Chapter 8 Root Locus and Magnitude-phase Representation § 8.1 Root Locus in Control Problem § 8.2...
-
Upload
godwin-bond -
Category
Documents
-
view
254 -
download
4
Transcript of Chapter 8 Root Locus and Magnitude-phase Representation § 8.1 Root Locus in Control Problem § 8.2...
Chapter 8 Root Locus and Magnitude-phase Representation
§ 8.1 Root Locus in Control Problem
§ 8.2 Magnitude-phase Representation
§ 8.3 Evan’s Root-locus Method
§ 8.4 Root-locus Method in Design
• Proportional Control Problem:
§ 8.1 Root Locus in Control Problem (1)
KG(s)Y(s)R(s)
+
H(s)
r(t) y(t)
b(t)
K: Can be controller parameter or plant parameter
)s(N)s(KN)s(D)s(D
)s(D)s(KNT(s)
)s(D
)s(NH(s) ,
)s(D
)s(NG(s) Let
G(s)H(s)K1
KG(s)T(s)
HGHG
HG
H
H
G
G
C-L System:
Zeros of T(s): Consist of the zeros of G(s) and poles of H(s)
Poles of T(s): Changes with K
Problem: Investigate the effect of varied K on controlling system dynamics
through the poles of closed-loop system.
m
lg
Kr
u(t)
b(t)
controller plant and sensor
• Fundamental of Root Locus: Definition of root locus:
§ 8.1 Root Locus in Control Problem (2)
Usage of root locus:
Graphical representation of the path of the closed-loop poles as one or more parameters of the open-loop transfer function are varied (usually ).
Provide relationship among open-loop system parameter, closed-loop pole location, and output transient response.
0
Root locus:
• Root Locus of 2nd-order Closed-loop System: § 8.1 Root Locus in Control Problem (3)
R(s) + Y(s)
)2s(s n
2n
1 ,i1
is 1,
- ,s 1,
1s 1, :Roots
s2s)s(T
2ndddn
nn
2nn
2nn
2
2n
parameter varied a as , fixed For (1) n parameter varied a as , fixed For (2) n
j
1
0
0
1
0 0
1
j
0n
ncos
n 1
• Vector Representation of Complex Numbers: § 8.2 Magnitude-phase Representation (1)
2 2 -1 bs = -a+bj= a +b -(tan ), a, b>0a
Cartesian
Coordinate
Polar Coordinate
s s
b
b
a 0
j
1s =-a+bj
2s =-a-bj
Part)y j(ImaginarPart) (Reals
Coord. Cartesian
0
1s
2s
(phase))(Magnitudes
Coord. olarP
12
12
ss
ss
1s2s1s
2s
1s
• Magnitude-phase Representation(1) Absolute magnitude and phase
sjessss
§ 8.2 Magnitude-phase Representation (2)
(2) Relative magnitude and phase As a free vector
s s
o
s
s s
s
Any horizontal reference line
a
a 0
ja
b
a a
a a
c
point a: - +jpoint b: - -jpoint c: -
a
c1
1
1
1
c
a
ac
a
b
a1
1
1
1
c0
j
a
c
c/a
1j1
11
aac
caa
e
j)(
)()j( :c to relative a point
)j(1
1o
111
1e
)360( or :c to relative b point
Ex: Find residues in partial fraction expansion for
Y(s) 1.25T(s)
R(s) s(s 2.5)(s 4.5)
§ 8.2 Magnitude-phase Representation (3)
by pole-zero representation.
5.4s
K
5.2s
K
s
K)s(T 321
Sol:
9
1
)05.4)(05.2(
25.1
)5.4s)(5.2s(
25.1s)s(TK
0s0s1
4
1
)02)(1805.2(
25.1
)5.4s(s
25.1)5.2s()s(TK
5.2s5.2s2
9
25.1
)1802)(1805.4(
25.1
)5.2s(s
25.1)5.4s()s(TK
5.4s5.4s3
pole-zero representation
5.4s9
25.1
5.2s4
1
s9
1)s(T
j
5.25.4
05.2
05.4
)K( 1
j
5.4
1805.2
02
)K( 2
j
1805.4
1802
)K( 3
Ex: If s is a complex variable, solve 0)1s(s
25.01
§ 8.2 Magnitude-phase Representation (4)
(1) 1)1s(s
25.00
)1s(s
25.01
Sol:
poles: s=0, s=-1In complex plane
From eq.(1)
3, 1,n ,e11
)e1s)(es(
25.0 o180nj)1s(jsj
3, 1,n ,180n))1s(s(:Phase
)a2(11ss
25.0:Mag
o
)b2(1801)s(s ,1n If o
0
j
11
j
j
)e( )0(j
)e( )270(j
)e( )90(j
)e( )180(j
je
circle Unit
2
)1s(
1s
1s
21s 3
seq.(1)satisfy to ?s
Where is the “s” to satisfy eqs. (2a) and (2b) in complex plane?
