2.04A Lecture Notes – Lecture 12 - MIT …2.04A Spring ’13 Lecture 12 – Tuesday, March 5 Today’s goal • Root Locus examples and how to apply the rules – single pole –

Lecture 12 – Tuesday, March 52.04A Spring ’13

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

1

X(s)Y (s)

✓X(s)

Y (s)

◆

OL

=1

s+ 2

✓X(s)

Y (s)

◆

CL

=K

s+ 2 +K

As K varies from 0 to 1 . . .

K = 0K = 1

Lecture 12 – Tuesday, March 52.04A Spring ’13

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 ∞.

2

+

�K G(s) =

1

s+ 2G(s) =

1

s+ 2

Y (s) X(s)

�2

�

j!Open-Looppole

�2

�

j!Open-LooppoleRoot Locus

Lecture 12 – Tuesday, March 52.04A Spring ’13

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 real-

axis finite poles/zeros

3

−8 −7 −6 −5 −4 −3 −2 −1 0 1−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

✓X(s)

Y (s)

◆

OL

=1

s+ 2

−6 −5 −4 −3 −2 −1 0 1−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

✓X(s)

Y (s)

◆

OL

=s+ 5

s+ 2

1 branch

to the left of the single pole

1 branch

to the left of the pole

nothing to the leftof one zero +

one pole

✓X(s)

Y (s)

◆

CL

=K (s+ 5)

(K + 1) s+ (5K + 2)✓

closed–loop

pole

◆= �5K + 2

K + 1

! �5, as K ! 1

✓X(s)

Y (s)

◆

CL

=K

s+ (K + 2)✓

closed–loop

pole

◆= � (K + 2) ! �1, as K ! 1

Lecture 12 – Tuesday, March 52.04A Spring ’13

Root-locus sketching rules

• Rule 4: begins at poles, ends at zeros

4

−6 −5 −4 −3 −2 −1 0 1−0.2

−0.15

−0.1

−0.05

0

0.05

0.1

0.15

0.2

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

✓X(s)

Y (s)

◆

OL

=s+ 5

s+ 2

1 branch

to the left of the pole

nothing to the leftof one zero +

one pole

K = 0K = 1

−8 −7 −6 −5 −4 −3 −2 −1 0 1−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

✓X(s)

Y (s)

◆

OL

=1

s+ 2

1 branch

to the left of the single pole

K = 0K = 1

We say that this TF has a “zero at infinity”

−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

G(s) =1

(s+ 1) (s+ 3)

✓a = ⇡/2�a = �2

Lecture 12 – Tuesday, March 52.04A Spring ’13

Root Locus sketching rules

• Rule 5: Real-axis intercept and angle of asymptote

5

➡ Two branches (Rule 1)➡ Symmetric (Rule 2)➡ On real axis to the left of the

first pole at -1 (Rule 3)➡ Two zeros at infinity (Rule 4)

−8 −7 −6 −5 −4 −3 −2 −1 0 1−5

−4

−3

−2

−1

0

1

2

3

4

5

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

G(s) =1

(s+ 1) (s+ 2) (s+ 3)

➡ 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)

fi

�a =

Pnite poles�

Pfinite zeros

#finite poles�#

Pfinite zeros

✓a =

(2m+ 1)⇡

#finite poles�#

Pfinite zeros

�a = �2✓a = ⇡/3

Solve

X

n

1

�b � zn=

X

q

1

�b � pq

Lecture 12 – Tuesday, March 52.04A Spring ’13

Root Locus sketching rules

• Rule 6: Real axis breakaway and break-in points σb

6

−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

G(s) =1

(s+ 1) (s+ 2)

✓a = ⇡/2�a = �2

➡ Two branches (Rule 1)➡ Symmetric (Rule 2)➡ On real axis to the left of the

first pole at -1 (Rule 3)➡ Two zeros at infinity (Rule 4)

−8 −7 −6 −5 −4 −3 −2 −1 0 1−5

−4

−3

−2

−1

0

1

2

3

4

5

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

G(s) =1

(s+ 1) (s+ 2) (s+ 3)

➡ 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)

�a = �2✓a = ⇡/3

�b = �2�b ⇡ �1.423

Solve KG (j!x

) = �1

Lecture 12 – Tuesday, March 52.04A Spring ’13

Root Locus sketching rules

• Rule 7: Imaginary axis crossings

7

−8 −7 −6 −5 −4 −3 −2 −1 0 1−5

−4

−3

−2

−1

0

1

2

3

4

5

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

G(s) =1

(s+ 1) (s+ 2) (s+ 3)

➡ 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)

�a = �2✓a = ⇡/3

�b ⇡ �1.423

!x

⇡ 3.32

K = 60

Lecture 12 – Tuesday, March 52.04A Spring ’13

What else is the Root Locus telling us

• Gain = product of distances to the poles

8

−8 −7 −6 −5 −4 −3 −2 −1 0 1−5

−4

−3

−2

−1

0

1

2

3

4

5

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

G(s) =1

(s+ 1) (s+ 2) (s+ 3)

p 12p 15p 20

p11

K = 60

−3.5 −3 −2.5 −2 −1.5 −1 −0.5 0 0.5−2

−1.5

−1

−0.5

0

0.5

1

1.5

2

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

G(s) =1

(s+ 1) (s+ 2)

p2p 2

K = 2

Lecture 12 – Tuesday, March 52.04A Spring ’13

The zeros are “pulling” the Root Locus

• Because of Rule 4• Therefore, adding a zero makes the response

– faster– stable

9

−16 −14 −12 −10 −8 −6 −4 −2 0 2−3

−2

−1

0

1

2

3

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

−6 −5 −4 −3 −2 −1 0 1−15

−10

−5

0

5

10

15

Root Locus

m (seconds−1)

jt (s

econ

ds−1

)

G(s) =s+ 5

(s+ 1) (s+ 3)G(s) =

s+ 5

(s+ 1) (s+ 2) (s+ 3)

Lecture 12 – Tuesday, March 52.04A Spring ’13

Practice 1: Sketch the Root Locus

Nise Figure P8.2

10

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Lecture 12 – Tuesday, March 52.04A Spring ’13

Practice 2:Are these Root Loci valid? If not, correct them

Nise Figure P8.1

11

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