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1
Modern Control TheoryDigital Control
Lecture 4
By Kirsten Mølgaard Nielsen
Based on notes from
Jesper Sandberg Thomsen
2
OutlineThe Root Locus Design Method
Introduction Idea, general aspects. A graphical picture of how changes of
one system parameter will change the closed loop poles. Sketching a root locus
Definitions 6 rules for positive root locus (6 rules for negative root locus)
Selecting the parameter value
3
Introduction
0)()()(1
)()()(1
)()()(
)(
)(
sHsGsD
sHsGsD
sGsDsT
sR
sYClosed loop transfer function
Characteristic equation,roots are poles in T(s)
4
Introduction
)(,8.1
))((2)(
1
1)(
2
22
2
2
1
pnr
dd
n
nn
n
Mt
jsjssssH
ssH
Dynamic features depend on the pole locations.
For example, time constant , rise time tr, and overshoot Mp
(1st order)
(2nd order)
5
Introduction
Root locus Determination of the closed
loop pole locations under varying K.
For example, K could be the control gain.
Ksa
sbK
sL
sKbsa
sa
sbK
sKL
sHsGsD
1
)(
)(
1)(
0)()(
0)(
)(1
0)(1
0)()()(1
The characteristic equation can be written in various ways
6
IntroductionSome root loci examples (K from zero to infinity)
7
Introduction
Polynomials b(s) and a(s)
n
iim
mnn
m
iim
mmm
pspspsps
asassa
zszszszs
bsbssb
sbsasa
sbsL
121
11
121
11
)()())((
)(
)()())((
)(
als1)polynomi power highest thent to(coefficie
monic are )( and )(,)(
)()(
8
IntroductionRoot locus of a motor position control (example)
2
41
2
1,
0)()(
of roots of grapha is locusRoot
)(,1)(
,)1(
1)(,
have, We
)1()(
)(
)(
)(
)( positionmotor -DC
21
2
2
Krr
KsssKbsa
sssasb
sssLAK
ss
AsG
sU
sY
sV
s
a
m
9
Introduction
0.5) (here, somefor calculated be can
21part real ,conjugatedcomplex :),(
,41For
[-1;0] interval thein real :),(
,410For
1
0
2
1
2
1),(
loop)-(open,0For 2
41
2
1),(
21
21
21
21
K
rr
K
rr
K
rr
K
Krr
break-away point
10
Sketching a Root Locus
Definition 1 The root locus is the values of s for which 1+KL(s)=0
is satisfied as K varies from 0 to infinity (pos.).
Definition 2 The root locus is the points in the s-plane where the
phase of L(s) is 180°. Def: The angle to the test point from zero number i is i.
Def: The angle to the test point from pole number i is i. Therefore,
In def. 2, notice,
integer,an is l ),1(360180 lii
positive and realK for
180)/1(,/1)( KKsL
11
110
10
100
0
1
)(
where
1
)(
1)(
spoint Test
)(
1
)(
1)(
1
ps
Rps
eRpssG
psassG
j
Sketching a Root Locus
s0
p1
What is meant by a test point? For example
12
)(exp
)()(
)()()(general, In
1121
21
)()(1
)()(1
010
0100 1
1
nmaa
bb
jan
ja
jbm
jb
n
m
jrr
rr
erer
erer
psps
zszssG
n
m
Sketching a Root Locus
s0
p
p*
13
Sketching a Root LocusHow do we find the root
locus?
We could try a lot of test points s0
Sketch by hand (6 rules)
Matlab
s0
p1
14
Sketching a Root Locus
Rule 1 (of 6) The n branches of the locus start at the poles L(s) and
m of these branches end on the zeros of L(s).
