me 534 - 04 controller design ii (rev. 1.3).pptme.metu.edu.tr/courses/me534/protected/me 534 -...

Outline – Root Locus

• Root Locus TechniqueRoot Locus Technique– Review– Examplesp– Calculation of gain

• Dominant Poles– Case studies– Bandwidth– Bode plot

• Design Exampleg p

Chapter 4b ME 534 2

Root Locus of a Negative Feedback CTS

Consider the TF of the CLcontrol system w/o D(s) :

)()()(1)()(

)()(

sHsGsGsGsG

sRsY

pc

pc

+=

)()()()( pc

Characteristic equation is 0)(11 =+=+ sGHGG OCharacteristic equation is 0)(11 =+=+ sGHGG OLpc

)()(b

zsM

i+∏or 0)(1 =+ sKG where

)()(

)(

)()( 1

sasb

pssG N

i

ii

=+

=

∏

∏=

Chapter 4b ME 534 3

1i=

Root Locus (Cont’d)( )Roots of a(s) are the open loop (OL) poles while those of b(s) are calledthe OL zeros Hence the characteristic equation becomesthe OL zeros. Hence, the characteristic equation becomes

0)()()( =+= sKbsasA

where the roots of A(s) give the CL poles (or eigenvalues). To satisfy thecharacteristic equation, we obtain the following conditions:

Magnitude Condition:

1)(1)(0)(1 =∴−=⇒=+ sGsGsG OLOLOL

Angle Condition:

1)(1)(0)(1 −∠=∠∴−=⇒=+ sGsGsG OLOLOL

)12(180)()01()( +±=∠⇒+−∠=∠ ksGjsG oOLOL

1)(1)(0)(1 ∠∠∴⇒+ sGsGsG OLOLOL

where k ∈ {0 1 2 }Chapter 4b ME 534 4

where k ∈ {0, 1, 2, ...}

Root Locus (Cont’d)( )Our objective is to draw the loci of CL poles as K varies. In fact, the anglecondition is sufficient to form such a plot. Magnitude condition could beused to calibrate the resulting plot.

Test Point (so) As an illustration, consider a CTS( )

pole2

with 3 OL poles and a zero. A test point (so) is selected and theangle condition is queried:

l

pole2g q

)12(180)( 0 +±=∠ ksG oOL

?pole1

l

zero1 )12(1803211 +±=−−− koθθθϕ

If this condition is satisfied, s0 is onpole3If this condition is satisfied, s0 is onthe root loci. Unfortunately, thetechnique is very cumbersome!

Chapter 4b ME 534 5

Analytical Approach for Calculating Root Locus

• A straightforward method for calculating root loci isroot loci is– To increment K systematically

To calculate the roots of corresponding– To calculate the roots of corresponding A(K,s) = num{1 + GOL(K,s)}.

• MATLAB employs this simple algorithm to• MATLAB employs this simple algorithm to plot root locus of a control system with arbitrary complexityarbitrary complexity.

Chapter 4b ME 534 6

ExamplePlot the root locus of the following system using analytical techniques:

K)1(

)(+

=ssKsGOL

Solution:So u o

Characteristic polynomial of this system is

(K = 0 25)

))(()(

21

2

pspsKsssA++=

++= (K = 0) (K = 0)(K = 0.25)

-12))(( 21 pp

Hence,

411 K±

2

2411

2,1Kp −±−

=

Chapter 4b ME 534 7

Root Locus TechniqueRoot Locus Technique

• First developed by Evans in 1948.p y• Method can be applied to both CTS and

DTSDTS.• Procedure includes 6 steps to sketch the

root locus easily.

Chapter 4b ME 534 8

Procedure for Sketching Root LocusLocus

1. Plot the OL poles and zeros in s-plane. Root locus branches start from OL poles pand terminate either at zeros or at infinity:infinity:• If n is the number of OL poles and m is the

b f OL ( ≥ ) thnumber of OL zeros (n ≥ m), then mbranches end up at zeros while n-mb h i d tbranches wind up at ∞.

Chapter 4b ME 534 9

Procedure (Cont’d)( )

2. Determine the root loci on the real axis.

• Pick a test point on real axis. If total number of “ l” d “ l”“real” zeros and “real”poles at the right side of this point is an odd

1 (odd)

1 ( dd)

2 ( )

3 ( dd)this point is an odd

number (1, 3, 5, ...), then that point lies on the root

(odd)(odd)(even)(odd)

plocus.

