Process Control
Introduction
In the chemical industry,
the design of a control system is essential to
ensure:
 Good Process Operation
 Process Safety
 Product Quality
 Minimization of Environmental Impact
Introduction

What is the purpose of a control system?
“To maintain important process characteristics at
desired targets despite the effects of external
perturbations.”
Perturbations
Plant
Processing
objectives
Safety
Make $$$
Environment...
Market
Economy
Climate
Upsets...
Control
Introduction
What constitutes a control system?
Control
Combination of process
sensors, actuators and
computer systems
designed and tuned
to orchestrate
safe and profitable
operation.
Plant
Introduction

Process Dynamics:
 Study of the transient behavior of processes

Process Control
 the use of process dynamics for the
improvement of process operation and
performance
or
 the use of process dynamics to alleviate the
effect of undesirable (unstable) process
behaviors
Introduction
What do we mean by process?
A process, P, is an operation that takes an INPUT or a
DISTURBANCE and gives an OUTPUT
u
P
y
d
Information Flow
INPUT: (u) Something that you can manipulate
DISTURBANCE: (d) Something that comes as a result of
some outside phenomenon
OUTPUT: (y) An observable quantity that we want to
regulate
Examples

Stirred tank heater
M
Tin, w
T, w
Q
Inputs
Tin
w
Q
Output
Process
T
Examples

The speed of an automobile
Force of
Engine
Friction
Inputs
Output
Friction
Process
Engine
Speed
Examples
e.g. Landing on Mars
Examples
e.g. Millirobotics
Laparoscopic Manipulators
Introduction

Process
A process, P, is an operation that takes an INPUT or a
DISTURBANCE and gives an OUTPUT
u
P
y
d
Information Flow
INPUT: (u) Something that you can manipulate
DISTURBANCE: (d) Something that comes as a result of
some outside phenomenon
OUTPUT: (y) An observable quantity that we want to
regulate
Control
What is control?

To regulate of a process output despite the effect of
disturbances e.g.
 Driving a car
 Controlling the temperature of a chemical
reactor
 Reducing vibrations in a flexible structure

To stabilize unstable processes e.g.
 Riding a bike
 Flight of an airplane
 Operation of a nuclear plant
Benefits of Control

Economic Benefits
 Quality (waste reduction)
 Variance reduction (consistency)
 Savings in energy, materials, manpower

Operability, safety (stability)
 Performance
 Efficiency
 Accuracy
 robotics
 Reliability
 Stabilizability
 bicycle
 aircraft
 nuclear reactor
Control
What is a controller?
Process
Controller

A controller is a system designed to regulate a given
process
 Process typically obeys physical and chemical
conservation laws
 Controller obeys laws of mathematics and logic
(sometimes intelligent)
e.g.
- Riding a bike (human controller)
- Driving a car
- Automatic control (computer
programmed to control)
Block representations

Block diagrams are models of the physical systems
Input variables
System Physical
Boundary
Process
Output variables
Transfer of
fundamental
quantities
Physical
Mass, Energy and Momentum
Abstract
Operation
Control

A controlled process is a system which is
comprised of two interacting systems:
e.g. Most controlled systems are feedback controlled
systems
Disturbances
Outputs
Process
Action
intervene
Observation
Controller
monitor
The controller is designed to provide regulation of
process outputs in the presence of disturbances
Introduction
What is required for the development of a
control system?
1. The Plant (e.g. SPP of Nylon)
Nylon
Gas Make-up
Reheater
Relief
Pot
Dehumidifier
Steam
Heater
Blower
Water
Vent
Introduction
What is required?
1. Process Understanding
Required measurements
Required actuators
Understand design limitations
2. Process Instrumentation
Appropriate sensor and actuator selection
Integration in control system
Communication and computer architecture
3. Process Control
Appropriate control strategy
Example

Cruise Control
Friction
Engine
Process
Controller
Human or Computer
Speed
Classical Control

Control is meant to provide regulation of process
outputs about a reference, r, despite inherent
disturbances
d
r +
e
-
u
Controller
y
Process
Classical Feedback Control System

The deviation of the plant output, e=(r-y), from its
intended reference is used to make appropriate
adjustments in the plant input, u
Control

Process is a combination of sensors and actuators

Controller is a computer (or operator) that performs
the required manipulations
e.g. Classical feedback control loop
d
Computer
r+
Actuator
e
-
y
C
A
P
Process
M
Sensor
Examples
Driving an automobile

Driver
r +
Steering
e
-
y
C
A
P
Automobile
M
Visual and tactile measurement
Actual trajectory
Desired trajectory
r
y
Examples

Stirred-Tank Heater
Tin, w
Heater
Q
TC
T, w
Thermocouple
Tin, w
Controller
TR
Heater
e
+
-
y
C
A
P
Tank
M
Thermocouple
Examples

Measure T, adjust Q
Tin, w
Controller
+
TR -
Heater
e
T
C
A
P
Tank
M
Thermocouple
Feedback control
Controller:
where
Q=K(TR-T)+Qnominal
Qnominal=wC(T-Tin)
Q: Is this positive or negative feedback?
Examples

Measure Ti, adjust Q
Ti
M
C
A
+ DQ
Qi
+
Q
Feedforward Control
P
Control Nomenclature

Identification of all process variables
 Inputs
 Outputs

(affect process)
(result of process)
Inputs
 Disturbance variables
Variables affecting process that are due to
external forces
 Manipulated variables
Things that we can directly affect
Control Nomenclature

Outputs
 Measured
speed of a car
 Unmeasured
acceleration of a car
 Control variables
important observable quantities that we
want to regulate
can be measured or unmeasured
Disturbances
Manipulated
Other
Process
Controller
Control
Example
wi, Ti
L
Pc
wc, Tci
h
T
wc, Tco
Po
Variables
• wi, wo:
• Ti, To:
• wc:
• Pc:
• Po :
• Tci, Tco:
• h:
wo, To
T
Tank inlet and outlet mass flows
Tank inlet and outlet temperatures
Cooling jacket mass flow
Position of cooling jacket inlet valve
Position of tank outlet valve
Cooling jacket inlet and outlet
temperatures
Tank liquid level
Example
Variables
Inputs
Disturbances
Outputs
Manipulated Measured Unmeasured Control
wi
Ti
Tci
wc
h
wo
To
Pc
Po
Task: Classify the variables
Process Control and Modeling

In designing a controller, we must






Define control objectives
Develop a process model
Design controller based on model
Test through simulation
Implement to real process
Tune and monitor
d
r
e
u
Controller
y
Process
Model
Design
Implementation
Control System Development
Control development is usually carried out following these
important steps
Define Objectives
Develop a process
model
Design controller
based on model
Test by
Simulation
Implement and Tune
Monitor
Performance
Often an iterative process, based on performance we may
decide to retune, redesign or remodel a given control system
Control System Development

Objectives
 “What are we trying to control?”

Process modeling
 “What do we need?”
Mechanistic and/or empirical

Controller design
 “How do we use the knowledge of process
behavior to reach our process control
objectives?”
 What variables should we measure?
 What variables should we control?
 What are the best manipulated variables?
 What is the best controller structure?
Control System Development

Implement and tune the controlled process
 Test by simulation
 incorporate control strategy to the process
hardware
 theory rarely transcends to reality
 tune and re-tune

Monitor performance
 periodic retuning and redesign is often
necessary based on sensitivity of process or
market demands
 statistical methods can be used to monitor
performance
Process Modeling

Motivation:
 Develop understanding of process
a mathematical hypothesis of process
mechanisms
 Match observed process behavior
useful in design, optimization and control
of process

Control:
 Interested in description of process dynamics
Dynamic model is used to predict how
process responds to given input
Tells us how to react
Process Modeling
What kind of model do we need?

Dynamic vs. Steady-state
 Steady-state
Variables not a function of time
useful for design calculation
 Dynamic
Variables are a function of time
Control requires dynamic model
Process Modeling
What kind of model do we need?

Experimental vs Theoretical
 Experimental
Derived from tests performed on actual
process
Simpler model forms
Easier to manipulate
 Theoretical
Application of fundamental laws of physics
and chemistry
more complex but provides understanding
Required in design stages
Process Modeling

Dynamic vs. Steady-state
65
60
Steady-State 1
Output
55
Steady-State 2
50
45
40
0

50
100
150
Time
200
250
300
Step change in input to observe
 Starting at steady-state, we made a step change
 The system oscillates and finds a new steadystate
 Dynamics describe the transitory behavior
Process Modeling

Empirical vs. Mechanistic models
 Empirical Models
only local representation of the process
(no extrapolation)
model only as good as the data
 Mechanistic Models
Rely on our understanding of a process
Derived from first principles
Observing laws of conservation of
 Mass
 Energy
 Momentum
Useful for simulation and exploration of
new operating conditions
May contain unknown constants that must
be estimated
Process Modeling

Empirical vs Mechanistic models
 Empirical models
do not rely on underlying mechanisms
Fit specific function to match process
Mathematical French curve
1.3
1.2
1.1
Output
1
0.9
0.8
0.7
0.6
0.5
0.4
0
50
100
150
Time
200
250
300
Process Modeling

Linear vs Nonlinear
 Linear
basis for most industrial control
simpler model form, easy to identify
easy to design controller
poor prediction, adequate control
 Nonlinear
reality
more complex and difficult to identify
need state-of-the-art controller design
techniques to do the job
better prediction and control

In existing processes, we really on
 Dynamic models obtained from experiments
 Usually of an empirical nature
 Linear

In new applications (or difficult problems)
 Focus on mechanistic modeling
 Dynamic models derived from theory
 Nonlinear
Process Modeling

General modeling procedure
 Identify modeling objectives
end use of model (e.g. control)
 Identify fundamental quantities of interest
Mass, Energy and/or Momentum
 Identify boundaries
 Apply fundamental physical and chemical laws
Mass, Energy and/or Momentum balances
 Make appropriate assumptions (Simplify)
ideality (e.g. isothermal, adiabatic, ideal
gas, no friction, incompressible flow,
etc,…)
 Write down energy, mass and momentum
balances (develop the model equations)
Process Modeling

Modeling procedure
 Check model consistency
do we have more unknowns than equations
 Determine unknown constants
e.g. friction coefficients, fluid density and
viscosity
 Solve model equations
typically nonlinear ordinary (or partial)
differential equations
initial value problems
 Check the validity of the model
compare to process behavior
Process Modeling

For control applications:
 Modeling objectives is to describe process
dynamics based on the laws of conservation of
mass, energy and momentum

The balance equation
Rate of Accumulation
of fundamental quantity
=
Flow
In
+
-
Flow
Out
Rate of
Production
1. Mass Balance (Stirred tank)
2. Energy Balance (Stirred tank heater)
3. Momentum Balance (Car speed)
Process Modeling

Application of a mass balance
Holding Tank
Fin
h
F

Modeling objective: Control of tank level

Fundamental quantity: Mass

Assumptions: Incompressible flow
Process Modeling
Total mass in system = rV = rAh
Flow in = rFin
Flow out = rF
Total mass at time t = rAh(t)
Total mass at time t+Dt = rAh(t+Dt)
Accumulation
rAh(t+Dt) - rAh(t) = Dt(rFin-rF ),
rAh(t + tD ) - rAh(t )
Dt
lim
 r ( Fin - F ),
rAh(t + tD ) - rAh(t )
Dt
Dt 0
rA
 r ( Fin - F ),
dh
 r ( Fin - F ).
dt
Process Modeling
Model consistency
“Can we solve this equation?”
Variables: h, r, Fin, F, A
5
Constants: r, A
2
Inputs: Fin, F
2
Unknowns: h
1
Equations
1
Degrees of freedom
0
There exists a solution for each value of the
inputs Fin, F
Process Modeling
Solve equation
 Specify initial conditions h(0)=h0 and integrate
h( t ) 
t  Fin ( ) h(0) + 0 

F ( ) 
 d

A
2
Fin
flow
1.5
F
1
0.5
0
0
10
20
30
40
50
60
70
80
90
100
0
10
20
30
40
50
60
70
80
90
100
1.3
h
1.2
1.1
1
0.9
Process Modeling

Energy balance
M
Tin, w
T, w
Q
Objective:
Control tank temperature
Fundamental quantity: Energy
Assumptions: Incompressible flow
Constant hold-up
Process Modeling