3) Rule()(
1) Rule)((0)(,(end) if
(poles) 0)(,(start)0 if
1
)(
)(0)()(
,order of is )( and ,order of is )(
Notice,
sa
zerossbK
saK
Ksa
sbsKbsa
mnmsbnsa
15
Sketching a Root Locus
Rule 2 (of 6) The loci on the real axis (real-axis part) are to the left
of an odd number of poles plus zeros.
Notice, if we take a test point s0 on the real axis :
• The angle of complex poles cancel each other.• Angles from real poles or zeros are 0° if s0 are to the right.• Angles from real poles or zeros are 180° if s0 are to the left.• Total angle = 180° + 360° l
16
Sketching a Root Locus
Rule 3 (of 6) For large s and K, n-m branches of the loci are
asymptotic to lines at angles l radiating out from a point s = on the real axis.
mn
zp
mnlmn
l
ii
l
,,2,1,)1(360180
l
For example
17
Sketching a Root Locus
180 havemust weNotice,
,0)(
11
pages),next (see ionApproximat
01
for ionapproximat an derive can we largeFor
3) Rule()(
1) Rule(0)(0)(, If
,/1)()(
)(have, We
1
11
11
n-m
ii
mn
nnn
mmm
α)(s
Rs
K
asas
bsbsK
s
sa
sbsLK
mnKsLsa
sb
s0
°
Example,n-m = 4
Thus, l°
18
Sketching a Root Locus(approximation)
)1()()()(
)(
so ,for small are )( and )( Notice,
)()(
)())((
2 that assuming , find usLet
,)()()(
)()()(
)(
)(
)()()()( Notice,
,)(
)(1
)(
)(
11
11
2
111111
21
11
11
1
11
11
11
11
sqssqssQsb
sa
ssRsO
baqqba
sRsOsqsbssas
sRsbssqssas
n-mQ(s)
ssqssQsb
sRsbsQ
sb
sa
sRsbsQsa
mnKsb
sa
Ksa
sb
mnmnmn
n
nnnnnn
mmmnmnnn
mnmn
19
Sketching a Root Locus (approximation)
0)(
11
1
)(
1 have weSo,
)( where,
)(
)()(11
1)(
)(
have weNow,
0 around,1)1(Taylor notice First,
11
11
1
11
1
mnmn
mn
mnmn
mn
r
sK
Ks
mn
q
mn
qsK
mn
qs
mns
qs
s
qsK
s
qsK
sb
sa
xxrx
20
Sketching a Root Locus
mn
zp
mn
ba
mn
qα
zzzzb
s
zszszsbsbssb
ppppa
s
pspspsasassa
ii
im
m
nmmm
in
n
nnnn
)(
have weNow,
: is oft coefficien The
)())(()(
: is oft coefficien The
)())(()(
Notice, .at look usLet
111
211
1
211
1
211
1
211
1
21
Sketching a Root Locus
n-m=1 n-m=2 n-m=3 n-m=4
mn
zp
mnlmn
l
ii
l
,,2,1
,)1(360180
3 Rule
For ex.,n-m=3
22
Sketching a Root LocusRule 4 (of 6)
The angle of departure of a branch of the locus from a pole of multiplicity q is given by (1)
The angle of departure of a branch of the locus from a zero of multiplicity q is given by (2)
)1(360180)2(
)(
)()(
)(
)( mult. of Zero
)1(360180)1(
)(
)(
)(
1
)(
)( mult. of Pole
,
0
,
0
lq
sa
sbzs
sa
sbq,
lq
sa
sb
pssa
sbq,
liiidepl
q
liiidepl
q
Notice, the situation is similar to the approximation in Rule 3
23
Sketching a Root Locus
arrldepl
liiarrli
liidepli
ii
q
q
l
,,
,
,
and find can weSo,
Notice,
)1(360180
have weBecause
s0
Departure and arrival?For K increasing from zero to infinity poles go towards zeros or infinity. Thus,• A branch corresponding to a pole departs a some angle.• A branch corresponding to a zero arrives at some angle.
