root locus – contents...2009 spring me451 - ggz week 10-11: root locus page 28 • examples for...

2009 Spring ME451 - GGZ Week 10-11: Root Locus

• Root locus, sketching algorithm

• Root locus, examples

• Root locus, proofs

• Root locus, control examples

• Root locus, influence of zero and pole

• Root locus, lead lag controller design

Root Locus Root Locus –– ContentsContents


• W. R. Evans developed in 1948.

• Pole location characterizes the feedback system stability

and transient properties.

• Consider a feedback system that has one parameter (gain) K > 0 to be designed.

• Root locus graphically shows how poles of CL system varies

as K varies from 0 to infinity.

LL((ss):): openopen--loop TFloop TF

Root Locus Root Locus –– What is it?What is it?

K )(sL


• Characteristic eq.

• K = 0: s = 0,-2

• K = 1: s = -1, -1

• K > 1: complex numbers

ClosedClosed--loop polesloop poles

ReRe

ImIm

Root Locus Root Locus –– A Simple ExampleA Simple Example

0)2(

11 =

++

ssK

)2(

1)(

+=

sssLK )(sL

022 =++ Kss Ks −±−= 11


• Characteristic eq.

• It is hard to solve this analytically for each K.

• Is there some way to sketch a rough root locus by hand? (In Matlab, you may use command “rlocus.m”.)

Root Locus Root Locus –– A Complicated ExampleA Complicated Example

)3)(2(

1)(

++

+=

sss

ssLK )(sL

0)3)(2(

11 =

++

++

sss

sK

0)1()3)(2( =++++ sKsss ???=s


• Root locus is symmetric w.r.t. the real axis.

• The number of branches = order of L(s)

• Mark poles of L with “x” and zeros of L with “o”.

ReRe

ImIm

Root Locus Root Locus –– Rule 0Rule 0

)3)(2(

1)(

++

+=

sss

ssL


• RL includes all points on real axis to the left of an odd number of real poles/zeros.

• RL originates from the poles of L and terminates at the

zeros of L, including infinity zeros.

ReRe

ImIm

Indicate the direction Indicate the direction

with an arrowhead.with an arrowhead.



• Number of asymptotes = relative degree (r) of L:

• Angles of asymptotes are

Root Locus Root Locus –– Rule 2 (Asymptotes)Rule 2 (Asymptotes)

1,,2,1,0 ),12( −=+× rkkr

Lπ

{ {)deg(deg(den)

:num

mnr −=


• Intersections of asymptotes

Asymptotes Asymptotes

(Not root locus)(Not root locus)

ReRe

ImIm

Root Locus Root Locus –– Rule 2 (Asymptotes) Rule 2 (Asymptotes) (cont(cont’’d)d)

)3)(2(

1)(

++

+=

sss

ssL

r

∑ ∑− zeropole

22

)1())3()2(0(zeropole−=

−−−+−+=

−∑ ∑r


• Breakaway points are among roots of

For each candidate s, check the positivity ofFor each candidate s, check the positivity of

Points where two or Points where two or

more branches meet more branches meet

and break away.and break away.


)3)(2(

1)(

++

+=

sss

ssL

0)(

=ds

sdL

2

23

)]3)(2([

3542

)(

++

+++−=

sss

sss

ds

sdL

js 7925.07672.0 ,4656.2 ±−−=((MatlabMatlab CMD: roots([1 4 5 3])CMD: roots([1 4 5 3])

)(

1

sLK −=

jK 2772.47907.1 ,4186.0 ±=


Quotient rule:

Root Locus Root Locus –– Rule 3 Rule 3 (cont(cont’’d d -- 1)1)

2))((

))(()()(

))(()(

)(

sD

ds

sDdsNsD

ds

sNd

ds

sD

sNd −

=

2

23

2

22

))3)(2((

61082

))3)(2((

)6103)(1()65()3)(2(

1

++

−−−−=

++

+++−++=

++

+

sss

sss

sss

ssssss

ds

sss

sd


ReRe

ImIm

Breakaway pointBreakaway point

Root Locus Root Locus –– Rule 3 Rule 3 (cont(cont’’d d -- 2)2)


