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Chapter 4

Root-Locus Analysis

and Design

Control Automático

3º Curso. Ing. IndustrialEscuela Técnica Superior de Ingenieros

Universidad de Sevilla

Depto. Ing. de Sistemas y Automática. Control Automático. 3º Ing. Industrial. ESI.US.

Outline of the presentation

� Introduction to the Root Locus

� Root-Loci Plot for positive gains

� Controller gain design

� Improving the closed-loop behaviour:

� Transitory response

� Steady state response

� Controller design

� Extensions:

� Root-Loci Plot for negative gains

� Generalized Root-Locus


Introduction to the Root Locus


� The canonical problem

� Example

� Characterization of roots: Magnitude and angle conditions







� Extensions:




The canonical Problem

� Objective: analysis of the effect of a single parameter variation on the location of the closed-loop poles (Evans, 1948)

RRG(s)C(s)

-

+ E U YG(s)C(s)

-

+

Controller

G(s)G(s)C(s)K

System-

+ E U Y

How does the location of closed-loop poles changeas K varies?


The canonical Problem

� Purpose

� Analysis: How system behaviour changes with the parameter

� Synthesis: How to select K to obtain the desired behaviour

G(s)C(s)-

+R E U YG(s)C(s)

-

+

Controller

G(s)G(s)C(s)K

System-

+R E U Y

• Analitically: impossible for higher order systems

• Graphically: Parameterized plot in K


Example

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Closed-loop poles:

K=0

-2 -1

K=1 K>1

x

x

x

K<0

xx x x

Root-Loci

Controller System


The Canonical Problem

What if ?

� Impossible to compute analytically the location of closed-loop poles depending on K

� Graphic Plot (approximate plot)

Root-Loci Construction(Evans, 1948)


Characterization of Roots

� Angle Condition

� Magnitude Condition


Characterization of Roots

� Angle Condition

G(s) takes the form

o x

x

x

x


Example

G(s)C(s)-

+R E U YG(s)C(s)

-

+

Controller

G(s)G(s)C(s)K

System-

+R E U Y

-2 -1xx x


Example

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

-2 -1

x

x

x x

Controller System


Example

-2xx x

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Controller System


Example

-2

x

x x

x

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Controller System


Example

K=0

-2 -1

K<0

x x

K>0

Branch for the pole s=0

Branch for the pole s=-2

Break-away point

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Controller System


Root-Loci Plot for positive gains








� Extensions:




Root-Loci Plot: important facts

� Number of branches:

� If G(s)=n(s)/d(s) where n(s) and d(s) are polynomials of degree m and n respectivelly then the closed-loop roots are the solutions to

� The number of roots (branches) is constant and equal to n if the polynomial d(s)+Kn(s) has constant degree n for every K (this is satisfied, for example, if n>m).

�Moreover, if the degree of the characteristic polynomial does not change with K, the roots vary continuously with K.


Root-Loci Plot: important facts

� Symmetry

� The root locus is symmetrical about the real axis.

� This is because complex closed-loop poles appear always in conjugate pairs.

� Starting points (K=0)

� For small values of K>0, the characteristic equation d(s)+Kn(s)=0 can be approximated by d(s)=0.

� Consequence: The root locus starts at the poles of G(s)=n(s)/d(s).

� End points (K→∞)� The characteristic equation can be rewritten as (1/K)d(s)+n(s)=0.

Which, for large values of K>0, is approximated by n(s)=0.

� Consequence: From the n branches, m converge to the m zeros of G(s) (the n-m remaining branches converge to n-m asymptotes).

� .


-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xis

RL example

� 4 open loop poles (n=4)

� 1 zero. m=1

� 4 branches

� 1 break away point

� 1 break in point

1 zero

3 asymptotesxx

x

x


Root-loci plot for K>0

� We start from the following canonical form

� The Root Loci consists of n branches

� The branches start from the open-loop poles

�m branches tend to the open-loop zeros

� n-m branches tend to infinity in an asymptotic way.


Root-Loci plot for K>0

� Root-Loci plot

� There exists a systematic procedure to obtain the plot

� The obtained plot is an approximation

� Important qualitative information

� Illustrative example


Steps for the Root-Loci plot

Step 1: Place open-loop poles and zeros

-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xisx → pole

o → zero



Step 2: Root-Loci Plot on the real axis

so ∈ RL if it is to the left of an odd number of real poles and zeros.

-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xis



Step 3: Determination of the break away and break in points

-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xis

Break away point

Break in point



Computation of the break away and break in points

They are roots with multiplicity of the characteristic polynomial

• They are roots of the closed-loop denominator

• They are roots also of the derivative of the denominator



� Example

Roots

-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xis



Step 4: Asymptotes computationm branches tend to the zerosn-m branches converge to the asymptotesn branches

Centroid

Asymptote angles



� Example

-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xis



Step 5: Computation of the possible crossings with theImaginary axis

Routh-Hurwitz Table

Second order even factor



� Example

-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xis

Second order even factor

Subsidiary equation



Step 6: Angle of departure from complex poles and angle of arrival at complex zeros

Departure angle from a complex pole

Arrival angle at a complex zero

o x

x

x



� Example

-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xis


-7 -6 -5 -4 -3 -2 -1 0 1 2 3-6

-4

-2

0

2

4

6Root Locus

Real Axis

Imag

inar

y A

xis



Controller gain design� Introduction to the Root Locus


� Controller gain design:

� Hypothesis for the design

� Control specifications and admisible regions

� Computing the controller gain

� Examples





� Extensions:




General assumption

Im

Re

<=>

:1

:1

:1

δδδ Overdamped

Critically damped.

Underdamped.


Underdamped system

00

Tiempo

y(t)

)(

)()(..

∞∞−

=y

ytyOS p

)(∞y

etptst


Underdamped system

00

Tiempo

y(t)

)(

)()(..

∞∞−

=y

ytyOS p

)(∞y

etptst

Settling time T5%: The time required to reach and stay within 5% of the steady-state value.


Normalized rise time

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.91.5

2

2.5

3

3.5

4

4.5

5

5.5

6

6.5

00

Tiempo

y(t)

)(

)()(..

∞∞−

=y

ytyOS p

)( ∞y

etptst


Underdamped system


Example 1:

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

-2

-1

0

1

2

3

Real Axis

Imag

Axi

s

Root Locus Editor (C)

xx

x

xK=1.38

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Controller System

Which is the overshoot and the rise time for a proportional controller with K=1.38 ?

From the closed-loop characteristic polynomial we obtain that the closed-loop poles are placed at -1.5+2.57j and -1.5-2.57j

-1.5

2.57


Example 1:

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

-2

-1

0

1

2

3

3

0.64 0.5 0.38 0.28 0.17 0.08

0.38 0.08

0.94

0.8

2

2.5

1

1.5

0.28 0.17

0.8

2.5

0.5

2

0.50.64

1.5

0.94

0.5

1

Real Axis

Imag

Axi

s


xx

x

x

K=1.38A


Example 1:

Step Response

Time (sec)

Am

plitu

de

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

ts=0.84 tp

� Since the open-loop transfer function has no integrators, the steady state position error is different from zero

� The expressions for the overshoot and rise time are exact in this case because the closed-loop system is a second-order system without zeros.


• Dominant dynamics: poles with the slowest response

• In practice, dominant poles are determined through their relative distance to the imaginary axis.

Re

Imp1

p’1

p2

p’2

d2

d1

p1 are dominant if d2/d1>5

Re

Im

p1

p2

p’2

d2

d1

The static gain must be the same

Closed-loop dominant poles


Closed Loop Dominant Poles

-1 is dominant. The remaining poles are neglected

Re

Im

-1-16

-17

Tiempo(s)0 1 2 3 4 5 6

0

0.5

1

1.5

2

2.5

y(t)

1

2

)17)(16)(1(

544

)17)(16)(1(

544)(

+=

+≈

+++=

ssssssG


Proportional controller

� The root locus illustrates how the transient response varies with the proportional gain� The overshoot and the rise time corresponding to each point of the root locus can be estimated (assuming the existence of dominant poles)� Design process:� Given a set of control specifications, it is possible to determine

if there is a point (belonging to the root locus) satisfying all the specifications� If such a point exits, then the corresponding gain is obtained by

means of the magnitude condition: If A belongs to the root locus and satisfies all the specifications, then the corresponding gain is obtained from:


Specifications on the complex plane

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Controller System

xx

Im

Rex

The region in green corresponds to the simultaneous satisfaction of SO and settling time specifications.

Similar regions are obtained for simultaneous SO and wn specifications.


Example 2:

� Design a proportional controller in such a way that the system is as fast as possible and the SO is not greater than 20%

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Controller System


Example 2:

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

-2

-1

0

1

2

3

Real Axis

Imag

Axi

s


xx62.87º

A

Step Response

Time (sec)

Am

plitu

de

0 0.5 1 1.5 2 2.5 3 3.5 40

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1


Example 3:

Proportional control of servo system

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Controller System

Is it possible to obtain a proportional controller such that SO and is not greater than 10% and the rise time is not greater than 5 seconds ? xx

Im

Rex

-5 -0.05


G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K

-

+R E U Y

Controller System

The characteristic polynomial can be approximated by a second order one: Im

Re In the region of interest, one can make the approximation:

region of interest

xx x-5 -0.05

Example 3:



In the region of interest, the root locus can be approximated by the root locus corresponding to:

Re

That is,

Im

x x-0.05 -0.025

Example 3:



In order to satisfy the SO specification:

Re

Im

x x-0.05 -0.025

A

53.75º

Example 3:



Step Response

Time (sec)

Am

plitu

de

0 50 100 150 200 2500

0.2

0.4

0.6

0.8

1

1.2

1.4

ts

Since the assumption of dominant poles is valid, the transient behaviour of the closed loop third order system is the one predicted.

The proportional controller fails to meet the rise time specification. In order to make the closed loop faster, one requires to increase the gain K, however, this increases the SO beyond its specified value.

Example 3:



Improving the closed-loop behaviour








� Extensions:



Depto. Ing. de Sistemas y Automática. Control Automático. 3º Ing. Industrial. ESI.US. 55

Design for transient specifications

� The root locus illustrates how the transient response varies with the proportional gain.

� The overshoot and the rise time corresponding to each point of the root locus can be estimated (assuming the existence of dominant poles)

� Design process (First case):

� Given a set of control specifications, it is possible to determine if there is a point A (belonging to the root locus) satisfying all the specifications

� If such a point exits, then the corresponding gain is obtained by means of the magnitude condition:


Design for transient specifications

� Design process (second case):

� If there is no point in the root locus satisfying all the transient specifications then it is necessary to modify the root locus (a simple proportional controller is not able to meet the transient specifications).� One should determine a point A in the complex plane with

damping ratio and natural frequency satisfying the transient specifications.� Once this point is chosen, the root locus is modified by means of

the inclussion of a compensator.� The compensator (PD or Lead compensator) is designed in such

a way that the compensated root locus goes through the desired point A (the parameters of the compensator can be obtained from the angle condition at A)� Once the zeros and poles of the compensator are obtained, the

gain of the compensator is obtained from the application of the magnitude condition at A


G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)K(s+c)

-

+R E U Y

Controller System

Example 4:

Proportional Derivative control of servo system

Rex

-5x x

x

-0.05

A

Im


Rex

-5x x

x

-0.05

A

Im

� Including a zero (PD controller) it is possible to modify the angle condition in such a way that the desired point A belongs to the new root locus� The gain of the PD controller is used to make the point satisfy the magnitude condition

� The uncompensated root locus does not go through the desired point ( A does not satisfy the angle condition)� This means that it is impossible to meet the specifications with a proportional controller

Example 4:



Rex

-5x x

x

-0.05

A

Im

Example 4:



Rex

-5x x

x

-0.05

A

Im

o-c

Example 4:



Rex

-5x x

x

-0.05

A

Im

o-c

� Root locus of the compensated system� The gain of the controller is obtained from the magnitude condition

Example 4:



� The closed loop behaviour of the compensated system does not correspond to the predicted overshoot and rise time.