Under constant hold-up and constant mean
pressure (small pressure changes)
 Balance equation can be written in terms of the
enthalpies of the various streams
dH 
 Hin - H out + Q + Ws
dt
 Typically work done on system by external
forces is negligible
dH 
 Hin - H out + Q
dt
 Assume that the heat capacities are constant
such that
H  r CPV (T - Tref )
H in  r C P w(Tin - Tref )
H out  r C P w(T - Tref )
Process Modeling
After substitution,
d ( rCPV (T - Tref ))
dt
 rCP w(Tin - Tref ) - rCP w(T - Tref ) + Q
Since Tref is fixed and we assume constant r
,Cp
rCPV
d (T - Tref )
dt
 rCP w(Tin - Tref ) - rCP w(T - Tref ) + Q
Divide by r CpV
dT w
Q
 (Tin - T ) +
dt V
rCPV
Process Modeling
Resulting equation:
dT F
Q
 (Tin - T ) +
dt V
rVC P
Model Consistency
Variables: T, F, V, Tin, Q, Cp, r
7
Constants: V, Cp, r
Inputs: F, Tin, Q
Unknown: T
3
3
1
Equations
1
There exists a unique solution
Process Modeling
Assume F is fixed
T (t )  T (0)e
- t /
t
+  e( - t )/ (
0
Tin ( )

+ Q( )
rC pV )d
where V/F is the tank residence time (or
time constant)
If F changes with time then the differential
equation does not have a closed form
solution.
dT (t ) F (t )
Q(t )

(Tin (t ) - T (t )) +
dt
V
rVCP
Product F(t)T(t) makes this differential
equation nonlinear.
Solution will need numerical integration.
Process Modeling
A simple momentum balance
Rate of
Accumulation
=
Momentum
In
+
-
Momentum
Out
Sum of forces
acting on system
Speed (v)
Friction
Force of
Engine (u)
Objective:
Control car speed
Quantity:
Momentum
Assumption: Friction proportional to speed
Process Modeling
Forces are:
Force of the engine = u
Friction = bv
Balance:
Total momentum = Mv
d ( Mv (t ))
dv (t )
M
 u(t ) - bv (t )
dt
dt
Model consistency
Variables:
Constants:
Inputs:
Unknowns
M, v, b, u
M, b
u
v
4
2
1
1
Process Modeling

Gravity tank
Fo
h
Objectives: height of liquid in tank L
Fundamental quantity: Mass, momentum
Assumptions:
 Outlet flow is driven by head of liquid in the
tank
 Incompressible flow
 Plug flow in outlet pipe
 Turbulent flow
F
Process Modeling
From mass and momentum balances,
dh Fo AP v

dt
A
A
dv hg K F v 2

dt
L
rA P
A system of simultaneous ordinary
differential equations results
Linear or nonlinear?
Process Modeling
Model consistency
Variables
Fo, A, Ap, v, h, g, L, KF, r
9
Constants
A, Ap, g, L, KF, r
6
Inputs
Fo
1
Unknowns
h, v
2
Equations
Model is consistent
2
Solution of ODEs

Mechanistic modeling results in nonlinear
sets of ordinary differential equations

Solution requires numerical integration

To get solution, we must first:
 specify all constants (densities, heat capacities,
etc, …)
 specify all initial conditions
 specify types of perturbations of the input
variables
For the heated stirred tank,
dT F
Q
 (Tin - T ) +
dt V
rVC P
 specify r, CP, and V
 specify T(0)
 specify Q(t) and F(t)
Input Specifications

Study of control system dynamics
 Observe the time response of a process
output in response to input changes

Focus on specific inputs
1. Step input signals
2. Ramp input signals
3. Pulse and impulse signals
4. Sinusoidal signals
5. Random (noisy) signals
Common Input Signals
1. Step Input Signal: a sustained
instantaneous change
e.g. Unit step input introduced at time 1
1.5
Input
1
0.5
0
0
1
2
3
4
5
Time
6
7
8
9
10
0
1
2
3
4
5
6
7
8
9
10
1.4
1.2
1
0.8
0.6
0.4
0.2
0
Common Input Signals
2. Ramp Input: A sustained constant rate of
change
e.g.
9
8
7
Input
6
5
4
3
2
1
0
0
1
2
3
4
5
Time
6
7
8
9
10
8
7
6
Output
5
4
3
2
1
0
-1
0
1
2
3
4
5
Time
6
7
8
9
10
Common Input Signals
3. Pulse: An instantaneous temporary change
e.g. Fast pulse (unit impulse)
100
90
80
70
Input
60
50
40
30
20
10
0
0
1
2
3
4
5
Time
6
7
8
9
10
0
1
2
3
4
5
Time
6
7
8
9
10
0.45
0.4
0.35
0.3
Output
0.25
0.2
0.15
0.1
0.05
0
-0.05
Common Input Signals
3. Pulses:
e.g. Rectangular Pulse
1.5
Input
1
0.5
0
0
1
2
3
4
5
Time
6
7
8
9
10
0
1
2
3
4
5
Time
6
7
8
9
10
1.2
1
Output
0.8
0.6
0.4
0.2
0
-0.2
Common Input Signals
4. Sinusoidal input
1.5
1
Input
0.5
0
-0.5
-1
-1.5
0
5
10
15
Time
20
25
30
0
5
10
15
Time
20
25
30
0.8
0.6
0.4
Output
0.2
0
-0.2
-0.4
-0.6
-0.8
Common Input Signals
5. Random Input
1.5
1
Input
0.5
0
-0.5
-1
-1.5
0
5
10
15
Time
20
25
30
0.6
0.4
Output
0.2
0
-0.2
-0.4
-0.6
-0.8
0
5
10
15
Time
20
25
30
Solution of ODEs using Laplace
Transforms
Process Dynamics and Control
Linear ODEs

For linear ODEs, we can solve without
integrating by using Laplace transforms

- st
F ( s)  [ f (t )]   f (t )e dt
t 0

Integrate out time and transform to
Laplace domain
dy (t )
 ay (t ) + bu(t )
dt
y (0) c
Integration
Multiplication
Y(s) = G(s)U(s)
Common Transforms
Useful Laplace Transforms
1. Exponential
f (t )  e - bt
[e
- bt

] e
- bt - st
e
dt   e - ( s+ b) t dt
0
[e
- bt
] -

0
e
- ( s+ b) t  

s + b 
0
1

s+b
2. Cosine
e - jt + e jt
f (t )  cos(t ) 
2

1  - ( s- j ) t
- ( s + j ) t 
[cos(t )]    e
dt +  e
dt 
2 0

0
1 1
1 
s
 
+
 2
2  s - j s + j  s +  2
Common Transforms
Useful Laplace Transforms
3. Sine
e jt - e - jt
f (t )  sin(t ) 
2j

1  - ( s- j ) t
- ( s + j ) t 
[sin(t )]    e
dt -  e
dt 
2 j 0

0
1  1
1 

 
 2
2 j  s - j s + j  s +  2
Common Transforms
Operators
1. Derivative of a function f(t)
df (t )
dt
du  df
v  e - st

df

[ ]  uv  0 -  udv  f (t )e - st
dt
0


0

-  ( - sf (t )e - st )dt
0

df
[ ]  s  f (t )e - st dt - f (0)  sF ( s) - f (0)
dt
0
2. Integral of a function f(t)
t
  - st t
F ( s)
  f ( )d    e (  f ( )d )dt 
s
0
 0
0
Common Transforms
Operators
3. Delayed function f(t-)
t 
 0
g (t )  
 f (t -  ) t  

 g (t )   e
0
- st

(0)dt +  e - st f (t -  )dt

 g (t )  e - s F ( s)
Common Transforms
Input Signals
1. Constant

[a ]   ae
f (t )  a
- st
- st
ae
dt  (
0
s
) 0  a
s
0 t  0
f (t )  
a t  0
2. Step

[ f (t )]   ae
- st
- st
ae
dt  (
0
s
) 0  a
s
0 t 0
3. Ramp function f (t )  
at t  0

 f (t )   ate - st dt  0
e
- st


at 
ae - st
a
dt  2
 +
s 0 0 s
s
Common Transforms
Input Signals
4. Rectangular Pulse
t0
0

f ( t )  a 0  t  t w
0
t  tw

tw
a
 f (t )   ae - st dt  (1 - e - t w s )
s
0
5. Unit impulse
 (t )  lim
1
tw  0 tws
(1 - e - t w s )
se - t w s
 (t )  lim
1
tw  0 s
Laplace Transforms
Final Value Theorem
lim  y (t )  lim  sY ( s)
t 
s 0
Limitations:
y (t )  C1 ,
lim  sY ( s) exists s Re( s)  0
s 0
Initial Value Theorem
y(0)  lim  sY ( s)
s
Solution of ODEs
We can continue taking Laplace transforms
and generate a catalogue of Laplace
domain functions. See SEM Table 3.1
The final aim is the solution of ordinary
differential equations.
Example
Using Laplace Transform, solve
dy
5 + 4y  2 ,
dt
y (0)  1
Result
y (t )  0.5 + 0.5e -0.8t
Solution of Linear ODEs
Stirred-tank heater (with constant F)
dT F
Q
 (Tin - T ) +
dt V
rVC P
T (0)  T0
taking Laplace
V  dT 
1

 Tin (t ) - T (t ) +
Q(t )
F  dt 
r FCP
 (T ( s) - T (0))  Tin ( s) - T ( s) + K P Q( s)

1
K
T ( s) 
T (0) +
Tin ( s) + P Q( s)
s + 1
s + 1
s + 1
To get back to time domain, we must
 Specify Laplace domain functions Q(s), Tin(s)
 Take Inverse Laplace
Linear ODEs
Notes:
 The expression
T ( s) 

s + 1
T (0) +
1
K
Tin ( s) + P Q( s)
s + 1
s + 1
describes the dynamic behavior of the process
explicitly
 The Laplace domain functions multiplying T(0),
Tin(s) and Q(s) are transfer functions
Tin(s)
Q(s)
T(0)
1
s + 1
KP
s + 1

s + 1
+
+
+
T(s)
Laplace Transform
Assume Tin(t) = sin(t) then the transfer
function gives directly
1

Tin ( s)  2
s + 1
( s +  2 )(s + 1)
Cannot invert explicitly, but if we can find A
and B such that
B

+
 2
2
2 s + 1
s +
( s +  2 )(s + 1)
we can invert using tables.
A
Need Partial Fraction Expansion to deal with
such functions
Linear ODEs
We deal with rational functions of the form
r(s)=p(s)/q(s) where degree of q > degree
of p
q(s) is called the characteristic polynomial of
the function r(s)
Theorem:
Every polynomial q(s) with real
coefficients can be factored into the
product of only two types of factors
 powers of linear terms (x-a)n and/or
 powers of irreducible quadratic terms,
(x2+bx+c)m
Partial fraction Expansions
1. q(s) has real and distinct factors
n
q ( s)   ( s + bi )
i 1
expand as
n
r ( s)  
i
i 1s + bi
2. q(s) has real but repeated factor
q ( s)  ( s + b) n
expanded
r ( s) 
1
+
2
s + b ( s + b)
2
++
n
( s + b) n
Partial Fraction Expansion
Heaviside expansion
For a rational function of the form
n 
p( s)
p( s)
i
r ( s) 
 n
 
q ( s)
( s + bi )
i

1
 ( s + bi )
i 1
Constants are given by
p ( s) 
 i  ( s + bi )
q ( s)  s- b
i
Note: Most applicable to q(s) with real and
distinct roots. It can be applied to more
specific cases.
Partial Fraction Expansions
3. Q(s) has irreducible quadratic factors of the
form
q ( s)  ( s2 + d1s + d 0 ) n
where d 2
 d0
4
Algorithm for Solution of ODEs





Take Laplace Transform of both sides of ODE
Solve for Y(s)=p(s)/q(s)
Factor the characteristic polynomial q(s)
Perform partial fraction expansion
Inverse Laplace using Tables of Laplace
Transforms
Transfer Function Models
of Dynamical Processe
Process Dynamics and Control
Transfer Function

Heated stirred tank example

1
KP
T ( s) 
T (0) +
Tin ( s) +
Q( s)
s + 1
s + 1
s + 1
Tin(s)
Q(s)
T(0)
e.g. The block
1
s + 1
KP
s + 1
+
+
T(s)
+

s + 1
KP
s + 1
is called the transfer function relating Q(s)
to T(s)
Process Control
Time Domain
Laplace Domain
Process Modeling,
Experimentation and
Implementation
Transfer function
Modeling, Controller
Design and Analysis
Ability to understand dynamics in Laplace and
time domains is extremely important in the
study of process control
Transfer function
Order of underlying ODE is given by
degree of characteristic polynomial
e.g. First order processes

KP
Y ( s) 
U ( s)
s + 1
Second order processes
KP
Y ( s)  2 2
U ( s)
 s + 2s + 1
Steady-state value obtained directly
e.g. First order response to unit step function
Kp
Y ( s) 
s(s + 1)
Final value theorem

lim sY ( s)  limG ( s)  K P
s 0

s 0
Transfer functions are additive and
multiplicative
Transfer function

Effect of many transfer functions on a
variable is additive

1
KP
T ( s) 
T (0) +
Tin ( s) +
Q( s)
s + 1
s + 1
s + 1
Tin(s)
Q(s)
T(0)
1
s + 1
KP
s + 1

s + 1
+
+
+
T(s)
Transfer Function

Effect of consecutive processes in series in
multiplicative
U(s)

KP
s + 1
Y1(s)
KP
s + 1
Transfer Function
KP
U ( s)
s + 1
K
Y 2( s)  P Y1 ( s)
s + 1
KP   KP 

Y1 ( s)  

 U ( s)
 s + 1  s + 1
Y1 ( s) 
Y2(s)
Deviation Variables
To remove dependence on initial condition
e.g.


