Lec 3 desgin via root locus

Control Systems
LECT. 3 DESIGN VIA ROOT LOCUS
BEHZAD FARZANEGAN
3/1/2020 PROVIDED BY: BF(B.FARZANEGAN@AUT.AC.IR)

Given the control plant, the procedure of controller design to satisfy the
requirement is called system compensation.
What is system compensation?
 The closed-loop system has the function of self-tunning. By selecting a
particular value of the gain K, some single performance requirement
may be met.
Is it possible to meet more than one performance requirement?
 Sometimes, it is not possible.K
e
Process+
-
Setpoint u CV

Improving transient response
Given Sample root locus, showing possible design point via gain
adjustment (A) and desired design point that cannot be met via simple
gain adjustment (B).

Improving steady-state error
Compensation techniques
o Cascade
o feedback

Improving steady-state error via
cascade compensation
Pole at A is:
o on the root locus without compensator
o not on the root locus with compensator pole added

PI controller
 A method to implement an Ideal integral compensator is shown.
𝐺𝑐(𝑠) = 𝐾1 +
𝐾2
𝑠
=
𝐾1 𝑠 +
𝐾2
𝐾1
𝑠

Ideal Integral compensation (PI)
approximately on the root locus with compensator pole and zero
added

Example
Problem: The given system
operating with damping ratio
of 0.174. Add an ideal integral
compensator to reduce the ss
error.
𝐾 = 164.6 𝐾 𝑝 = 8.23→
𝜉 = 0.174
)𝜃 = cos−1(𝜉
→ 𝜃 = 78.98∘
⇒ 𝑒(∞) =
1
1 + 𝐾 𝑝
= 0.108

Example (cont.)
Solution: We compensate the
system by choosing a pole at
the origin and a zero at -0.1.
 Almost same transient response and gain,
but with zero ss error since we have a type
one system.

Lag Compensator
 Using passive networks, the compensation pole and zero is moved to
the left, close to the origin.
 The static error constant for uncompensated system is
 Assuming the compensator is used, then the static error is
𝐾𝑣𝑜 =
𝐾𝑧1 𝑧2. . . .
𝑝1 𝑝2. . . . . . .
𝐾𝑣𝑁 =
𝐾𝑧1 𝑧2. . . . )(𝑧 𝑐
𝑝1 𝑝2. . . . . . . )(𝑝𝑐
𝑮 𝒄(𝒔) =
𝒔 + 𝒛 𝒄
𝒔 + 𝒑 𝒄

Effect on transient response
Almost no change on the transient response and same gain K. While
the ss error is effected since
𝑮 𝒄(𝒔) =
𝒔 + 𝒛 𝒄
𝒔 + 𝒑 𝒄
𝑲 𝒗𝑵 = 𝑲 𝒗𝒐
𝒛 𝒄
𝒑 𝒄
> 𝑲 𝒗𝒐

Example
Problem: Compensate the shown system to improve the ss error by a
factor of 10 if the system is operating with a damping ratio of 0.174.
Solution: the uncompensated system error from previous example is
0.108 with Kp= 8.23.
A ten fold improvement means
→ 𝑲 𝒑 = 𝟗𝟏. 𝟓𝟗𝑒𝑠𝑠 = 0.0108
𝒛 𝒄
𝒑 𝒄
=
𝑲 𝑷𝑵
𝑲 𝑷𝒐
=
𝟗𝟏. 𝟓𝟗
𝟖. 𝟐𝟑
= 𝟏𝟏. 𝟏𝟑⇒ ⇒
𝑃𝑐 = 0.01
𝑍 𝑐 = 11.13𝑃𝑐 ≃ 0.111
𝑲 𝒔 + 𝒛 𝒄
𝒔 + 𝒑 𝒄

Predicted characteristics of uncompensated
and lag-compensated systems

PD controller implementation
 K2 is chosen to contribute to the required loop-gain value. And K1/K2 is
chosen to equal the negative of the compensator zero.
1
2 1 2
2
( ) ( )c
K
G s K s K K s
K
   

Improving Transient response
via Cascade Compensation
Ideal Derivative compensator is called PD controller.
 When using passive network it’s called lead compensator Using ideal
derivative compensation
uncompensated compensator zero at –2
𝑮 𝒄(𝒔) = 𝒔 + 𝒛 𝒄

