn  cn 1
 ken  cn
T
cn 
(1 

T

T
cn 1  kcn  cn
) cn 

T
cn 1  ken
Taking z-transform
(1 

)  cˆ( z ) 

z 1  cˆ( z )  keˆ( z )
T
T
cˆ( z )
k
or

e( z ) (1   )   z 1
T
T
Z-transform for a given continuous
system with transfer function G(s)
and a ZOH
1  e Ts

Z  H ( s )G ( s )   Z 
G (s) 
 s

G s 
 Ts G  s  
Z
  Z e

s 
 s 

 G  s   1  G  s  
Z
z Z 

 s 
 s 
G s 
 1  z  Z 

 s 
1
Example: Pure Integrator with Hold
c*(s)
1  e  ST
s
Kp
s
1  e Ts K p
HG p ( s ) 

s
s
Kp
K p e Ts
1  e Ts K p
Z HG ( s )  Z [
.
]  Z[ 2 ]  Z[ 2 ]
s
s
s
s
1
Kp
K pTz
K pT  z 1
-1
1
 [1 - z ]  Z [ 2 ]  [1  z ]

s
(1  z 1 ) 2 (1  z 1 )
Step response
Hence of
c( s ) 
1
s
cˆ( z ) 
1
1  z 1
K p  z 1
yˆ ( z ) 
K p  z 1
1


(1  z 1 ) (1  z 1 ) (1  z 1 ) 2
which impulse a ramp response
y*(s)
Example：First order lag system
1  e  ST K p

H ( s )G p ( s ) 
 ps 1
s
Kp
1  e  ST
]

Z [ HG ( s )]  Z [
s ( p s  1)
s
Kp
1
 [1  z ]Z [
s ( p s  1)
]
1
1
]
 K p [1  z 1 ]Z [ 
s s 1  p
 K p [1  z 1 ][
T 
1
1
]

T  p 1
1
1 z
z
1 e
(1  e p ) z 1
 Kp
T 
1  e p z 1
Step Response for 1st order
lag system
From tables, for
c( s ) 
yˆ (z) 
1
s
cˆ( z ) 
K p (1  e
  p
1
1  z 1
) z 1
 
(1  z 1 )(1  e p z 1 )
Kp
Kp


1
(1  z ) (1  e   p z 1 )
y (nT )  K p [1  e
 nT  p
y (t )  K p to t  
y(t)
＊
＊
＊
＊
＊
＊
＊
＊
＊
＊
]
＊
＊
time
Note: Compare with discrete approximation to
First-order system
Generalization
cn  a 0en  a1en  1  ...  aken  k  b1cn  1
 b 2cn  2  ...  bmcn  m
cˆ( z )  a 0eˆ( z )  a1 z 1eˆ( z )  ...  ak eˆ( z ) z  k
 b1cˆ( z ) z 1  b2 cˆ( z ) z  2  ...  bm cˆ( z ) z  m
or
cˆ( z ) a0  a1 z 1  a2 z 2  ...  ak z  k

 D( z )
1
2
m
eˆ( z ) 1  b1 z  b2 z  ...  bm z
D (z)=Transfer function relating e and c
Analogous to Laplace transfer
Discrete time input/output model
Remark：
Note that D(z) is the z-transform of the response of the system
to an impulse input eˆ( z )  1
Z-transform of a continuous
process with Sample and Hold
Hold
H (s)
discrete
input
Process
Gp (s)
c*(s)
y (s)
continuous
variables
y*(s)
discrete
output
we seek a relationship (Z-transfer function) between c and y.
Consider a impulse input
Then
y ( s )  H ( s )Gp ( s )c  ( s )
c*(z)=1
c*(s)=1
 H ( s )Gp ( s )
yˆ ( z )  Z  y s 
 Z [ H s Gp ( s )]
 HGP ( z )
HGp(z) called the pulse transfer function
(since it represents the z-transform of the pulse response of Gp (s) )
Properties of pulse Transfer Function
1.
Z G1 (s)G2 (s)  Z G1 (s) Z G2 (s)
2.An impulse input is converted into a pulse input by the first
order hold element . Hence HG(z) is the pulse response of
G(s) sampled at z internals of T.
3.The pulse transfer function of two systems in series can be
combined if there is a sample and hold in between.
c2
c1
T
c3 ( z )
 G3 ( z )
c1 ( z )
G1(z)
G2(z)
c2 ( z )
 G1 ( z )
c1 ( z )
c3
c3 ( z )
 G2 ( z )
c 2( z )
Closed-Loop System
disturbance
T
set
point
eˆ ( z )
+
D (z)
ysp (z)
-
ˆ ( z)
y
Hold
T
Process
H (s)
Gp (s)
m (s)
ˆ( z )
c
y2 ( s )
y(2)
y1(s)
sampled output
yˆ ( z )  Z  y ( s )  Z  y1 ( s )  y2 ( s )
 Z  y1 ( s )  Z  y2 ( s )
 HG p ( z )eˆ( z )  yˆ 2 ( z )
 HG p ( z ) D ( z )eˆ( z )  yˆ 2 ( z )


