Download - Course Outline - AMME Current Student Information Pagesweb.aeromech.usyd.edu.au/AMME3500/Course_documents/material... · 3 Dr. Stefan B. Williams AMME 3500 : Root Locus Slide 9 Example:

1

Amme 3500 : System Dynamics and Control

Root Locus

Dr. Stefan B. Williams

Slide 2 Dr. Stefan B. Williams Amme 3500 : Introduction

Course Outline Week Date Content Assignment Notes 1 1 Mar Introduction 2 8 Mar Frequency Domain Modelling 3 15 Mar Transient Performance and the s-plane 4 22 Mar Block Diagrams Assign 1 Due 5 29 Mar Feedback System Characteristics 6 5 Apr Root Locus Assign 2 Due 7 12 Apr Root Locus 2 8 19 Apr Bode Plots No Tutorials 26 Apr BREAK 9 3 May Bode Plots 2 Assign 3 Due 10 10 May State Space Modeling 11 17 May State Space Design Techniques 12 24 May Advanced Control Topics 13 31 May Review Assign 4 Due 14 Spare

Slide 3 Dr. Stefan B. Williams AMME 3500 : Root Locus

Designing Control Systems •  We have had a quick look at a number of

methods for specifying system performance •  We have examined some methods for designing

systems to meet these specifications for first and second order systems

•  We will now look at a graphical approach, known as the root locus method, for designing control systems

•  As we have seen, the root locations are important in determining the nature of the system response


Proportional Controller •  When the feedback control signal is made to be

linearly proportional to the system error, we call this proportional feedback

•  We have seen how this form of feedback is able to minimize the effect of disturbances

K -

+

R(s) E(s) C(s) G(s)

2


Proportional Controller

•  The closed look transfer function is given by

•  Assuming we have two poles, G(s)=1/(s+a)(s+b)

( )( )1 ( )KG sT sKG s

=+

2( )( )

KT ss a b s ab K

=+ + + +


Proportional Controller

•  We can also look at the system parameters as a function of the gain, K

•  Given fixed values for the roots of the plant, we can find K to meet performance specifications

2

2 2

( )( )

2n

n n

KT ss a b s ab K

Cs

!"! !#

=+ + + +

=+ +

2

2

n

ss

ab Ka bab K

a b

abeab K

!

"

#

= ++=+

+=

=+


Root Location

•  The location of the roots, and hence the nature of the system performance, are a function of the system gain K

•  In order to solve for this system performance, we must factor the denominator for specific values of K

•  We define the root locus as the path of the closed-loop poles as the system parameter varies from 0 to ∞


Example: Second order system •  A system to

automatically track a subject in a visual image can be modelled as follows

•  We can solve for the closed loop transfer function as a function of the system parameter, K

3


Example: Second order system

•  We can also determine the closed loop poles as a function of the gain for the system



The individual pole locations The root locus


Properties of the Root Locus

•  We can easily derive the root locus for a second order system

•  What about for a general, possibly higher order, control system?

•  Poles exist when the characteristic equation (denominator) is zero

( )( )1 ( ) ( )

KG sT sKG s H s

=+

1 ( ) ( ) 0KG s H s+ =



•  How do we find values of s and K that satisfy the characteristic equation?

•  This holds when

1 ( ) ( ) 0KG s H s+ =

( ) ( ) 1

( ) ( ) (2 1)180

KG s H s

KG s H s k

=

! = + !

(2 1)180

1( ) ( )

zero angles pole angles k

pole lengthK

G s H s zero length

! = +

= =

" "##

!

4



•  The preceding angle and magnitude criteria can be used to verify which points in the s-plane form part of the root locus

•  It is not practical to evaluate all points in the s-plane to find the root locus

•  We can formulate a number of rules that allow us to sketch the root locus


Basic Root Locus Rules •  Rule 1 : Number of Branches – the n branches of the

root locus start at the poles

•  For K=0, this suggests that the denominator must be zero (equivalent to the poles of the OL TF)

•  The number of branches in the root locus therefore equals the number of open loop poles

1 ( ) ( ) 0( ) ( ) 0KG s H s

Den s KNum s+ =

+ =


Basic Root Locus Rules •  Rule 2 : Symmetry - The root locus is symmetrical

about the real axis. This is a result of the fact that complex poles will always occur in conjugate pairs.


Basic Root Locus Rules •  Rule 3 – Real Axis Segments – According to the angle

criteria, points on the root locus will yield an angle of (2k+1)180o.

•  On the real axis, angles from complex poles and zeros are cancelled.

•  Poles and zeros to the left have an angle of 0o. •  This implies that roots will lie to the left of an odd number

of real-axis, finite open-loop poles and/or finite open-loop zeros.

5


Basic Root Locus Rules

•  Rule 4 – Starting and Ending Points – As we saw, the root locus will start at the open loop poles

•  The root locus will approach the open loop zeros as K approaches ∞

•  Since there are likely to be less zeros than poles, some branches may approach ∞

1 ( ) ( ) 0( ) ( ) 0KG s H s

Den s KNum s+ =

+ =


Example •  Consider the system

at right •  The closed loop

transfer function for this system is given by

•  Difficult to evaluate the root location as a function of K

2

( 3)( 4)( )(1 ) (3 7 ) (2 12 )

K s sT sK s K s K

+ +=+ + + + +


Example •  Open loop poles and

zeros •  First plot the OL poles

and zeros in the s-plane

•  This provides us with the likely starting (poles) and ending (zeros) points for the root locus


Example

•  Real axis segments •  Along the real axis,

the root locus is to the left of an odd number of poles and zeros

6


Example

•  Starting and end points •  The root locus will start

from the OL poles and approach the OL zeros as K approaches infinity

•  Even with a rough sketch, we can determine what the root locus will look like


Basic Root Locus Rules •  Rule 5 – Behaviour at infinity – For large s and K, n-m

of the loci are asymptotic to straight lines in the s-plane •  The equations of the asymptotes are given by the real-

axis intercept, sa, and angle, qa

•  Where k = 0, ±1, ±2, … and the angle is given in radians relative to the positive real axis

(2 1)

a

a

finite poles finite zeroesn m

kn m

!

"#

$=

$+=$

% %


Basic Root Locus Rules •  Why does this hold? •  We can write the characteristic equation as

•  This can be approximated by

•  For large s, this is the equation for a system with n-m poles clustered at s=σ

11

11

1 0m m

mn n

n

s b s bKs b s b

!

!

+ + ++ =+ + +

!!

11 0( )n mKs ! "+ ="


Example

•  Here we have four OL poles and one OL zero

•  We would therefore expect n-m = 3 distinct asymptotes in the root locus plot

7


Example •  We can calculate the

equations of the asymptotes, yielding

( 1 2 4) ( 3) 43 3

(2 1)

/ 3 ( 0)( 1)

5 / 3 ( 2)

a

a

finite poles finite zeroesn m

kn mk

kk

!

"#

"""

$=

$$ $ $ $ $= = $

+=$

= == == =

% %


Angles of Departure and Arrival

•  For poles on the real axis, the locus will depart at 0o or 180o

•  For complex poles, the angle of departure can be calculated by considering the angle criteria


Angles of Departure and Arrival

•  A similar approach can be used to calculate the angle of arrival of the zeros


Imaginary Axis Crossing

•  We may also be interested in the gain at which the locus crosses the imaginary axis

•  This will determine the gain with which the system becomes unstable

8


Using Available Resources

•  All of this probably seems somewhat complicated

•  Fortunately, Matlab provides us with tools for plotting the root locus

•  It is still important to be able to sketch the root locus by hand because – This gives us an understanding to be applied

to designing controllers –  It will probably appear on the exam


Root Locus as a Design Tool

•  As we saw previously, the specifications for a second order system are often used in designing a system

•  The resulting system performance must be evaluated in light of the true system performance

•  The root locus provides us with a tool with which we can design for a transient response of interest



•  We would usually follow these steps – Sketch the root locus – Assume the system is second order and find

the gain to meet the transient response specifications

– Justify the second-order assumptions by finding the location of all higher-order poles

–  If the assumptions are not justified, system response should be simulated to ensure that it meets the specifications


Root Locus as a Design Tool •  Recall that for a second order system with

no finite zeros, the transient response parameters are approximated by

– Rise time :

– Overshoot :

– Settling Time (2%) :

5%, 0.716%, 0.520%, 0.45

pM!!!

="#$ =%# =&

1.8r

n

t!

"

4st !"

9


Example: Second order system •  Recall the system

presented earlier •  Determine a value of

the gain K to yield a 5% percent overshoot

•  For a second order system, we could find K explicitly



•  Examining the transfer function

•  Solve for K given the desired damping ratio specified by the desired overshoot

2

2 2

( )10

2n

n n

KT ss s K

Cs s

!"! !#

=+ +

=+ +

2

2

2 105% , 0.7

50.751

n

n

K

for overshoot

therefore K

!"!"

==#

$ %= & '( )=



•  Alternatively, we can examine the Root Locus 1 0

1

1( 10)

KGHKGH

Ks s

+ ==

=+

x x

Im(s)

Re(s)

10 5 0

x

x

S=5+5.1j

1( 5 5.1 )( 5 5.1 10)

51.01

Kj j

K

=! + ! + +=

θ=sin-1ζ



•  We can use Matlab to generate the root locus

!!!% define the OL system!

sys=tf(1,[1 10 0])!% plot the root locus!rlocus(sys)!

Root Locus

Real Axis

Imag Axis

-10 -8 -6 -4 -2 0-5

-4

-3

-2

-1

0

1

2

3

4

5 System: sys2 Gain: 52.5

Pole: -5 + 5.24i Damping: 0.69

Overshoot (%): 5 Frequency (rad/sec): 7.24

10



•  We also need to verify the resulting step response !% set up the closed loop TF!cl=51*sys/(1+51*sys)!% plot the step response!step(cl)!

Step Response

Time (sec)

Amplitude

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20

0.2

0.4

0.6

0.8

1

1.2

1.4

System: cl Time (sec): 0.6 Amplitude: 1.05



•  Consider this system •  This is a third order

system with an additional pole

•  Determine a value of the gain K to yield a 5% percent overshoot



•  With the higher order poles, the 2nd order assumptions are violated

•  However, we can use the RL to guide our design and iterate to find a suitable solution

Root Locus

Real Axis

Imag Axis

-10 -8 -6 -4 -2 0

-10

-8

-6

-4

-2

0

2

4

6

8

10

System: sys Gain: 51.2 Pole: -4.65 + 4.88i Damping: 0.69 Overshoot (%): 5 Frequency (rad/sec): 6.74



•  The gain found based on the 2nd order assumption yields a higher overshoot

•  We could then reduce the gain to reduce the overshoot

Step Response

Time (sec)

Amplitude

0 1 2 3 4 5 6 7 8 9 100

0.2

0.4

0.6

0.8

1

1.2

1.4

System: untitled1 Time (sec): 0.604 Amplitude: 1.12

11


Generalized Root Locus

•  The preceding developments have been presented for a system in which the design parameter is the forward path gain

•  In some instances, we may need to design systems using other system parameters

•  In general, we can convert to a form in which the parameter of interest is in the required form



•  Consider a system of this form

•  The open loop transfer function is no longer of the familiar form KG(s)H(s)

•  Rearrange to isolate p1

•  Now we can sketch the root locus as a function of p1

21 1

10( )( 2) 2 10

T ss p s p

=+ + + +

21

2

12

10( )2 10 ( 2)102 10( 2)12 10

T ss s p s

s sp ss s

=+ + + +

+ += +++ +



•  This results in the following root locus as a function of the parameter p1


Conclusions

•  We have looked at a graphical approach to representing the root positions as a function of variations in system parameters

•  We have presented rules for sketching the root locus given the open loop transfer function

•  We have begun looking at methods for using the root locus as a design tool

12


Further Reading

•  Nise – Sections 8.1-8.6

•  Franklin & Powell – Section 5.1-5.3

Download - Course Outline - AMME Current Student Information Pagesweb.aeromech.usyd.edu.au/AMME3500/Course_documents/material... · 3 Dr. Stefan B. Williams AMME 3500 : Root Locus Slide 9 Example:

Top Related