Review: Today's Topic:: Pay Attention in Class!!
Review: Today's Topic:: Pay Attention in Class!!
Review: Today's Topic:: Pay Attention in Class!!
Note!! The way I teach the Root Locus differs a bit from what
the textbook does (good news: it is simpler). Still, pay
attention in class!!
Course structure so far:
modeling — examples
↓
analysis — transfer function, response, stability
↓
design — some simple examples given
where
I K is a constant gain
b(s)
I L(s) = , where a(s) and b(s) are some polynomials
a(s)
The Root Locus Design Method
+
R K L(s) Y
Y KL(s) b(s)
Closed-loop transfer function: = , L(s) =
R 1 + KL(s) a(s)
Closed loop poles are solutions of:
1
1 + KL(s) = 0 ⇔ L(s) = −
K
m
Kb(s)
1+ =0
a(s)
m
a(s) + Kb(s) = 0 characteristic equation
| {z }
characteristic
polynomial
A Comment on Change of Notation
q(s) b(s)
from H(s) or G(s) = to L(s) =
p(s) a(s)
Im
settling
time
ove
rsh
oot
rise
time
Re
0
+
R K L(s) Y
Y KL(s) b(s)
Closed-loop transfer function: = , L(s) =
R 1 + KL(s) a(s)
For what values of K do we best satisfy given design specs?
1 + KL(s) = 0,
as K varies from 0 to ∞.
A Simple Example
1
L(s) = b(s) = 1, a(s) = s2 + s
s2 +s
Characteristic equation: a(s) + Kb(s) = 0
s2 + s + K = 0
√
1 − 4K
1
Root locus = − ± :0≤K<∞ ⊂C
2 2
Example, continued
√
1 − 4K
1
Root locus = − ± :0≤K<∞ ⊂C
2 2
Let’s plot it in the s-plane:
I start at K = 0 the roots are − 12 ± 12 ≡ −1, 0
note: these are poles of L (open-loop poles)
Im
x x Re
1 0
Example, continued
√
1 − 4K
1
Root locus: − ± :0≤K<∞ ⊂C
2 2
I as K increases from 0, the poles start to move
1 − 4K > 0 =⇒ 2 real roots
K = 1/4 =⇒ 1 real root s = −1/2
Im
x x x Re
1 1 0
2
Example, continued
√
1 − 4K
1
Root locus: − ± :0≤K<∞ ⊂C
2 2
I as K increases from 0, the poles start to move
K > 1/4 =⇒ 2 complex roots with Re(s) = −1/2
Im
x x x Re
1 1 0
2
Im
increase K
x x x Re
1 1 0
2
Im
increase K
x x x Re
1 1 0
2
Thus, the root locus helps us visualize the trade-off between all
the specs in terms of K.
However, for order > 2, there will generally be no direct formula
for the closed-loop poles as a function of K.
Our goal: develop simple rules for (approximately) sketching
the root locus in the general case.
Equivalent Characterization of RL: Phase Condition
Recall our original definition: The root locus for 1 + KL(s) is
the set of all closed-loop poles, i.e., the roots of
1 + KL(s) = 0,
as K varies from 0 to ∞.
A point s ∈ C is on the RL if and only if
1
L(s) = − for some K > 0
K
|{z}
negative and real
There are six rules for sketching root loci. These rules are
mainly qualitative, and their purpose is to give intuition about
impact of poles and zeros on performance.
These rules are:
I Rule A — number of branches
I Rule B — start points
I Rule C — end points
I Rule D — real locus
I Rule E — asymptotes
I Rule F — jω-crossings
Today, we will cover mostly Rules A–C (and a bit of D).
Rule A: Number of Branches
Rule A:
#(branches) = deg(a)
Rule B: Start Points
a(s) + Kb(s) = 0
1
b(s) = − a(s)
K
— as K → ∞,
I branches end at the roots of b(s) = 0, or
I branches end at zeros of L(s)
+ 1
R KP + K D s
s2 1
Y
Gc
Gp
Y G c Gp
= poles: 1 + Gc (s)Gp (s) = 0
R 1 + G c Gp
1
1 + (KP + KD s) 2 =0
s −1
+ 1
R KP + K D s
s2 1
Y
Gc
Gp
z1
o x x Re
1 0
breakaway point
s − z1
L(s) =
s2 − 1
Im
z1
o x x Re
1 0
breakaway point
z1
o x x Re
1 0
breakaway point
+ 1
R KP + K D s
s2 1
Y
Gc
Gp
Gc (s) = 10 + 5s
u = 10e + 5ė, e=r−y
s+2
Characteristic equation: 1+5 =0
s2 − 1
s2 + 5s + 9 = 0
But: Rules A–C cannot tell the whole story. How do we know
which way the branches go, and which pole corresponds to
which zero?
Rules D–F!!
Example
s+1
Let’s consider L(s) =
s(s + 2) s + 1)2 + 1
(
m=1
I Rule A: =⇒ 4 branches
n=4
I Rule B: branches start at open-loop poles
s = 0, s = −2, s = −1 ± j
I Rule C: branches end at open-loop zeros s = −1, ±∞
Im
p3
x
x o x Re
p2 z1 p1
x
p4
Example, continued
b(s)
∠L(s) = ∠
a(s)
(s − z1 )(s − z2 ) . . . (s − zm )
=∠
(s − p1 )(s − p2 ) . . . (s − pn )
m
X X n
= ∠(s − zi ) − ∠(s − pj )
i=1 j=1
— this sum must be ±180◦ for any s that lies on the RL.
Rule D: Real Locus
So, we try test points:
Im
p3
x
∠(s1 − z1 ) = 0◦ (s1 > z1 )
◦
∠(s1 − p1 ) = 180 (s1 < p1 )
s1
Re ◦
p2
x o
z1
x
p1 ∠(s1 − p2 ) = 0 (s1 > p2 )
∠(s1 − p3 ) = −∠(s1 − p4 )
x (conjugate poles cancel)
p4
x o x Re
p2 z1 p1
x
p4