Improving Transient response
via Cascade Compensation
1. uncompensated
2. compensator zero at –2
𝑮 𝒄(𝒔) = 𝒔 + 𝒛 𝒄

Predicted characteristics for
the systems of previous slides

Ex: Feedback control system
Problem: Given the system in the figure, design an ideal derivative
compensator to yield a 16% overshoot with a threefold reduction in
settling time.
The settling time for the uncompensated system
𝑻 𝒔 =
𝟒
𝝃𝝎 𝒏
𝑻 𝒔𝒏𝒆𝒘 =
𝑻 𝒔
𝟑
= 𝟏. 𝟏𝟎𝟕
→ 𝝈 =
4
𝑇𝑠𝑛𝑒𝑤
𝝎 𝒅 = 3.613tan(180∘ − 120.26∘) = 6.193→
→=
𝟒
𝟏. 𝟐𝟎𝟓
= 𝟑. 𝟑𝟐𝟎
=
4
1.107
= 3.613

Evaluating the location of the
compensating zero
The sum of angles from all poles to the desired compensated pole -
3.613+j6.193 is −275.6
The angle of the zero to be on the root locus is 275.6-180=95.6
The location of the compensator zero is calculated as
6.193
tan(180 95.6 )
Thus
3.613
3.006


 

o o

Uncompensated and compensated
system characteristics for Example

Geometry of lead compensation
 Advantages of a passive lead network over an active PD controller:
 no need for additional power supply
 noise due to differentiation is reduced
 Three of the infinite possible lead compensator solutions
2 1 3 4 5 (2 1)180k          o
2 1Note ( ) c   

Lead compensator design
Problem: Design 3 lead compensators for the system in figure that will
reduce the settling time by a factor of 2 while maintaining 30%
overshoot.
Solution: The uncompensated settling time is
new settling time is
4 4
3.972
1.007
s
n
T

  
3.972
1.986
2
sT  
𝝎 𝒅 = 𝟐. 𝟎𝟏𝟒𝐭𝐚𝐧(𝟏𝟏𝟎. 𝟗𝟖∘) = 𝟓. 𝟐𝟓𝟐→
→ 𝝈 =
𝟒
𝑻 𝒔
=
𝟒
𝟏. 𝟗𝟖𝟔
= 𝟐. 𝟎𝟏𝟒

Lead compensator design(cont.)
S-plane picture used to calculate the location of the compensator pole
Arbitrarily assume a compensator zero at -5 on the real axis as
possible solution. Then we find the compensator pole location as shown
in figure.
Note sum of angles of compensator zero and all uncompensated poles
and zeros is -172,69 so the angular contribution of the compensator
pole is -7.31.
𝟓. 𝟐𝟓𝟐
𝒑 𝒄 − 𝟐. 𝟎𝟏𝟒
= 𝐭𝐚𝐧𝟕. 𝟑𝟏∘ and 𝒑 𝒄 = 𝟒𝟐. 𝟗𝟔

Comparison of lead compensation designs

Improving Steady-State Error
and Transient Response
PID controller or using passive network it’s called lag-lad compensator
𝑮 𝒄(𝒔) = 𝑲 𝟏 +
𝑲 𝟐
𝒔
+ 𝑲 𝟑 𝒔 =
𝑲 𝟏 + 𝑲 𝟐 + 𝑲 𝟑 𝒔 𝟐
𝒔
=
𝑲 𝟑(𝒔 𝟐
+
𝑲 𝟏
𝑲 𝟑
𝒔 +
𝑲 𝟐
𝑲 𝟑
)
𝒔

PID controller design
Evaluate the performance of the uncompensated system to determine
how much improvement is required in transient response
Design the PD controller to meet the transient response specifications. The
design includes the zero location and the loop gain.
Simulate the system to be sure all requirements have been met.
Redesign if the simulation shows that requirements have not been met.
Design the PI controller to yield the required steady-sate error.
Determine the gains, K1, K2, and K3 shown in previous figure.
Simulate the system to be sure all requirements have been met.
Redesign if simulation shows that requirements have not been met.

