Transcript Document

7.1 Root Locus (RL) Principle
We introduce the RL through an example.
Consider servo motor system shown bellow
A(s) +
–
compensator
K
motor
1
s( s  2)
R(s)
To have more understanding how K affect the
system characteristic we will plot the locus of
the roots in the s plane as K is varied from 0
to infinity. The roots are
s
 2  4  4K
 1  1  K
2
The closed loop transfer function is
K
K
s ( s  2)
T (s) 
 2
K
s  2s  K
1
s ( s  2)
The CE is
s2 + 2s + K = 0
The system is stable for K > 0.
It is not evident how K affect the transient
response.
K=2
450
K=0.
–2
–1
K=2
s
j
K=0.
K=1
–j
RL is a plot of the closed loop poles as some
parameters of the system is varied.
7.1 Root Locus Principle
The CE of this system is
s
2nd
For
order
system the RL
appears as a
family of 2 paths
or branches
traced out by 2
roots of CE.
We generally consider a system as in the
following figure
+
K
G(s)
–
1+KG(s)H(s)=0
(1)
A value s1 is a point in the RL if and only if
satisfies (1) for real 0<K<.
For this system we call KG(s)H(s) the open
loop function (OLF)
Equation (1) can be written as
K= –1/G(s)H(s)
Since G(s) and H(s) are generally complex and
K is real then there are 2 condition must be met
|G(s)H(s)|=1/|K| and
 G(s)H(s) = r, r = ±1, ±3, ±5 …
H(s)
G(s) includes plant and compensator TF.
The closed loop TF is
T ( s) 
KG ( s)
1  KG ( s) H ( s)
(2)
(3)
(4)
We call (3) the magnitude criterion of the RL,
and (4) the angle criterion.
Any value s on the RL must meet both
criterions.
7.1 Root Locus Principle
To illustrate this criterion let us consider
KG ( s) H ( s) 
K ( s  z1 )
( s  p1 )(s  p2 )
(all zero angles) – (all pole angles) = r
s1
s1- p2
s1- z1
s1- p1
1
z1
s
2
p2
From the preceding discussion, it is seen that the
condition of a point to be on the RL is that
3
p1
Suppose that s1on the RL. Thus the angle
condition becomes
1 – 2 – 3= ± 
The locus of all points meet this condition
forms the complete RL of the system
where zero angle is the angle of s – zk and pole
angle is the angle of s – pk . zk and pk are zero and
pole of the OLF
Calculating and plotting RL digitally is available
and it is a convenient way to do it.
However a good knowledge of the rule for
plotting the RL will offer insight effect of
changing parameter and adding poles and/or zeros
in the design process.
From mason gain formula, the CE is
(s) = 1 + F(s) = 0
where F(s) is the OLF.
Many design procedures are based on the OLF.
7.2 Root Locus Rules
Since the TF is real function then its complex
roots must exist in conjugate pairs, then RL is
symmetrical with respect to the real axis
Rule 1.
RL is symmetrical with respect to the real
axis
With the zk and pk are zeros and poles of the
OLF. The CE may be expressed as
1  KG ( s) H ( s)  1 
Kbm ( s  z1 )(s  z1 ) 
( s  p1 )(s  p2 ) 
(s  p1 )(s  p2 ) Kbm (s  z1 )(s  z1 )  0
Therefore for K = 0 the roots of CE are
simply the poles of OLF  RL start at the
poles of G(s)H(s).
It is always that the number of zeros, NZ,,
equals the number of poles, NP.
If the number of finite zeros < NP then the rest
of zeros lies at infinity.
If K  but s remain finite then the branches
of RL must approach the zeros of the OLF, else
there must be zeros at infinity.
This mean that RL finish at zeros of G(s)H(s).
Rule 2.
RL originates on the poles of G(s)H(s) for
K=0 and terminates on the zeros of G(s)H(s)
as K, including zeros at infinity.
When there are zeros at infinity then the RL
will extend to infinity as well. The asymptotes
can be determined. If there is  zeros at infinity
then the OLF can be written as
K (bm s m  bm1s m1 
KG(s) H (s)  m
s
 an1s m1 
7.2 Root Locus Rules
K (bm s m  bm1s m1 
KG(s) H (s)  m
s
 an1s m1 
As s , the polynomial becomes
Kbm
s  s 
lim KG ( s ) H ( s )  lim
s 
Remember that the values of s must satisfy
the CE 1-KG(s)H(s)= 0 regardless the value
of s hence we have
Kbm
0
s  s 
1  lim KG ( s) H ( s)  1  lim
s 
asymptote angle
0
No asymptote
1
1800
2
±900
3
±600,1800
4
±450, ± 1350
The asymptote intersect the real axis at
a
hence we can write for large s
s + Kbm=0
The angle of the roots are principal values of
angle (asymptote angles)
 = r/