1
KP
T ( s) 
T (0) +
Tin ( s) +
Q( s)
s + 1
s + 1
s + 1
Remove dependency on T(0)
1
KP
T  ( s) 
T  ( s) +
Q ( s)
s + 1 in
s + 1
Transfer functions express extent of deviation
from a given steady-state

Procedure




Find steady-state
Write steady-state equation
Subtract from linear ODE
Define deviation variables and their derivatives
if required
 Substitute to re-express ODE in terms of
deviation variables
Example

Jacketed heated stirred tank
F, Tin
Fc, Tcin
h
Fc, Tc
F, T
Assumptions:
 Constant hold-up in tank and jacket
 Constant heat capacities and densities
 Incompressible flow
Model
dT F
hc Ac
 (Tin - T ) +
(Tc - T )
dt V
rC PV
dTc Fc
h A
 (Tcin - Tc ) - c c (Tc - T )
dt Vc
rcC PcVc
Nonlinear ODEs
Q: If the model of the process is nonlinear,
how do we express it in terms of a transfer
function?
A: We have to approximate it by a linear one
(i.e.Linearize) in order to take the Laplace.
f(x)
f
( x0 )
x
f(x0)
x0
x
Nonlinear systems

First order Taylor series expansion
1. Function of one variable
f ( x )  f ( xs ) +
 f ( xs )
( x - xs )
x
2. Function of two variables
f ( x, u)  f ( xs , us ) +
 f ( xs ,us )
 f ( xs ,us )
( x - xs ) +
(u - us )
x
u
3. ODEs
x  f ( x)  f ( xs ) +
 f ( xs )
( x - xs )
x
Transfer function

Procedure to obtain transfer function from
nonlinear process models
 Find steady-state of process
 Linearize about the steady-state
 Express in terms of deviations variables about
the steady-state
 Take Laplace transform
 Isolate outputs in Laplace domain
 Express effect of inputs in terms of transfer
functions
Y ( s)
 G1 ( s)
U1 ( s)
Y ( s)
 G2 ( s)
U 2 ( s)
First order Processes
Examples, Liquid storage
Fi
h
F
rA
dh
 rFi - rF  rFi -  h
dt
rA dh r
 Fi - h
 dt 
dh
 + h  K p Fi
dt
dh 

+ h   K p Fi 
dt
First Order Processes
Examples: Speed of a Car
dv 
 u - bv 
dt
M dv  1
 u - v 
b dt b
dv 

 K pu - v 
dt
Stirred-tank heater
M
dT 
rC pV
 - rC p FT  + Q
dt
V dT 
1

Q - T 
F dt rC p F

dT 
 K pQ - T 
dt
Note:
Tin (t )  0
Kp
v  ( s)

u ( s)  s + 1
Kp
T  ( s)

Q ( s)  s + 1
First Order Processes
Kp

Liquid Storage Tank
r/
rA/
Speed of a car
M/b
1/b
Stirred-tank heater
1/rCpF
V/F
First order processes are characterized by:
1. Their capacity to store material, momentum
and energy
2. The resistance associated with the flow of
mass, momentum or energy in reaching their
capacity
First order processes
Liquid storage:
 Capacity to store mass : rA
 Resistance to flow : 1/
Car:
 Capacity to store momentum: M
 Resistance to momentum transfer : 1/b
Stirred-tank heater
 Capacity to store energy: rCpV
 Resistance to energy transfer : 1/ rCpF
Time Constant =  = (Storage capacitance)*
(Resistance to flow)
First order process
Step response of first order process

Kp M
Y ( s) 
s + 1 s
Step input signal of magnitude M
1
0.9
0.8
0.7
0.632
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
t/
4
5
6
First order process

What do we look for?
 Process Gain: Steady-State Response
 Kp 
Overall Change in y Dy
lim 
 Kp 


s 0 s + 1
Overall Change in u Du
 Process Time Constant:
 = Time Required to Reach
63.2% of final value

What do we need?




Process at steady-state
Step input of magnitude M
Measure process gain from new steady-state
Measure time constant
First order process
Ramp response:
Kp a
Y ( s) 
s + 1 s2
Ramp input of slope a
5
4.5

y(t)/Kpa
4
3.5
3
a
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
2.5
t/
3
3.5
4
4.5
5
First order Process
Sinusoidal response

KP
A
Y ( s) 
s + 1 s2 +  2
Y ( s) 
KP A
-1
lim 
1+  
Sinusoidal input Asin(t)
t 
2
2
sin(t + f )
2
1.5
1
AR
0.5
0
-0.5
f
-1
-1.5
0
2
4
6
8
10
t/
12
14
16
18
20
First order Processes
Bode Plots
10
0
AR/Kp
High Frequency
10
10
Asymptote
Corner Frequency
-1
-2
10
-2
-1
10
10
0
p
10
1
10
2
0
-20
f
-40
-60
-80
-100
-2
10
-1
10
Amplitude Ratio
K
AR 
1 +  2 2
10
p
0
10
1
Phase Shift
f  - tan-1( )
10
2
Integrating Processes
Example: Liquid storage tank
Fi
h
F
dh
 rFi - rF
dt
dh
A  Fi - F
dt
rA
H  ( s)
1/ A

s
F ( s)
Fi   Fi - Fis , F   F - Fs
dh 
A
 Fi  - F 
dt
H  ( s) 1 / A


s
F ( s)
i
Process acts as a pure integrator
Process Modeling

Step input of magnitude M
K M KM
Y ( s) 
 2
s s
s
Slope = KM
Time
Time
 0
y (t )  
 KMt
t0
t0
Integrating processes

Unit impulse response
K
KM
Y ( s)  M 
s
s
KM
Time
Time
 0
y (t )  
 KM
t0
t0
Integrating Processes

Rectangular pulse response
KM
KM
- tws
Y ( s) 
(1 - e
)  2 (1 - e- tws )
s s
s
Time
 KMt
y (t )  
 KMt w
Time
t  tw
t  tw
Second Order Processes
Three types of second order process:
1. Multicapacity processes: processes that
consist of two or more capacities in series
e.g. Two heated stirred-tanks in series
2. Inherently second order processes: Fluid
or solid mechanic processes possessing
inertia and subjected to some acceleration
e.g. A pneumatic valve
3. Processing system with a controller:
Presence of a controller induces oscillatory
behavior
e.g. Feedback control system
Second order Processes

Multicapacity Second Order Processes
 Naturally arise from two first order processes in
series
U(s)
U(s)
K P1
 1s + 1
K P2
 2s + 1
K P1 K P 2
( 1s + 1)( 2 s + 1)
Y(s)
Y(s)
 By multiplicative property of transfer functions
K P1 K P 2
Y ( s) 
U ( s)
( 1 s + 1)( 2 s + 1)
Second Order Processes
Inherently second order process:
e.g. Pneumatic Valve

p
x
Momentum Balance
M
d dx
dx
 pA - Kx - C
dt dt
dt
M d 2 x C dx
A
+
+x p
K dt K dt
K
A
x  ( s)
K

p ( s) M s2 + C s + 1
K
K
Second order Processes

Second order process:
 Assume the general form
KP
Y (S )  2 2
U ( s)
 s + 2s + 1
where KP = Process steady-state gain
 = Process time constant
 = Damping Coefficient

Three families of processes
1
=1
>1

Underdamped
Critically Damped
Overdamped
Note: Chemical processes are typically
overdamped or critically damped
Second Order Processes

Roots of the characteristic polynomial
-2  4 2 2 - 4 2
2 2
 1 2
- 
 -1


Case 1) >1: Two distinct real roots
System has an exponential
behavior
Case 2) =1: One multiple real root
Exponential behavior
Case 3) 1: Two complex roots
System has an oscillatory
behavior
Second order Processes

Step response of magnitude M
KP
M
Y (S )  2 2
 s + 2s + 1 s
2
0
1.8
0.2
1.6
1.4
1.2
1
0.8
0.6
2
0.4
0.2
0
0
1
2
3
4
5
6
7
8
9
10
Second order process
Observations
 Responses exhibit overshoot (y(t)/KM >1)
when <1
 Large  yield a slow sluggish response
 Systems with =1 yield the fastest response
without overshoot
 As  (with 1) becomes smaller system
becomes more oscillatory
 If 0, system oscillates without bounds
(unstable)
Second order processes

Example - Two Stirred tanks in series
M
Tin, w
Response of T2 to
Tin is an example of an
overdamped second
order process
Q
M
T1, w
T2, w
Q
Second order Processes
Characteristics of underdamped second order
process
1. Rise time, tr
2. Time to first peak, tp
3. Settling time, ts
4. Overshoot:


a

OS   exp  
2
b
 1-  
5. Decay ratio:

c
2 

DR   exp 2
b
 1-  
Second order Processes
1.8
1.6
P
1.4
b
1.2
+5%
c
1
-5%
0.8
0.6
a
0.4
0.2
0
tp
0 tr
5
10
15
20
ts
25
30
35
40
45
50
Second Order Process

Sinusoidal Response
Y ( s) 
y (t ) 
 2 s2 + 2s + 1 s2 +  2
Kp A

1 - ( )
where
A
Kp

2 2
sin(t + f )
+ (2 )2
 2
1
f  - tan 


2
1 - ( ) 
ARn 

1 - ( )
1

2 2
+ (2 )2
Second Order Processes
Bode Plots

1
10
0.1
0
10
=1
-1
10
-1
10
0
1
10
10
0
-50
=1
-100
-150
0.1
-1
10
0
10
1
10
More Complicated processes
Transfer function typically written as rational
function of polynomials
r
a
+
a
s
+

+
a
s
r ( s)
0
1
r
G ( s) 

q ( s) b0 + b1s++b s
where r(s) and q(s) can be factored as
q ( s)  b0 (1s + 1)( 2 s + 1)( s + 1)
r ( s)  a0 ( a1s + 1)( a 2 s + 1)( a s + 1)
s.t.
G ( s)  K
( a1 s + 1) ( ar s + 1)
(1s + 1) ( s + 1)
Poles and zeroes
Definitions:
 the roots of r(s) are called the zeros of G(s)
z1  -
1
 a1
, , zr  -
1
 ar
 the roots of q(s) are called the poles of G(s)
p1  -
1
1
,, p  -
1

Poles: Directly related to the underlying
differential equation
If Re(pi)<0, then there are terms of the
form e-pit in y(t) - y(t) vanishes to a unique
point
If any Re(pi)>0 then there is at least one
term of the form epit - y(t) does not vanish
Poles
e.g. A transfer function of the form
K
s(1s + 1)( 2 s2 + 2 2 s + 1)
with 0    1 can factored to a sum of
 A constant term from s
 A e-t/ from the term (1s+1)
 A function that includes terms of the form
- t
e
- t
e
 2 sin( 1 -  2 t
2)
 2 cos( 1 -  2 t
2 )
Poles can help us to describe the qualitative
behavior of a complex system (degree>2)
The sign of the poles gives an idea of the
stability of the system
Poles

Calculation performed easily in MATLAB
Function ROOTS
e.g.