Example:
Problem: Using the system in the Figure, Design a PID controller so that
the system can operate with a peak time that is 2/3 that of the
uncompensated system at 20% overshoot and with zero steady-state
error for a step input.
Solution: The uncompensated system operating at 20% overshoot has
dominant poles at -5.415+j10.57 with gain 121.5, and a third pole at -
8.169. The complete performance is shown in next table.

Example:
To compensate the system to reduce the peak time to 2/3 of original,
we must find the compensated system dominant pole location.
𝝎 𝒅 =
𝝅
𝑻 𝒑
=
𝝅
𝟐 𝟑)(𝟎. 𝟐𝟗𝟕
= 𝟏𝟓. 𝟖𝟕
⇒ 𝝈 =
𝝎 𝒅
𝐭𝐚𝐧𝟏𝟏𝟕. 𝟏𝟑∘
= −𝟖. 𝟏𝟑

Calculating the PD compensator
To design the compensator, we find the sum of angles from the
uncompensated system’s poles and zeros to the desired compensated
dominant pole to be -198.37. Thus the contribution required from the
compensator zero is 198.37-180=18.37. Then we calculate the location
of the zero as:
𝟏𝟓. 𝟖𝟕
𝒛 𝒄 − 𝟖. 𝟏𝟑
= 𝐭𝐚𝐧𝟏𝟖. 𝟑𝟕∘
and 𝒛 𝒄 = 𝟓𝟓. 𝟗𝟐
⇒ 𝑮 𝑷𝑫 𝒔 = 𝒔 + 𝟓𝟓. 𝟗𝟐
⇒ 𝑲 = 𝟓. 𝟑𝟒

and PD- compensated systems

Calculating the PI compensator
Choosing the ideal integral compensator to be
Finally to implement the compensator and find the K’s, using the PD
and PI compensators
we find K1= 259.5, K2=128.6, and K3=4.6
𝑮 𝑷𝑰(𝒔) =
𝒔 + 𝟎. 𝟓
𝒔
⇒ 𝑮 𝑷𝑰𝑫(𝒔) =
)𝑲(𝒔 + 𝟓𝟓. 𝟗𝟐)(𝒔 + 𝟎. 𝟓
𝒔
=
𝟒. 𝟔(𝒔 𝟐 + 𝟓𝟔. 𝟒𝟐𝒔 + 𝟐𝟕. 𝟗𝟔
𝒔

Predicted characteristics of uncompensated,
PD and PID- compensated systems

Lag-Lead Compensator Design
Problem: Using the system in the Figure, Design a lag-lead compensator
so that the system can operate with a twofold reduction in settling time,
and 20% overshoot and a tenfold improvement in steady-state error for
a ramp input
Solution: The uncompensated system operating at 20% overshoot has
dominant poles at -1.794+j3.501 with gain 192.1, and a third pole at -
12.41. The complete performance is shown in next table.

Example
To compensate the system to realize a twofold reduction in settling
time, the real part of the dominant poles must be increased by a factor
of 2, thus,
And the imaginary part is
−𝝃𝝎 𝒏 = −𝟐(𝟏. 𝟕𝟗𝟒) = −𝟑. 𝟓𝟖𝟖
𝝎 𝒅 = 𝝃𝝎 𝒏 𝒕𝒂𝒏𝟏𝟏𝟕. 𝟏𝟑∘
= 𝟑. 𝟓𝟖𝟖𝒕𝒂𝒏𝟏𝟏𝟕. 𝟏𝟑∘
= 𝟕. 𝟎𝟎𝟑

Evaluating the compensator
pole for the Example
Now to design the lead compensator, arbitrarily select a location for
the lead compensator zero at -6, to cancel the pole.
To find the location of the compensator pole. Using program sum the
angles to get -164.65. and the contribution of the pole is -15.35 we find
the location of the pole from the figure as
𝟕. 𝟎𝟎𝟑
𝒑 𝒄 − 𝟑. 𝟓𝟖𝟖
= 𝐭𝐚𝐧𝟏𝟓. 𝟑𝟓∘
and p 𝒄 = −𝟐𝟗. 𝟏

Predicted characteristics of uncompensated, lead-compensated
compensated systems of the Example