Root Locus

Real Axis

Ima

gin

ary

Ax

is

-3 -2.5 -2 -1.5 -1 -0.5 0-8

-6

-4

-2

0

2

4

6

8

Root Locus Root Locus –– MatlabMatlab Command Command ““rlocus.mrlocus.m””

)3)(2(

1)(

++

+=

sss

ssL


• Asymptotes

– Relative degree 2

– Asymptote intersection

• Breakaway point

ReRe

ImIm

Root Locus Root Locus –– A Simple Example RevisitedA Simple Example Revisited

)2(

1)(

+=

sssLK )(sL

0))2((

)22()(2

=+

+−=

ss

s

ds

sdL

12

)2(0−=

−+

1−=s


• Root locus

– What is root locus

– How to roughly sketch root locus

• Sketching root locus relies heavily on experience.

PRACTICE!

• To accurately draw root locus, use Matlab.

• Next, more examples

Root Locus Root Locus –– Rule SummaryRule Summary


Root Locus Root Locus –– Practices Practices (hand drawing and (hand drawing and MatlabMatlab))

ssL

1)( =

2

1)(

ssL =

3

1)(

ssL =

)4(

1)(

+=

sssL

)2(

1)(

+

+=

ss

ssL

)5)(1(

1)(

++=

ssssL


• Consider a feedback system that has one parameter (gain)

K > 0 to be designed.

• Root locus graphically shows how poles of a CL system

varies as K varies from 0 to infinity.

LL((ss):): openopen--loop TFloop TF

Root Locus Root Locus –– What Is It (Review)?What Is It (Review)?

K )(sL


ReRe

ImIm

Root Locus Root Locus –– Rule 0 (Mark Pole/Zero)Rule 0 (Mark Pole/Zero)




)5)(1(

1)(

++=

ssssL


ReRe

ImIm

Indicate the direction Indicate the direction

with an arrowhead.with an arrowhead.

Root Locus Root Locus –– Rule 1 (Real Axis)Rule 1 (Real Axis)

• RL includes all points on real axis to the left of an odd number of real poles/zeros.

• RL originates from the poles of L and terminates at the

zeros of L, including infinity zeros.


Root Locus Root Locus –– Rule 2 (Asymptotes)Rule 2 (Asymptotes)



1,,2,1,0 ),12( −=+× rkkr

Lπ

{ {)deg(deg(den)

:num

mnr −=


Asymptote (Not Asymptote (Not

root locus)root locus)

ReRe

ImIm

Root Locus Root Locus –– Rule 2 (Asymptotes) Rule 2 (Asymptotes) (cont(cont’’d)d)


)5)(1(

1)(

++=

ssssL

r

∑ ∑− zeropole

23

)0())5()1(0(zeropole−=

−−+−+=

−∑ ∑r


For each candidate s, check the positivity ofFor each candidate s, check the positivity of

Root Locus Root Locus –– Rule 3 (Breakaway)Rule 3 (Breakaway)


Points where two or Points where two or

more branches meet more branches meet

and break away.and break away.

)5)(1(

1)(

++=

ssssL

0)(

=ds

sdL

2

2

)]5)(1([

5123)(

++

++−=

sss

ss

ds

sdL

3

212 ±−=s

)(

1

sLK −=

−=−≈−−=

≈−≈+−=

1.1352.33

212

13.147.03

212

Ks

Ks


What is this value? What is this value?

For what K?For what K?

ReRe

ImIm


RouthRouth--Hurwitz!Hurwitz!

Root Locus Root Locus –– Rule 3 (Breakaway) Rule 3 (Breakaway) (cont(cont’’d d -- 1)1)

)5)(1(

1)(

++=

ssssL

)13.1(

47.0

=

−

K


• Characteristic equation

• Routh array

• When K = 30

Stability conditionStability condition

Root Locus Root Locus –– Find Find KK for critical stabilityfor critical stability

0560)5)(1(

1 23 =+++⇔=++

+ Kssssss

K

300 << K

Ks

s

Ks

s

K

0

6

301

2

3

6

51

−

jss 50306 2 ±=⇒=+


ReRe

ImIm


)5)(1(

1)(

++=

ssssL

)13.1(

47.0

=

−

K

Root Locus Root Locus –– Root Locus ExampleRoot Locus Example

)30(

5

=K

j

)30(

5-

=K

j


ReRe

ImIm


After Steps 0,1,2,3, we obtainAfter Steps 0,1,2,3, we obtain

How to compute How to compute

angle of departure?angle of departure?

