lecture10(partial)

(Problem 4.5, p72)
Re
C K = R s 2 + 10 s + K
s1, 2 = −5 ± 25 − K
− 10
−5
− 5i
Root Locus Method: Magnitude & Phase Equations
Characteristic equation:
Geometric explanation:
N ( s ) D ′( s ) − N ′( s ) D ( s ) = 0 , G (s) H (s) = KN ( s ) D (s)
Breakaway point
G
C
2
Im
1
Imag Axis
0
GH ( s ) = G ( s ) = K ( s + 1) s ( s + 4 )( s 2 + 2 s + 2 )
Im Im Re Im Re
2
Im Axis mag
( 0 − 1 − 2) − 0 σa = 3 = −1
N ( s ) D ′( s ) − N ′ ( s ) D ( s ) = 0 ⇒ 3s 2 + 6 s + 2 = 0
1
Re
0
(a)
-1
(b)
Im Re
(c)
Im Re
Im
-2
s1, 2 = − 0 .423 , − 1 .577
How to understand this rule:
K∏| s − zj |
m j =1 n
Rule 2: (number of segments, continuity & symmetry) • Number of segments ＝ Number of closed-loop poles ＝ Number of open-loop poles = n (order of the system). • The loci are continuous. • The loci are symmetrical about the real axis.
( 2k + 1)π ϕa = , k = 0,1,2 L ( n − m − 1). n−m
p3
Re
p4
p2
z1
p1
4
Example:
R + _
3
Rule5: (Breakaway points & breakaway angles) • Location of breakaway points are determined by:
G( s) H ( s) = K s( s + 10)
− 10
=1
Im
s
1 + G( s) H ( s) = 0
⇒ G ( s ) H ( s ) = −1 + j 0
l2Leabharlann ⎧| GH |= 1 ⇒⎨ o ⎩∠GH = −180
Magnitude & phase equations:
K∏ | s − zj |
Rule 1: (Start points & end points) Every locus starts at an open-loop pole for K=0 and finishes at an open-loop zero or infinity for K=∞. If n>m, (n-m) segments of the locus will go infinite. We can also say that there are (n-m) infinite open-loop zeros.
∏ | s − pi |
i =1
=1⇒
For the magnitude equation to hold: • If K→0, s→ pi • If K→∞ , s→ zj or s→∞
1
Rule 3: (Asymptotes) • Number of asymptotes — (n-m)！ • All the asymptotes intersect the real axis at:
(Review your homework：)
Module 9:
Root Locus: Magnitude & Phase Equations
R
+ _
K s( s + 10)
C
K = 0, 10, 25, 50, 100
Im
5i
Module 10
Rules for Plotting the Root Locus
-3 -3 -2 -1 0 1 2
Re
Real Axis
(d)
(e)
(f)
→Continue on next page!
2
Note: We only concern about breaking-away on the real axis, which is the most common case.
Example:
K GH ( s ) = s ( s + 1)( s + 2 )
3
Homework: 1.Sketch the approximate root loci for the following arrangements of open-loop poles and zeros:
Rule 4: (root locus on the real axis) Segments of the real axis to the left of an odd number of open-loop poles/zeros are segments of the root locus.
Im
j =1
m
j
) − ∑ ∠ ( s − pi ) = −180 o ⇒ −α1 − α 2 = −180
i =1
∏ (s − p )
i i =1
j =1 n
∑ ∠( s − z
j =1
m
j
) − ∑ ∠ ( s − pi ) = − 180 o
i =1
(Phase)
• Root locus is determined by the phase equation alone! • The magnitude equation can help determine K at specific points on the locus.
(0 − 4 − 2) − (−1) 3 = −1.67
Re
-1 1
or:
∑
m
-2
j =1
1 = s − zj
∑
n
i =1
1 s − pi
-3
σa =
-4 -6
-5
-4
-3
-2 Real Axis
-1
0
1
2
• Breakaway angles (outgoing directions with respect to an arbitrary incoming one): ( 2k + 1)π ϕ= , k = 0,1,L (l − 1), l : number of breakaway (outgoing) segments. l
m
α2
0
α1
l1
Re
K∏ | s − zj |
m
Suppose:
G( s) H ( s) =
Root locus gain
K∏ (s − zj )
m
∏|s− p
i =1
j =1 n
= 1 (Magnitude)
∏|s− p
i =1
j =1 n
⇒
i
|
n
K =1 l1 ⋅ l 2
o
i
|
n
∑ ∠( s − z
σa =
∑ pi − ∑ z j
i =1 j =1
n
m
（Open-loop poles/zeros off the real axis don’t matter.）
Im Re
n−m
σa
ϕa
and their directions (with respect to the positive real axis) are: