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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
Slide 4 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 5 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
=+ + + +
Slide 6 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
!
"
#
= ++=+
+=
=+
Slide 7 Dr. Stefan B. Williams AMME 3500 : Root Locus
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 ∞
Slide 8 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 9 Dr. Stefan B. Williams AMME 3500 : Root Locus
Example: Second order system
• We can also determine the closed loop poles as a function of the gain for the system
Slide 10 Dr. Stefan B. Williams AMME 3500 : Root Locus
Example: Second order system
The individual pole locations The root locus
Slide 11 Dr. Stefan B. Williams AMME 3500 : 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+ =
Slide 12 Dr. Stefan B. Williams AMME 3500 : Root Locus
Properties of the Root Locus
• 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
Slide 13 Dr. Stefan B. Williams AMME 3500 : Root Locus
Properties of the Root Locus
• 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
Slide 14 Dr. Stefan B. Williams AMME 3500 : 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+ =
+ =
Slide 15 Dr. Stefan B. Williams AMME 3500 : Root Locus
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.
Slide 16 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 17 Dr. Stefan B. Williams AMME 3500 : Root Locus
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+ =
+ =
Slide 18 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
+ +=+ + + + +
Slide 19 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 20 Dr. Stefan B. Williams AMME 3500 : 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
Slide 21 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 22 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
!
"#
$=
$+=$
% %
Slide 23 Dr. Stefan B. Williams AMME 3500 : Root Locus
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 ! "+ ="
Slide 24 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 25 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
!
"#
"""
$=
$$ $ $ $ $= = $
+=$
= == == =
% %
Slide 26 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 27 Dr. Stefan B. Williams AMME 3500 : Root Locus
Angles of Departure and Arrival
• A similar approach can be used to calculate the angle of arrival of the zeros
Slide 28 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 29 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 30 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 31 Dr. Stefan B. Williams AMME 3500 : Root Locus
Root Locus as a Design Tool
• 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
Slide 32 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 33 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 34 Dr. Stefan B. Williams AMME 3500 : Root Locus
Example: Second order system
• 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
!"!"
==#
$ %= & '( )=
Slide 35 Dr. Stefan B. Williams AMME 3500 : Root Locus
Example: Second order system
• 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
Slide 36 Dr. Stefan B. Williams AMME 3500 : Root Locus
Example: Second order system
• 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
Slide 37 Dr. Stefan B. Williams AMME 3500 : Root Locus
Example: Second order system
• 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
Slide 38 Dr. Stefan B. Williams AMME 3500 : Root Locus
Root Locus as a Design Tool
• 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
Slide 39 Dr. Stefan B. Williams AMME 3500 : Root Locus
Root Locus as a Design Tool
• 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
Slide 40 Dr. Stefan B. Williams AMME 3500 : Root Locus
Root Locus as a Design Tool
• 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
Slide 41 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 42 Dr. Stefan B. Williams AMME 3500 : Root Locus
Generalized Root Locus
• 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
=+ + + +
+ += +++ +
Slide 43 Dr. Stefan B. Williams AMME 3500 : Root Locus
Generalized Root Locus
• This results in the following root locus as a function of the parameter p1
Slide 44 Dr. Stefan B. Williams AMME 3500 : Root Locus
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
Slide 45 Dr. Stefan B. Williams AMME 3500 : Root Locus
Further Reading
• Nise – Sections 8.1-8.6
• Franklin & Powell – Section 5.1-5.3