0 2 4 6 8 10 12 14 16 180

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

� This is because the closed transfer function has a closed loop zero at –c which is not cancelled by any closed-loop pole

� Closed loop transfer function:

Example 4:



Example 4 (revisited):

Proportional derivative control of servo system

Rex

-5x x

x

-0.05

A

Im

o-c

� Notice that point Bcorresponds to the same overshoot as point A, that is, same δ.� However, it has faster response (wn is larger)

� Also, point B has associated a greater value of controller gain. � This implies better steady state behaviour.

xB




Rex

-5x x

x

-0.05

A

Im

o-c

� Approximating the vertical branch by the asymptote one obtains:

xB




Rex

-5x x

x

-0.05

A

Im

o-c

� Applying the magnitude condition: xB

� Predicted values for overshoot and rise time:


� The closed loop behaviour of the new compensated system (in red) is much faster than the original one.

� The closed transfer function has a closed loop zero at –c which is partially cancelled by a closed loop pole.

� Closed loop transfer function:



0 2 4 6 8 10 12 14 16 180

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de


-

Example 5:

Lead compensator+R E U Y

Rex x

xA

Im

x-6 -2 -1


Example 5:

Lead compensator

Rex x

xA

Im

x-6 -2 -1

o-3


Example 5:

Lead compensator

-18 -16 -14 -12 -10 -8 -6 -4 -2 0

-6

-4

-2

0

2

4

6

Root Locus

xxxx

Ax


Example 5:

Lead compensator

-150

-100

-50

0

50

Mag

nitu

de (

dB)

10-2

10-1

100

101

102

103

-270

-180

-90

0

Pha

se (

deg)

Bode DiagramGm = 16.4 dB (at 10.3 rad/sec) , Pm = 55.6 deg (at 3.35 rad/sec)

Frequency (rad/sec)


Example 5:

Lead compensator

0 0.5 1 1.5 2 2.5 30

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de


Design for steady state

specifications

� Suppose that a compensator (proportional controller, PD, lead compensator) has been designed for the system taking into account ONLY control transient specifications (the closed loop system has a (dominant) pole at a desired location A).

� It is possible to improve (a posteriori) the steady state specifications including (in series) with the initial compensator a PI or a Lag compensator

� The idea is very simple: if the zero and pole of the compensator are close to the origin then the new compensated root locus will be almost identical to the original one in the proximity of A.

� For that purpose, it suffices to take the zero of the compensator smaller than |A|/10.

� The ratio between the zero and the pole determine the improvement in the steady state response


-

Example 6:

Proportional Integral controller +R E U Y


Example 6:

Proportional Integral controller


Example 6:


-20 -15 -10 -5 0

-8

-6

-4

-2

0

2

4

6

Root Locus

Real Axis

Imag

inar

y A

xis

-18 -16 -14 -12 -10 -8 -6 -4 -2 0

-6

-4

-2

0

2

4

6

Root Locus (Lead compensator)

xxxx

Ax

xA

The PI does not modify the root locus around A


Example 6:


10-2

10-1

100

101

102

103

-270

-225

-180

-135

-90

-45

Pha

se (

deg)

-150

-100

-50

0

50

Mag

nitu

de (

dB)

Bode DiagramGm = 15.4 dB (at 9.76 rad/sec) , Pm = 47.5 deg (at 3.37 rad/sec)

Frequency (rad/sec)


Example 6:


0 1 2 3 4 5 6 70

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de


Example 6:


0 1 2 3 4 5 6 70

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

The high settling time is due to the slow pole close to the origin


-

Example 7:

Lag compensator +R E U Y


Example 7:

Lag compensator


-18 -16 -14 -12 -10 -8 -6 -4 -2 0 2

-6

-4

-2

0

2

4

6

Root Locus

Real Axis

Imag

inar

y A

xis

Example 7:

