Lecture 2_3 - College of Engineering, Purdue University

Introduction to Feedback Control
• Control System Design
– Why Control?
– Open-Loop vs Closed-Loop (Feedback)
– Why Use Feedback Control?
• Closed-Loop Control System Structure
– Elements of a Feedback Control System
– Closed-Loop Transfer Functions (CLTF)
• Performance Specifications
– Steady State Specifications
– Transient (Dynamic) Specifications
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ME375 Feedback Control - 1
Control System Design
Control is the process of causing a system to behave in a prescribed manner.
Specifically, control system design is the process of causing a system variable
(output) to conform to some desired input (reference).
Reference
Input
Input
R (s)
U (s)
System
(Plant)
GP(s)
Output
Y(s)
The objective of the control system is to control the output y by using the input u,
such that the output y follows a set of reference inputs r.
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Open-Loop vs Closed-Loop
• Open-Loop Control
The control input u(t) (or U(s)) is synthesized based on the a priori knowledge
of the system (plant) and the reference input r(t) (or R(s)). The control system
does not measure the output, and there is no comparison of the output to make it
conform to the desired output (reference input).
Reference Input
(Command)
R(s)
System Output
C(s)
U(s)
Control Input
GP(s)
Y(s)
Plant or System
Q: Ideally, if we want Y(s) to follow R(s) (i.e. want Y(s) = R(s)), how would you design the controller
C(s) ?
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Open-Loop Control Example
Open loop Cruise Control
W
The vehicle speed model can be
approximated by a static gain between
the throttle angle (input) and the vehicle
speed (output). From experiment, on
level road, at 55 mph, 1o of throttle angle
causes 10 mph change in speed. When
the road grade changes by 1%, 1o of
throttle angle will only change vehicle
speed by 5 mph. Design an open-loop
cruise controller for this vehicle.
R
U
Y
W
: reference speed, mph
: throttle angle, degree
: actual speed, mph
: road grade, %
R
U
−
Y
+
Cruise
Controller
Speed
Model
Q: What are potential problems with this cruise control ?
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Open-Loop vs Closed-Loop
• Closed-Loop (Feedback) Control
The control input u(t) (or U(s)) is synthesized based on the a priori knowledge
of the system (plant), the reference input r(t) (or R(s)) and the measurement of
the actual output y(t) (or Y(s)). For example the temperature control of this
classroom:
Disturbance D(s)
Heater
Actuator
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Room
Temperature
Room
Y(s)
Plant or System
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Closed-Loop Control Example
Closed Loop Cruise Control
Same vehicle system as the previous example. The vehicle speed is measured and fed back.
Design a closed-loop cruise control that uses the measured vehicle speed and the reference
speed.
W
R
−
U
Y
+
Speed Model
Q: How would road grade, plant uncertainty affect the closed-loop performance ?
Q: How is the steady state performance ? Will you have any steady state error ?
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Closed-Loop Control Example
Cruise Control (Closed-Loop Control)
(a) Find the actual vehicle speed when the reference speed is 50 mph and the road grade is 1%
and 10%, respectively.
(b) If the actual vehicle speed model is 1o of throttle angle corresponds to 9 mph change in
speed, what is the actual vehicle speed with the same cruise controller.
(c) When there is no grade and the vehicle speed model is accurate, what is the actual output
speed when a reference speed of 50 mph is desired.
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Why Feedback ?
Using feedback, we can change the closed-loop system’s dynamic behavior (the
Closed-Loop Transfer Function (CLTF) will be different from the original
system’s (open-loop) transfer function). By using feedback to change the CLTF,
we can achieve the following:
– Stabilize Unstable Systems
For example, unstable plants such as inverted pendulum can be stabilized
using feedback.
– Improve System Performance (Achieve Performance Specifications)
• Steady State Performance -- For example, reduce steady state error ...
• Transient Performance -- For example, reduce rise time, reduce settling
time, reduce overshoot …
– Reduce (attenuate) the effect of modeling uncertainty (error) and
external disturbances
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Example
More Realistic Cruise Control Problem
The relationship between a vehicle’s speed y and the throttle angle u is described by
a first order system with a steady state gain KC and a time constant of 3 sec. The
gain KC is affected by various operating conditions like the temperature and
humidity. Due to these effects, the actual value of KC is between 5 and 15. The
objective of the cruise control is to design a control law (strategy) to determine the
throttle angle u such that the vehicle’s steady state speed will stay within 2% of the
desired reference speed set by the driver.
Cruise
Controller
R
Speed Model
U
Y
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Use a simple “proportional” feedback
control, i.e. the control input u(t) is
proportional to the regulation error
e(t) = r(t) − y(t). The control design
parameter is the proportional constant
between the input and the error. This
constant KP is usually called the
feedback gain or the proportional
gain.
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Example
Calculate Closed-Loop Transfer Function (CLTF):
Select an appropriate feedback gain KP to satisfy the performance specification :
Q: Will this proportional control law work for attenuating external disturbances ?
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Elements of Feedback Control
Elements of a Feedback Control System:
• Plant (Process) GP(s) -- The plant is the system (process) whose output is to be
controlled, e.g., the room in the room temperature control example.
• Actuator -- An actuator is a device that can influence the input to the plant, e.g. the heater
(furnace) in the room temperature control example.
• Disturbance d(t) -- Disturbances are uncontrollable signals to the plant that tend to
adversely affect the output of the system, e.g., opening the windows in the room
temperature control example.
• Sensor (Measurement System) H(s) -- The transfer function (frequency response
function) of the device (system) that measures the system output, e.g., a thermocouple.
• Controller GC (s) -- The controller is the device that generates the controlled input that is
to affect the system output, e.g., the thermostat in the room temperature control example.
Controller
Reference
Input
Disturbance D(s)
Output
GC (s)
R(s)
Heater
Actuator
Sensor
GP(s)
Y(s)
Plant
(Process)
H(s)
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Closed-Loop Transfer Function
Disturbance D(s)
Control
Input
U(s)
Output
GP(s)
Y(s)
Plant
Plant Equation (Transfer function model that we all know how to obtain ?!):
Control Law (Algorithm) (we will try to learn how to design):
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Closed-Loop Transfer Function
Disturbance D(s)
Reference
Input
R(s)
+
Error
− E(s)
GC (s)
Control
Input +
+
U(s)
Output
GP(s)
Y(s)
Plant
H(s)
Y (s) =
⋅ R( s) +



