Transcript chapter8-note

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
§ 8.1 Root Locus in Control Problem (1)
• Proportional Control Problem:
r(t) +
KG(s)
R(s)
m
y(t)
u(t)
Y(s)
r
b(t)
l
g
K
b(t)
H(s)
controller
plant and sensor
K: Can be controller parameter or plant parameter
C-L System:
T(s) 
KG(s)
1  KG(s)H(s)
N ( s)
N ( s)
Let G(s)  G , H(s)  H
DG ( s)
DH ( s )
 T(s) 
KNG (s)DH (s)
DG (s)DH (s)  KNG (s)NH (s)
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.
§ 8.1 Root Locus in Control Problem (2)
• Fundamental of Root Locus:
Definition 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 0   ).
Usage of root locus:
Provide relationship among open-loop system parameter, closedloop pole location, and output transient response.
§ 8.1 Root Locus in Control Problem (3)
• Root Locus of 2nd-order Closed-loop System:
n
s( s  2n )
2
R(s) +
T(s) 
n
Y(s)
2
s2  2ns  n
2
Roots :   1, s  n  n  2  1
  1, s  n , - n
1
  1, s  n  id    id , d  n 1   2

Root locus:
(1) For fixed n,  as a varied parameter (2) For fixed , n as a varied parameter
j
0

 1
     
0
0
j
1
n  
1


  1   

n  0
   0
n  
cos   
§ 8.2 Magnitude-phase Representation (1)
• Vector Representation of Complex Numbers:
s = -a+bj= a2 +b2 -(tan-1 b a ), a, b>0
Cartesian
Coordinate
s
s
Polar Coordinate
j
s=-a+bj
1
a
s=-a-bj
2
v
s1
b
0
v
s1

s2
s1
0  s
1
b
Cartesian Coord.
r
s  (Real Part)  j(Imaginar y Part)
v
v
s2  s1
v
s2
v
s2
Polar Coord.
r
s  (Magnitude)(phase)
s2  s1
§ 8.2 Magnitude-phase Representation (2)
• Magnitude-phase Representation
(1) Absolute
magnitude and phase
v
s  s s  s e js
v
v
s
s
s
s
s
s
o
As a free vector
Any horizontal reference line
(2) Relative magnitude and phase
j
a
l1
c
 c
q1
 q1
 a
l1
b
point a: -+j

aa
point b: --j

aa
point c: -c
a
0
a
v

l1
a
a/c

j
v
c

0
a
q1
c
 q1
a
l1
 a
point a relative to c : (   a  j  a )  (   c )
b
 (  c   a )  j a

l 1 q 1
l 1e j q
o
point b relative to c : l 1  q 1 or l 1 ( 360  q 1 )


1
l 1e j(  q
1)
c  a
§ 8.2 Magnitude-phase Representation (3)
Ex: Find residues in partial fraction expansion for
Y(s)
1.25
T(s) 

by pole-zero representation.
R(s) s(s  2.5)(s  4.5)
K3
K
K2
Sol: T(s)  1 

s s  2 . 5 s  4 .5
1.25
1.25
1
K 1  T(s)  s s0 


(s  2.5)( s  4.5) s0 (2.50)( 4.50) 9
1.25
1.25
1
K 2  T(s)  (s  2.5) s 2.5 


s(s  4.5) s 2.5 (2.5180)( 20)
4
1.25
1.25
1.25
K 3  T(s)  (s  4.5) s 4.5 


s(s  2.5) s 4.5 ( 4.5180)( 2180)
9
pole-zero representation
j
j
j
4.50
 4 .5

 2 .5
2180
20

 4 .5
2.50
(K 1 )
1.25
1
1
9
4 
T(s)  9 
s s  2.5 s  4.5
2.5180
(K 2 )

4.5180
(K 3 )
§ 8.2 Magnitude-phase Representation (4)
0.25
0
Ex: If s is a complex variable, solve 1 
s(s  1)
0.25
0.25
Sol: 1 
0
 1     (1)
s(s  1)
s(s  1)
poles: s=0, s=-1
In complex plane
s  ? to satisfy eq.(1)
s
s 1  l2
j
j (e j( 90 ) )
s  l1
s  q1
q3
j(180 )
(e
)  1 (s  1) 0
 q2
From eq.(1)
q
Unit circle
e jq