q( s)  s3 + s2 + s + 1
» ROOTS([1 1 1 1])
ans =
-1.0000
0.0000 + 1.0000i
0.0000 - 1.0000i
»
MATLAB
Poles
Plotting poles in the complex plane
1
0.8
0.6
Imaginary axis
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-1.2
-1
-0.8
-0.6
-0.4
Real axis
-0.2
q( s)  s3 + s2 + s + 1
Roots: -1.0, 1.0j, -1.0j
0
0.2
Poles
Process Behavior with purely complex poles
Unit Step Response
1.8
1.6
1.4
1.2
y(t)
1
0.8
0.6
0.4
0.2
0
0
5
10
15
20
25
t
30
35
40
45
50
Poles
1
0.8
0.6
Imaginary axis
0.4
0.2
0
-0.2
-0.4
-0.6
-0.8
-1
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
Real axis
0.2
0.3
2s3 + 2.5s2 + 3s + 1
Roots: -0.4368, -0.4066+0.9897j,
-0.4066-0.9897j
0.4
0.5
Poles
Process behavior with mixed real and
complex poles
Unit Step Response
1
0.9
0.8
0.7
y(t)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
2
4
6
8
10
t
12
14
16
18
20
Poles
1.5
1
Imaginary axis
0.5
0
-0.5
-1
-1.5
-0.8
-0.6
-0.4
-0.2
0
Real axis
0.2
2s4 + 2.5s3 + 3s2 + s - 0.5
Roots: -0.7441, -0.3805+1.0830j,
-0.3805-1.0830j, 0.2550
0.4
0.6
Poles
Process behavior with unstable pole
Unit Step Response
160
140
120
100
y(t)
80
60
40
20
0
-20
0
2
4
6
8
10
t
12
14
16
18
20
Zeros
Transfer function:
G ( s) 
K p ( a s + 1)
(1s + 1)( 2 s + 1)
  - - t
- t 


y ( t )  K p M  1 + a 1 e 1 + a 2 e  2 
 1 -  2



1
2


Let 3
1 is the dominant time constant
1 >  2
16
2.5
y(t)/KM
2
8
1.5
4
1
2
1
0.5
0
-1
-2
0
-0.5
0
2
4
6
8
10
Time
12
14
16
18
20
Zeros
Observations:
 Adding a zero to an overdamped second order
process yields overshoot and inverse response
 Inverse response is observed when the zeros lie
in right half complex plane, Re(z)>0
 Overshoot is observed when the zero is
dominant ( a > 1)
 Pole-zero cancellation yields a first order
process behavior
 In physical systems, overshoot and inverse
response are a result of two process with
different time constants, acting in opposite
directions
Zeros
Can result from two processes in parallel
K1
1s + 1
U(s)
Y(s)
K2
 2s + 1
( a s + 1)
G( s)  K
(1s + 1)( 2 s + 1)
K  K1 + K2
K  + K21
a  1 2
K1 + K2
If gains are of opposite signs and time
constants are different then a right half
plane zero occurs
Dead Time
Fi
Control loop
h
Time required for the
fluid to reach the valve
usually approximated as
dead time
Manipulation of valve does not lead to immediate
change in level
Dead time
Delayed transfer functions
U(s)
e
- d s
G ( s)
Y ( s)  e - d sG ( s)U ( s)
e.g. First order plus dead-time
G ( s) 
e - d s K p
 s+1
Second order plus dead-time
G ( s) 
e - d s K P
 2 s2 + 2s + 1
Y(s)
Dead time

Dead time (delay)
G ( s)  e - d s
 Most processes will display some type of lag
time
 Dead time is the moment that lapses between
input changes and process response
1
0.9
0.8
0.7
y/KM
0.6
0.5
0.4
0.3
0.2
0.1
0
0
D 0.5
1
1.5
2
2.5
3
t/tau
3.5
4
4.5
5
5.5
Step response of a first order plus dead time process
Dead Time

Problem
 use of the dead time approximation makes
analysis (poles and zeros) more difficult
G ( s) 

e - d s K p
 s+1
Approximate dead-time by a rational
(polynomial) function
 Most common is Pade approximation

1- s
2
e - s  G1( s) 

1+ s
2

2


1 - s + s2
2
12
e -s  G2 ( s) 
2
1 + s + s2
2
12
Pade Approximations

In general Pade approximations do not
approximate dead-time very well

Pade approximations are better when one
approximates a first order plus dead time
process
- s

1- s K
p
2
G ( s) 

 s + 1 1 +  s s + 1
2
e
Kp

Pade approximations introduce inverse
response (right half plane zeros) in the
transfer function

Limited practical use
Process Approximation

Dead time
 First order plus dead time model is often used
for the approximation of complex processes

Step response of an overdamped second
order process
1
0.9
0.8
0.7
0.6
- First Order plus dead time
o Second Order
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
8
Process Approximation

Second order overdamped or first order
plus dead time?
1.2
1
0.8
-- First order plus dead time
- Second order overdamped
o Actual process
0.6
0.4
0.2
0
-0.2

0
1
2
3
4
5
6
7
8
Second order process model may be more
difficult to identify
Process Approximation

Transfer Function of a delay system
 First order processes
KPe-  s
Y ( s) 
U ( s)
s + 1
D
 Second order processes
KP e-  s
Y (S )  2 2
U ( s)
 s + 2s + 1
D
U(s)
Y(s)
e- 
Ds
G(s)
Process Approximation

More complicated processes
 Higher order processes (e.g. N tanks in series)
U(s)
K P1K P 2 K PN
(1s + 1)( 2 s + 1)( N s + 1)
Y(s)
 For two dominant time constants 1 and 2
process well approximated by
G ( s) 
e - s K p
(1s + 1)( 2 s + 1)

N
  i
i3
 For one dominant time constant 1, process well
approximated by
N
e - s K p
G ( s) 
   i
(1s + 1)
i2
Process Approximation

Example
G ( s) 
1
(10s + 1)(25s + 1)( s + 1)2
1.2
1
0.8
0.6
e -12 s
G1( s) 
25s + 1
e -2 s
G2 ( s) 
(10s + 1)(25s + 1)
0.4
0.2
0
-0.2
0
20
40
60
80
100
120
140
160
180
200
Empirical Modeling
Objective:
 To identify low-order process dynamics (i.e.,
first and second order transfer function models)
 Estimate process parameters (i.e., Kp,  and )
Methodologies:
1. Least Squares Estimation
more systematic statistical approach
2. Process Reaction Curve Methods
quick and easy
based on engineering heuristics
Empirical Modeling
Least Squares Estimation:
Simplest model form
E[ y ]  0 + 1x
Process Description
y  0 + 1x + 
where
y
x
1 , 0
vector of process measurement
vector of process inputs
process parameters
Problem:
Find 1, 0 that minimize the sum of squared
residuals (SSR)
n
SSR   ( yi - 0 - 1xi ) 2
i 1
Empirical Modeling
Solution
Differentiate SSR with respect to parameters
n
 SSR
 -2  ( yi - 0 - 1xi )  0
 0
i 1
n
 SSR
 -2  xi ( yi - 0 - 1xi )  0
 1
i 1
These are called the normal equations.
Solving for parameters gives:
0  y - 1x
1 
where
n
 xi yi - nxy
i 1
n 2
2
 xi - nx
i 1
n x
x  i,
i 1 n
n y
y  i
i 1 n
Empirical Modeling
Compact form
Define
 y1 
1
y 
1
Y   2 , X  


y 
1
 n

x1 
x2 
0 

,  
 

 1
xn 
Then
 y1 - 0 - 1x1 
y -  -  x 
0
1 2
E 2
 Y - X



y -  -  x 
 n
0
1 n
Problem
find value of  that minimize SSR
SSR  E T E
Empirical Modeling
Solution in Compact Form
Normal Equations can be written as
 ET E
0

which can be shown to give
X T X  X T Y
or
(
  X T X
)
-1
X TY
In practice
 Manipulations are VERY easy to perform in
MATLAB
 Extends to general linear model (GLM)
E[ y ]  0 + 1x1 ++  p x p
 Polynomial model
E[ y ]  0 + 1x1 + 11x12
Empirical Modeling
Control Implementation:
 previous technique applicable to process model
that are linear in the parameters (GLM,
polynomials in x, etc…)
 ei
i.e. such that, for all i, the derivatives

a function of 
are not
 typical process step responses
first order
E[ y(t )]  K p M (1 - e - t / )
Nonlinear in Kp and 
overdamped second order
 1e - t /1 -  2e - t / 2 

E[ y (t )]  K p M  1 1 -  2


Nonlinear in Kp, 1 and 2
 nonlinear optimization is required to find the
optimum parameters
Empirical Modeling
Nonlinear Least Squares required for control
applications
 system output is generally discretized
y(t )  [ y(t1), y(t2 ),, y(tn )]
or, simply
y(t )  [ y1, y2 ,, yn ]
First Order process (step response)
E[ yi ]  K p M (1 - e - ti / )
Least squares problem becomes the minimization
of
n
SSR   ( yi - K p M (1 - e - ti / )) 2
i 1
This yields an iterative problem solution best
handled by software packages: SAS, Splus,
MATLAB (function leastsq)
Empirical Modeling
Example
Nonlinear Least Squares Fit of a first order process
from step response data
Model
E[ y (t )]  3.0 K p (1 - e - t / )
Data
Step Response
4.5
4
3.5
3
y(t)
2.5
2
1.5
1
0.5
0
-0.5
0
10
20
30
40
t
50
60
70
80
Empirical Modeling
Results:
Using MATLAB function “leastsq” obtained
K p  13432
.
,   118962
.
Resulting Fit
Step Response
4.5
4
3.5
3
y(t)
2.5
2
1.5
1
0.5
0
-0.5
0
10
20
30
40
t
50
60
70
80
Empirical Modeling
Approximation using delayed transfer
functions
 For first order plus delay processes
0
0 t 

E[ yi ]  
- (ti - )/
K
M
(
1
e
)
t 
 p
Difficulty
 Discontinuity at  makes nonlinear least
squares difficult to apply
Solution
1. Arbitrarily fix delay or estimate using
alternative methods
2. Estimate remaining parameters
3. Readjust delay repeat step 2 until best value of
SSR is obtained
Empirical Modeling
Example 2
Underlying “True” Process
G ( s) 
1
(10s + 1)(25s + 1)( s + 1)2
Data
3.5
3
2.5
y(t)
2
1.5
1
0.5
0
-0.5
0
20
40
60
80
t
100
120
140
Empirical Modeling
Fit of a first order plus dead time
10000
.
e -11s
G1( s) 
(27.3899 s + 1)
Second order plus dead time
0.9946e -2 s
G2 ( s) 
(24.9058s + 1)(101229
.
s + 1)
3.5
3
2.5
y(t)
2
1.5
1
0.5
0
-0.5
0
20
40
60
80
t
100
120
140
Empirical Modeling
Process reaction curve method:
 based on approximation of process using first
order plus delay model
D(s)
M/s
Y*(s)
Gp
Gc
U(s)
Gs
Ym(s)
Manual Control
1. Step in U is introduced
2. Observe behavior ym(t)
3. Fit a first order plus dead time model
KMe -s
Ym ( s) 
s( s + 1)
Y(s)
Empirical Modeling
First order plus dead-time approximations
1.2
1
0.8
0.6
KM
0.4
0.2
0

-0.2
0
1

2
3
4
5
6
7
 Estimation of steady-state gain is easy
 Estimation of time constant and dead-time is
more difficult
8
Empirical Modeling
Estimation of time constant and dead-time
from process reaction curves
 find times at which process reaches 35.3% and
85.3%

1
0.9
0.8
0.7
y(t)
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
t1
 Estimate
60
t2
80
t
100
  13
. t1 - 0.29t2
  0.67(t2 - t1)
120
140
160
Empirical Process
Example
For third order process
G ( s) 
1
(10s + 1)(25s + 1)( s + 1)2
Estimates:
t1  23, t2  62.5
   1178
. ,   26.46
Compare:
Least Squares Fit
10000
.
e -11s
G1( s) 
(27.3899 s + 1)
Reaction Curve
100
. e -11.78s
G1( s) 
(26.46s + 1)
Empirical Modeling
Process Reaction Curve Method
 based on graphical interpretation
 very sensitive to process noise
 use of step responses is troublesome in normal
plant operations
frequent unmeasurable disturbances
difficulty to perform instantaneous step
changes
maybe impossible for slow processes
 restricted to first order models due to reliability
 quick and easy
Least Squares
 systematic approach
 computationally intensive
 can handle any type of dynamics and input
signals
 can handle nonlinear control processes
 reliable
Feedback Control

Steam heated stirred tank
Fin,Tin
TC
TT
IP
LT
Ps
Steam
Condensate
LC
IP
F,T
Feedback control system: Valve is manipulated to
increase flow of steam to control tank temperature
Closed-loop process: Controller and process are
interconnected
Feedback Control
Control Objective:
 maintain a certain outlet temperature and tank
level
Feedback Control:
 temperature is measured using a thermocouple
 level is measured using differential pressure
probes
 undesirable temperature triggers a change in
supply steam pressure
 fluctuations in level trigger a change in outlet
flow
Note:
 level and temperature information is measured
at outlet of process/ changes result from inlet
flow or temperature disturbances
 inlet flow changes MUST affect process before
an adjustment is made
Examples
Feedback Control:
 requires sensors and actuators
e.g. Temperature Control Loop
Tin, F
Controller
+
Valve
e
T
C
TR -
A
P
Tank
M
Thermocouple
Controller:
 software component implements math
 hardware component provides calibrated signal for
actuator
Actuator:
 physical (with dynamics) process triggered by
controller
 directly affects process
Sensor:
 monitors some property of system and transmits
signal back to controller
Closed-loop Processes