Chapter 4b ME 534 10

Procedure (Cont’d)( )3(a). Determine the asymptotes of root loci:( ) y p

}...,,1,0{)12(180 mnkka −∈

+±=Φ

mn −

3(b) Determine the abscissa of intersection of the3(b). Determine the abscissa of intersection of the asymptotes:

zerospoles −∑ ∑mn

zerospoles iia −

−= ∑ ∑σ

Thi i l t th t ti f t fThis is analogous to the computation for center of gravity.



4 Find the breakaway and break in points on4. Find the breakaway and break-in points onthe root loci:

)()(0)()(

sbsaKsbKsa −

=⇒=⋅+

02 =⋅−⋅

=b

abdsdK ds

dbdsda

Real roots of this equation yield the locations of either

bds

break-away or break-in points on the real axis.



5(a) Determine the angle of departure of the root locus5(a). Determine the angle of departure of the root locus from a complex pole:

∠departure = 180o – (∑ angles of vectors to the complex ∠departure = 180 – (∑ angles of vectors to the complex pole in question from other poles)

+ (∑ angles of vectors to complex pole inquestion from other zeros)question from other zeros)

5(b). Determine the angle of arrival at a complex zero:

∠arrival = 180o – (∑ angles of vectors to the complex pole in question from other zeros) f

+ (∑ angles of vectors to complex pole inquestion from other poles)


Procedure (Cont’d)( )As an illustration, consider the CTSwith three OL poles. Let us find the

2

angle of departure from the complexpole2. From geometry, the angles ofvectors can be directly determined: 2 y

θ1 = 135 oθ3 = 90 o

1

3ϕ1 = 0 o (no zero)

Therefore,3

o

oooo

450)90135(1802

=

++−=

θ

θTherefore,

452 −=θThat is, the locus is leaving off thispole towards the first one.


p


6. Find the break-away point at s = jω(imaginary axis)(imaginary axis). • From this point on, the CL poles enter the

i t bilit i Th t binstability region. The system becomes unstable.

• For DTS, one needs to determine the breakaway points on the unit circle.


Example 1pSketch the root locus of the following system:

1+s)4(

1)( 2 ++

=ss

sKsGOL

Solution: tes

So u o

1. Poles: p1 = p2 = 0 (repeated poles); p3 = -4 (n = 3)

Z 1 ( 1) asym

ptot

Zeros: z1 = -1 (m = 1).

2. Root loci on the real axis: -4 ≤ s ≤ -1pole3

pole

2zero

1

a

5.1)1()400(−=

−−−+=σ

3. Asymptotes:pole1,2

5.113 −aσ

oo

a 9013

180±=

−±

=Φ


13 −

Solution of Example 1 (Cont’d)p ( )4. Break-away / break-in points:

0=−dsdba

dsdab

y p

0)4()83)(1( 232 =+−++ sssssdsds

0

5. Branches depart from poles @ 0

s = 0 s = -1.75 ± j0.97 (nonsense!)

2

5. Branches depart from poles @ 0with ∠±90o.

6. No crossing of imaginary axis.g g y


Solution of Example 1 (Cont’d)

3

4Root Locus

p ( )

L t if lt i

2

3Let us verify our results usingMatlab:

% d fi l f

0

1

ary

Axi

s

% define open-loop TF of CTS% (K is excluded in OLTF)G = tf([1 1],[1 4 0 0]); % plot root locus

-1

0

Imag

ina% p ot oot ocus

rlocus(G);

-3

-2

-4 -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0-4

3

Real Axis


Real Axis

Example 2pSketch the root locus of the following system:

1)2)(1(

1)(++

=sss

KsGOL

Solution:t t

So u o

1. Poles: p1 = 0; p2 = -1; p3 =-2 (n = 3)Zeros: none (m = 0).

asymptote

2. Root loci on the real axis: -1 ≤ s ≤ 0 and s ≤ -2

pole3 pole1pole2 60o

-60o

10)210(−=

−−−=σ

3. Asymptotes:

asymptote

60

103 −aσ

oooo

ak 300,180,6003

)12(180=

+±=Φ


03 −


0=−dsdba

dsdab

y p

0)263( 2 =++− ssdsds

0 4226-0.4226

5 No departure from poles

s = -0.4226 s = -1.5774 (nonsense!)

5. No departure from poles.

6. Locus is to cross imaginary axis.


Solution of Example 2 (Cont’d)4

Root Locus

p ( )Let us verify this result with

2

3y

Matlab again:

% define open-loop TF of CTS

0

1

ry A

xis

p pG = tf(1,poly([0 -1 -2]));% plot root locusrlocus(G);

-1

0

Imag

inar

-3

-2

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

3

Real Axis


Real Axis

Example 3 – DTS pSketch the root locus of the following system:

50z2)1(5.0)(

−−

=zz

zKzGOL

Solution:

Imz-plane

jasymptote

So u o

1. Poles: p1 = 0; p2 = p3 = 1 (n = 3)Zeros: z1 = 0.5 (m = 1).

2. Root loci on the real axis: 0 ≤ z ≤ 0.5 Re

2pole1 zero1 pole2,3-10.50

75.03

5.0)011(=

−++=aσ

3. Asymptotes:

13 −a

oo

ak 9013

)12(180±=

+±=Φ

-j


13 −


Imz-plane

0=−dzdba

dzdab

y pj

05.135.32 23 =−+− zzzdzdz

z = 12-1

5 Branches depart from poles

z1 = 1 z2,3 = 0.375 ± j0.7806 (nonsense)

Re0.50

5. Branches depart from poles@ 1 with ±90o

6. Locus does not cross the unit -j6. Locus does not cross the unitcircle.

NOT POSSIBLE TO STABILIZE!


Solution of Example 3 (Cont’d)0 8

1

0 7 /T

0.6π/T0.5π/T

0 3 /T

0.4π/T

0 1

Root Locus

p ( )Let us verify this result with

0 4

0.6

0.8

0.8π/T

0.7π/T

0.2π/T

0.3π/T0.1

0.2

0.3

0.4

0.5

0 6

yMatlab again:

% define open-loop TF of DTS

0

0.2

0.4

0.9π/T 0.1π/T

0.9

π/T

0.7

0.8

0.6

ry A

xis

G = tf([1 -.5],poly([0 1 1]),-1);% plot root locusrlocus(G);zgrid;

-0.2

0π/T

0.1π/T0.9π/T

π/T

Imag

inar

-0.6

-0.4

0 3 /T

0.2π/T0.8π/T

0 7 /T

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1

-0.8

0.6π/T0.5π/T

0.4π/T

0.3π/T0.7π/T

Real Axis


Obtaining Gain from the Plotg

• Once the root locus plot is obtained, appropriate CL pole locations are selected.

• Gain (K) yielding the desired pole locations are determined from the plot using the magnitude condition:

|GOL(s0)| = |KG(s0)| = 1K = 1 / |G(s0)|

where s0 is the desired location on the plot.0 p


ExampleConsider the root locus of the followingsystem:

K]16)4[(

)( 2 ++=

ssKsGOL

D t i K f th CL l ( ) hpole2 s0

|s0 - s2|

Determine K for the CL poles (■) shown on the plot.

pole1

|s0 - s1||s0 - s3|

Solution: pole1

pole3

Solution:

1||||||

)|(|302010

0 =−−−

=ssssss

KsGOL302010

51|||||| 302010 =−−−=434214342143421ssssssK/8.5/2.2/0.4 sradsradsrad

Be careful with the scale and units. After the vector length is measured(say in mm) one needs to convert it to [rad/s] for the calculationChapter 4b ME 534 26

(say in mm), one needs to convert it to [rad/s] for the calculation.

CL Pole and Gain Calculation with Matlab

• Matlab uses an interactive function to calculate the gain yielding the CL poles selected by the user:>> rlocus(G);

[K l ] l fi d(G)>> [K,poles] = rlocfind(G)• Once rlocus plots the root locus, Matlab

t th t i t th d i d lprompts the user to point the desired pole locations on the graph (No need to be exact!).

• Hence, rlocfind returns not only the gain but also the selected CL poles.


Example – Gain Calculation8

10Root Locus

pRevisit the previous example.

4

6G = tf([1],[1 8 32 0])); rlocus(G);[K,poles] = rlocfind(G)

0

2

4

ary

Axi

s

Matlab’s Output:Select a point in the Graphics window

Selected CL poles

4

-2

0

Imag

ina

pselected_point =-2.6588 + 3.3075i

-6

-4K =47.5909

poles =

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

-8

Real Axis

poles -2.6581 + 3.2660i-2.6581 - 3.2660i-2.6839


Dominant Poles

A typical control system may have several poles• A typical control system may have several poles (eigenvalues). – The system might be decomposed into several paralel y g p p

subsystems with different “weights” and “operating speeds.”• In controls literature, the dominant pole of a control

system refers to the slowest pole (“subsystem”) thatsystem refers to the slowest pole ( subsystem ) that dominates the overall dynamics of the system.

• To use root locus tools efficiently, one needs to y,understand:– How zeros and poles interact with each other,

How the location of closed loop poles affect the system dynamics– How the location of closed loop poles affect the system dynamics,– How to simplify the overall system.


Dominant Poles – Case 1

Im

To see the effect of pole locations a unit step to thislocations, a unit step to this system. At each test, the location of the pole at δ is ReResponsepchanged. Response

gets faster


Dominant Poles - Case 1 (Cont’d)

0.9

1Step Response

0.7

0.8

δ ∈ {0.6, 0.7, 0.8, 0.9}

0.5

0.6

plitu

de

A th l t δ t d th

0.3

0.4

Am

p As the pole at δ moves towards the origin, the response gets faster.

0.1

0.2

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.1

Time (sec)


( )


0

A h d dIm

As the next case, a second-order system with two real roots is considered Slowest pole at δ isconsidered. Slowest pole at δ is fixed while the location of the other pole is incrementally Rep ymoved towards the origin at each test.

ReResponse does not

change much!


Dominant Poles – Case 2 (Cont’d)

0.9

1Step Response

0.7

0.8

0 5

0.6

0.7

itude

δ = 0.9 ε ∈ {0.6, 0.7, 0.8, 0.9}

0 3

0.4

0.5

Am

pli

Slowest pole at δ dominates the system dynamics Thus

0 1

0.2

0.3 the system dynamics. Thus, changing the location of the fastest pole at ε does not havethat much effect

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.1

Time (sec)

that much effect.


Time (sec)


ImA zero (at ξ), which is adjacent to the slowest pole at δ, is added to the previous system. Pole at δ as

ell as the ero are all stationar

ReResponse gets

well as the zero are all stationary while the location of the other pole is incrementally moved Response gets

noticeably faster!pole is incrementally moved towards the origin at each test.


Dominant Poles – Case 3 (Cont’d)

0.9

1Step Response

0 7

0.8δ = 0.9 and ξ = 0.85ε ∈ {0 6 0 7 0 8 0 9}

0 5

0.6

0.7

tude

ε ∈ {0.6, 0.7, 0.8, 0.9}

0 3

0.4

0.5

Am

plit

Effect of the slowest pole at δ iscompensated by the adjacent zero!

0.2

0.3 compensated by the adjacent zero!Thus, changing the location of the other pole at ε does have a noticeable i fl th

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.1 influence on the response.


Time (sec)

System Simplificationy p

• If the pole pi is close to the unit circle pi is relativelyIf the pole pi is close to the unit circle, pi is relatively more “dominant” over the other poles since the transient response associated with it decays slowly.– After a while, the system dynamics will be completely dominated

by that pole.

If l i l t it ff t th t i t• If a pole is very close to a zero, its effect on the transient response can be neglected. – That pole and corresponding zero can be excluded from the CL– That pole and corresponding zero can be excluded from the CL

transfer function.– This omission is more justified when

• That pole is close to other poles and far from zeros (large residue) • That pole is far from other poles and close to zeros (small residue)


Simplification (Cont’d)p ( )Im Im

System might becharacterized by the slowest pole!

Reslowfast Reslow

the slowest pole!

)( 0KG )(' 0KG)9.0(

)( 0

−=

zzzG

9.0)(' 0

−=

zzG


Simplification (Cont’d)

0 9

1G'(z)G(z)

Step Response

p ( )

0 7

0.8

0.9 ( )

0 5

0.6

0.7

tude

D it th i lifi ti

0 3

0.4

0.5

Am

plit Despite the simplification,

the difference in responsesis quite small...

0 1

0.2

0.3

0 10 20 30 40 50 600

0.1

Time Index k (sec)


Time Index k (sec)

Bandwidth Frequencyq ySometimes, the location of dominant pole is indirectly specified in terms of a bandwidth frequency or break frequency (fb):

sfbf

ee aTπ

δ2−− ==

in terms of a bandwidth frequency or break frequency (fb):

where fs =1/T is the sampling frequency.

B k f f th t ll d

max

Break frequency of the controlledsystem is to meet the specifiedbandwidth frequency.

f sm

This could be verified by checking the frequency which correspondsq y pto -3 dB point in the Bode plot.