1 ( e j( 0  ) )
j
(e
j( 270  )
)
0.25
jn180 o
 1  1 e
, n  1,  3,   
( s e js )( s  1e j( s1) )
0.25

Mag
:
 1   (2a)

s s 1

Phase : (s  (s  1))  n  180o , n  1,  3,   

If n  1, s  (s  1)  180o    (2b)
§ 8.2 Magnitude-phase Representation (5)
Where is the “s” to satisfy eqs. (2a) and (2b) in complex plane?
(A) From phase relationship (2b)
q1  q2  180
 q1  q3  180
(B) From magnitude relationship (2a)
0.25
1
l1  l 2
 l 1  l 2 (eq.3)
 q2  q3
 l 1  l 2    (3)
 l 1  l 2  0.25
2
i.e. s is on the bisection line
j
2
 l 1  l 2  0 .5
j
s
1
0

Conclusions: From (A) and (B)
s
 1 l1
s=-0.5, -0.5
l2
0

§ 8.2 Magnitude-phase Representation (6)
Ex: Find the root locus of a DC-servo position control system with varied K.
plant
R(s) +
u
K
u
1
s(s  1)
-
Y(s)
,K  0
y
r
y
Plant
K
controller
C
G
K


Sol: C-L system,
R 1 2GH s2  s  K
characteristic eq. s  s  K  0
(A) Direct solution method
s  0.5  0.25  K
pole-zero representation
j
K 0
K  1.25
1 .0
K  0 .5
0 .5
K  0.25
K 0
 0 .5
1
0
K  0 .5
 0 .5
K  1.25
 1 .0
Root locus with varied K
Controller
Gain, K
Poles, s
0
0,  1
0.25
 0.5,  0.5
0.5
 0 . 5  0 .5 j
1.25
 0 .5  j
O-L System

C-L system is stable with varied K.
C-L system is underdamping when K>0.25.
§ 8.2 Magnitude-phase Representation (7)
(B) Use magnitude-phase representation
2
characteristic eq. s  s  K  0
1
1
i.e. 2

s s
K
Mag : s s  1  K
K
or
 1  
s(s  1)
Phase : (s  0)  (s  ( 1))  180
From the example in section 8.2:
(a) Phase relationship  s is on the bisection line between 0 and –1.
(b) Magnitude relationship
j
K  0.25  l 1  l 2  l  0.5
1
K  0.5  l  0.5, i.e. l 
2
K  0.25, s  0.5,  0.5
From (a) and (b)  
K  0.5, s  0.5  j0.5
2
K  0.5
l
l
K  0.25
1 K  0
 0 .5
K  0.5
Root locus
0
K 0

§ 8.2 Magnitude-phase Representation (8)
Step response
r(t)
K=1.25
K=0.5
shaft pointer
1
K=0.25
K=0.25
K=0.5
Start
Start
Rotation and then stop
t
Key parametric values of K in root locus:
(1) Intersection between root locus and j -axis
Critical K of Instability
(2) Intersection between root locus and  -axis
Critical K of Oscillation
Oscillation and then stop
§ 8.3 Evan’s Root-locus Method (1)
• Method
Open-loop
pole-zero diagram
Evan's R-L
Method
Closed-loop
pole locations
Adjust parameter
K
1. Parametrilization of closed-loop system with K varied as:
+
KG(s)
sm  bm1sm1      b1s  b0
G(s)H(s)  n
,n  m
n1
s  an1s      a1s  a0
H(s)
The highest power in both the numerator and denominator of G(s)H(s) are
normalized to unity.
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:
Magnitude criterion : KG(s)H(s)  1
Phase criterion : KG(s)H(s)  180  k  360  (2k  1)  180, k  0,  1,  2,   
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.
§ 8.3 Evan’s Root-locus Method (2)
• Terminologies and Symbols About Root Locus
j
Asymptotic line
Branch(1)
Centroid of
asymptotes
Angles of
departure
o
qd
Breakaway point
Break-in point

Branch(3)
Angles of
asymptotes
qk
Branch(2)
1
 0, K  0, a  c)
(s  c)(s  a  bi)(s  a - bi)
K(s  z1 )(s  z 2 )    (s  zm )
KG(s)H(s) 
,n  m
(s  p1 )( s  p 2 )    (s  pn )
# p : number of poles  pi : Algebraic sum of the values of poles
(Ex : 1  K
# z : number of zeros
 z : Algebraic sum of the values of zeros
i
§ 8.3 Evan’s Root-locus Method (3)
• Five Rules for Sketching the Root Locus
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 K   , 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.
Equation of asymptotes
0 
(  pi   zi )
#p  # z
( 2k  1)180
qk 
, k  0, 1, 2,   
#p  # z
§ 8.3 Evan’s Root-locus Method (4)
• Refining the Sketch
Root locus calibration:
1
G(st )H(st ) 
for points s t on the locus
K
 st  pi
K 
 s t  zi
Breakaway (Break-in) points:
K attains local maximum (minimum) on the real axis.
dK
i.e.
 0 for necessary condition.
ds sst
Points of imaginary-axis crossings:
Found by using Routh-Hurwitz criterion or substituting s  j 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.
§ 8.3 Evan’s Root-locus Method (5)
Ex: Find root locus for a DC motor position servo as
+
K
s(s  1)
Sol: C-L poles: 1 
K
 0, s(s  1)+k=0, Two branches
s(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.
(2k  1)180
Asymptotes: qk 
, k  0, 1, 2,  
20
 q0  90, q1  270
(0  ( 1))  (0)
1

20
2
dK
1
1

0
,
s



K

Breakaway point:
ds
2
4
0 
§ 8.3 Evan’s Root-locus Method (6)
j
real axis
segment
270 
90 
K0
s  1

1
2
K
K0
1
4
s0

Breakaway
point
Asymptotes
The closed-loop position servo is stable for any positive proportional gain.
§ 8.3 Evan’s Root-locus Method (7)
Ex: Find stability conditions for various K in the characteristic equation:
s3  3s2  2s  K  0, K  0
K
Sol: Standard form 1 
0
s(s  1)(s  2)
C-L poles: s3  3s2  2s  K  0, Three branches of loci
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.
(2k  1)180
, k  0, 1, 2,  
Asymptotes: qk 
30
 q0  60, q1  180, q2  300
0 
( 0  1  2)  ( 0 )
 1
30
3
2
Imaginary axis crossing: set s  j ⇒ K  ( s  3s  2s)
s  j
Real part, K  32
  0, K  0
Imaginary part, j(2 - 2 )  0  

   2, K  6
§ 8.3 Evan’s Root-locus Method (8)
s  0.423  K  0.385
dK
0
Breakaway point:
ds
s  1.58 (Not feasible )
Root locus and stability conditions
j
2, K  6
K=0
180
-2
K=0
60
-1
300
-0.423
K=0.385
0
K=0
 2, K  6
Asymptotes
Stability conditions
(1) 0  K  0.385
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
§ 8.3 Evan’s Root-locus Method (9)
Ex: Consider a system includes lightly damped flexible modes near
imaginary axis, find the root for 1  KG(s)H(s)  0
(s  0.1)2  25
(s  0.1) 2  36
(1) G(s)H(s) 
(2) G(s)H(s) 
2
s(( s  0.1)  36)
s(( s  0.1)2  25)
j
j

Sol: (1)
(2)
2
1
1
st
1
2
1


3
2
2
3
Find angle of arrival 1 (s t close to zero)
Find angle of departure (s t close to pole)
 1  2  3  1   2  180  360k, k  0  1  2  3  1   2  180  360k, k  0
1   2  3  90
2  90  1  180
Stability insensitive
to varied K
j
1   2  3  90

1  90  2  0
Stability sensitive
to varied K
j

§ 8.4 Root-locus Method in Design (1)
• Design Problems
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.
• Configurations of Compensation
+
Gc(s)
G(s)
+
+
G(s)
Hc(s)
Cascade Compensation Feedback Compensation
O-L Transfer Function: Gc (s)G(s)
Static Compensator: Gc (s)  K p
Gc(s)
G(s)
Hc(s)
Cascade and Feedback
Compensation
G(s)Hc (s)
Gc (s)G(s)Hc (s)
Hc (s)  K b
Gc (s)  K p , Hc (s)  K b
§ 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
sz
z
PD controller
PI controller
D(s) 
, DC Gain 
K
sp
p
Gc (s)  K P  K Ds Gc (s)  K P  I
s
Phase lead
Phase lag
j
j
Compensator