GYR ( s )
⋅ D( s)



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GYD ( s )
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Closed-Loop Transfer Function
The closed-loop transfer functions relating the output y(t) (or Y(s)) to the reference
input r(t) (or R(s)) and the disturbance d(t) (or D(s)) are:
Y (s) =
GYR ( s)

⋅ R( s) +
Closed-Loop Transfer Function
From R( s ) to Y ( s )
⋅ D( s)
GYD ( s)

Closed-Loop Transfer Function
From R( s ) to Y ( s )
The objective of control system design is to design a controller GC (s), such that
certain performance (design) specifications are met. For example:
• we want the output y(t) to follow the reference input r(t), i.e.,
for certain frequency range. This is equivalent to specifying that
,
• we want the disturbance d(t) to have very little effect on the output y(t) within the
frequency range where disturbances are most likely to occur. This is equivalent
to specifying that
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Performance Specifications
Given an input/output representation, GCL (s), for which the output of the system
should follow the input, what specifications should you make to guarantee that the
system will behave in a manner that will satisfy its functional requirements?
Input
R(s)
Output
GCL (s)
r(t)
Y(s)
y(t)
Time
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Time
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Unit Step Response
1.6
yMAX
1.4
Unit Step Response
OS
1.2
± X%
1
0.8
0.6
0.4
0.2
0
tP
Time
tS
tr
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Performance Specifications
• Steady State Performance → Steady State Gain of the Transfer Function
Specifies the tracking performance of the system at steady state. Often it is
specified as the steady state response, y(∞) (or ySS(t)), to be within an X% bound of
the reference input r(t), i.e., the steady state error eSS(t) = r(t) − ySS(t) should be
within a certain percent. For example:
r (t ) − ySS (t )
≤ 2%= 0.02
r (t )
⇔
ySS (t )
≥ 98%= 0.98
r (t )
⇔
To find the steady state value of the output, ySS(t):
– Sinusoidal references: use frequency response, i.e.
– General references: use FVT, provided that
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is stable, ...
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Performance Specifications
• Transient Performance (Transient Response)
Transient performance of a system is usually specified using the unit step response
of the system. Some typical transient response specifications are:
– Settling Time (tS): Specifies the time required for the response to reach and
stay within a specific percent of the final (steady-state) value. Some typical
settling time specifications are: 5%, 2% and 1%. For 2nd order systems, the
specification is usually:  4
 ζω for 2% bound
 n

