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III. CONTROL SYSTEM DESIGN
Desired
Performance ControllerPlant /
ProcessComparison
Sensor
Output
Feedback Loop
InputIII.
II.
I.Errorsignals
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III. A Root Locus Design Method
III.A.1 Root Locus Concepts
Definition
R(s) K G(s) Y(s)+-
- Open loop transfer function G(s) (usually given)
- Gain controller: gain K to be designed
- Closed loop transfer function T(s)
)(1)()(
)()(
sKGsKGsT
sRsY
a(s) )=1+KG(s) (Characteristic Equation or CE)
- Closed loop system poles given by roots of
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- Example:
- Root locus: trajectories of closed loop poles, i.e., roots ofa(s) on s-plane as K:
Trajectories of r1, r2as K:
)4(
1)(
sssG
)4(1
)4()(
ss
K
ss
K
sT
roots of a(s) = 0 042 Kss
2
4164, 21
Krr
)4(1)(
ss
Ksa
0
0
CE:
How about root locus for general transfer function?
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Phase and Magnitude Requirements:
a(so)=1+KoG(so) = 0( a(s) may have many roots
at K=Ko, so is one of them)
- Let sobe on root locus at K=Ko, i.e.,
so on root locus
at K=Ko
.
s
Branches of root locus
of closed loop system
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* Phase requirement (PR):
l = 0,1,2,3,
* Magnitude requirement (MR):
o
o
KsG
1)( or
)(
1
o
o
sGK
)360180()(
0
11)(0)(1
lj
oo
sGjoo e
KKesGsGK
o
)()()(
osGjoo esGsG L
G(so) complex- Generally, so complex
- With complex G(so) expressed as ,
- Comparing the phase and magnitude on both sides:
then
360180)( lsG o
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- is complex
number going from
pi to so
Graphical Interpretation of PR
If (sum of angles from zeros of G(s) to so
- sum of angles from poles of G(s) to so) = 180 l360,
then so is on root locus !
test point so
L(so+z2)
L(so+z1)
-z2
-z1
so+z2
so+z1
-p1 -p2
L(so+p1) L(so+p2)
so+p1 so+p2
.
x x
)( io
n
i
ps
1
)( io
M
i
zs
1
o denotes zeros of G(s)x denotes poles of G(s)
s)( io zs - is complex
number going fromzi to so
)( io ps
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- If so on root locus, use MR to determine corresponding value
MR:
Ko of K at so
Moo
noo
o
o
zszsC
psps
sGK
...
...
)( 111
o
o
KsG 1)(
(Note: MR will produce a value of Ko
even if so
is not on
root locus in that case the value is meaningless)
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- Example:
C = 1, M = 0
n = 2, p1 = 0, p2 = 4
)4(1)( ss
sG
* Is so = -2+2j on root locus?
PR check - L(so+p1) - L(so+p2) = 180 l360 ?
LHS: -135 -45 = -180 = RHS forl = -1
YES! so is on root locus!
* Corresponding Ko value?844
1 )()(
)(oooo
o
o sssssG
K
* Conclusion: so is on root locus at K=Ko = 8
-p2 -2
2j
-p1
so
MR
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-2 -1 -z1
-p2 -p1
soj- Example:
C = 1, M = 1, z1 = 0
n = 2, p1 = 1, p2 = 2
)2)(1()(
ss
ssG
* Is so = -1+j on root locus?
PR check L(so+z1) - L(so+p1) - L(so+p2) = 180 l360 ?
LHS = RHS for any l
Conclusion: so is not on root locus
04590135
* Meaningless to apply MR to calculate Ko
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III.A.2 General Procedures in Drawing Root Locus
Given Characteristic Equation (CE):
0
11
1
1
)(
)()()(
i
n
i
i
M
i
ps
zsKsKPsa
- Assume C=1 for P(s) or absorbed C into K
))...((
))...(()(
n
M
psps
zszsCsP
1
1where
- npoles of P(s):
Mzzz ,,, 21
nppp ,,, 21
-Mzeros of P(s):
R(s)K P(s)
Y(s)+
-
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Starting and ending points of root loci?
011
)()( iM
ii
n
izsKps- Roots of a(s) = 0
- K 0, -pi are roots of a(s) = 0
- K , -zi are roots of a(s) = 0
Step 1: Denote x and o on s-plane as pole and zero of P(s).