Example (cont.)
Since the uncompensated system’s open-loop transfer function is
The static error constant of the uncompensated system is 3.201
Since the open-loop transfer function of the lead-compensated system is
the static error constant of the lead-compensated system is 6.794, so we
have improvement by a factor of 2.122.
To improve the original system error by a factor of 10, the lag compensator
must be designed to improve the error by a factor of 10/2.122 = 4.713
𝑮(𝒔) =
𝟏𝟗𝟐. 𝟏
)𝒔(𝒔 + 𝟔)(𝒔 + 𝟏𝟎
𝑮 𝑳𝑪(𝒔) =
𝟏𝟗𝟕𝟕
)𝒔(𝒔 + 𝟏𝟎)(𝒔 + 𝟐𝟗. 𝟏

Example (cont.)
We arbitrarily choose the lag compensator
pole at 0.01, which then places the zero at
0.04713 yielding
as a lag compensator
and
as lag-lead-compensated system open-loop
transfer function
𝑮𝒍𝒂𝒈(𝒔) =
𝒔 + 𝟎. 𝟎𝟒𝟕𝟏𝟑
𝒔 + 𝟎. 𝟎𝟏
𝑮 𝑳𝑳𝑪(𝒔) =
)𝑲(𝒔 + 𝟎. 𝟎𝟒𝟕𝟏𝟑
)𝒔(𝒔 + 𝟏𝟎)(𝒔 + 𝟐𝟗. 𝟏)(𝒔 + 𝟎. 𝟎𝟏

Predicted characteristics of uncompensated, lead-compensated, and
lag-lead- compensated systems of Example

Types of cascade compensators

Active-Circuit Realization
Operational amplifier configured for transfer function realization
𝑽 𝒐 𝒔
𝑽𝒊 𝒔
= −
𝒁 𝟐 𝒔
𝒁 𝟏 𝒔

Physical System Realization
PI Compensator
Lag Compensator
PD Compensator
Lead Compensator
PID Compensator
22
1
1
( )
s
R CR
C s
R s
 
 
  
2 2
1 2
1 2
1
( )
1
( )
c
s
R R C
G s
R R s
R R C


 

R1
R2
C
Vi(s) Vo(s)
2
1
1
( )cG s R C s
R C
 
   
 
1
1 2
1
( )
1 1
,
c
c
c c
c
s
R C
G s
s
R C R C
s z
z p
s p


 

 

R1
R2
C2
Vi(s) Vo(s)
C1
2 1 1 2
2 1
1 2
1
( )c
R C R C
G s R C s
R C s
 
  
      
  
  

Lag-lead compensator implemented
with operational amplifiers

Example
PROBLEM: Implement the controller of PID Example.
SOLUTION: The transfer function of the PID controller is
Comparing the PID controller in previous Table with above equation we
obtain the following three relationships:
we arbitrarily select a practical value for one of the elements.
𝑮 𝒄 𝒔 =
𝒔 + 𝟓𝟓. 𝟗𝟐 𝒔 + 𝟎. 𝟓
𝒔
𝑅2
𝑅1
+
𝐶1
𝐶2
= 56.42
𝑅2 𝐶1 = 1
1
𝑅1 𝐶2
= 27.96

Example
PROBLEM: Implement the controller of Lead Example.
SOLUTION: The transfer function of the Lead controller is
Comparing the lead controller in previous Table with above equation we
obtain the following three relationships:
we arbitrarily select a practical value for one of the elements.
𝑮 𝒄 𝒔 =
s+4
𝒔 + 𝟐𝟎. 𝟎𝟗
1
𝑅1C
+
1
𝑅2 𝐶
= 20.09
1
𝑅1 𝐶
= 4

Lag-Lead Realization
The lag-lead transfer function can be put in the following form:
Thus, the terms with T1 form the lead compensator, and the terms with
T2 form the lag compensator. We see that the ratio of the lead
compensator zero to the lead compensator pole must be the same as
the ratio of the lag compensator pole to the lag compensator zero.
α < 1

Cruise Control: System Modeling
Physical setup
Automatic cruise control is an excellent example of a feedback control
system found in many modern vehicles. The purpose of the cruise
control system is to maintain a constant vehicle speed despite external
disturbances, such as changes in wind or road grade. This is
accomplished by measuring the vehicle speed, comparing it to the
desired or reference speed, and automatically adjusting the throttle
according to a control law.