r = ±1, ±3,…
PZ


N p  Nz
Where P is the finite poles and Z is the
finite zeros
7.2 Root Locus Rules
7.3 Additional Techniques
Rule 3
If OLF has  zeros at infinity, the RL
approach  asymptote as K approach
infinity. The asymptote intersect real axis
at a with angle .
K
s( s  2)
There are 2 zeros at infinity, hence there are
2 asymptotes with angle ±900.
The asymptote intersect the real axis at
a
P   Z [0  (2)]  0



 1
N p  Nz
-
a
z1
Example.
Consider the OLF
KG ( s) H ( s) 
s
20
p2
p1

Consider point a to the right of p1. The angle
condition states that if a is on the RL then
(all zero angles) – (all pole angles) = r.
Since this not the case then any points on the
real axis to the right of p1 are not on the RL.
By the same argumentations we conclude that
the points between p1 and p2 are on the RL,
the points between p2 and z1 are not on the RL,
the points to the left of z1 are on the RL.
Rule 4
The RL includes all points to the left of an
odd number of real critical frequencies (poles
and zeros)
7.3 Additional Techniques
Multiple roots on the real axis.
And (2) can be written as
If CE has multiple roots a on the real axis then
we can write the characteristic polynomial as
1+KG(s)H(s) =(s-a)kQ(s)
(1)
Differentiating (1) with respect to s gives
(3)
The multiple points are breakaway points,
they are points at which branches of RL leave
or enter the real axis.
Rule 5.
The breakaway points on RL will appear
among the roots of polynomial obtained
either from (2) or (3).
 k (s  a)k 1Q(s)  (s  a)k Q' (s)
For s =a it must be
d
d
KG ( s) H ( s)  G ( s) H ( s)  0
ds
ds
N(s)D’(s)- N’(s)D(s)=0
Since on the RL
(2)
Hence the multiple roots a can be found by
solving (2).
Since G(s)H(s) is rational we can represent
G(s)H(s)= N(s)/D(s)
K = –1/ G(s)H(s) = – D(s)/N(s)
we conclude that
K’= N(s)D’(s)- N’(s)D(s)=0
Hence at breakaway points K is either
maximum or minimum.
7.3 Additional Techniques
Angle of Departure from Complex Poles
Suppose the open loop poles are as shown in the
figure below, and s1 is on the RL.
s1
1
 p
1
For very small , 2=900, hence
1= 1800 – 900 – 3
Rule 6
Loci will depart from a pole pj (arrive at
zj) at angle of d(a) where
d = zi   pi + r
a = pi   zi + r
3
p3
2
p2
We are interested in finding the angle 1 when
 is very small. This is the angle of departure.
From angle criterion we know that
– 1 – 2 – 3=1800
7.3 Additional Techniques
Example 1
Routh-Hurwitz method can be used to
find values of K for stability, this is
found to be 6 < K< 10.
We consider a system with OLF
KG ( s) H ( s) 
K
( s  1)(s  2)(s  3)
(1)
with 3 finite poles and 3 infinite zeros.
K=10
Rule 2. The RL start from pole s =1, s = – 2, and s = – 3.
Rule 3. The asymptote intersect the real axis
-4/3
at (1-2-3)/(3-0) = – 4/3, with angle ±600, and 1800.
Rule 4. The RL occurs on the real axis for – 2< s <1
and s<–3.
As K is increased p1 and p3 move left, but p2 move right.
Rule 5. Determine the breakaway point by differentiating
(1) and set the result to zero.
3s2+8s+1=0  s1 = -0.132 and s2 = -2.54 (ignore s2)
p3
-3
j
-0.132
p2
-2
p1
-1
1
-j
K=6
7.3 Additional Techniques
Example 2
KG ( s ) H ( s ) 
K ( s  1)
s2
Example 3
A simplified ship steering system modeled as a
system of order 2
KG ( s) H ( s) 
K
s 2  0.1s
The RL start from double pole s =0 and
stop at zero s = 1 and s = 
The RL start from s =0 and s =0.1 and stop at
s = .
One asymptote at 1800.
Two asymptotes at 900 intersect the real axis at
s =0.05
The RL occurs on real axis for s < 1
Breakaway point at s = 0 and s = 2
-2
-1
The RL occurs on real axis for 0.1< s <1
Breakaway point at s = 0.05
 0.1
 0.05
1
7.3 Additional Techniques
Example 4
The order-3 model of ship steering system is
KG ( s) H ( s) 
0.01K a
K