Root Locus Root Locus –– Example with Complex PolesExample with Complex Poles

1)(

2 ++=

ss

ssL

2

3-:poles

0:zero

2

1 j±

1

0)1(

)12(1)(22

2

±=⇒

=++

+−++=

s

ss

ssss

ds

sdL


• Angle condition: For “s” to be on RL,

ReRe

ImIm

Root Locus Root Locus –– Rule 4 (Angle of Departure)Rule 4 (Angle of Departure)

)1(180

)()()(

))(()(

211

211

21

1

−∠==−−=

−∠−−∠−−∠=

−−

−∠=∠

o

pspszs

psps

zssL

θθφ2

3

2

11

jp +−=

2

3

2

12

jp −−=

""s 1θ

2θ

1φ

01 =z

o

o

150

90,120

, toclose is s"" If

1

21

1

−=

≈≈

θ

θφ o

p



ReRe

ImIm

Root Locus Root Locus –– Rule 4 (Angle of Departure)Rule 4 (Angle of Departure)


• Examples for root locus.

– Gain computation for marginal stability, by using Routh-Hurwitz criterion

– Angle of departure (Angle of arrival can be obtained by a

similar argument.)

• Next, sketch of proofs for root locus algorithm

Root Locus Root Locus –– SummarySummary


Root Locus Root Locus –– Exercises 1Exercises 1

1)(

2 ++=

ss

ssL

2

1)(

s

ssL

+=

)1)(1()(

2 ++=

ss

ssL

)3)(2)(1(

1)(

++−=

ssssL


)2)(1(

1)(

++=

ssssL

)22)(2(

1)(

2 +++=

sssssL


)22)(1(

1)(

2 +++=

ssssL

)52)(54(

1)(

22 ++++=

sssssL



)22)(3(

1)(

2 +++=

sssssL

)22)(2)(1(

1)(

2 ++++=

ssssssL

54

1)(

2 ++

+=

ss

ssL

)54)(1(

3)(

2 +++

+=

sss

ssL



)54)(3)(1(

4)(

2 ++++

+=

ssss

ssL

)1(

)3)(2()(

+

++=

ss

sssL

)106)(54(

2)(

22 ++++

+=

ssss

ssL

)3)(2(

)22()(

2

++

++=

sss

sssL



• Root locus is obtained by

– for a fixed K > 0, finding roots of the characteristic

equation, and

– sweeping K over real positive numbers.

• A point “s” is on the root locus, if and only if L(s) evaluated for that “s” is a negative real number.

Root Locus Root Locus –– Characteristic equation and RLCharacteristic equation and RL

0)(1 =+ sKL)(

1

sLK −=

KsL

1)( −=


• Characteristic eq. can be split into two conditions.

– Angle condition

– Magnitude condition

Odd numberOdd number

For any point For any point ““ss””, ,

this condition holds this condition holds

for some positive K.for some positive K.

Root Locus Root Locus –– Angle and Magnitude ConditionsAngle and Magnitude Conditions

Ko

,2 ,1 ,0 ),12(180)( ±±=+×=∠ kksL

KsL

1)( =


• Select a point s=-2+j

ReRe

ImIm

s is on root locus.s is on root locus.

• Select a point s=-1+j

s is NOT on root locus.s is NOT on root locus.

Root Locus Root Locus –– A Simple ExampleA Simple Example

)2(

1)(

+=

sssL

2

1

)1)(1(

1

)2(

1)2(

1

−=++

=

+=+−

+−=

jj-

ssjL

js

180)1( =+−∠ jL

2)(

1==

sLK

)2(1

))(2(1

2)2(

1

)2(

jj

jj

jsss

jL

−

+−

+−=+

−=

=

=+−

180)( ≠∠ sL



– Characteristic equation is an equation with real

coefficients. Hence, if a complex number is a root, its

complex conjugate is also a root.