§ 8.2 Magnitude-phase Representation (5)
(A) From phase relationship (2b)
(3)
180
180
21
32
31
21
5.0
25.0
(eq.3)
125.0
21
2
2
2
1
21
21
(B) From magnitude relationship (2a)
i.e. s is on the bisection line
Conclusions: From (A) and (B) s=-0.5, -0.5
0
j
s
1 0
j
s
1 21
§ 8.2 Magnitude-phase Representation (6)Ex: Find the root locus of a DC-servo position control system with varied K.
Sol: C-L system,
Ku+
-
R(s) Y(s)
)1s(s
1
plant
controller
K
r
Controller
y
u
y
Plant
0K ,
Kss
K
GH1
G
R
C2
characteristic eq. 0Kss2 (A) Direct solution method
K25.05.0s pole-zero representation
Root locus with varied K
C-L system is stable with varied K.
C-L system is underdamping when K>0.25.
0
j
1 5.0
5.0
0.1
0.1
5.0
25.1K
5.0K
25.0K 0K 0K
5.0K
25.1K
Gain, K Poles, s
0
0.25
1.25
0.5
1 ,0
5.0 ,5.0
j5.05.0
j5.0
O-L System
§ 8.2 Magnitude-phase Representation (7)(B) Use magnitude-phase representation
characteristic eq. 0Kss2
180))1(s(0)(s :Phase
K1ss :Mag1
1)s(s
K or
K
1
ss
1 .e.i
2
From the example in section 8.2:
(a) Phase relationship s is on the bisection line between 0 and –1.
(b) Magnitude relationship
2
1 i.e. ,5.05.0K
5.025.0K
2
21
j0.5 0.5s 0.5,K
0.5 0.5,s 0.25,K (b) and (a) From
0
j
1
5.0 0K 0K
25.0K
5.0K
5.0K
Root locus
§ 8.2 Magnitude-phase Representation (8)Step response
Key parametric values of K in root locus:
(1) Intersection between root locus and -axis
Critical K of Instability
(2) Intersection between root locus and -axis
Critical K of Oscillation
j
1
r(t)
t
K=1.25
K=0.5
K=0.25K=0.25
Start
Rotation and then stop
shaft pointer
K=0.5
Start
Oscillation and then stop
§ 8.3 Evan’s Root-locus Method (1)
1. Parametrilization of closed-loop system with K varied as:
• MethodEvan's R-L
MethodOpen-loop
pole-zero diagramClosed-loop
pole locations
Adjust parameterK
2. Obtain characteristic equation of the closed-loop poles: 1+KG(s)H(s)=0
3. From magnitude-phase representation to obtain criteria for poles and zeros:
KG(s)+
H(s)
4. Sketch shape with scale and parameter K:
The shape of root locus is determined entirely by the phase criterion.
The magnitude criterion is used only to assign scale and parameter K of the
locus.
1KG(s)H(s) :criterion Magnitude
mn ,asasas
bsbsbsG(s)H(s)
011n
1nn
011m
1mm
The highest power in both the numerator and denominator of G(s)H(s) are
normalized to unity.
2, 1, 0,k ,180)1k2(360k180KG(s)H(s) :criterion Phase
§ 8.3 Evan’s Root-locus Method (2)• Terminologies and Symbols About Root Locus
mn ,)ps()ps)(ps(
)z(s)z)(szK(sKG(s)H(s)
c)a 0,K ,0bi)-abi)(sac)(s(s
1K1 :Ex(
n21
m21
zeros of values the of sum Algebraic:z zeros of number :z#
poles of values the of sum Algebraic:p poles of number :p#
i
i
j
Breakaway point
Branch(1)
Branch(2)
Branch(3)
Break-in point
Asymptotic line
Angles ofasymptotes
k
Centroid ofasymptotes
Angles ofdeparture
d
o
§ 8.3 Evan’s Root-locus Method (3)• Five Rules for Sketching the Root Locus
2 1180
i i0
k
Equation of asymptotes
( p z )
#p # z
( k ) , k 0, 1, 2,
#p # z
1. Number of branches: The number of branches of the locus is equal to
the order of the characteristic polynomial.
