Bài giảng Fundamentals of Control Systems - Chapter 8: Analysis of discrete control systems - Huỳnh Thái Hoàng

ransient performances
Method 1: Analyzing the transient performance based on
the time response y(k) of discrete systems.
yy  Percentage of overshoot: %100
ss
ssmax
y
POT 
y and y are the maximum and steady state values of y(k)max ss - 
 Settling time: Tkt ss 
where ks satisfying the condition:
kkyyky  .)( ss s ,100ss
kkykyy   1)(1 
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 43
s  ,100100 ssss
Transient performances
Method 2:Analyzing the transient performances based 
on the dominant poles.
  ln r
 The dominant poles: jrez *2,1 

 

22
1
)(ln  r
 
22)(ln  r
Tn
 Percentage of overshoot: %100
1
exp
2



 
POT 
 Settling time: t 3 (according to 5% criterion)
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 44
ns
Steady state error
GC(z)
Y(s)+ T G(s)ZOH
R(s) E(z)
H(s)
 E i )(zRrror express on:
)()(1
)(
zGHzG
zE
C

 Steady state error: )()1(lim)(lim 1
1s
zEzkee
zks

 
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Performances of discrete system – Example 1
Y( )R( )
10)( sG
s+ G(s)ZOH
s
1.0T
)3)(2(  ss
1. Calculate the closed-loop transfer function of the system.
2. Calculate the time response of the system to step input.
3. Evaluate the performance of the system: POT, settling
time, steady-state error.
 Solution:
1. The closed-loop TF of the system: )()( zGzGcl 
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)(1 zG
Performance of discrete system – Example 1 (cont’)
 10


s
sGzzG )()1()( 1 Z

)3)(2(
)(  sssG
 

)3)(2(
10)1( 1
sss
z Z
))()(1(
)()1(10 1.031.02
1




ezezz
BAzzz
))()(1(
)(
))((
1
ezezz
BAzz
bsass bTaT 




 Z
)7410)(8190(
036.0042.0)(  zzG
)1()1(
)(
)1()(
ebeeae
abab
eaebA
aTbTbTaT
bTaT



..  zz
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 47
)( abab
B 

Performance of discrete system – Example 1 (cont’)
03600420 z
)(1
)()( 
zG
zGzGcl  )741.0)(819.0(
..)(  zzzG
)741.0)(819.0(
036.0042.0


zz
z
)741.0)(819.0(
036.0042.01 


zz
z

64305181
036.0042.0)( 2 

zz
zzGcl ..
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Performance of discrete system – Example 1 (cont’)
Th ti f th t t t i t
036.0042.0)( zzG)()()( zRzGzY k
2. e me response o e sys em o s ep npu .
643.0518.12  zzk
)(
643.0518.1
036.0042.0
2 zRzz
z


)(
643051811
036.0042.0
21
21
zR
zz
zz




..
 )()036.0042.0()()643.0518.11( 2121 zRzzzYzz  
 )2(036.0)1(042.0)2(643.0)1(518.1)(  krkrkykyky
)2(0360)1(0420)2(6430)1(5181)(  krkrkykyky
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.... 
Performance of discrete system – Example 1 (cont’)
01)( kkU it t i t , rn s ep npu :
Initial condition: 0)2()1(  yy
Substitute the initial condition to the recursive equation
f (k) h
ky ...0.5003; 0.3909; 0.2662; 0.1418; 0.0420; ;0)( 
o y , we ave:
;....; .; .; .; .; .
;.; .; .;.; .; .
619106251063410646106606067600 
...689806985069750681706459058600 
)2(0360)1(0420)2(6430)1(5181)(  krkrkckckc
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.... 
Performance of discrete system – Example 1 (cont’)
Step Response
0.6
0.7
0.5
A
m
p
l
i
t
u
d
e
0.3
0.4
0.2
0
0.1
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 51
Time (sec)
0 0.5 1 1.5 2 2.5 3
03600420 z
Performance of discrete system – Example 1 (cont’)
3 Transient performances:
2
1)(
643.0518.1
..)( 
R
zz
zGk
.
)()1(li 1 Y
The steady state response:
11  zzm1 zzy zss  
)()()1(lim 1
1
zRzGz kz

 







  12
1
1 1
1
643.0518.1
036.0042.0)1(lim
zzz
zz
z
6240.ssy
The maximum value: 6985.0max y
 Percentage of overshoot:
%9411%100624.06985.0%100ssmax  yyPOT
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 52
.
624.0

ssy
Performance of discrete system – Example 1 (cont’)
 Settling time (5% criterion): 6240
First, we need to find ks satisfying:
    kkk  1)(1 
05.0%5
.



xlc
syyy ,ssss
skkky  ,655.0)(593.0
From the time response calculated before 14sk
1.014 Tkt ss

sec4.1st
 St d t t

;.; .; .;.; .; .
kc
...689806985069750681706459058600 
...0.5003;0.3909; 0.2662;0.1418;0.0420; ;0)( ea y s a e error:
Since the system is unity negative feedback, we have:
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 53
;....; .; .; .; .; . 619106251063410646106606067600 624.01ssssss  yre 376.0ss e
Performance of discrete system – Example 1 (cont’)
 Note: It is possible to calculate POT and t based on the dominant poless
The poles of the closed-loop system are the roots of the equation:
0643.0518.12  zz
3285.08019.02587075900*2,1  . j .z
557908019.0lnln  r .
3285.0)8019.0(ln)(ln 2222 r
3958032850)80190(ln1)(ln1 2222   r