– If L(s) =n(s)/d(s), then Characteristic eq. is d(s)+Kn(s)=0,

which has roots as many as the order of d(s).


ReRe

ImIm

Root Locus Root Locus –– StepStep--byby--Step: Step 0Step: Step 0

))(()(

21

1

psps

zssL

−−

−=

1z1

p2

p


• RL includes all points on real axis to the left of an odd

number of real poles/zeros.

ReRe

ImIm

Test pointTest point

ReRe

ImIm

Not satisfy angle condition!Not satisfy angle condition!

Satisfy angle condition!Satisfy angle condition!

Root Locus Root Locus –– StepStep--byby--Step: Step 1Step: Step 1--11

1z1

p2p

1z1

p2p

0

)()()()(

0

2

0

1

0

1

=

−∠−−∠−−∠=∠434214342143421pspszssL

180

)()()()(

0

2

0

1

180

1

=

−∠−−∠−−∠=∠434214342143421pspszssL


Root Locus Root Locus –– StepStep--byby--Step: Step 1Step: Step 1--1 1 (cont(cont’’d)d)

• RL includes all points on real axis to the left of an odd

number of real poles/zeros.

ReRe

ImIm

Test pointTest point

ReRe

ImIm

Not satisfy angle condition!Not satisfy angle condition!

Satisfy angle condition!Satisfy angle condition!

1z1

p2p

1z1

p2p

0

)()()()(

0

2

180

1

180

1

=

−∠−−∠−−∠=∠434214342143421pspszssL

180

)()()()(

180

2

180

1

180

1

−=

−∠−−∠−−∠=∠434214342143421pspszssL

s

s


• RL originates from the poles of L, and terminates at the zeros of L, including infinity zeros.

s: Poles of s: Poles of LL((ss)) s: Zeros of s: Zeros of LL((ss))


∞=K0=K

}

0)(

)(10)()(0

)(

)(1

)(

=+⇔=+⇔=+sd

sn

KsKnsd

sd

snK

sL

0)( =sd 0)(0)(

)(=⇒= sn

sd

sn





)numdeg()dendeg(: −=r

K ,1 ,0 ),12( =+× kkr

π

1=r 2=r 3=r 4=r


• For a very large s,

• Characteristic equation is approximately

Root Locus Root Locus –– StepStep--byby--Step: Step 2Step: Step 2--1 1 (cont(cont’’d)d)

rn

rn

s

n

s

snsL 00)( ≈

+

+=

−

L

L

0010)(10

0 =+⇔=+⇔=+ Knss

nKsKL r

r

)0 assuming( 000

><−=⇒ nKnsr

... 2, 1, ,0 ),12( =+×=∠⇒ kksr π

... 2, 1, ,0 ),12( =+×=∠⇒ kkr

sπ



• Proof for this is omitted and not required in this course.

• Interested students should read page 363 in the book by

Dorf & Bishop.


r

zp

r

zerospoles ii∑+∑

=∑−∑



Suppose that Suppose that ss = = bb is a breakaway point.is a breakaway point.

Root Locus Root Locus –– StepStep--byby--Step: Step 3 Step: Step 3

0)(

=ds

sdL

=+

=+

0)()(

0)()(

bnKbd

bKnbd

&&0)(

)(

)()( =− bn

bn

bdbd &&

0)()(

)()(

)(

)(

)(

)()()()()(

2

2

=

−−=

−=

=

bnbn

bdbd

bd

bn

bd

bdbnbdbn

ds

sdL

bs

&&

&&


• RL departs from a pole pj with angle of departure

• RL arrives at a zero zj with angle of arrival

(No need to memorize these formula.)(No need to memorize these formula.)