 HG p ( z ) D ( z ) y sp ( z )  yˆ ( z )  yˆ 2 ( z )
or
yˆ ( z ) 
HG p ( z ) D ( z ) y sp ( z )
1  HG p ( z ) D( z )

yˆ 2 ( z )
1  HG p ( z ) D ( z )
1. Roots of the Characteristic equation
1+HGp(z)D(z)=0
Determine stability of the closed-loop system
2. Note similarity to continuous system.
Example: closed-loop response of a
first-order system
G p (s) 
kp
 pS 1
HG p ( z )  k p
(1  e
T / 
p
) z 1
1 e
T / 
p
z 1
For proportional control
D( z )  kc
T / 
p
(1  e
) z 1
ˆ sp ( z )
kp
kc y
T /  p
1
1 e
z
ˆ ( z) 
y
T /  p
(1  e
) z 1
1 k p
kc
T /  p
1
1 e
z
k p k c (1  b) z 1
ˆ sp ( z )

y
1
1  k p k c  b 1  k p k c  z

where
be
T / p

For a unit step change in set point
and
yˆ ( z ) 
k p k c (1  b) z 1
1  [ k p k c  b(1  k p k c )] z
y ( nT ) 

p 
ˆ sp ( z ) 
y
k p kc
1  k p kc
p
1  k p kc
1  e
 nT /  p 

1
1  z 1

1
1
(1  z 1 )

The response is very similar to continuous control.
The steady state value of y(t) is
y
Kc K p
1  K p Kc
Hence the offset is
Kc K p
1
offset  1 

1  Kc K p 1  K p Kc
Stability of Discrete Systems
A system is consider to be stable if output remains bounded for bounded
Inputs. Consider a discrete system with transfer function

a0  a1 z 1      am z  m
D(z)   
1
n
1

b
z





b
z
1
n
e (z)
cn
c1
c2


  
1
1
1  P1 z
1  P2 z
1  Pn z 1
c (z)
Where P1,P2,…,Pn are n roots of:
1  b1 z -1  b2 z -2  .....  bn z -n  0

For an impulse imput e(z)  1 and C(z)  D(z)
consider t he k th term
Ck
f K (z) 
1
1 - PK z
f k (nT)  C K e
nlnPK
Let PK  α  jβ | PK | e jw
ln Pk  ln(P K )  jw
e
lnPK
e
ln(PK )
e
jw
e nlnPK  e nln(PK ) e njw
The second term is always bounded between(-1 and 1)
IF
| PK | 1 then ln | PK | 0 and e nln|PK |  0 as n  
| PK | 1 then ln | PK | 0 and e nln|PK |  1
| PK | 1 then ln | PK | 0 and e nlnPK   as n  
Hence : if PK lies within th e unit - circle on the complex
plane, PK will remain bounded.
" A discrete system is stable if all its poles lies inside
or on the unit circle in the complex plane"
Im
Unstable
roots
real
STABLE
REGION
Unit
circle
Location of poles
Consider t he impule response of a system with
1
D(z) 
1 - P1 z 1

1
2 -2
-1
y(z) 

1

P
z

P
 ....
1
1 z
1
1 - P1 z
y(1)  P1
y(nT)  P1
if
na
| P1 | 1, y(t)   as t  
0  Pi  1 then y(t)  0 : exponentia l decay
- 1  P1  0 then y(t)  0 : oscillator y
Example: Stability of closed-loop
Proporhona l control of First - order Process characteri stic equation
1  [K P K C - b(1  K P K C )]z -1  0
P1  (1  K P K C )b - K P K C
 (1  K P K C )e
e
-T
τP
-T
τP
 K P K C [e
-T
τP
- KPKC
 1]
Example: Stability of closedloop - Continued
p1 will cross the unti circle where
-T
τP
-T
τP
 1  e - K P K C [1 - e ]
-T
τP
-T
τP
or( 1 - e ) - K P K C [1 - e ]
or since e
-T
τP
1
K P K C  -1
or K P K C 
1 e
-T
τP
-T
τP
1- e
note that instabity may be caused by high values of K C or high values of T
Digital Feedback - Control
1.Digital Apperoximation of classical controllers
Displacement form
t
1
de
C(t)  K C [e(t)   e(t)dt  τ 0 ]  Cs
τt 0
dt
τ
T
(e k )  D (e n -e n-1 )]  Cs
T
k 0 τ I
n
C n  K C [e n  
Alternative form :
Velocity form
n 1
τ
T
e k  D (e n-1 -e n-2 )]  Cs
T
k 0 τ I
C n-1  K C [e n-1  
ΔC n  K C [1 
2τ
τ
T τD
 ]e n -K C [1  D ]e n-1  K C D e n-2
τI T
T
T

D(z) 
Δ τ(z)

e (z)
 K C [(1 
2τ
τ
T τD
 )-(1  D )z -1  D z -2 ]
τI T
T
T

1.
2.
3.

No initialization is necessary. [ Cs is not needed ]
 Bumpless transfer from manual / automatic
Automatic ‘reset-windup’ protection.
Protection in case of computer failure
Disadvantages:
1.
2.

Advantages of velocity Form
Since different modes are indistinguishable, on-line tuning
methods will not work.
Difficult to put constraints on integral and / or derivative term.
Tuning Digital Controllers:
1.
2.
3.
Ziegler – Nichols
Cohen – Coon settings
Time - integral performance criteria
2.Deadbeat or Minimal Prototype controller
This control - law is designed such that the output(sam pled values)
no error at the frist sampling following a set point change :
we have
HG(z)D(z) 
y sp (z)
y(z) 
1  HG(z)D(z)

1
step - change
y sp 
Let
-1
1- z
we want


z -1
y( z ) 
1 - z -1
solving for D(z) :
z -1 Y(t)
1
D(z) 
HG(z) 1 - z -1
Possible responses :
ex.y(t) goes to 1 after one sampling
actual
ideal
time
Derivation of Deadbeat
Controller- Continued

HG(z)D(z) 
y (z) 
ysp (z)
1  HG(z)D(z)
1
z 1
ysp  z  
; yˆ  z  
1
1 z
1  z 1
yˆ ( z )
z 1
HG(z)D(z)


ysp  z  1 1  HG(z)D(z)
z -1
HG(z)D(z)

-1
1-z
1
z -1
1
D  z   -1
1-z HG(z)
Deadbeat
Deadbeat
0~10 sec
Deadbeat control for (1/(s+1)3)
DZ  


1
Z 1
Z 3  0.406Z 2  0.05495Z  0.002479
Z 1



HG ( Z ) 1  Z 1
0.3233Z 2  0.3073Z  0.01584
1  Z 1
1  0.406Z 1  0.05495Z  2  0.002479Z 3
0.3233  0.016Z 1  0.2915Z  2  0.01584Z 3
Sampling time: 2
Ringing and Pole-placement
Ringing
refers to excessive value movement caused by a widely
oscillating controller output.
Caused by negative poles in D(z).
Hence avoid poles near -1.
Change controller design such that poles are on the side or near
zero on negative side

SYS = TF(1,[1 3 3 1])





Transfer function:
1
--------------------s^3 + 3 s^2 + 3 s + 1


>> sysd=c2d(SYS,2)





Transfer function:
0.3233 z^2 + 0.3073 z + 0.01584
-------------------------------------z^3 - 0.406 z^2 + 0.05495 z - 0.002479