Study of process dynamics focused on uncontrolled
or Open-loop processes
 Observe process behavior as a result of specific input
signals
U(s)
Gp
Y(s)
In process control, we are concerned with the
dynamic behavior of a controlled or Closed-loop
D(s)
process
process
controller
actuator
R(s)+
+ Y(s)
+
Gc
Gp
Gv

sensor
Gm
 Controller is dynamic system that interacts with the
process and the process hardware to yield a specific
behaviour
Closed-Loop Transfer Function
Block Diagram of Closed-Loop Process
D(s)
controller
R(s)+
-
Gc
actuator
Gv
process
Gp
sensor
Gm
Gp(s)
- Process Transfer Function
Gc(s)
- Controller Transfer Function
Gm(s)
- Sensor Transfer Function
Gv(s)
- Actuator Transfer Function
+
+ Y(s)
Closed-Loop Transfer Function
For control, we need to identify closed-loop
dynamics due to:
- Setpoint changes
- Disturbances
Servo
Regulatory
1. Closed-Loop Servo Response
 transfer function relating Y(s) and R(s) when
D(s)=0
Y ( s)  G p ( s)V ( s)
Y ( s)  G p ( s)Gv ( s)U ( s)
Y ( s)  G p ( s)Gv ( s)Gc ( s) E ( s)
Y ( s)  G p ( s)Gv ( s)Gc( s)  R( s) - Ym ( s)
Y ( S )  G p ( s)Gv ( s)Gc ( s) R( s) - Gm ( s)Y ( s)
 Isolate Y(s)
Y ( s) 
G p ( s)Gv ( s)Gc ( s)
1 + G p ( s)Gv ( s)Gc ( s)Gm ( s)
R ( s)
Closed-Loop Transfer Function
2. Closed-loop Regulatory Response
 Transfer Function relating D(s) to Y(s) at
R(s)=0
Y ( s)  D( s) + G p ( s)V ( s)
Y ( s)  D( s) + G p ( s)Gv ( s)U ( s)
Y ( s)  D( s) + G p ( s)Gv ( s)Gc ( s) E ( s)
Y ( s)  D( s) + G p ( s)Gv ( s)Gc ( s)0 - Ym ( s)
Y ( s)  D( s) + G p ( s)Gv ( s)Gc ( s)0 - Gm ( s)Y ( s)
 Isolating Y(s)
Y ( s) 
1
D( s)
1 + G p ( s)Gv ( s)Gc ( s)Gm ( s)
Closed-loop Transfer Function
2. Regulatory Response with Disturbance
Dynamics
Gd ( s)
Y ( s) 
D( s)
1 + G p ( s)Gv ( s)Gc ( s)Gm ( s)
Gd(s)
Disturbance (or load) transfer
function
3. Overall Closed-Loop Transfer Function
Servo
Y ( s) 
G p ( s)Gv ( s)Gc ( s)
1 + G p ( s)Gv ( s)Gc ( s)Gm ( s)
R( s) +
Gd ( s)
D( s)
1 + G p ( s)Gv ( s)Gc ( s)Gm ( s)
Regulatory
PID Controllers
The acronym PID stands for:
P
I
D
- Proportional
- Integral
- Derivative
PID Controllers:
greater than 90% of all control
implementations
dates back to the 1930s
very well studied and understood
optimal structure for first and second order
processes (given some assumptions)
always first choice when designing a
control system
PID controller equation:

1 t
de 
u(t )  Kc e(t ) +  e( )d +  D  + u R
I 0
dt 

PID Control
PID Control Equation
Proportional
Action
Derivative
Action

1 t
de 
u(t )  Kc e(t ) +  e( )d +  D  + u R
I 0
dt 

Integral
Action
Controller
Bias
PID Controller Parameters
Kc
I
D
uR
Proportional gain
Integral Time Constant
Derivative Time Constant
Controller Bias
PID Control
PID Controller Transfer Function


1
u(t ) - u R   U  ( s)  Kc  1 +
+  D s E ( s)
 Is

or:
I
U  ( s)   P + + Ds E ( s)


s
Note:
 numerator of PID transfer function cancels
second order dynamics
 denominator provides integration to remove
possibility of steady-state errors
PID Control
Controller Transfer Function:


1
Gc ( s)  Kc  1 +
+  D s
 Is

or,
I

Gc ( s)   P + + Ds


s
Note:
Many variations of this controller exist
Easily implemented in SIMULINK
each mode (or action) of controller is better
studied individually
Proportional Feedback
Form:
u(t ) - uR  Kce(t )
Transfer function:
or,
U '( s)  Kc E ( s)
Gc ( s)  Kc
Closed-loop form:
Y ( s) 
G p ( s)Gv ( s) Kc
1 + G p ( s)Gv ( s) KcGm ( s)
R( s) +
1
D( s)
1 + G p ( s)Gv ( s) KcGm ( s)
Proportional Feedback
Example:
Given first order process:
G p ( s) 
Kp
s + 1
, Gv ( s)  1, Gm ( s)  1
for P-only feedback closed-loop dynamics:
K p Kc
Y(s) 
1 + K p Kc




 s + 1
 1 + K p Kc 
R (s)



1

 s +
 1 + K p Kc  1 + K p Kc
+
D(s)




 s + 1
 1 + K p Kc 
Closed-Loop
Time Constant
Proportional Feedback
Final response:
K p Kc
1
lim yservo ( t ) 
, lim y reg ( t ) 
1
+
K
K
1 + K p Kc
t 
p c t 
Note:
 for “zero offset response” we require
lim yservo (t )  1,
t 
Tracking Error
lim yreg (t )  0
t 
Disturbance rejection
 Possible to eliminate offset with P-only
feedback (requires infinite controller gain)
 Need different control action to eliminate offset
(integral)
Proportional Feedback
Servo dynamics of a first order process under
proportional feedback
1
10.0
0.9
5.0
0.8
y(t)/KM
0.7
0.6
1.0
0.5
0.4
0.5
0.3
0.2
Kc
0.1
0
0
1
2
3
4
5
t/
6
7
8
0.01
9
- increasing controller gain eliminates off-set
10
Proportional Feedback
High-order process
e.g. second order underdamped process
1.5
y(t)/KM
1
5.0
2.5
1.0
0.5
0.5
0.01
0
0
5
10
15
20
25
 increasing controller gain reduces offset, speeds
response and increases oscillation
Proportional Feedback
Important points:
 proportional feedback does not change the order
of the system
started with a first order process
closed-loop process also first order
order of characteristic polynomial is
invariant under proportional feedback
 speed of response of closed-loop process is
directly affected by controller gain
increasing controller gain reduces the
closed-loop time constant
 in general, proportional feedback
reduces (does not eliminate) offset
speeds up response
for oscillatory processes, makes closedloop process more oscillatory
Integral Control
Integrator is included to eliminate offset
 provides reset action
 usually added to a proportional controller to
produce a PI controller
PID controller with derivative action turned
off
PI is the most widely used controller in
industry
optimal structure for first order processes
PI controller form

1
u ( t )  Kc  e ( t ) +
I

t

e
(

)
d


 + uR

0
Transfer function model

1 
U  ( s)  K c  1 +
 E ( s)
  I s
PI Feedback
Closed-loop response
  s + 1
G p ( s)Gv ( s) Kc  I

 Is 
Y ( s) 
R ( s) +
  I s + 1
1 + G p ( s)Gv ( s) Kc 
 Gm ( s)
 Is 
1
  I s + 1
1 + G p ( s)Gv ( s) Kc 
 Gm ( s)
 Is 
 more complex expression
 degree of denominator is increased by one
D( s)
PI Feedback
Example
PI control of a first order process
G p ( s) 
Kp
s + 1
, Gv ( s)  1, Gm ( s)  1
Closed-loop response
Y ( s) 
Is+1
  I   2  1 + Kc K p 

s +
 I s + 1
 Kc K p 
 Kc K p 
R( s) +
  I  2   I 

s +
s
 Kc K p 
 Kc K p 
D( s)
  I   2  1 + Kc K p 

s +
 I s + 1
 Kc K p 
 Kc K p 
Note:
offset is removed
closed-loop is second order
PI Feedback
Example (contd)
effect of integral time constant and controller gain
on closed-loop dynamics
natural period of oscillation
 cl 
 I
Kc K p
damping coefficient
1 K p

2 
 Kc K p + 1

Kc I 
 Kc K p 
 integral time constant and controller gain can
induce oscillation and change the period of
oscillation
PI Feedback
Effect of integral time constant on servo
dynamics
1.8
0.01
Kc=1
1.6
1.4
0.1
y(t)/KM
1.2
0.5
1
1.0
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
6
7
8
9
10
PI Feedback
Effect of controller gain
10.0
5.0
1
0.9
1.0
0.5
0.8
y(t)/KM
0.7
0.6
0.1
0.5
0.4
0.3
0.2
I=1
0.1
0
0
1
2
3
4
5
6
7
8
 affects speed of response
 increasing gain eliminates offset quicker
9
10
PI Feedback
Effect of integral action of regulatory
response
0.4
0.35
0.3
y(t)/KM
0.25
0.2
0.15
0.1
0.05
0
-0.05
-0.1
0
1
2
3
4
5
6
7
8
9
10
 reducing integral time constant removes effect
of disturbances
 makes behavior more oscillatory
PI Feedback
Important points:
 integral action increases order of the system in
closed-loop
 PI controller has two tuning parameters that can
independently affect
speed of response
final response (offset)
 integral action eliminates offset
 integral action
should be small compared to proportional
action
tuned to slowly eliminate offset
can increase or cause oscillation
can be de-stabilizing
Derivative Action
Derivative of error signal
 Used to compensate for trends in output
measure of speed of error signal change
provides predictive or anticipatory action
 P and I modes only response to past and current
errors
 Derivative mode has the form
D
de
 D
Kc
dt
if error is increasing, decrease control
action
if error is decreasing, decrease control
action
 Always implemented in PID form

1 t
de 
u(t )  Kc e(t ) +  e( )d +  D  + u R
I 0
dt 

PID Feedback
Transfer Function


1
U  ( s )  Kc  1 +
+  D s E ( s)
 Is

Closed-loop Transfer Function
   s2 +  s + 1
I

G p ( s)Gv ( s) Kc  D I

Is


Y ( s) 
R( s) +
2
   s +  s + 1
I
 Gm ( s)
1 + G p ( s)Gv ( s) Kc  D I

Is


1
   s +  s + 1
I
 Gm ( s)
1 + G p ( s)Gv ( s) Kc  D I

Is


2
 Slightly more complicated than PI form
D( s)
PID Feedback
Example:
PID Control of a first order process
G p ( s) 
Kp
s + 1
, Gv ( s)  1, Gm ( s)  1
Closed-loop transfer function
Y ( s) 
 D I s2 +  I s + 1
  I
 2  1 + Kc K p 

  I s + 1
+  D I  s + 
 Kc K p

 Kc K p 
R( s) +
  I  2   I 

 s + 
 s
 Kc K p 
 Kc K p 
D( s)
  I
 2  1 + Kc K p 

  I s + 1
+  D I  s + 
 Kc K p

 Kc K p 
PID Feedback
Effect of derivative action on servo dynamics
1.6
1.4
y(t)/KM
1.2
1
0.1
0.8
0.5
0.6
1.0 2.0
0.4
0.2
0
0
1
2
3
4
5
6
7
8
9
10
PID Feedback
Effect of derivative action on regulatory
response
0.25
0.2
0.15
0.1
0.05
0.1
2.0
1.0
0.5
0
-0.05
-0.1
0
1
2
3
4
5
6
7
8
9
10
 increasing derivative action reduces impact of
disturbances on control variable
 slows down servo response and affects
oscillation of process
Derivative Action
Important Points:
 Characteristic polynomial is similar to PI
 derivative action does not increase the order of
the system
 adding derivative action affects the period of
oscillation of the process
good for disturbance rejection
poor for tracking
 the PID controller has three tuning parameters
and can independently affect,
speed of response
final response (offset)
servo and regulatory response
 derivative action
should be small compared to integral action
has a stabilizing influence
difficult to use for noisy signals
usually modified in practical
implementation
Closed-loop Stability
Every control problem involves a
consideration of closed-loop stability
General concepts:
BIBO Stability:
“ An (unconstrained) linear system is said to
be stable if the output response is bounded
for all bounded inputs. Otherwise it is
unstable.”
Comments:
Stability is much easier to prove than
unstability
This is just one type of stability
Closed-loop Stability
Closed-loop dynamics
Y ( s) 
GcGv G p
1 + GcGv G pGm
Y * ( s) +
1
D( s)
1 + GcGv G pGm
GOL
 if GOL is a rational function then the closed-loop
transfer functions are rational functions and
take the form
r
a
+
a
s
+