Root Locus Root Locus –– StepStep--byby--Step: Step 4 Step: Step 4

∑ ∑≠

+−−−=i jii

ijijdppzp

,

180)()(θ

∑ ∑≠

+−−−=i jii

ijijazzpz

,

180)()(θ


• Sketch of proof for angle of departure

ImIm

ReRe

For For ss to be on root locus,to be on root locus,

due to due to angle conditionangle condition

Root Locus Root Locus –– StepStep--byby--Step: Step 4 Step: Step 4 (cont(cont’’d)d)

180211

+−= θφθd

180)()(21111

=−∠−−−∠ ppzp θ


• Sketch of proof for angle of arrival

ImIm

ReRe

For s to be on root locus,For s to be on root locus,

due to due to angle conditionangle condition

Root Locus Root Locus –– StepStep--byby--Step: Step 4 Step: Step 4 (cont(cont’’dd--1)1)

18023211

+−++= φθθθφA

01 →− zs

180)()(1

3

1

211=−∠−−∠+ ∑

=

i

i

pzzzφ


• Four step drawing process:– Root locus is symmetric w.r.t. the real axis; the number of branches

equals to the order of L(s); and mark poles of L with “x” and zeros of L with “o”.

– RL includes all points on real axis to the left of an odd number of real poles/zeros; and it originates from the poles of L, and terminates at the zeros of L, including infinity zeros.

– Number of asymptotes equals to relative degree (r) of L; and angles of asymptotes are π(2k+1)/r (k=0, 1, …)

– Breakaway points are among roots of dL(s)/ds=0

– RL departs from a pole pj with angle of departure

– RL arrives at a zero zj with angle of arrival

• Next, we will move on to root locus applications to control examples.

Root Locus Root Locus –– Summary (How to Draw) Summary (How to Draw)

∑ ∑≠

+−−−=i jii

ijijdppzp

,

180)()(θ

∑ ∑≠

+−−−=i jii

ijijazzpz

,

180)()(θ


a) Set Kt = 0. Draw root locus for K > 0.

b) Set K = 10. Draw root locus for Kt > 0.

c) Set K = 5. Draw root locus for Kt > 0.

Root Locus Root Locus –– Control Example 1 Control Example 1

)(sR )(sE )(sY

)5(

12 +ss

K

sKt


ReRe

ImIm

There is no There is no

stabilizing gain K!stabilizing gain K!

Root Locus Root Locus –– (a) (a) KKtt = 0 = 0 (Control Example 1)(Control Example 1)

)5(

1)(

2 +=

sssL


Characteristic eq.Characteristic eq.

Root Locus Root Locus –– (b) (b) KK = 10 = 10 (Control Example 1)(Control Example 1)

)(sR )(sE )(sY

)5(

12 +ss

10

sKt

0

)5(1

)5(

1

101

2

2

=

++

++

ss

sK

ss

t

0105

1

)(

23=

+++

4434421sL

tss

sK0105 23 =+++ sKss

t


ReRe

ImIm

By increasing Kt, we By increasing Kt, we

can stabilize the CL can stabilize the CL

system.system.

Root Locus Root Locus –– (b) (b) KK = 10 = 10 (Control Example 1)(Control Example 1)

105)(

23 ++=

ss

ssL



• Routh array

• When Kt = 2

Stability conditionStability condition

Root Locus Root Locus –– (b) (b) KK = 10 = 10 (Control Example 1) (Find (Control Example 1) (Find KKtt for MS)for MS)

01050105

1 23

23=+++⇔=

+++ sKss

ss

sK

tt

10

2

105

1

0

1

2

2

s

Ks

s

Ks

t

t

−

jss 20105 2 ±=⇒=+


-6 -5 -4 -3 -2 -1 0 1-15

-10

-5

0

5

10

150.040.0850.130.190.260.38

0.52

0.8

0.040.0850.130.190.260.38

0.52

0.8

2

4

6

8

10

12

14

2

4

6

8

10

12

14

Root Locus

Real Axis

Imag

inary

Axis

Damping ratioDamping ratio

If If K K = 10= 10, we , we

cannot achievecannot achieve

for any for any KKtt > 0> 0. .

Root Locus Root Locus –– (b) (b) KK = 10 = 10 (Control Example 1) ((Control Example 1) (MatlabMatlab ““rlocus.mrlocus.m””))


Characteristic eq.Characteristic eq.

