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8/3/2019 Lecture Wk07 No1
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ENGN3223-Control System:
Root-Locus Design
Department of Engineering
Australian National University
8/3/2019 Lecture Wk07 No1
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ENGN3223, 2009
Controller Design using Root locus
We will now look at a graphical approach, known as the rootlocus method, for designing control systems
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ENGN3223, 2009 3/37
This slide is also from Week 4
What will happen in pole locationsas K increases?
Lets try some cases, K=0, s=0, -2
K=1, s=-1, -1
K=2, s=-1j1
K= , s=-1j
So we can predict the timeresponses now.
H(s) =K
s2+ 2s+ K
=
K
(s+1)2+ (K1)
-2
more overshoot
faster rise time
0
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Selecting the proportional feedback gain
1. The time-domainspecifications can be
converted into s-plane ones:
2. The locations of the poles are
3. Thus,
tr 1.8/n
Mp 15 (%)
ts 4.6/
n1.8/ t
r=1.8/1.2 =1.5 (rad/s)
0.5
4.6/ ts= 4.6/5 = 0.92
n =1.5
=0.92
=0.5
-1-2
1 K1 3
1
1 i K1
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Selecting the proportional feedback gain
1. The time-domainspecifications can be
converted into s-plane ones:
2. Draw a root locus (thisis the goal of this week)
3. Look at overlap of 1 and 2.That gives you an appropriatefeedback gain K.
tr 1.8/n
Mp 15 (%)
ts 4.6/
n1.8/ t
r=1.8/1.2 =1.5 (rad/s)
0.5
4.6/ ts= 4.6/5 = 0.92
n =1.5
=0.92
=0.5
-1-2
1
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Rewrite the system
+-
G(s)
KH(s)
Closed-loop transfer function Y(s) =L(s)K
1+ L(s)KR(s)
+-
G(s)H(s)
= L(s)
K
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What is a root locus?
T(s) =L(s)K
1+ L(s)K
+-
G(s)H(s)
= L(s)
K
The root locus is defined by the characteristic equation of T(s)
1+KL(s) = 0
Now we want to see the behavior of the closed-loop transferfunction T(s) as a function ofK
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Example
T(s) =L(s)K
1+ L(s)K=
K
s2+ 2s+K
+-
G(s)H(s)
= L(s)
K
G =1
s+ 2
H=1
s
1st order
Integral
Gain K
Closed loop transfer function
The root locus is defined by
s2+ 2s+K= 0
s = 1 1K
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ENGN3223, 2009 9/37
This slide is also from Week 4
What will happen in pole locationsas K increases?
Lets try some cases, K=0, s=0, -2
K=1, s=-1, -1
K=2, s=-1j1
K= , s=-1j
So we can predict the timeresponses now.
H(s) =K
s2+ 2s+ K
=
K
(s+1)2+ (K1)
2
more overshoot
faster rise time
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ENGN3223, 2009
Remark
+-
G(s)
KH(s)
Closed-loop transfer function Y(s) =G(s)
1+ L(s)KR(s)
The transfer function is different from the unit feedback.
However, the denominator is the same.
The system has the same poles and the same root locus.
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Lets see the following properties
of root locus
(1)K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots(5) Symmetry
(6) Locus on the real axis
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Properties of root locus
T(s) =KL
(s
)1+KL(s)
L(s) B(s)
A(s)
The characteristic equation (denominator=0)
1+ KB(s)
A(s)= 0
The root locus is defined by this equation as a function of K.
We want to see how to draw the root locus when K:0
Closed loop transfer function
Let us defineA(s) = s
n+ a
1sn1
+ = (s p1)(s p2)(s pn )
B(s) = sm+ b
1sm1
+ = (sz1)(sz
2)(szm )
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Property (1) (2)
(K
0) the root locus starts from the poles of L(s)=G(s)H(s).
(K)the root locus goes to zeros of L(s)=G(s)H(s).