+
a
s
r ( s)
0
1
r
G ( s) 

q ( s) b0 + b1s++b s
and factor as
G ( s)  K
( a1 s + 1) ( ar s + 1)
(1s + 1) ( s + 1)
Closed-loop stability
General Stability criterion:
“ A closed-loop feedback control system is stable
if and only if all roots of the characteristic
polynomial are negative or have negative real
parts. Otherwise, the system is unstable.”
 Unstable region is the right half plane of the
complex plane.
 Valid for any linear systems.
 Underlying system is almost always nonlinear
so stability holds only locally. Moving away
from the point of linearization may cause
instability.
Closed-loop Stability
Problem reduces to finding roots of a
polynomial
Easy (1990s) way : MATLAB function ROOTS
Traditional:
1. Routh array:
 Test for positivity of roots of a
polynomial
2. Direct substitution
 Complex axis separates stable and
unstable regions
 Find controller gain that yields purely
complex roots
3. Root locus diagram
 Vary location of poles as controller
gain is varied
 Of limited use
Closed-loop stability
Routh array for a polynomial equation
an sn + an-1sn-1++ a1s + a0  0
is
an - 2
an - 4 
an - 1 an - 3
b1
b2
an - 5 
an
1
2
3
4
c1


n+1
z1
c2
b3


where
a a
-a a
a a
-a a
b1  n -1 n - 2 n - 3 n , b2  n -1 n - 4 n -5 n ,
an -1
an -1
ba
-b a
ba
-b a
c1  1 n - 3 2 n -1 , c2  1 n -5 3 n -1 ,
b1
b1
Elements of left column must be positive to
have roots with negative real parts
Example: Routh Array
Characteristic polynomial
2.36s5 + 149
. s4 - 0.58s3 + 121
. s2 + 0.42 s + 0.78  0
Polynomial Coefficients
a5  2.36, a4  149
. , a3  -058
. , a2  121
. , a1  0.42, a0  0.78
Routh Array
a5 (2.36)
a4 (149
. )
a3 ( -0.58) a1(0.42)
a2 (121
. )
b1( -2.50) b2 ( -0.82)
c1(0.72)
c2 (0.78)
d1(189
. )
d2 (0)
e1(0.78)
 Closed-loop system is unstable
a0 (0.78)
b3 (0)
Direct Substitution

Technique to find gain value that de-stabilizes the
system.

Observation:
Process becomes unstable when poles appear on
right half plane
Find value of Kc that yields purely complex
poles

Strategy:
 Start with characteristic polynomial
1 + KcGv ( s)G p ( s)Gm ( s)  1 + Kc
 Write characteristic equation:
q ( s) + Kcr ( s)  0
 Substitute for complex pole (s=j)
q ( j ) + Kcr ( j )  0
 Solve for Kc and 
r ( s)
q ( s)
Example: Direct Substitution
Characteristic equation
1 + Kc
s+1
3
2
s + 0.5s - 0.5s - 0.75
0
s3 + 0.5s2 - 0.5s - 0.75 + Kc s + Kc  0
s3 + 0.5s2 + ( Kc - 0.5) s + ( Kc - 0.75)  0
Substitution for s=j
( j ) 3 + 0.5( j ) 2 + ( Kc - 0.5) j + ( Kc - 0.75)  0
- j 3 - 0.5 2 + ( Kc - 0.5) j + ( Kc - 0.75)  0
Real Part
Complex Part
-0.5 2 + Kc - 0.75  0
( Kc - 0.5) -  3  0
 Kc  0.5 2 + 0.75  (0.5 2 + 0.75 - 0.5) -  3  0
 -0.5 2 + 0.25  0
    2 / 2, Kc  1
 System is unstable if Kc > 1
Root Locus Diagram
Old method that consists in plotting poles of
characteristic polynomial as controller gain is
changed
e.g.
s3 + 0.5s2 + ( Kc - 0.5) s + ( Kc - 0.75)  0
1.5
1
Kc-0
1
Imaginary Axis
0.5
0
-0.5
Kc-0
-1
-1.5
-1.5
-1
-0.5
0
Real Axis
0.5
1
1.5
Stability and Performance

Given plant model, we assume a stable closed-loop
system can be designed

Once stability is achieved - need to consider
performance of closed-loop process - stability is not
enough

All poles of closed-loop transfer function have
negative real parts - can we place these poles to get
a “good” performance
C
Space of all
Controllers
S
P
S: Stabilizing Controllers for a given plant
P: Controllers that meet performance
Controller Tuning
Can be achieved by
 Direct synthesis : Specify servo transfer
function required and calculate required
controller - assume plant = model
 Internal Model Control: Morari et al. (86)
Similar to direct synthesis except that plant and
plant model are concerned
 Tuning relations:
Cohen-Coon - 1/4 decay ratio
designs based on ISE, IAE and ITAE
 Frequency response techniques
Bode criterion
Nyquist criterion
 Field tuning and re-tuning
Direct Synthesis
From closed-loop transfer function
GcG p
C

R 1 + GcG p
Isolate Gc
1  C R
Gc 
G p  1 - C



R
For a desired trajectory (C/R)d and plant
model Gpm, controller is given by
( )
( )
 C

1 
Rd 
Gc 

G pm  1 - C

R d
 not necessarily PID form
 inverse of process model to yield pole-zero
cancellation (often inexact because of process
approximation)
 used with care with unstable process or
processes with RHP zeroes
Direct Synthesis
1. Perfect Control
 C  1
 
 R d
 cannot be achieved, requires infinite gain
2. Closed-loop process with finite settling
time
 C  1
 
 R d  cs + 1
 For 1st order Gp, it leads to PI control
 For 2nd order, get PID control
3. Processes with delay 
e - c s
 C
  
 R d  cs + 1
 requires c  
 again, 1st order leads to PI control
 2nd order leads to PID control
IMC Controller Tuning
D
R
+
Gc*
+
Gp
+
C
-
Gpm
+
Closed-loop transfer function
C
Gc*G p
1 + Gc* (G p - G pm )
R+
1 - Gc*G p
1 + Gc* (G p - G pm )
In terms of implemented controller, Gc
Gc 
Gc*
1 - Gc*G pm
D
IMC Controller Tuning
1. Process model factored into two parts
+
G pm  G pm
G -pm
where + contains dead-time and RHP
G pm
zeros,
steady-state gain scaled to 1.
2. Controller
Gc* 
1
G -pm
f
where f is the IMC filter
f 
1
( c s + 1) r
 based on pole-zero cancellation
 not recommended for open-loop unstable
processes
 very similar to direct synthesis
Example
PID Design using IMC and Direct synthesis
for the process
e -9 s 0.3
G p ( s) 
30s + 1
Process parameters: K=0.3, 30, 9
1. IMC Design: Kc=6.97, I=34.5, d=3.93
 Filter
f 
1
12 s + 1
2. Direct Synthesis: Kc=4.76, I=30
 Servo Transfer function
-9 s
 C  e
 
 R  d 12 s + 1
Example
Result: Servo Response
 IMC and direct synthesis give roughly same
results
25
IMC
20
Direct
Synthesis
15
y(t)
10
5
0
0
50
100
150
200
250
t
 IMC not as good due to Pade approximation
300
Example
Result: Regulatory response
40
35
30
y(t)
25
Direct Synthesis
20
IMC
15
0
50
100
150
200
250
t
 Direct synthesis rejects disturbance more
rapidly (marginally)
300
Tuning Relations
Process reaction curve method:
 based on approximation of process using first
order plus delay model
D(s)
1/s
Y*(s)
Gp
Gc
U(s)
Gs
Ym(s)
Manuel Control
1. Step in U is introduced
2. Observe behavior ym(t)
3. Fit a first order plus dead time model
Ke - s
Ym ( s) 
 s+1
Y(s)
Tuning Relations
Process response
1.2
1
0.8
0.6
KM
0.4
0.2
0

-0.2
0
1

2
3
4
5
6
7
8
4. Obtain tuning from tuning correlations
Ziegler-Nichols
Cohen-Coon
ISE, IAE or ITAE optimal tuning relations
Ziegler-Nichols Tunings
Controller
P-only
Kc
Ti
PI
(0.9 / K p )( /  )
3.3
PID
(1.2 / K p )( /  )
(1 / K p )( /  )
2.0
Td
0.5
- Note presence of inverse of process gain in controller
gain
- Introduction of integral action requires reduction in
controller gain
- Increase gain when derivation action is introduced
Example:
PI:
PID:
e -9 s 0.3
G p ( s) 
30s + 1
Kc= 10
Kc= 13.33
I=4.5
I=29.97
I=18
Example
Ziegler-Nichols Tunings: Servo response
50
Z-N PI
45
Z-N PID
40
y(t)
35
Direct Synthesis
30
25
20
0
50
100
150
t
200
250
300
Example
Regulatory Response
40
35
30
Direct Synthesis
25
Z-N PI
20
Z-N PID
15
10
0
50
100
150
200
250
Z-N tuning
 Oscillatory with considerable overshoot
 Tends to be conservative
300
Cohen-Coon Tuning Relations
Designed to achieve 1/4 decay ratio
 fast decrease in amplitude of oscillation
Controller
P-only
Kc
Ti
PI
(1 / K p )( /  )[ 0.9 +  / 12 ]
 [30 + 3( /  )]
9 + 20( /  )
PID
(1 / K p )( /  )[
(1 / K p )( /  )[1 +  / 3 ]
3 + 16
]
12
 [32 + 6( /  )]
13 + 8( /  )
Example:
PI: Kc=10.27
Kc=15.64
d=3.10
I=18.54
I=19.75
Td
4
11 + 2( /  )
Tuning relations
Cohen-coon: Servo
55
50
C-C PID
45
40
35
C-C PI
30
25
20
0
50
100
150
200
250
 More aggressive/ Higher controller gains
 Undesirable response for most cases
300
Tuning Relations
Cohen-Coon: Regulatory
40
35
30
C-C PI
25
y(t)
20
C-C PID
15
10
5
0
50
100
150
 Highly oscillatory t
 Very aggressive
200
250
300
Integral Error Relations
1. Integral of absolute error (IAE)

IAE   e(t ) dt
0
2. Integral of squared error (ISE)

ISE   e(t ) 2 dt
penalizes large errors
0
3. Integral of time-weighted absolute error
(ITAE)

ITAE   t e(t ) dt
0
penalizes errors that persist
 ITAE is most conservative
 ITAE is preferred
ITAE Relations
Choose Kc, I and d that minimize the ITAE:
 For a first order plus dead time model, solve
for:
 ITAE
 ITAE
 ITAE
 0,
 0,
0
 Kc
 I
 d
 Design for Load and Setpoint changes yield
different ITAE optimum
Type of
Input
Load
Type of
Controller
PI
Load
PID
Set point
PI
Set point
PID
Mode
A
B
P
I
P
I
D
P
I
P
I
D
0.859
0.674
1.357
0.842
0.381
0.586
1.03
0.965
0.796
0.308
-0.977
-0.680
-0.947
-0.738
0.995
-0.916
-0.165
-0.85
-0.1465
0.929
ITAE Relations
From table, we get
Load Settings:
( )
Y  A 
B
 KKc      d 
I
Setpoint Settings:
( )
( )
B

Y  A   KKc   d  ,    A + B  
I
Example
0.3e -9 s
Gs 
, GL  1
30s + 1
ITAE Relations
Example (contd)
Setpoint Settings
( )
KKc  0.965 9 30
- 0.85
 2.6852
Kc  2.6852 K  2.6852 0.3  8.95

( )
9

0
.
796
01465
.
30  0.7520
I
 I   0.7520  30 0.7520  39.89
 d  0.308 9 0.929  01006
.

30
 d  01006
.
  3.0194
( )
Load Settings:
- 0.947
9
KKc  1357
.
 4.2437
30
( )
Kc  4.2437

K
 4.2437
0.3
 14.15
- 0.738
9

0
.
842
 2.0474
30
I
( )
 I   2.0474  30 2.0474  14.65
 d  0.381 9 0.995  01150
.