Root Locus Root Locus –– (c) (c) KK = 5 = 5 (Control Example 1) (Control Example 1)

)(sR )(sE )(sY

)5(

12 +ss

5

sKt

0

)5(1

)5(

1

51

2

2

=

++

++

ss

sK

ss

t

055

1

)(

23=

+++

43421sL

tss

sK055 23 =+++ sKss

t


-6 -5 -4 -3 -2 -1 0 1-10

-8

-6

-4

-2

0

2

4

6

8

100.060.120.20.280.38

0.52

0.68

0.88

0.060.120.20.280.380.52

0.68

0.88

2

4

6

8

2

4

6

8

10

Root Locus

Real Axis

Imag

inary

Ax

is

Real axisReal axis

Ima

gin

ary

ax

isIm

ag

ina

ry a

xis

Root LocusRoot Locus

Root Locus Root Locus –– (c) (c) KK = 5 = 5 (Control Example 1) ((Control Example 1) (MatlabMatlab ““rlocus.mrlocus.m””))


a) Set T = 0. Draw root locus for K > 0.

b) Vary T to see the effect of a zero on root locus.

Root Locus Root Locus –– Control Example 2 Control Example 2

)2)(1(

1)(

++

+=

sss

TssL

)(sLK

)2)(1(

1)(

++=

ssssL


• Root locus for

ReRe

ImIm


Root Locus Root Locus –– (a) (a) (Control Example 2)(Control Example 2)

3

31+−

)2)(1(

1)(

++=

ssssL

=

2

33K

j2

j2−)6( =K


• When K is fixed and T is a positive parameter, the

characteristic equation can be written as

Root Locus Root Locus –– (b) (b) (Control Example 2)(Control Example 2)

0)2)(1(

11 =

++

++

sss

TsK

{ 0)2)(1(Twith

T without Term

=++++ TKsKsss444 3444 21

0)2)(1(

1 =+++

+Ksss

KsT


• Root locus for

various K & T

• Zero of L(s):

• Generally, addition of

a zero improves

stability.

Root Locus Root Locus –– (b) (b) (Control Example 2) (cont(Control Example 2) (cont’’d)d)

Ts

1−=


• Multiple parameter design examples

• Next, lead compensator design based on root locus

• More Example

– For the feedback system,

• Set a = 0, and draw RL for K > 0.

• Set K = 9, and draw RL for a > 0.

Root Locus Root Locus –– Control Example Summary Control Example Summary

)( ass

K

+


• Place closed-loop poles at desired location

– by tuning the gain C(s) = K.

• If root locus does not pass the desired location, then

reshape the root locus

– by adding poles/zeros to C(s). (How?)

PlantPlantControllerController

CompensationCompensation

Fixed!Fixed!Design Design

Target!Target!

(for time domain specs)(for time domain specs)

Root Locus Root Locus –– Closed Loop Design using RL Closed Loop Design using RL

)(sG)(sC


• Pulling root locus to the RIGHT

– Less stable

– Slow down the settling

ReRe

ImIm

ReRe

ImIm

ReRe

ImIm

Add a poleAdd a pole Add a poleAdd a pole

Root Locus Root Locus –– Effect of Adding Poles Effect of Adding Poles


• Pulling root locus to the LEFT

– More stable

– Speed up the settling

ReRe

ImIm

Add a zeroAdd a zero

ReRe

ImIm

ReRe

ImIm

ReRe

ImIm

Root Locus Root Locus –– Effect of Adding Zeros Effect of Adding Zeros


• Adding only zero

– often problematic because such controller amplifies the

high-frequency noise.

• Adding only pole

– often problematic because such controller generates a less stable system (by moving the closed-loop poles to

the right).

• These facts can be explained by using frequency response analysis.

• Add both zero and pole!

Root Locus Root Locus –– Adding Poles/Zeros Remarks Adding Poles/Zeros Remarks

0)( ),()( >+= zzsKsC

)0( ),/()( >+= ppsKsC


• Lead compensator

ReRe

ImIm

• Lag compensator

ReRe

ImIm

Why these are called Why these are called ““leadlead”” and and ““laglag””??

We will see that from frequency response in this class.We will see that from frequency response in this class.