Root loci begin at the poles (x-location) of P(s) and endat the zeros (o-location) of P(s) as Kgoes from 0 to
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Real axis portion of root loci?- PR L(so+zi ) - L(so+pi ) = 180 l360
All zeros All poles
- For so on real axis,
* Contribution from pole and zero to the left of so are zero
* PR not satisfied if even number of poles and zeros on the
right of so
* PR satisfied only if odd number of poles and zeros on the
right of so
Step 2: Real axis portion of locus Any point on real axis
having an odd number of real poles and zeros to its
right is on the root locus
* Contribution from complex pole and zero pair cancel out
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- P(s) has npoles andMzeros with n > M
- For large K,
*Mloci go fromMof the npoles toMzeros (Step 1)
* (n-M) loci from the remaining (n-M) poles will go to
following an asymptotespattern
- Asymptotespattern defined by angle and centroid A :
Asymptotes going to ?
Mn
MnkMn
k
M
A
A
iiz
n
iip
11
121018012
)(,,,,,)(
Step 3: Draw asymptotes according to the computed angle Aand centroid A, to be followed by root loci whenfar
away from origin as K becomes large
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Points, if any, where loci cross the imaginary axis?
- Method (a) : Put s=jinto a(s) = 0, equate real and
imaginary part to solve for and Kvalues
- Method (b) : Routh-Hurwitz Criterion:
set Kto have roots on imaginary axis, and
determine from Axillary Polynomial
Step 4: Compute cross-over points, i.e., where loci cross
imaginary axis
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Example:
Method (a)
0)328(
12
sss
K0328 23 Ksss
s = j 0328 23 Kjj
Real part
Imaginary part
082
K
0323
32Crossover: at K = 256
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Method (b)
K = 256
0
1
2
3
s
s
s
s 1 32
8 K
)256(8
1K
K
roots on imaginary axis
(Special Case II)
Axillary polynomial:
02568)( 2 ssAP
32js +crossover at
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- Breakaway point: where real roots meet and breakaway into complex roots
- To determine breakaway point
- Example:
* Then
)()()(sp
sPKsKP
101
0 )(spds
d
ds
dKBreakaway point occurs at
0263 2
bb
ss
ssdsdK
b
Breakaway point, if any, on real axis?
Some function of s
423.0or58.1 bs
423.0bs* Breakaway point between s=0 and 1:
* s= -1.58 gives rise to K
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* Why ? Plotting Kvs s (real number)
-- At K=K1, 3 real roots at
-- At K=K2, 3 real roots at
-- At K=K3, only 1 real root at , remaining 2 roots complex!
-- Occurrence of complex roots at K=Kb , where 2 real roots
meet at
-- occurs at local maximum of K, i.e.,
K
s-2 -1 0
K1
K2
K3
Kb
)1(1s
)2(1s )3(1s
)1(
2s)2(
2s)3(
2s
)1(
3s
)1(
bs)3()2( , bb ss
)3(
1
)2(
1
)1(
1 ,, sss)3(
2
)2(
2
)1(
2 ,, sss)1(
3s
)3()2(
bbb sss
0
ds
dK
bs
0ds
dK
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- Solving sb from analytically possible for simple
transfer function P(s).
- For complicated P(s), use rough estimation method:
bs
(i) Compute Kat discrete locations between s=0 and s=-1
(separation depends on desired accuracy for )
(ii) Pick maximum Klocation as approximate bs
)2)(1( sssK
0.4 (to an accuracy of 0.2)sb
0192.0336.0384.0288.00
0.18.06.04.02.00
K
s
0ds
dK
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- Note: Also Breakin point, where complex roots meet andbecome real roots.
))(()(
)(
))(()(
32
1
1
32
ss
ssK
ss
sssPExample:
* Breakaway point occurs between s = 0 and 1,0
ds
dKcorresponding to local max. of K
,0ds
dK
* Breakin point occurs between s = 2 and 3
corresponding to local min. of K
Again, focusing on :0K
Step 5: Compute/estimate locations of Breakaway and
Breakin points, if any.
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Angle of departure from a complex pole / Angle of arrival
- For angle of departure from complex pole -pj
Let test point so very close to complex -pj:
to a complex zero
so = -pj +
- By PR:
36018011
lppzp ijn
iij
M
i
)()(
- is angle of departure from pole -pj
36018011
lppzp ij
n
jii
ij
M
i
)()(
360180
11lppzp ij
n
ji
iij
M
i
)()(
36018011
lpszs io
n
iio
M
i
)()(
- PR becomes:
or
complex quantity with small magnitude
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- Generalize, for any single complex pole
* i is angle of vector from zero -zi to the complex pole
* i is angle of vector from pole -pi to the complex pole
* i,i known from given zero and pole locations -zi and pi
* For convenience, select integer l to keep -180 o
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Step 6: Compute angle of departures dep and angle of arrival
arr, if any.
Finally,
Step 7: Complete root locus sketch with information
obtained in previous steps
- Similarly, angle of arrival for any single complex zero
* i is angle of vector from pole -pi to the complex zero
360180
zeros
remainingpolesallliiarr
* i is angle of vector from zero -zi to the complex zero