2. Symmetry: The locus is symmetrical about the real axis.
3. Real-axis segments: For K>0, the locus on the real axis exists to the
left of an odd number of real-axis open-loop poles
plus zeros.
4. Starting and ending points: The locus begins at the poles of G(s)H(s)
with K=0 and terminate with , either at
the zeros of G(s)H(s) or at infinity.
5. Behavior at infinity: The locus approaches straight line asymptotes as the
locus approaches infinity.
K
§ 8.3 Evan’s Root-locus Method (4)• Refining the Sketch
Points of imaginary-axis crossings:
Found by using Routh-Hurwitz criterion or substituting into the equation of root locus and solving the equations.
Angles of departure (arrival):
Obtain by choosing an arbitrary point infinitesimally close to the pole (zero)
and applying the angle criterion.
Root locus calibration:
Breakaway (Break-in) points:
K attains local maximum (minimum) on the real axis.
t t t
t i
t i
1 G(s )H(s ) for points s on the locus
Ks p
Ks z
condition.necessary for 0ds
dK .e.i
tss
js
§ 8.3 Evan’s Root-locus Method (5)Ex: Find root locus for a DC motor position servo as
+
)1s(s
K
Sol: C-L poles:K
1 0, s(s 1)+k=0, Two branchess(s 1)
O-L poles: s=0, s=-1, # p=2
zeros: none, # z=0
Real-axis segment: lies between the poles of s=0 and s=-1.
,2 ,1 ,0k ,
02
180)1k2(kAsymptotes:
Breakaway point:4
1K
2
1s ,0
ds
dK
270 ,90 10
2
1
02
)0())1(0(0
§ 8.3 Evan’s Root-locus Method (6)
The closed-loop position servo is stable for any positive proportional gain.
j
1s 2
1
0K 0K
4
1K 0s
90
270
Breakaway point
Asymptotes
real axis segment
§ 8.3 Evan’s Root-locus Method (7)Ex: Find stability conditions for various K in the characteristic equation:
Sol: Standard form 02)1)(ss(s
K1
C-L poles:
O-L poles: s=0, s=-1, s=-2, # p=3
zeros: none, # z=0
Real-axis segment: lies between the poles of s=0 and s=-1.
lies to the left of pole s=-2.
,2 ,1 ,0k ,
03
180)1k2(kAsymptotes:
0K ,0Ks2s3s 23
loci of branches Three ,0Ks2s3s 23
Imaginary axis crossing:
js
23 )s2s3s(K js set ⇒
0)-(2j part,Imaginary 3K part, Real
2
2
6K ,2 0K ,0
300 ,180 ,60 210
103
)0()210(0
§ 8.3 Evan’s Root-locus Method (8)
Breakaway point:
)feasible Not( 58.1s
385.0K 423.0s 0
ds
dK
Root locus and stability conditions
Stability conditions
(1)
three negative real roots
stable
(2) 0.385<K<6
two complex roots, one negative real root
stable
(3) K>6
two complex roots in RHP, one negative
real root
unstable
385.0K0
0-1-2K=0
K=0
K=0
-0.423K=0.385
Asymptotes
6K ,2
6K ,2
60180
300
j
§ 8.3 Evan’s Root-locus Method (9)Ex: Consider a system includes lightly damped flexible modes near
imaginary axis, find the root for 0)s(H)s(KG1
)36)1.0s((s
250.1)(sG(s)H(s) (1)
2
2
)25)1.0s((s
360.1)(sG(s)H(s) (2)
2
2
Sol: (1) (2)
18090
90
0k ,k360180
zero) to close (s arrival of angle Find
12
321
21321
t1
090
90
0k ,k360180
pole) to close (s departure of angle Find
21
321
21321
t
j
j
j
ts
j
Stability insensitive
to varied KStability sensitive
to varied K
§ 8.4 Root-locus Method in Design (1)• Design Problems
• Configurations of Compensation
Adjust parameter and / or introduce the pole(s) and zero(s) of the dynamic
compensator to alter the root locus so that the performance specifications
can be satisfied.
Performance specs: Stability Margin, Transient Response, Steady State Error.