)(sG)(sC

Root Locus Root Locus –– Lead and Lag Compensators Lead and Lag Compensators

)0,0( ,)( >>+

+= pz

ps

zsKsC


• Positive angle contribution

ReRe

ImImTest pointTest point

ss

--zz11--pp11

Root Locus Root Locus –– Lead Compensators Lead Compensators

0)( >=∠LeadLead

sC θ

0

)()()(

z

11

1

1

>=−=

+∠−+∠=+

+∠=∠

Leadp

Leadpszs

ps

zssC

θθθ


• Negative angle contribution

ReRe


ss

--zz22 --pp22

Root Locus Root Locus –– Lag Compensators Lag Compensators

0)( <=∠LagLag

sC θ

0

)()()(

z

22

2

2

<=−=

+∠−+∠=+

+∠=∠

Lagp

Lagpszs

ps

zssC

θθθ



– Improve transient response

– Improve stability

• Lag compensator

– Reduce steady state error

• Lead-lag compensator

– Take into account all the above issues.

Root Locus Root Locus –– Rules of Lead/Lag Compensators Rules of Lead/Lag Compensators

1

1

1)(ps

zsKsC

Lead+

+=

2

2

2)(ps

zsKsC

Lag+

+=

)()()( sCsCsCLagLeadLL

=


Root Locus Root Locus –– Example: Radar Tracking System Example: Radar Tracking System

)2(

4

+ss


• Consider a system

• Analysis of CL system for C(s) = 1

– Damping ratio ζ = 0.5

– Undamped natural freq. ωn = 2 rad/s

• Performance specification

– Damping ratio ζ = 0.5

– Undamped natural freq. ωn = 4 rad/s

ReRe

ImIm

Desired poleDesired pole

CL pole with CL pole with

C(sC(s) = 1) = 1

Root Locus Root Locus –– RTS: Lead Compensator Design RTS: Lead Compensator Design (1)(1)


)(sG)(sC

)2(

4)(

+=

sssG

j32


• A point s to be on root locus �� it satisfies

– Angle condition

• For a point on root locus, gain K is obtained by

– Magnitude condition

Odd numberOdd number

Root Locus Root Locus –– RTS: Angle and RTS: Angle and MagMag Conditions Conditions

,...2,1,0 ),12(180)( ±±=+×=∠ kksLo

KsL

1)( =


Evaluate G(s) at the desired pole.

o If angle condition is satisfied,

compute the corresponding K.

o In this example,

Angle condition is not satisfied.

ReRe

ImImDesired poleDesired pole

Angle deficiencyAngle deficiency


jjjjG

33

1

32)322(

4)322(

+

−=

+−=+−

j32

210)322( −=+−∠ jG

30=φ


To compensate angle deficiency, design a lead compensator

satisfying

ReRe


There are many ways to design such There are many ways to design such C(sC(s)!)!

):(30)322( φ==+−∠ jC

ps

zsKsC

+

+=)(


180)322( −=+−∠ jGC

j32


• Positive angle contribution

• Triangle relationsReRe


ss

--zz11--pp11


0)( >=∠LeadLead

sC θ

πθπθθ =−++ )(zLeadp

Leadpzθθθ =−


How to select pole and zero:

• Draw horizontal line PA

• Draw line PO

• Draw bisector PB

• Draw PC and PD

• Pole and zero of C(s) are shown in the figure.

ReRe


PPAA

OOBBCC

DD

--z(=z(=--2.9)2.9)--p(=p(=--5.4)5.4)

APOBPOAPB ∠=∠=∠2

1

j32


2

φ=∠=∠ BPDCPB


Compensator realization:

• One example, using operational amplifiers

--

++--

++


4R

+

+⋅−

−=

1

1

)(

)(

22

11

1

2

3

4

sCR

sCR

R

R

R

R

sV

sV

i

o

3R

2R

1R

1C

2C

)(tvi )(tv

o


)(

)(

)(

)(

)(

)()(

22

11

22

11

1

1

23

14

3

4

1221

1112

CR

CR

CR

CR

i

o

s

s

CR

CR

R

R

sCRR

sCRR

sV

sVsC

+

+⋅=⋅

+

+==

• Transfer function


ReRe

ImIm

• Lag compensator

ReRe

ImIm


22

1

CR−

K

z

p

22

1

CR−

11

1

CR−

11

1

CR−


0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4Uncompensated system (Uncompensated system (C(sC(s)=1))=1)Compensated systemCompensated system