Sampling time: 2
0.3233 0.6306 0.3231

>> p1=[1 -1];p2=[0.3233 0.3073 0.01584]

p2 =


0.3233
0.3073

>> c=conv(p1,p2)

c=

0.0158
0.0158
0.3233 -0.0160 -0.2915 -0.0158
Canceling the ringing pole at
z=-0.8958




ans =
1.0000
-0.8958
-0.0547

>> p1=[1 0];p2=[1 -0.99];p3=[1 0.0547];
>> c=conv(p1,p2)

c=


1.0000 -0.9900

>> c=conv(c,p3)

c=



0
1.0000 -0.9353 -0.0542
0
Warning: Using a default value of 1 for maximum step size. The simulation step size will be limited to be
less than this value.
>>
Smoothing the Control Action







p=[0.3233 -0.016 -0.2915 -0.01584];
r=roots(p)
r=
1.0001
-0.8959
-0.0547
Delete the unstable pole z=-0.8959

p1=[1 -0.99]; p2=[1 0.0547];
c=conv(p1,p2)

c=


1.0000 -0.9353 -0.0542
Reconstruct the Control Loop
Scope1
Step
Subtract
num(z)
1
den(z)
den(s)
Discrete
Transfer Fcn
Zero-Order
Hold
Transfer Fcn
Scope
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
30
35
40
45
50
The treatment of unstable
poles

sysd=c2d(SYS,1)





Transfer function:
0.0803 z^2 + 0.1544 z + 0.01788
----------------------------------z^3 - 1.104 z^2 + 0.406 z - 0.04979


>> p2=[ 0.0803 0.1544 0.01788]

p2 =

0.0803
0.1544

>> p1=[1 -1]

p1 =

1
-1

>> c=conv(p1,p2)

c=

0.0179
0.0803
0.0741 -0.1365 -0.0179
The treatment of unstable
poles

>> roots(c)

ans =



-1.7990
1.0000
-0.1238
1
0.9
0.8
0.7

>> p1=[1 0];p2=[1 -0.99];p3=[1 0.1238];
>> c=conv(p1,p2)

c=

0.6
0.5
0.4
0.3


1.0000 -0.9900
0
0.2
0.1
>> c=conv(c,p3)
0


c=
1.0000 -0.8662 -0.1226
0
0
5
10
15
20
25
30
35
40
45
50
3.Dahlin’s Method

Require that the closed–loop system behave like a first-order
system with dead-time.
e- S 1
y(S) 

S  1 S
we want
for
Solving for D
(step response)
-T/ 
-1
(1
e
)z
ŷ(z)  z -k
(1 - z -1 )(1 - e-T/  z -1 )
1
ŷ sp (z) 
1 - z -1
1
(1 - e-T/  )z -k -1
D(z) 

HG(z) 1 - e-T/  z -1 - (1 - e-T/  )z -k -1
1. Choose ,  such that D(z) is realizable
2. Lot of algebra
Dahlin’s Method
Dahlin’s Method
0~10sec
0~50 sec
Dahlin’s Method for for
(1/(s+1)3)
1
DZ  

HG ( Z )

(1  e
1 e


T

) Z k 1
T

Z 1  (1  e

T

) Z  k 1
Z 3  0.406Z 2  0.05495Z  0.002479


0.3233Z 2  0.3073Z  0.01584
0.6321Z 1  0.2566Z  2  0.03474Z 3  0.001567 Z  4
0.3233  0.1884Z 1  0.3016Z  2  0.2001Z 3  0.01001Z 4

Sampling time: 2
(1  e
1 e


T

) Z  k 1
T

Z 1  (1  e

T

) Z k 1
Regulatory Control
Consider a process described by
yn = a1yn-1 + a2yn-2 + … + b1mn-1 + … + bk mn-k
In regulatory control, we want to keep y close to zero in presence of disturbances.
Ideally choose mn such that
yn ≡ ysp

ysp = a1yn + a2yn-2 + … + bkyn-k+1 + b1 mn + b2mn-1 + bkmn-k+1
Or mn = -1/b1 [ ysp – a1yn – a yn-1 - akyn-k+1 – b2mn-1 + … - bkmn-k+1 ]
Remark
1.
Problems can arise in practice if model parameters are not known
2.
The above choice is equivalent to minimizing
3.
If dead-time is present control will be unrealizable
1 N
P   (y n - y sp ) 2
N n 0