1+KB(s)
A(s)= 0
A(s)+KB(s) = 0 K0 A(s) = 0
1
KA(s)+ B(s) = 0
K B(s) = 0
A(s) = sn+ a
1sn1
+ = (s p1)(s p2)(s pn )
B(s) = sm + b1sm1 +
= (sz1)(sz2)
(szm )
pi : pole
zi : zero
For causal systems, mn
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Property (1) (2)
Check the real-line whether it is a part of the root-locus withthe starting and ending points of the CL poles
K=0K=0
K=K=
pole
zero
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Properties of root locus
A(s) = sn+ a
1
sn1
+ = (s p1
)(s p2
)(s pn
)
B(s) = sm+ b
1sm1
+= (sz1)(sz
2)(szm )
pi:
polezi : zero
For causal systems, mn
s 0 = A + KB sn + a1sn1
+ K(sm+ b
1sm1
)
K=sn+ a1s
n1
s
m
+ b1s
m1
= snm s
m+ a1s
m1
sm+ b1s
m1
= snm
1+a1
s
1+b1
s
snm 1+a1 b1s
s+a1 b1n m
nm
1
1+ 1
(1+ )n
1+ n
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Asymptotes of root locus
K= s+ a1 b1n m
nm
s = a1 b
1
n m+ (K)
1
nm
=
a1 b
1
n m +K
1
nm
e
i2l+1
nm
Equation (1) give the asymptotes of the root locus.
In fact, Equation (1) represents (n-m) lines from
l= 0,1,
n m 1
a1 b
1
n m
(1)
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Asymptotes of root locus
Equation (1) give the asymptotes of the root locus.
In fact, Equation (1) represents (n-m) lines from a1 b1n m
=
(poles) (zeros)n m
A(s) = sn+ a
1sn1
+ = (s p1)(s p
2)(s p
n
)
B(s) = sm+ b
1sm1
+= (sz1)(sz
2)(szm )
pi: pole
zi : zero
From this expression, we have
a1 = (p1 + p2 ++ pn ) = (poles)
b1 = (z1 + z2 ++ zm ) = (zeros)
Thus, we can also express as
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Lets see the following properties
of root locus
(1)K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots(5) Symmetry
(6) Locus on the real axis
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Multiple root
(s p)2= 0 s = p
d
ds(s p)
2= 0 s = p
A simple exampleLet p be a multiple root
A(s)+KB(s) = 0K= A(s)
B(s)
A'(s)+KB'(s) = 0
A'(s)A(s)
B(s)B'(s) = 0
A'(s)
A(s)B
'(s)
B(s)= 0
1
s pll=1
n
1
szll=1
m
= 0
The multiple root can be foundfrom these two equations
We apply this idea to
the characteristic
equation
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Lets see the following properties
of root locus
(1)K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots(5) Symmetry
(6) Locus on the real axis
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Symmetry
The root locus is always symmetric with respect to the realaxis.
This is trivial because the roots of the characteristic
equation are of the form
s =Re iIm
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Lets see the following properties
of root locus
(1)K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots(5) Symmetry
(6) Locus on the real axis
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Root Locus on the real axis
From the characteristic equation we have1+KB(s)
A(s)= 0
L(s) = K( ) =
-
-
L(s) =(s0+ z)(s
0+ p)
pole
zero
L(s) = () ()
=
contribution from symmetric poles & zeros = 0
|#{zeros}-#{poles}|=odd
=L(s)
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Root Locus on the real axis
If there are more poles than zeros, (n-m) > 0 is called the poleexcess, then there are (n-m) branches of the root locus that
diverge to infinity (zeros at infinity).
K=0K=0
K=K=
n-m=1 (there is a zero at infinity)
K=
pole
zero
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Example
L(s) = (s+ 3)s(s+1)(s+ 2)(s+ 4)+-
pole
zero
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Example
L(s) = (s+ 3)s(s+1)(s+ 2)(s+ 4)
pole
zero
(1) K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots
(5) Symmetry
(6) Locus on the real axis
s =
a1 b
1
n m+ K
1
nm
e
i2l+1
nm
= 7 3
4 1+ K
1
41ei2l+1
41
= 4
3+K
1
3ei2l+1
3
=
sm+ b
1sm1
+
sn+ a
1sn1
+
l= 0,1,2where
This represents three lines (red)
-4/3
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Example
L(s) = (s+ 3)s(s+1)(s+ 2)(s+ 4)
pole
zero
(1) K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots
(5) Symmetry
(6) Locus on the real axis
-4/3
The root locus starts from the four poles.
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Example
L(s) = (s+ 3)s(s+1)(s+ 2)(s+ 4)
pole
zero
(1) K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots
(5) Symmetry
(6) Locus on the real axis
-4/3
The root locus ends at the four zeros.
(three are at infinity)
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Example
(1) K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots
(5) Symmetry
(6) Locus on the real axis
|#{zeros}-#{poles}|=odd (to the right)
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Example
(1) K=0(2) K
(3) |s|Asymptotes
(4) Multiple roots
(5) Symmetry
(6) Locus on the real axis
|#{zeros}-#{poles}|=odd (to the right)