Cruise Control: System Modeling
System equations
With these assumptions we are left with a first-order mass-damper
system. Summing forces in the x-direction and applying Newton's 2nd
law, we arrive at the following system equation:
Since we are interested in controlling the speed of the vehicle, the
output equation is chosen as follows
For this example, let's assume that the parameters of the system are:
(m) vehicle mass 1000 kg
(b) damping coefficient 50 N.s/m
(u) nominal control force 500 N
𝒎 𝒗 + 𝒃𝒗 = 𝒖
𝒚 = 𝒗

Cruise Control: System Analysis
Performance specifications
The next step is to come up with some design criteria that the
compensated system should achieve. When the engine gives a 500
Newton force, the car will reach a maximum velocity of 10 m/s (22 mph).
An automobile should be able to accelerate up to that speed in less than
5 seconds. In this application, a 10% overshoot and 2% steady-state error
on the velocity are sufficient.
HW? Design a Cruise Controller to achieve the performance specifications.

Notch Filter
Root locus before cascading notch filter
typical closed-loop step response before cascading notch filter
root locus after cascading notch filter

Generic control system with
feedback compensation

Approach 1
The first is similar to cascade compensation. Assume a typical feedback
system, where G(s) is the forward path and H(s) is the feedback. Now
consider that a root locus is plotted from G(s)H(s).
With cascade compensation we added poles and zeros to G(s). With
feedback compensation, poles and zeros are added via H(s).

Transfer function of a tachometer
rate feedback
𝑮 𝒔 𝑯 𝒔 = 𝑲 𝒇 𝑲 𝟏 𝑮 𝟏 𝒔 (𝒔 + 𝑲 𝑲 𝒇

Example: Compensating Zero
via Rate Feedback
PROBLEM: Given the system of Figure (a), design rate feedback
compensation, as shown in Figure (b), to reduce the settling time by a
factor of 4 while continuing to operate the system with 20% overshoot

Example: Compensating Zero
via Rate Feedback
SOLUTION: First design a PD compensator. For the uncompensated
system, search along the 20% overshoot line (ξ = 0.456) and find that
the dominant poles are at -1.809 ±j3.531, as shown in Figure.
The settling time is 2.21 seconds and must be
reduced by a factor of 4 to 0.55 second.
𝑻 𝒔 = 𝟐. 𝟐𝟏 𝑻 𝒔𝑵 =
𝑻 𝒔
𝟒
= 𝟎. 𝟓𝟓
𝝈 = 𝟒 −𝟏. 𝟖𝟎𝟗 = − 𝟕. 𝟐𝟑𝟔
𝝎 𝒅 = − 𝟕. 𝟐𝟑𝟔 𝒕𝒂𝒏 𝟏𝟏𝟕. 𝟏𝟑° = 𝟏𝟒. 𝟏𝟐

Finding thecompensator zero in Example
The geometry shown in the Figure leads to the calculation of the
compensator's zero location.
𝟏𝟒. 𝟏𝟐 (𝟕. 𝟐𝟑𝟔 – 𝒛 𝒄) = 𝒕𝒂𝒏 𝟏𝟖𝟎° − 𝟗𝟕. 𝟑𝟑°
𝒛 𝒄 = 𝟓. 𝟒𝟐
𝐾1 𝐾𝑓 = 256.7
⇒ 𝐾𝑓 = 0.185
⇒ 𝐾1 = 1388
𝐾𝑣 = 𝐾1 ( 75 + 𝐾1 𝐾𝑓) = 4.18

and compensated systems of Example

Approach 2
The second approach allows us to use feedback compensation to design
a minor loop’s transient response separately from the closed-loop
system response.

Example
PROBLEM: For the system of Figure (a), design minor-loop feedback
compensation, as shown in Figure (b), to yield a damping ratio of 0.8 for
the minor loop and a damping ratio of 0.6 for the closed-loop system.

Example(cont.)
The transfer function of the minor loop
The open loop transfer function
Is there pole-zero cancellation at the origin?

Predicted characteristics of the uncompensated and
compensated systems of Example

Lec 3 desgin via root locus

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