 5 for 1% bound

 ζω n
≤ Desired Settling Time (tS )
⇒
– % Overshoot (%OS):(2nd order systems)
−π
ζω n
ω n 1−ζ
2
=
%OS 100=
e
100e
−π
ζ
1−ζ 2
≤ X%
Q: How can we link this performance specification to the closed-loop transfer function?
(Hint) What system characteristics affect the system performance ?
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Performance Specifications
Transient Performance Specifications and CLTF Characteristic Poles
Recall that the positions of the system characteristic poles directly affect the system
output. For example, assume that the closed-loop transfer function of a feedback
Kω n 2
control system is:
GCL ( s ) = 2
s + 2ζω n s + ω n 2
The characteristic poles are:
s1, 2 = −ζω n ± jω n 1 − ζ 2 = −ζω n ± jω d = −
± j⋅
Settling Time (2%): → Puts constraint on the real part of the dominating closed-loop poles.
tS (2%) =
4
ζω n
=
4
%OS: → Puts constraint on the imaginary part of the dominating closed-loop poles.
−
πζ
− π⋅
ζω n
ω n 1−ζ
=
%OS 100
=
e 1−ζ 100e=
100e
2
2
− π⋅
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Performance Specification → CL Pole Positions
Transient Performance Specifications and CLTF Pole Positions
Transient performance specifications can be interpreted as constraints on the
positions of the poles of the closed-loop transfer function. Let a pair of closedloop poles be represented as: p1, 2 = −σ ± jω
Img.
Transient Performance Specifications:
jω
−σ + jω
– Settling Time (2 %) ≤ TS
tS (2%) =
4
σ
– %OS ≤ X %
%OS =100e
σ
ω
− π⋅
≤ TS
⇒
σ ≥
4
TS
Real
σ
π⋅
100
100
= πσ ≤ X % ⇒ e ω ≥
X
e ω
−σ − jω
⇒
− jω
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Performance Specification → CL Pole Positions
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Example
A DC motor driven positioning system can be
modeled by a second order transfer function:
GP ( s ) =
Find closed-loop transfer function:
3
s ( s + 6)
A proportional feedback control is proposed
and the proportional gain is chosen to be 16/3.
Find the closed-loop transfer function, as well
as the 2% settling time and the percent
overshoot of the closed loop system when
given a step input.
Draw block diagram:
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Example
Find closed-loop poles:
2% settling time:
%OS:
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ME375 Feedback Control - 23
Example
A DC motor driven positioning system can be
modeled by a second order transfer function:
GP ( s ) =
Find closed-loop transfer function:
3
s ( s + 6)
A proportional feedback control is proposed.
It is desired that:
– for a unit step response, the steady state
position should be within 2% of the
desired position,
– the 2% settling time should be less than 2
sec, and
– the percent overshoot should be less than
10%.
Find (1) the condition on the proportional gain
such that the steady state performance is
satisfied; (2) the allowable region in the
complex plane for the closed-loop poles.
Write down the performance specifications:
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Example
Steady state performance constraint:
Percent Overshoot (%OS)
Img.
jω
Transient performance constraint:
2% Settling Time
Real
− jω
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ME375 Feedback Control - 25

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