Today’s goal
•
Root Locus examples and how to apply the rules
– single pole
– single pole with one zero
– two real poles
– two real poles with one zero
– three real poles
– three real poles with one zero
•
Extracting useful information from the Root Locus
– transient response parameters
– limit gain for stability
2.04A Spring ’13
Lecture 12 – Tuesday, March 5
1
Root Locus definition
•
Root Locus is the locus on the complex plane of closed-loop poles
as the feedback gain is varied from 0 to ∞.
Y (s)
X(s)
1
G(s) =
s+2
✓
X(s)
Y (s)
◆
=
OL
Open-Loop
pole
1
s+2
j!
Y (s) +
1
G(s) =
s+2
K
✓
X(s)
Y (s)
Root Locus
K=1
2
◆
=
CL
K
s+2+K
Open-Loop
pole
j!
K=0
2
As K varies from 0 to 1
2.04A Spring ’13
X(s)
Lecture 12 – Tuesday, March 5
...
2
Root-locus sketching rules
•
•
•
Rule 1: # branches = # poles
Rule 2: always symmetric with respect to the real axis
Rule 3: real-axis segments are to the left of an odd number of realaxis finite poles/zeros
Root Locus
0.4
0.3
✓
X(s)
Y (s)
0.2
◆
OL
1
=
s+2
0.2
0.15
−1
0.1
0
−0.1
X(s)
Y (s)
◆
=
OL
s+5
s+2
1 branch
0.05
0
−0.05
to the left of
the single pole
−0.2
−0.1
−0.15
−0.3
−0.4
−8
✓
0.1
1 branch
j (seconds )
j (seconds−1)
Root Locus
−7
−6
−5
2.04A Spring ’13
−4
−3
(seconds−1)
−2
−1
0
1
−0.2
−6
nothing to the left
of one zero +
one pole
−5
Lecture 12 – Tuesday, March 5
−4
to the left of
the pole
−3
−2
(seconds−1)
−1
0
1
3
Root-locus sketching rules
•
Rule 4: begins at poles, ends at zeros
✓
◆
K
X(s)
=
Y (s) CL
s + (K + 2)
✓
◆
closed–loop
= (K + 2) ! 1, as
pole
✓
K!1
✓
X(s)
Y (s)
◆
closed–loop
pole
=
CL
◆
=
We say that this TF
has a “zero at infinity”
Root Locus
0.3
✓
X(s)
Y (s)
0.2
◆
OL
1
=
s+2
0.15
0.1
K=0
K=1
0
−0.1
5,
as
K!1
X(s)
Y (s)
◆
=
OL
s+5
s+2
1 branch
0.05
0
K=0
K=1
−0.05
to the left of
the single pole
−0.2
−0.1
−0.15
−0.3
−0.4
−8
✓
0.1
1 branch
−1
j (seconds−1)
0.2
5K + 2
!
K +1
Root Locus
j (seconds )
0.4
K (s + 5)
(K + 1) s + (5K + 2)
−7
−6
−5
2.04A Spring ’13
−4
−3
(seconds−1)
−2
−1
0
1
−0.2
−6
nothing to the left
of one zero +
one pole
−5
Lecture 12 – Tuesday, March 5
−4
to the left of
the pole
−3
−2
(seconds−1)
−1
0
1
4
Root Locus sketching rules
•
Rule 5: Real-axis intercept and angle of asymptote
a
=
✓a
=
Root Locus
2
1.5
P
P
finite poles
finite zeros
P
#finite poles # finite zeros
(2m + 1) ⇡
P
#finite poles # finite zeros
Root Locus
5
1
G(s)
=
4
(s + 1) (s + 2) (s + 3)
1
G(s) =
(s + 1) (s + 3)
3
2
2
−1
a
0.5
=
j (seconds )
j (seconds−1)
1
✓a = ⇡/2
0
−0.5
➡ Two branches (Rule 1)
➡ Symmetric (Rule 2)
➡ On real axis to the left of the
−1
−2
−3.5
−3
−2.5
−2
2.04A Spring ’13
−1.5
−1
(seconds−1)
−0.5
0
0.5
2
✓a = ⇡/3
−1
−3
➡ Two zeros at infinity (Rule 4)
=
0
−2
first pole at -1 (Rule 3)
−1.5
a
1
−4
−5
−8
➡ Three branches (Rule 1)
➡ Symmetric (Rule 2)
➡ On real axis to the left of the first pole
at -1 and the third pole at -3 (Rule 3)