p

z
0  z  p,   
D( j)  0


z

p
0  p  z,   
D( j)  0

Effects
Passive
Active
Improve transient speed
Increase stability margin
Lead
PD
Network Controller
Improve s.s.error
Reduce stability margin
Lag
PI
Network Controller
PID controller or lag-lead compensator can be used to improve transient
response, s.s. error and trade off stability margin.
§ 8.4 Root-locus Method in Design (3)
• Design Constraints
Feasible region of 1st / 2nd-order dominant poles
j
j
cos 1 


s 
% o.s.  a%
(2nd  order )
Optimal region of 2nd-order dominant poles
stable but
oscillatiory
stable but
sluggish
45°

 = 0.7, % o.s. < 5%
optimal
region

1 4

 ts
SettlingTime (Ts )  t s
(1st / 2nd  order )
j
j
%o.s.  a%

Ts  t s
§ 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.
j
j
K   j
K
K
K
1
K=0
K0
K0
1
1

2
0

K=0
K=0
K=0
-2
-1
0
K=0

-2
-1
K=0
0

K=0
-1
K
K
G(s)H(s) :
K
s(s  1)
K
K
K
s(s  1)( s  2)
K
s(s  1)( s  2  j)( s  2  j)
§ 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.
K
j
j
j
K
K0
K0
1
1

2
0

K  K0
K
-2
-1
K0

K0
-2
-1
K0
0
K
K
G(s)H(s) :
K
s(s  1)
K(s  2)
s(s  1)
K(s  2  j)( s  2  j)
s(s  1)

§ 8.4 Root-locus Method in Design (6)
• Practical Compensator Realized by OP:
1

 s
Eo (s)  R 4C1 
R1C1


Ei (s)  R3C2  s  1

R 2C 2


1

s

T
  Kc
1
s

T

RC
Eo(s) T  R1C1, T  R 2C2 , K c  4 1
R 3C2
C2
C1
R2
R1
R4
R3
-
-
+
Ei(s)
+
E(s)
(1) R1C1  R2C2 : Lead Compensato r
(  1)
(  1)
j

1
1

R 2C 2
R1C1
j
1
st

(2) R2C2  R1C1 : Lag Compensato r

st




1
1

R1C1 R 2C 2
1



§ 8.4 Root-locus Method in Design (7)
• Root Sensitivity
Robust K of characteristic roots
s
s  K ds
K
s dK
K
dK
0
Breakaway (breakin) point:
ds
s
 sK  
S 
s
K
Avoid selecting the value of K to operate at the breakaway (breakin) points.
m
For characteristic eq. 1+KG(s)H(s)=0, KG(s)H(s) 
K  (s  zi )
i1
n
 (s  p )
i
i1
sensitivity of C-L pole at s  - : sK
s

K
GH  s   

s
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.
§ 8.4 Root-locus Method in Design (8)
Ex: For a DC-servo with position and tacho feedbacks, find K1 and K2 to satisfy the
control specs: (1) settling time  3 sec, (2) dominant poles with   0.707,
(3) ramp-input steady state error  10%
R(s)
1  K 2s
+
Ea(s)
K1
s(s  2)
Y(s)
1  K 2s
Sol: G(s)H(s) 
K1(1  K 2s)
s(s  2)
Characteristic eq. 1 G(s)H(s)  0, s2  2s  K1K 2s  K1  0
or s2  2s  as  b  0, a  K1K 2 , b  K1
From ramp input
1
ess  lim
 0.1  K1  20
s0 sG( s)H( s)
From setting time
3
1 4
Ts  4  3 sec,   ,   
4
 3
Root locus for b=K1=20
as
1 2
0
s  2s  20
§ 8.4 Root-locus Method in Design (9)
Feasible region and root locus
Intersecti on of root locus and ξ  0.707 line :
  0 . 707
5
a  4 .3
4.36
sd
4
3
2
1
4
-6
-5
-4
-3
-2
3
a  4.3  20K 2  K 2  0.215
Intersecti on point
sd  -3.15  3.15j,
1

 3.15
0
-1
-1
-2
-3
-4
-4.36
-5
Check settling time
Ts  4  1.27 sec
K  20
 select  1
K 2  0.215