Lead compensatorLead compensator gives gives

•• faster transient responsefaster transient response

(shorter rise and settling time)(shorter rise and settling time)

•• improved stabilityimproved stability


System responses (uncompensated and compensated)System responses (uncompensated and compensated)


0 1 2 3 4 50

1

2

3

4

5

Error constants (after lead compensation)

• Step-error constant

• Ramp-error constant

Lag compensator can reduce steadyLag compensator can reduce steady--state error. state error.

Unit ramp inputUnit ramp input

Ramp responseRamp response

NOT SATISFACTORY!NOT SATISFACTORY!


)4.5(

)9.2(675.4

)2(

4)()(

+

+⋅

+=

s

s

sssCsG

Lead

∞==→

)()(lim:0

sCsGKLead

sp

02.5)()(lim:0

==→

sCssGKLead

sv


How to design lag compensator?

• Lag compensator

• We want to increase ramp-error constant

Take, for example, z =10p.

• We do not want to change CL pole location s1 so much

(already satisfactory transient).

Root Locus Root Locus –– RTS: Lag Compensator Design RTS: Lag Compensator Design (1)(1)

ps

zssC

Lag+

+=)(

5002.5)()()(lim:0

>⋅==→ p

zsCsCssGK

LagLeads

v

≈

=+

1)(

0)()(1

1

11

sC

sCsG

Lag

Lead

0)()()(1111

≈+ sCsCsGLagLead


Guidelines to choose z and p

• The zero and the pole of a lag compensator should be close

to each other, for

• The pole of a lag compensator should be close to the origin,

to have a large ratio z/p, leading to a large ramp-error

constant Kv.

• However, the pole of a lag compensator too close to the origin may be problematic:

– Difficult to realize (recall op-amp realization)

– Slow settling (due to closed-loop pole near the origin)


1)(1

≈sCLag


Root locus with lag compensator

• Without compensator

ss11

• With compensator

ss11


180321

=−−− θθθ 180321

≈−+−−−pz

θθθθθ


How to design lag compensator?

• For the desired CL pole

• Take a small p (by trial-and-error!)

• Lead-lag controller


js 3221

+−=

010

,110

1)(1

1

1

1

1≈

+

+∠≈

+

+⇔≈

ps

ps

ps

pssC

Lag

025.0=p o

ps

ps

ps

ps88.2

10 ,97.0

10

1

1

1

1 −≈

+

+∠=

+

+

025.0

25.0

4.5

9.2675.4)(

+

+⋅

+

+⋅=

s

s

s

ssC

LL


Root locusRoot locus

-6 -4 -2 0-1 5

-1 0

-5

0

5

1 0

1 5

-6 -4 -2 0-1 5

-1 0

-5

0

5

1 0

1 5Ro o t Lo c u s

Re a l A x is

Imagin

ary

Axis

Ro o t Lo c u s

Re a l A x is

Imagin

ary

Axis

With lead compensatorWith lead compensator With leadWith lead--lag compensatorlag compensator

Desired poleDesired pole



Comparison of step responses

0 1 2 3 4 50

0.2

0.4

0.6

0.8

1

1.2

1.4

UncompensatedUncompensated

With lead compensatorWith lead compensator

With leadWith lead--lag compensatorlag compensator



Comparison of ramp responses

0 1 2 3 4 50

1

2

3

4

5

UncompensatedUncompensated

With lead compensatorWith lead compensator

With leadWith lead--lag compensatorlag compensator

Unit ramp inputUnit ramp input



• Controller design based on root locus

– Lag compensator design

• Lag compensator improves steady state error.

– Lead-lag compensator design

• Lead-lag compensator improves stability, transient and steady-state responses.

• Next, frequency response and Bode plot

Root Locus Root Locus –– Summary Summary

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