In general, we must find as0 such thatangle(L(s))=180°
Vary s0 untilangle(L(s))=180°
24
Sketching a Root LocusRule 5 (of 6)
The locus crosses the j axis at points where the Routh criterion shows a transition from roots in the left half-plane to roots in the right half-plane.
Routh: A system is stable if and only if all the elements in the first column of the Routh array are positive.
0)Row(s
3)Row(s
2)Row(s
1)Row(s
1)Row(s
0
3213-n
3212-n
5311-n
42n
cccn
bbbn
aaan
aan
21
31
11
51
4
12
31
2
11
det1
1det
1
1det
1
bb
aa
bc
aa
a
ab
aa
a
ab
25
Sketching a Root Locus
Rule 6 (of 6) The locus will have multiplicative roots of q at points
on the locus where (1) applies. The branches will approach a point of q roots at
angles separated by (2) and will depart at angles with the same separation.
q
l
ds
dba
ds
dab
)1(360180)2(
0)1(
26
Sketching a Root LocusExampleRoot locus for double integrator with P-control
Rule 1: The locus has two branches that starts in s=0. There are no zeros. Thus, the branches do not end at zeros.
Rule 3: Two branches have asymptotes for s going to infinity. We get
02
0
270
90
2
)1(360180)1(360180
mn
zp
l
mn
l
ii
l
Eq.)(Char.01
1)(,1
)(22
skksD
ssG pp
27
Sketching a Root Locus
Rule 2: No branches that the real axis.
Rule 4: Same argument and conclusion as rule 3.
Eq.)(Char.01
1)(,1
)(22
skksD
ssG pp
Rule 5: The loci remain on the imaginary axis. Thus, no crossings of the j-axis.
Rule 6: Easy to see, no further multiple poles. Verification:
0021)(
10
1)(,)(
22
2
ssds
ds
ds
sd
ds
dba
ds
dab
sbssa
28
Sketching a Root Locus
Eq.)(Char.0)1(
101
)1(1
1and
Eq.)(Char.01
)(1
)(,1
)(
22
2
2
s
sK
ss
K
k
kkK
sskk
skksDs
sG
D
pD
Dp
Dp
ExampleRoot locus for satellite attitude control with PD-control
29
Sketching a Root Locus
90 areat 0 pole double thefrom departure of angles The
:4 Rule
axis real negative thealong asymptote one is there,1 Because
:3 Rule
locus theon is 1 ofleft the toaxis real The
:2 Rule
infinity. approachesother The
.1at zero theon s terminateOne
.0at start that branches twoare There
:1 Rule
1)1(0
)1(1
22
s
mn
s
s
s
Ks
s
s
sK
30
Sketching a Root Locus
2,002)1(
2,1,)(,1)(
from found are roots multiple of points The :6 Rule
1
axis.imaginary thecrossnot does locus theThus,
below.array thefind wecriterion, s Routh'Applying :5 Rule
Eq.)(Char.0)1(
1
2
2
2
issssds
dba
ds
dab
sds
da
ds
dbssassb
K
K
K
s
sK
31
Sketching a Root Locus
32
Selecting the Parameter ValueThe (positive) root locus
A plot of all possible locations of roots to the equation 1+KL(s)=0 for some real positive value of K.
The purpose of design is to select a particular value of K that will meet the specifications for static and dynamic characteristics.
For a given root locus we have (1). Thus, for some desired pole locations it is possible to find K.
LK
sLK
1
)(
1)1(
33
Selecting the Parameter Value
657.7*1.2*4)()()(
1
))((
1
then,5.0 want say we usLet
, ,at Poles
)44)(44(
1
)16)4((
1
32000
32000
321
2
ssssssL
K
sssss)L(s
sss
jsjss
ssL(s)
Example
34
Selecting the Parameter ValueMatlab
A root locus can be plotted using Matlab rlocus(sysL)
Selection of K [K,p]=rlocfind(sysL)