s 3  2.1s 2  0.2s s 2  2.1s  0.2s
The RL start from poles s =0, s =0.1, & s =2
and stop at 3 zeros at s = 
3 asymptotes intersect real axis at s =0.7
The real axis RL, for 0.1<s<0 and s<2
Breakaway point at s = 0.0494
The CE is
1  KGH  0  s3  2.1s 2  0.2s  0.01Ka
The Routh array is
s3 1
0.2
s2 2.1
0.01Ka
s1 (0.420. 01Ka)/2.1
Ka< 42
s0 0.01Ka
Ka> 42
For stability 0<Ka<42.
The auxiliary polynomial for Ka=42 is
Qa = 2.1 s2 + 0.01(42) = 2.1 (s2 + 0.2)
j0.4472
Ka=42
2
s =0.7  0.1
1
Setting Qa to zero we find s = ±j0.4472.
At this point the system is oscillating with
frequency = 0.4472 rad/s
7.3 Additional Techniques
Example 5: The order-4 model of ship steering system is
KG ( s) H ( s) 
0.2 K a
K
 4
s( s  20)(s  2)(s  0.1) s  22.1s 3  42.2s 2  4s
The RL start from poles s =0, s =0.1, s =2, & s =2 and stop at 4 zeros at s = 
4 asymptotes intersect real axis at s =5.525.
The real axis RL, for 0.1<s<0 and 20<s<2
Breakaway point at s = 0.0493, s = 15.19
For stability 0<Ka<38.03.
For Ka=38.03 the system is oscillating with frequency 0.4254.
1350
 20
 15.19
450
s =5.525  2
j0.4254
Ka=38.03
0
7.3 Additional Techniques
Table 1. Stabilities of the last 3 example
example
Ka for
marginal
stability
3 (order 2)
Always
stable
4 (order 3)
5 (order 4)
42
38.03
Oscillation
frequency
…
0.4472
0.4254
Table 2. Poles of example 3, 4, and 5.
ex
Ka =1
Ka =20
Ka =40
3
–0.05±j0.05 –0.05±j0.312
–0.05±j0.44
4
–0.04±j0.05 –0.025±j0.31
– 2.003
– 2.050
–0.002±j0.4
– 2.0096
5
–0.04±j0.05 –0.022±j0.31
– 2.002
– 2.055
–19.99997
–19.9994
–0.023±j0.4
– 2.106
–19.9990
The last 3 example illustrate RL construction and the reduction of order of system.
From table 1 and 2 we see that for small Ka (=1) model of order 2 is adequate.
For moderate Ka (= 20) model of order 3 is adequate.
For large Ka (= 40) model of order 4 is required.
Six rules for RL construction have been developed many additional rule are available for accurate
RL graph, however we should use digital computer to draw accurate RL.
7.4 Additional RL properties
Sketching RL relies on experiences, followings are some low order RL
( s  z1 )
( s  p1 )( s  p2 )
1
( s  p1 )( s  p2 )
1
s p1
p1
p2
1
( s  p1 )( s  p2 )( s  p3 )
p1
p2 z1
1
( s  p1 )( s  p2 )( s  p3 )
p2
p3 p2
p1
p1
z1 p2
1
( s  p1 )( s  p2 )( s  p3 )
p1
( s  z1 )
( s  p1 )( s  p2 )( s  p3 )
p2
p1
p3
( s  z1 )
( s  p1 )( s  p2 )
p1
p3
p2
p1 z1
p1
7.4 Additional RL properties
The CE can be written as
D(s) + KN(s) =0
For a given K1 we have
D(s) + K1 N(s) =0
Suppose that K is increased by K2 from K1
then the CE becomes
D(s) + (K1 + K2)N(s) =0
Example suppose we have OLF
KG ( s ) H ( s ) 
K ( s  1)
s2
(1)
The CE can be written as s2 + K(s+1)=0
For K =K1=2, we have s2+2s+2=(s+1)2+1=0
With roots s = 1±j.
Thus the RL of OLF
K ( s  1)
(2)
KG ( s) H ( s)  2
s  2s  2
which can be expressed as