Lag compensator

-18 -16 -14 -12 -10 -8 -6 -4 -2 0

-6

-4

-2

0

2

4

6

Root Locus (Lead compensator)

xxxx

Ax

xA

The Lag compensator does not modify the root locus around A


0 1 2 3 4 5 60

0.2

0.4

0.6

0.8

1

1.2

1.4Step Response

Time (sec)

Am

plitu

de

Example 7:

Lag compensator


Extensions








� Extensions:




Root Locus for K<0

K<0

In some situations, one is interested in a negative gain (K<0)

• Inverse action:

If ↑↑↑↑u � ↓↓↓↓y, then ↑↑↑↑e � ↓↓↓↓ u

(Negative controller gain )

G(s)C(s)-

+ E U YG(s)C(s)

-

+

Controller

G(s)G(s)C(s)K

System-

+ E U Y


Example

Negative Bode Gain !

0 10 20 30 40 50 60−10

−8

−6

−4

−2

0

2

Step Response

Time (sec)

Am

plitu

de

In order to guarantee the stability of the negative feedback we require negative controller gain


Characterization of the roots

� Angle condition

� Magnitude condition


Root Loci Plot for K<0

1- Placement of open loop poles and zeros

2- Root Loci on the real axis

3- Break away and break in points

4- Asymptotes and centroid

5- Crossing points at the imaginary axis

6- Departure and arrival angles at complex poles and zeros

s0 2 RL if it is to the left of an even number of real poles and zeros.


Example

-1

-1.5+j

-1.5-j

Step 1. Placement of open-loop poles and zeros

Step 2. Root-loci on the real axis


Example

-1

-1.5+j

-1.5-j

0.12

Step 3. Breaking away and breaking in points

Roots of:

Root-Loci on the real axis for K<0


Example

Step 4. Asymptotes and centroid

n=2,m=1 implies the existence of a unique asymptote

-1

-1.5+j

-1.5-j

0.12


Example

Step 5. Crossing points at the imaginary axis

Routh-Hurwitz

-1

-1.5+j

-1.5-j

0.12

Subsidiary equation K=-3

Subsidiary equation K=-3.25

0.5j

-0.5j


Example

Step 5. Departure angle from complex poles116.6º

90º

-1.5+j

-1.5-j

1

-1

-1.5+j

-1.5-j

0.12

0.5j

26.6º


Generalized Root-Locus








� Extensions:





G(s)C(s)-

+R E U YG(s)C(s)

-

+

Controller

G(s)G(s)C(s)K

System-

+R E U Y

Root Locus:

G(s)C(s)-

+R E U YG(s)C(s)

-

+G(s)G(s)C(s)C(s)

-

+R E U Y

Generalized Root Locus: Parameters different from K

Example: C(s) = PD, PI, PID...

Controller System



The closed loop system depends on a parameter αin such a way that

Ths is the same structure considered for the Root Locus:

Canonical problem with

We can use the same strategies !!


Exampe

+R E U Y

-

Objective: Analysis of the effect of the parameter T in the closed-loop poles.

1) Characteristic equation:

2) Computation of

T>0


Example

-1

-0.5+0.866j

-0.5-0.866j

Step 1. Placement of poles and zeros

Step 2. Root-Loci on the real axis

T>0

0


Example

Step 3. Breaking away and breaking in points

Roots of :

They do not belong to the real axis for T>0

-1

-0.5+0.866j

-0.5-0.866j

0

There are no breaking away points


Example

Step 4. Asymptotes and centroid

n=3,m=1 implies the existence of two asymptotes

-1

Centroid

270º

90º


Example

Step 5. Crossing point with the imaginary axis

Routh-Hurwitz The third entry of the first column vanishes for

T = -3/2 <0

The Root Locus for T>0 does notcross the imaginary axis.

-1

-0.5+0.866j

-0.5-0.866j

0


Example

Step 5. Departure angle from complex poles120º

90º

-0.5+0.866j

-0.5-0.866j

0-1

60º

-1

-0.5+0.866j

-0.5-0.866j

0

150º

chapter 4 root-locus analysis and design - …control-class.com/ch_4/slides/root_locus.pdf ·...

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