30
 d  01150
.
  3.4497
( )
ITAE Relations
Servo Response
60
55
ITAE(Load)
50
45
40
ITAE(Setpoint)
35
30
25
20
0
50
100
150
200
250
300
 design for load changes yields large overshoots
for set-point changes
ITAE Relations
Regulatory response
40
35
30
ITAE(Setpoint)
25
20
15
10
5
0
ITAE(Load)
0
50
100
150
200
250
 Tuning relations are based GL=Gp
 Method does not apply to the process
 Set-point design has a good performance for
this case
300
Tuning Relations

In all correlations, controller gain should be
inversely proportional to process gain

Controller gain is reduced when derivative action is
introduced



Controller gain is reduced as   increases
Integral time constant and derivative constant
should increase as   increases

In general, d   0.25
I

Ziegler-Nichols and Cohen-Coon tuning relations
yield aggressive control with oscillatory response
(requires detuning)

ITAE provides conservative performance (not
aggressive)
CHE 446
Process Dynamics and Control
Frequency Response of
Linear Control Systems
First order Process
Response to a sinusoidal input signal

lim 
t 
Y ( s) 
KP A
-1
1+  
2
10
14
2
sin(t + f )
2
1.5
1
AR
0.5
0
-0.5
f
-1
-1.5
0
2
4
6
8
12
16
18
20
t/
Recall: Sinusoidal input Asin(t) yields
sinusoidal output caharacterized by AR and
f
First order Processes
Bode Plots
10
0
AR/Kp
High Frequency
10
10
Asymptote
Corner Frequency
-1
-2
10
-2
-1
10
10
0
p
10
1
10
2
0
-20
f
-40
-60
-80
-100
-2
10
-1
10
Amplitude Ratio
K
AR 
1 +  2 2
10
p
0
10
1
Phase Shift
f  - tan-1( )
10
2
Second Order Process

Sinusoidal Response
Y ( s) 
y (t ) 
 2 s2 + 2s + 1 s2 +  2
Kp A

1 - ( )
where
A
Kp

2 2
sin(t + f )
+ (2 )2
 2
1
f  - tan 


2
1 - ( ) 
ARn 

1 - ( )
1

2 2
+ (2 )2
Second Order Processes
Bode Plot
1
10
AR
0.1
Amplitude reaches
a maximum at
resonance frequency
0
10
=1
-1
10
-1
10
0
1
10
10
0
f
-50
=1
-100
-150
0.1
-1
10
0
10

1
10
Frequency Response
Q: Do we “have to” take the Laplace inverse
to compute the AR and phase shift of a 1st
or 2nd order process?
No
Q: Does this generalize to all transfer function
models?
Yes
Study of transfer function model response to
sinusoidal inputs is called “Frequency
Domain Response” of linear processes.
Frequency Response
Some facts for complex number theory:
i) For a complex number:
w  a + bj
a  Re( w), b  Im( w)
Im
b
w

a
It follows that
a  w cos( ), b  w sin( )
2
w  Re( w) + Im( w)
such that
Re
w  w e j
2
where
 Im( w) 

 Re( w) 
  arg( w)  tan -1
Frequency Response
Some facts:
ii) Let z=a-bj and w= a+bj then
w  z and arg( z)  - arg( w)
iii) For a first order process
G( s) 
Let s=j
Kp
 s+1
Kp
K p
(1 -  j )
G ( j ) 

j
2
2
2
2
 j + 1 (1 -  j ) 1 +   1 +  
such that
Kp
G ( j ) 
(  AR)
1 +  2 2
Kp
arg(G ( j ))  - tan -1 ( )(  Phase Lag)
Frequency Response
Main Result:
The response of any linear process G(s) to
a sinusoidal input is a sinusoidal.
The amplitude ratio of the resulting signal
is given by the Modulus of the transfer
function model expressed in the frequency
domain, G(i).
The Phase Shift is given by the argument of
the transfer function model in the
frequency domain.
i.e.
AR  G ( j )  Re(G ( j )) 2 + Im(G ( j )) 2
-1 Im(G ( j )) 
Phase Angle    tan 

 Re(G ( j )) 
Frequency Response
For a general transfer function
r ( s) e - s ( s - z1)( s - zm )
G ( s) 

q ( s)
( s - p1)( s - pn )
Frequency Response summarized by
G ( j )  G ( j ) e j
where
is the modulus of G(j) and 
( j )
is theGargument
of G(j)
Note: Substitute for s=j in the transfer
function.
Frequency Response
The facts:
For any linear process we can calculate the
amplitude ratio and phase shift by:
i) Letting s=j in the transfer functionG(s)
ii) G(j) is a complex number. Its modulus is the
amplitude ratio of the process and its argument
is the phase shift.
iii) As , the frequency, is varied that G(j) gives
a trace (or a curve) in the complex plane.
iv) The effect of the frequency, , on the process
is the frequency response of the process.
Frequency Response
Examples:
1. Pure Capacitive Process G(s)=1/s
K  - j 
K
G ( j ) 

- j
j  - j 

AR 
K
-1  - K /  
, f  tan 


0
2. Dead Time G(s)=e-s
G ( j )  e - j
AR  1, f  -


2
Frequency Response
Examples:
3. n process in series
G( s)  G1( s)Gn ( s)
Frequency response of G(s)
G ( j )  G1 ( j ) Gn ( j )
 G1 ( j ) e jf1  Gn ( j ) e jfn
therefore
n
AR  G ( j )   Gi ( j )
i 1
n
n
i 1
i 1
f  arg(G ( j ))   arg(Gi ( j ))   fi
Frequency Response
Examples.
4. n first order processes in series
K1
Kn
G( s) 

1s + 1  n s + 1
AR 
K1
1 +  12 2

Kn
1 +  n2 2
f  - tan -1( 1 )-- tan -1( n )
5. First order plus delay
G ( s) 
AR 
K p (1)
1 +  2 2
K p e - s
 s+1
, f  - tan -1( ) - 
Frequency Response

To study frequency response, we use two
types of graphical representations
1. The Bode Plot:
Plot of AR vs.  on loglog scale
Plot of f vs.  on semilog scale
2. The Nyquist Plot:
Plot of the trace of G(j) in the
complex plane

Plots lead to effective stability criteria and
frequency-based design methods
Bode Plot
Pure Capacitive Process
AR 
K

-

2
2
AR
10
1
10
0
10
-2
10
-1
10
0
10
Phase Angle
-89
-89.5
-90
-90.5
-91
-2
10
-1
10
Frequency (rad/sec)
0
10
Bode Plot
G( s)  G1( s)G2 ( s)G3 ( s)
G1( s) 
1
1
1
, G2 ( s) 
, G3 ( s) 
10s + 1
5s + 1
s+1
0
10
G3
G2
-2
10
G1
-4
10
-4
10
-3
10
-2
10
-1
10
0
10
G
1
10
0
-100
-200
-300
-4
10
G ( j ) 
-3
10
-2
10
-1
10
0
10
1
10
1
(1 + 102  2 )(1 + 52  2 )(1 + 12  2 )
  - tan -1(10 ) - tan -1(5 ) - tan -1( )
Bode Plot
G ( s)  e - s
G( j )  1,   - 
Example: Effect of dead-time
Gd ( s)  e -2 sG1( s)G2 ( s)G3 ( s)
0
10
-2
10
G=Gd
-4
10
-4
10
-3
10
-2
10
-1
0
10
10
1
10
0
-100
-200
Gd
-300
-4
10
-3
10
-2
10
-1
10
G
0
10
1
10
Nyquist Plot
Plot of G(j) in the complex plane as  is
varied
Relation to Bode plot
 AR is distance of G(j) for the origin
 Phase angle,  , is the angle from the Real
positive axis
Example First order process (K=1, =1)

G( j )
Nyquist Plot
Dead-time

Second Order
1
1
Nyquist Plot
Third Order
1
G ( s)  3
s + 3s2 + 3s + 1
Effect of dead-time (second order process)
G ( s) 
1
s2 + 3s + 1
, Gd ( s)  e -2 s
Che 446: Process Dynamics and
Control
Frequency Domain
Controller Design
PI Controller
AR  Kc
1
 I
2
2
+1
  tan -1( -1 /  I )
3
10
2
10
AR101
0
10
-3
10
-2
10
-1
10
0
10
1
10
0
-20
-40
 -60
-80
-100
-3
10
-2
10
-1
10

0
10
1
10
PID Controller
2

1 
AR  Kc   D  +1

 I 
-1 
1 
  tan   D 

 I 
3
10
2
10
AR
1
10
0
10
-3
10
-2
10
-1
10
0
10
1
10
100
50

0
-50
-100
-3
10
-2
10
-1
10

0
10
1
10
Bode Stability Criterion
Consider open-loop control system
D(s)
+
R(s) +
-
Gp
Gc
U(s)
+
Gs
Ym(s)
Open-loop Response to R(s)
1. Introduce sinusoidal input in setpoint (D(s)=0)
and observe sinusoidal output
2. Fix gain such AR=1 and input frequency such
that f=-180
3. At same time, connect close the loop and set
R(s)=0
Q: What happens if AR>1?
Y(s)
Bode Stability Criterion
“A closed-loop system is unstable if the
frequency of the response of the open-loop
GOL has an amplitude ratio greater than
one at the critical frequency. Otherwise it
is stable. “
Strategy:
1. Solve for  in
arg(GOL ( j ))  -
2. Calculate AR
AR  GOL ( j )
Bode Stability Criterion
To check for stability:
1. Compute open-loop transfer function
2. Solve for  in f=-
3. Evaluate AR at 
4. If AR>1 then process is unstable
Find ultimate gain:
1. Compute open-loop transfer function
without controller gain
2. Solve for  in f=-
3. Evaluate AR at 
4. Let
1
Kcu 
AR
Bode Criterion
Consider the transfer function and controller
.s
5e - 01
G( s) 
( s + 1)(0.5s + 1)
1 
Gc ( s)  0.4 1 +

 01
. s
- Open-loop transfer function
.s
5e -01
1 

GOL ( s) 
0.4 1 +

( s + 1)(0.5s + 1)  01
. s
- Amplitude ratio and phase shift
AR 
5
1+ 
1
2
1 + 0.25
2
0.4 1 +
1
0.01 2
1 
f  -01
.  - tan ( ) - tan (0.5 ) - tan 

 01
. 
-1
-1
- At =1.4128, f=-, AR=6.746
-1 
Ziegler-Nichols Tuning
Closed-loop tuning relation
 With P-only, vary controller gain until system
(initially stable) starts to oscillate.
 Frequency of oscillation is c,
 Ultimate gain, Ku, is 1/M where M is the
amplitude of the open-loop system
 Ultimate Period
Pu 
2
c
Ziegler-Nichols Tunings
P
PI
PID
Ku/2
Ku/2.2
Ku/1.7
Pu/1.2
Pu/2
Pu/8
Nyquist Stability Criterion
“If N is the number of times that the
Nyquist plot encircles the point (-1,0) in
the complex plane in the clockwise
direction, and P is the number of openloop poles of GOL that lie in the right-half
plane, then Z=N+P is the number of
unstable roots of the closed-loop
characteristic equation.”
Strategy
1. Substitute s=j in GOL(s)
2. Plot GOL(j) in the complex plane
3. Count encirclements of (-1,0) in the
clockwise direction
Nyquist Criterion
Consider the transfer function
5e -0.1s
G ( s) 
( s + 1)(0.5s + 1)
and the PI controller
1 

Gc ( s)  0.4 1 +

 01
. s
Stability Considerations

Control is about stability

Considered exponential stability of
controlled processes using:





Routh criterion
Direct Substitution
Polynomial
Root Locus
(no dead-time)
Bode Criterion (Restriction on phse angle)
Nyquist Criterion

Nyquist is most general but sometimes
difficult to interpret

Roots, Bode and Nyquist all in MATLAB

MAPLE is recommended for some
applications.
CHE 446
Process Dynamics and
Control
Advanced Control Techniques:
1. Feedforward Control
Feedforward Control
Feedback control systems have the general
form:
D(s)
UR(s)
R(s) +
Gc
+ +
GD
Gv
Gp
U(s)
+
+
Y(s)
Gs
Ym(s)
where UR(s) is an input bias term.