N ( s)
N ( s)
1  K2 

1

K
0
2

D1 ( s)
 D( s)  K1 N ( s) 
With respect to K2 the locus appears to
originate on roots as placed by letting K = K1
but still terminates on the original zeros.
is the same with RL of (1) but start from 1±j.
-2
-1
7.4 Additional RL properties
The RL of OLF
The CE can be written as
1+ KG(s)H(s) =0
(1)
Consider replacing s with s1
1+ KG(s-s1)H(s-s1) =0
(2)
KG ( s ) H ( s ) 
K ( s  2)  1
( s  2) 2
(3)
is as shown below
then for K0, (s-s1) = s0 satisfied (2).
This is the efect of shifting all poles and
zeros of the OLF by a constant amount s1.
-3
-4
-3
-2
Example. The RL of OLF
KG ( s ) H ( s ) 
K ( s  1)
s2
is as shown below
-2
-1
(3)
The last property is that a RL has symmetry at
breakaway points. Suppose 4 branches come
together at a common breakaway point, the the
angle between 4 branches will be 3600/4=900.
7.5 Other Configuration
Now (3) is the same form as (1) with
So far we plot RL of CE
1+ KG(s)H(s) =0
and
by varying K from 0 to .
G(s)H(s) = s/(s2+5)
Other system configuration should be
converted to this form first.
As an example let us vary  from 0 to 
for the following CE
1
5
0
s ( s  )
K= 
(1)
In general the procedure is as follows
1. Write the CE in a polynomial of s
2. Grouped the terms that are multiplied by 
and that are not, that is we express
(2)
or
Dc(s) +  Nc(s)=0
Example
Let us design PI controller using RL.
s  s  5  0
2
We rewrite the equation to yield
and
compensator
(s 2  5)  s  0
s
1  2
0
s 5
+
(3)
–
KP 
KI
s
motor
0.25
( s  0.1)
7.5 Other Configuration
The system CE is
K  0.25 

1   K P  I 
0
s  s  0.1 

(1)
Supposed that KI =1, and we want to vary
KP and plot the RL. Eq. (1) becomes
s 2  0.1s  0.25  0.25K P s  0
We have to rewrite this equation as
s
1  0.25 K P 2
 0 (2)
s  0.1s  0.25
The sketch of th RL is as follow
s 2  0.1s  0.25
0.45
KP 

 3.6
0.25s
0.125
s  0.5
The required PI compensator is then
Gc(s) = 3.6 + 1/s
and the design is complete
j0.5
-0.5
Now the RL is available, it is easier to specify
the roots.
Suppose we want critical damping condition
then s =-0.5 and from (2) KP is
0
-j0.5