%1112%10014.35579.0exp%100exp 


  POT
...
1.0Tn
..
5579.01
.
1 22  
 
 
36133t
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 54
sec.
3958.05579.0
 nqñ
Performance of discrete system – Example 2
y(t)+ ZOH
r(t)
T
eR(t)e(kT)e(t)
)(sG
)5(2 s
with T = 0.1
1 F l t th t t ti d ibi th t
)3)(2(
)(  sssG
. ormu a e e s a e equa ons escr ng e sys em
2. Calculate the response of the system to unit step input
(assuming the initial conditions are zeros) using the state
equation formulated above.
3 Calculate POT settling time steady state error
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. , ,
Performance of discrete system – Example 2 (cont’)
 S l ti
1. Formulate the state equation:
o u on:
65
102
)3)(2(
)5(2
)(
)()( 2 


ss
s
ss
s
sE
sYsG
R
0)(10)( txtx  
 The state equation of the continuous plant:
 
)(
1)(56)( 2
1
2
1 te
txtx R
B

 
 
A
  


)(
)(
210)( 1
tx
ty
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2 txC
Performance of discrete system – Example 2 (cont’)
 The transient matrix:
 
11
1
56
1
56
10
10
01
)(



















s
s
sss -AI









 
 6
)3)(2(
1
)3)(2(
5
6
15
6)5(
1
s
ssss
s
s
 

)3)(2()3)(2( ssss
sss
  1123 11







 


 
  



3266
3232)]([)(
11
1 ssssst
LL
LL
L
  
 )()23()(
3232 tttt eeeet
  3232 ssss
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   )32()66( 3232 tttt eeee
Performance of discrete system – Example 2 (cont’)
 The state equation of the   )()(])1[( kTekTTk BxAx
discrete open-loop system:   )()( kTkTy d
Rdd
xC











5850.04675.0
0779.09746.0
)32()66(
)()23()(
1.0
3232
3232
T
TTTT
TTTT
d eeee
eeeeTA
 

 





 
TT
d deeee
eeeed
0
3232
3232
0 1
0
)32()66(
)()23()(  

BB





 

 

 

 07790
0042.0)
32
(
)(
)(
1.032
32
32




ee
dee
T
 210CC
   
 .)(32 0
320  eeee
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d
Performance of discrete system – Example 2 (cont’)
 The state equation of the discrete closed-loop system:
 

 
)()(
)()(])1[(
kTkT
kTrkTTk dddd
C
BxCBAx
y d x
with
    







 4292.02465.1
0695.09326.0
210
0779.0
0042.0
5850.04675.0
0779.09746.0
ddd CBA
)(
07790
0042.0
)(
)(
4292024651
0695.09326.0
)1(
)1( 11 kTr
kx
kx
kx
kx












  


)(
)(
.210)( 1
kx
kx
ky
 ... 22
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2
Performance of discrete system – Example 2 (cont’)
2 Time response of the system:.
  )(0042.0)(0695.0)(9326.0)1( 211 krkxkxkx
From the closed-loop state equations, we have:
With initial condition x1(1)=x2(1)=0, unit step input, we can
  )(0779.0)(4292.0)(2465.1)1( 212 trkxkxkx


 59.7;... 57.4; 54.0; 49.1; 42.6; 34.2; 24.2; 13.5; 4.2; ;010)( 31  kx
calculate the solution to the state equation:
...62.662.6;62.7;62.7; 62.8;62.8;62.7;62.5;62.0; 61.2; 

0 40 5;0 5;0 5;0 3;0 3;1 4;3 4;6 5;11 4;
 ... 18.5; 28.3; 41.2; 57.2; 75.4; 93.5; 106.6; 106.1; 77.9; ;010)( 32
-----
kx  
............ . 
 0.634;...0.631;0.622;0.606;0.577;0.529;0.455;0.348; 0.198; ;0)( ky
)(2)(10)( 21 kxkxky The closed-loop system response:
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...0.625 0.625; 0.626; 0.627; 0.627; 0.629; 0.630; 0.632; 0.634; 0.635; 
Performance of discrete system – Example 2 (cont’)
Step Response
0.6
0.7
0.5
A
m
p
l
i
t
u
d
e
0.3
0.4
0.2
0
0.1
6 December 2013 © H. T. Hoàng - www4.hcmut.edu.vn/~hthoang/ 61
Time (sec)
0 0.5 1 1.5 2 2.5
Performance of discrete system – Example 2 (cont’)
3 Performances of the system:.
 Percentage of overshoot:
630 5.max y
625.0ssy
%6.1%100
ss
ssmax 
y
yyPOT
 The settling time:
    skkykyy  ,05.01)(05.01
6,656.0)(594.0  kky 
According to the response of the system:
 0.634;...0.631;0.622;0.606;0.577;0.529;0.455;0.348; 0.198; ;0)( kc
S d 375062501
6sk sec6.0 Tkt ss
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...0.625 0.625; 0.626; 0.627; 0.627; 0.629; 0.630; 0.632; 0.634; 0.635;  tea y state error: ..ssssss  yre