➡ Three zeros at infinity (Rule 4)
−7
−6
Lecture 12 – Tuesday, March 5
−5
−4
−3
(seconds−1)
−2
−1
0
1
5
Root Locus sketching rules
•
Rule 6: Real axis breakaway and break-in points σb
Solve
X
1
b
n
zn
=
X
q
1
b
pq
Root Locus
Root Locus
2
1.5
5
1
G(s)
=
4
(s + 1) (s + 2) (s + 3)
1
G(s) =
(s + 1) (s + 2)
3
2
2
−1
a
0.5
=
j (seconds )
j (seconds−1)
1
✓a = ⇡/2
0
−0.5
b
=
2
➡ Two branches (Rule 1)
➡ Symmetric (Rule 2)
➡ On real axis to the left of the
−1
−2
−3.5
−3
−2.5
−2
2.04A Spring ’13
−1.5
−1
(seconds−1)
−0.5
0
0.5
2
✓a = ⇡/3
−1
−3
➡ Two zeros at infinity (Rule 4)
=
0
−2
first pole at -1 (Rule 3)
−1.5
a
1
−4
−5
−8
b
➡ Three branches (Rule 1)
➡ Symmetric (Rule 2)
➡ On real axis to the left of the first pole
⇡
1.423
at -1 and the third pole at -3 (Rule 3)
➡ Three zeros at infinity (Rule 4)
−7
−6
Lecture 12 – Tuesday, March 5
−5
−4
−3
(seconds−1)
−2
−1
0
1
6
Root Locus sketching rules
Rule 7: Imaginary axis crossings
Solve
KG (j!x ) =
1
Root Locus
5
1
G(s)
=
4
(s + 1) (s + 2) (s + 3)
K = 60
!x ⇡ 3.32
3
2
−1
j (seconds )
•
=
2
✓a = ⇡/3
0
−1
−2
−3
−4
−5
−8
2.04A Spring ’13
a
1
b
➡ Three branches (Rule 1)
➡ Symmetric (Rule 2)
➡ On real axis to the left of the first pole
⇡
1.423
at -1 and the third pole at -3 (Rule 3)
➡ Three zeros at infinity (Rule 4)
−7
−6
Lecture 12 – Tuesday, March 5
−5
−4
−3
(seconds−1)
−2
−1
0
1
7
What else is the Root Locus telling us
•
Gain = product of distances to the poles
Root Locus
Root Locus
1
j (seconds−1)
1
G(s)
=
4
(s + 1) (s + 2) (s + 3)
p 2
0.5
K = 60
3
K = 2p
2
p
2
−1
1.5
5
1
G(s) =
(s + 1) (s + 2)
j (seconds )
2
0
−0.5
p 0
2 p 5
1
1
0
11
p 12
−1
−2
−1
−3
−1.5
−2
−3.5
−4
−3
−2.5
−2
2.04A Spring ’13
−1.5
−1
(seconds−1)
−0.5
0
0.5
−5
−8
−7
−6
Lecture 12 – Tuesday, March 5
−5
−4
−3
(seconds−1)
−2
−1
0
1
8
The zeros are “pulling” the Root Locus
•
•
Because of Rule 4
Therefore, adding a zero makes the response
– faster
– stable
Root Locus
Root Locus
3
15
G(s) =
2
10
1
5
j (seconds−1)
−1
j (seconds )
s+5
G(s) =
(s + 1) (s + 3)
0
0
−1
−5
−2
−10
−3
−16
−14
−12
−10
2.04A Spring ’13
−8
−6
(seconds−1)
−4
−2
0
2
s+5
(s + 1) (s + 2) (s + 3)
−15
−6
−5
Lecture 12 – Tuesday, March 5
−4
−3
−2
(seconds−1)
−1
0
1
9
Practice 1: Sketch the Root Locus
Nise Figure P8.2
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Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
2.04A Spring ’13
Lecture 12 – Tuesday, March 5
10
Practice 2:
Are these Root Loci valid? If not, correct them
Nise Figure P8.1
© John Wiley & Sons. All rights reserved. This content is excluded from our Creative
Commons license. For more information, see http://ocw.mit.edu/help/faq-fair-use/.
2.04A Spring ’13
Lecture 12 – Tuesday, March 5
11
MIT OpenCourseWare
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2.04A Systems and Controls
Spring 2013
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