Bode Plot of a Discrete-time System

Consider CL transfer function of a discrete-time control system:

)()( FzY )()()( zF

zRzY

=

Bode plot of this system is obtained byBode plot of this system is obtained by

TjzFjF ω = )()(' TjezzFjF ωω

=)()(

Magnitude (in dB) = 20 log |F ’(jω)|

where 0 ≤ ω ≤ 2πfs/2 = π/T. By default, the sampling frequency must be at least twice the specified bandwidth frequency.


p q y

ExampleConsider the following discrete-time plant:

)( zzY2)1(

)()()(

−==

zzzG

zMzY

p

where sampling period (T) is selected as 0.001 seconds. Design a PD controller to achieve a (CL) bandwidth frequency of at least 100 Hz while the maximum overshoot is not to exceed 10%while the maximum overshoot is not to exceed 10%.

Solution:

TF function of the PD controller is

bzKzKKG −− −11)( TK

pKK d+=̂

zbzK

TzKKzG dpc =+=

1)( where

dp

d

p

KTKKb+

=̂


dp

Solution (Cont’d)OL transfer function becomes

)( −bzK2)1()()(

−−

=z

bzKzGOL

I t l d i t h i th OL (b) i l t d fi t ThIn root-locus design technique, the OL zero (b) is selected first. Thenthe CL poles are placed to desired location in the z-plane so as to achieve the desired CL bandwidth frequency. For instance,

5335.0001./110022

=== −− ππδ ee sf

bf (points the location ofdominant pole!)

Accompanying Matlab script is utilized in the design. Hence, the gains are

⎫02590K⇒

⎪⎭

⎪⎬⎫

×=

=-4104.9225

0259.0

d

p

KK fb = 127 [Hz] > 100 Hz.

Mp = 8% < 10%


Solution (Cont’d)1

0 6π/T0.5π/T

0 4π/T

Root Locus

( )

Matlab’s Output:

0.6

0.8

0.8π/T

0.7π/T

0.6π/T

0.2π/T

0.3π/T

0.4π/T

0.2

0.1

0.3

0.40 5

Matlab s Output:

Select a point in the graphics window

0.2

0.4

0.9π/T 0.1π/T

π/T

0.8

0.9

0.7

0.50.6

y A

xis

selected_point =0.5379 - 0.0031i

-0.4

-0.2

0 π/T

0.1π/T0.9π/T

π/T

Imag

inar

y

K =0.5182

poles =

-0.8

-0.6

0.4

0.7π/T 0.3π/T

0.2π/T0.8π/T

poles 0

0.94390.5379

-1 -0.8 -0.6 -0.4 -0.2 0 0.2 0.4 0.6 0.8 1-1 0.5π/T

0.4π/T0.6π/T

Real Axis


Solution (Cont’d)0

2

System: Gcl

Bode Diagram

-4

-2

gnitu

de (

dB)

y Frequency (rad/sec): 796 Magnitude (dB): -3

-10

-8

-6

Mag

-45

0

g)

-10

-180

-135

-90

Pha

se (

deg

101

102

103

104

-225

-180

Frequency (rad/sec)


Frequency (rad/sec)

Solution (Cont’d)1.4

S t G l

Step Response

1

1.2 System: Gcl Time (sec): 0.00847 Amplitude: 1.08

0.8

itude

0.6Am

pli

An overshoot is observed as the effectof zero @ 0.95 is not perfectly cancelled by that of the pole sitting at 0 9439!

0.2

0.4 by that of the pole sitting at 0.9439!

0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050

Time (sec)


Time (sec)

Matlab ScriptpT = 0.001; b = .95; % ParametersGp = tf([1 0],[1 -2 1],T); % Define plant's TFGc = tf([1 -b],[1 0],T); % Now PD controllerG = Gc*Gp; % OL TF w/o Krlocus(G); zgrid; % Plot root locusrlocus(G); zgrid; % Plot root locus[K,poles] = rlocfind(G) % Place the CL polesGcl = feedback(K*G,1); % CL TFfigure(2); bode(Gcl); % Check the bandwidthfigure(3); step(Gcl); % and transient resp.Kd = b*K*T; Kp = K - Kd/T; % Calculate gains


me 534 - 04 controller design ii (rev. 1.3).pptme.metu.edu.tr/courses/me534/protected/me 534 -...

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