Feedback controllers
 output of process must change before any action
is taken
 disturbances only compensated after they affect
the process
Feedforward Control

Assume that D(s)
 can be measured before it affects the process
 effect of disturbance on process can be
described with a model GD(s)
Feedforward Control is possible.
Feedforward
Controller
D(s)
Gf
GD
R(s) +
Gc
+ +
Gv
U(s)
Gp
+
+
Gs
Ym(s)
Feedback/Feedforward Controller
Structure
Y(s)
Feedforward Control
Heated Stirred Tank
F,Tin
TT
TT
TC1
Ps
Steam
Condensate
F,T
 Is this control configuration feedback or
feedforward?
 How can we use the inlet stream thermocouple
to regulate the inlet folow disturbances
 Will this become a feedforward or feedback
controller?
Feedforward Control
A suggestion:
TC2
TT
F,Tin
TT +
+
TC1
Ps
Steam
Condensate
How do we design TC2?
F,T
Feedforward Control
The feedforward controller:
D(s)
Gf
GD
UR(s)+ +
Gv
Gp
U(s)
+
+
Y(s)
Transfer Function
Y ( s)  GD ( s) D( s) + G P ( s)Gv ( s)U ( s)
Y ( s)  GD ( s) D( s) + G P ( s)Gv ( s)(U R ( s) + G f ( s) D( s))
Y ( s)  (GD ( s) + G p ( s)Gv ( s)G f ( s)) D( s) + G p ( s)Gv ( s)U R ( s)
Y ( s)  (GD ( s) + G p ( s)Gv ( s)G f ( s)) D( s) + YR ( s)
Tracking of YR requires that
GD ( s) + G p ( s)Gv ( s)G f ( s)  0
 G f ( s)  -
GD ( s)
G p ( s)Gv ( s)
Feedforward Control
Ideal feedforward controller:
G f ( s)  -
GD ( s)
G p ( s)Gv ( s)
 Exact cancellation requires perfect plant and
perfect disturbance models.
GD ( s) + G p ( s)Gv ( s)G f ( s)  0
 Feedforward controllers:
very sensitive to modeling errors
cannot handle unmeasured disturbances
cannot implement setpoint changes
 Need feedback control to make control system
more robust
Feedforward Control
Feedback/Feedforward Control
D(s)
Gf
GD
R(s) +
Gc
+ +
Gv
U(s)
Gp
+
+
Gs
Ym(s)
What is the impact of Gf on the closed-loop
performance of the feedback control system?
Y(s)
Feedforward Control
Regulatory transfer function of
feedforward/feedback loop
C ( s) GD ( s) + G f ( s)Gv ( s)G p ( s)

D( s) 1 + Gc ( s)Gv ( s)G p ( s)Gm ( s)
Perfect control requires that (as above)
GD ( s)
G f ( s)  Gv ( s)G p ( s)
Note:
 Feedforward controllers do not affect closedloop stability
 Feedforward controllers based on plant models
can be unrealizable (dead-time or RHP zeroes)
 Can be approximated by a lead-lag unit or pure
gain (rare)
G f ( s)  K f
(1s + 1)
( 2 s + 1)
KD
G f ( s)  Kv K p
Feedforward Control
Tuning: In absence of disturbance model
lead-lag approximation may be good
(1s + 1)
G f ( s)  K f
( 2 s + 1)
 Kf obtained from open-loop data
Kf  -
KD
Kv K p
- 1 and 2
from open-loop data
1   p ,  2   D
from heuristics
1
1
 0.5
 2.0
2
2
Trial-and-error
1 -  2  c
Feedforward Control
Example:
Plant:
10
(10s + 1)(5s + 1)( s + 1)
1
GD ( s) 
(2.5s + 1)( s + 1)
G p ( s) 
Plant Model:
10e -6s
e- s
G pm ( s) 
GDm ( s) 
10s + 1
2.5s + 1
Feedback Design from plant model: IMC PID
tunings
Kc  0.26,  I  13,  D  2.31
Feedforward Control
Possible Feedforward controllers:
1. From plant models:
e5s (10s + 1)
G f ( s)  10 (2.5s + 1)
Not realizable
2. Lead-lag unit
 1  10,  2  2.5
1
Kf  3. Feedforward gain10
controller:
Kf  -
1
10
Feedforward Control
For Controller 2 and 3
Disturbance Controller with Feedforward
1.2
1
0.8
.. - Gain Controller
0.6
-- - Lead-Lag Controller
- - No FF Controller
0.4
0.2
0
-0.2
-0.4
-0.6
0
20
40
60
80
100
120
140
160
180
200
 Some attenuation observed at first peak
 Difficult problem because disturbance dynamic
are much faster
Feedforward Control

Useful in manufacturing environments if
good models are available
 outdoor temperature dependencies can be
handle by gain feedforward controllers
 scheduling issues/ supply requirements can be
handled

Benefits are directly related to model
accuracy
 rely mainly on feedback control

Disturbances with different dynamics
always difficult to attenuate with PID
 may need advanced feedback control approach
(MPC, DMC, QDMC, H4-controllers, etc…)

Use process knowledge (and intuition)
CHE 446:
Process Dynamics and
Control
Advanced Control Techniques
2. Cascade Control
Cascade Control
Jacketed Reactor:
TT
F,Tin
TT
TC1
Ps
Steam
FT
Condensate
F,T
Conventional Feedback Loop:
 operate valve to control steam flow
 steam flow disturbances must propagate
through entire process to affect output
 does not take into account flow measurement
Cascade Control
Consider cascade control structure:
TT
F,Tin
TT
TC1
FC
FT
Ps
Steam
Condensate
F,T
Note:
 TC1 calculates setpoint cascaded to the flow
controller
 Flow controller attenuates the effect of steam
flow disturbances
Cascade Control
Cascade systems contain two feedback loops:
 Primary Loop
regulates part of the process having slower
dynamics
calculates setpoint for the secondary loop
e.g. outlet temperature controller for the
jacketed reactor
 Secondary Loop
regulates part of process having faster
dynamics
maintain secondary variable at the desired
target given by primary controller
e.g. steam flow control for the jacketed
reactor example
+
-
Gc1
+
-
Block Diagram
Gc2
Gm1
Gm2
Gv2
Gp2
+
D2
Gp1
D1
+
Cascade Control
Cascade Control
Closed-loop transfer function
1. Inner loop
G p2 Gv 2 Gc2
C2

 Gcl 2
R2 1 + G p2 Gv 2 Gc2 Gm2
2. Outer loop
G p1Gcl 2Gc1
C1

R1 1 + G p1Gcl 2Gc1Gm1
Characteristic equation
1 + G p1Gcl 2Gc1Gm1  0
1 + G p1
G p2Gv 2Gc2
1 + G p2Gv 2Gc2Gm2
Gc1Gm1  0
1 + G p2Gv 2Gc2Gm2 + G p1G p2Gv 2Gc2Gc1Gm1  0
Cascade Control
Stability of closed-loop process is governed by
1 + G p2Gv 2Gc2Gm2 + G p1G p2Gv 2Gc2Gc1Gm1  0
Example
G p1 
K p1
1s + 1
G p2 
1 + Kc 2
, Gc1  Kc1, Gv1  Gm1  1
K p2
 2s + 1
, Gc2  Kc2 , Gm2  1
K p2
 2s + 1
+ Kc1
K p2
K p1
 2 s + 1 1s + 1
0
(1s + 1)( 2 s + 1) + Kc2 K p2 (1s + 1) + Kc1K p2 K p1  0
1 2 s2 + (1 +  2 + Kc2 K p21) s + 1 + Kc2 K p2 +
Kc1K p2 K p1  0
Cascade Control
Design a cascade controller for the following
system:
1. Primary:
e -0.1s
G p1( s) 
, Gm1  1,
(0.5s + 1)( s + 1)

1 
Gc1  Kc1 1 +

  I s
2. Secondary:
1
, Gv 2  Gm2  1
01
. s+1
Gc2  Kc2
G p2 
Cascade Control
1. PI controller only
-0.1s
1
1
e
GOL1  Kc1 1 + 
 s 01
. s + 1 (0.5s + 1)( s + 1)
1
AR  Kc1 1 + 2

1
1
1
0.01 2 + 1 0.25 2 + 1  2 + 1
-1  1 
  - tan   - tan -1 (01
. )
 
- tan -1 (0.5 ) - tan -1 ( ) - 01
.
Critical frequency
c  2.99, AR  0178
.
Maximum gain
Kc1  561
.
Cascade Control
Bode Plots
AR


ln()
Cascade Control
2. Cascade Control
 Secondary loop
GOl 2  Kc2
1
01
. s+1
no critical frequency gain can be large
Let Kc2=10.
 Primary loop
10
-0.1s
1
e
. s+1
GOL1  Kc1 1 +  01
1 (0.5s + 1)( s + 1)
 s
1 + 10
01
. s+1
10
e -0.1s
 Kc1
.
11 s(0.5s + 1)( 01
s + 1)
11
Cascade Control
Closed-loop stability:
AR 10 1

Kc1 11 

1
1
2
2
1
+
0
.
25

01
.
1 +    2
 11
. 
-1  01
f  - - 01
.  - tan 
  - tan -1 (0.5 )
 11 
2
 Bode
 c  413
. , AR  0.0958
Maximum gain Kc1=10.44
 Secondary loop stabilizes the primary loop.
Cascade Control
Use cascade when:
 conventional feedback loop is too slow at
rejecting disturbances
 secondary measured variable is available which
responds to disturbances
has dynamics that are much faster than
those of the primary variable
can be affected by the manipulated variable
Implementation
 tune secondary loop first
 operation of two interacting controllers requires
more careful implementation
switching on and off
CHE 446
Process Dynamics and Control
Advanced Control Techniques
3. Dead-time Compensation
Dead-time Compensation
Consider feedback loop:
D
R
Gc
Gp
e-s
 Dead-time has a de-stabilizing effect on closedloop system
 Presence of dead-time requires detuning of
controller
 Need a way to compensate for dead-time
explicitly
C
Dead-time Compensation
Motivation
e - s
G ( s)  2
, 01
.    0.75
s + 3s + 2
1
Gc ( s)  4 1 + 
 s
0.75
0.5 0.25 0.1
Dead-time Compensation
Use plant model to predict deviation from
setpoint
D
R
Gc
Gp
e-s
Gpm
Result:
 Removes the de-stabilizing effect of dead-time
Problem:
 Cannot compensate for disturbances with just
feedback (possible offset)
 Need a very good plant model
C
Dead-time Compensation
Closed-loop transfer function
- s
C ( s)
C ( s) GcG p e
 1,

D( s)
R( s) 1 + GcG pm
Characteristic Equation becomes
1 + GcG pm  0
 Effect of dead-time on closed-loop
stability is removed
 Controller is tuned to stabilize
undelayed process model
 No disturbance rejection
Dead-time Compensation
D
R
1
4 1 + 
 s
1
C
e-0.5s
2
s + 3s + 2
1
s2 + 3s + 2
Servo Response
1.5
1
0.5
0
0
1
2
3
4
5
6
7
8
9
10
7
8
9
10
Regulatory Response
1
0.8
0.6
0.4
0.2
0
0
1
2
3
4
5
6
Dead-time Compensation
Include effect of disturbances using model
predictions
D ( s)  Y ( s) - Y ( s)
- s
-s

D( s)  G p e U ( s) - G pme U ( s)
Adding this to previous loop gives
D
R
+
+
+
Gc
Gpm
Gp
Gpm
e-s
+
C
+
e-s
+
-
Dead-time Compensation
Closed-loop transfer function
- s
- s
1 + (e
-e
)GcG pm
C ( s)

D( s) 1 + G G + G (G e - s - G e - s )
c pm
c
p
pm
GcG p e - s
C ( s)

R( s) 1 + G G + G (G e - s - G e - s )
c pm
c
p
pm
Characteristic Equation
1 + GcG pm + Gc (G pe
- s
- G pme
- s
)0
Slow
Fast
Dynamics
Dynamics
 Effect of dead-time on stability is
removed
 Disturbance rejection is achieved
 Controller tuned for undelayed
dynamics
Dead-time Compensation
D
R
+ -
s2 + 3s + 2
1
1
s2 + 3s + 2
s2 + 3s + 2
+
+
1
1
4 1 + 
 s
+ C
e-0.5s
+
+
e-0.5s
-
D ( s)
Servo Response
1.5
1
0.5
0
0
1
2
3
4
5
6
7
8
9
10
7
8
9
10
Regulatory Response
1
0.5
0
-0.5
0
1
2
3
4
5
6
Dead-time Compensation
Alternative form
D
R
+
+
+
Gc
Gp
e-s
+
+
Gpm(1-e-s)
Reduces to classical feedback control system
with
Gc* ( s) 
Gc ( s)
1 + G pm (1 - e -ms )
called a Smith-Predictor
C
Dead-time compensation
Smith-Predictor Design
1. Determine delayed process model
Y ( s)  G pm ( s)e -ms
2. Tune controller Gc for the undelayed
transfer function model Gpm
3. Implement Smith-Predictor as
Gc* ( s) 
Gc ( s)
1 + G pm (1 - e -ms )
4. Perform simulation studies to tune
controller and estimate closed-loop
performance over a range of modeling
errors (Gpm and m)
Dead-time Compensation
Effect of dead-time estimation errors:
R
+ -
+
+
1
1

4 1 + 
 s
s2 + 3s + 2
1
1
s2 + 3s + 2
s2 + 3s + 2
e-0.5s
D
+ C
+
e-s
+
-
D ( s)