G(s)+
Hc(s)
Gc(s)+
G(s) Gc(s)+
G(s)
Hc(s)
Cascade Compensation Feedback Compensation Cascade and Feedback Compensation
O-L Transfer Function:
Static Compensator: pc
c
K)s(G
)s(G)s(G
bc
c
K)s(H
)s(H)s(G
bcpc
cc
K)s(H ,K)s(G
)s(H)s(G)s(G
§ 8.4 Root-locus Method in Design (2)• Dynamic Compensators
Passive compensator --- Mainly RC type network
Active compensator --- Mainly OP and RC circuit
Require external power
Basic Gain and Phase Compensation Passive Compensator Active Compensator
PD controller PI controller
Phase lead Phase lag
PID controller or lag-lead compensator can be used to improve transient
response, s.s. error and trade off stability margin.
p
zGain DC ,
ps
zs)s(D
sKK)s(G DPc
s
KK)s(G I
Pc
p z
j
,pz00)j(D
pz
j
,zp0
0)j(D
CompensatorEffects Passive Active
LeadNetwork
LagNetwork
PDController
PIController
Improve transient speedIncrease stability margin
Improve s.s.errorReduce stability margin
§ 8.4 Root-locus Method in Design (3)• Design Constraints
Feasible region of 1st / 2nd-order dominant poles
Optimal region of 2nd-order dominant poles
j
1cos
j
1 4
sst
j
%a.s.o % )ordernd2(
s sSettlingTime (T ) t)ordernd2/st1(
s s
%o.s. a%
T t
j
°45
stable butoscillatiory
5%<o.s. % ,7.0=
stable butsluggish
optimalregion
§ 8.4 Root-locus Method in Design (4)• Addition of Poles to G(s)H(s)
The effect of adding a pole to G(s)H(s) is to push the root loci toward the
RHP.
:)s(H)s(G )2s)(1s(s
K
)j2s)(j2s)(1s(s
K
j
12
1 0
0K 0K
K
K
0-1-2
K=0 K=0 K=0
j
K
K
)1s(s
K
0-1-2
K=0
K=0 K=0
j
K
K
K=0-1
1
K
K
§ 8.4 Root-locus Method in Design (5)• Addition of Zeros to G(s)H(s)
The effect of adding a left-half plane zero to G(s)H(s) is to move and bend
the root loci toward the LHP.
:)s(H)s(G )1s(s
)2s(K
)1s(s
)j2s)(j2s(K
j
12
1 0
0K 0K
K
K
)1s(s
K
j
0K
-2 -1K
0K K
j
0K
-2 -1
K
0K
K
0
§ 8.4 Root-locus Method in Design (6)• Practical Compensator Realized by OP:
1)(
rCompensato Lead :CRCR (1) 2211
1)(
rCompensato Lag :CRCR (2) 1122
j
22CR
1
11CR
1
ts 1
j
22CR
1
11CR
1
ts 1
+
--
+
R1
R2
R3
R4
Eo(s)Ei(s)
C1
C2
E(s)
T1
s
T1
sK
CR1
s
CR1
s
CR
CR
)s(E
)s(Ec
22
11
23
14
i
o
23
14c2211 CR
CRK ,CRT ,CRT
§ 8.4 Root-locus Method in Design (7)• Root Sensitivity Robust K of characteristic roots
Breakaway (breakin) point:
dK
ds
s
K
KK
ss
SsK
0ds
dK
sKs
Avoid selecting the value of K to operate at the breakaway (breakin) points.
Root locus is less sensitive to changes in gain at the lower value of K.
Root sensitivity provides information of changes in both magnitude and direction of specific root for designer.
For characteristic eq. 1+KG(s)H(s)=0,
n
1ii
m
1ii
)ps(
)zs(K)s(H)s(KG
K ss
Ksensitivity of C-L pole at s - : s GH s
§ 8.4 Root-locus Method in Design (8)
From ramp input
Characteristic eq.
)2s(s
)sK1(K)s(H)s(G 21
20K1.0)s(H)s(sG
1lime 1
0sss
Ex: For a DC-servo with position and tacho feedbacks, find K1 and K2 to satisfy the
control specs: (1) settling time sec, (2) dominant poles with ,
(3) ramp-input steady state error 3 707.0
%10
Y(s)R(s) + Ea(s)
sK1 2
)2s(sK1
sK1 2
Sol:
0KsKK2ss0,G(s)H(s)1 1212
1212 Kb ,KKa ,0bass2s or
From setting time
3
41 ,
4
3 sec,34Ts
Root locus for b=K1=20
020s2s
as1
2
§ 8.4 Root-locus Method in Design (9)
0.215K20K4.3a
:line 0.707ξ and locus root of onIntersecti
22
Feasible region and root locus
3.151
3.15j,-3.15s
point onIntersecti
d
sec27.14T
time settling Check
s
215.0K
20K select
2
1
-6 -5 -4 -3 -2 -1
-5
-4
-3
-2
-1
0
1
2
3
4
5
-4.36
4.36
707.0
ds3.4a
34