Winter 2006
Instructor:
M.Guay
TA:
V. Adetola

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

 What is the purpose of a control system?
“To maintain important process characteristics at desired targets despite the effects of external perturbations.”
Perturbations
Market
Economy
Climate
Upsets...
Plant
Processing objectives
Safety
Make $$$
Environment...
Control

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 d
P y
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
T in
, w
Q
Inputs
T in w
Q
Process
T, w
Output
T

 The speed of an automobile
Examples
Friction
Inputs
Friction
Engine
Process
Force of
Engine
Output
Speed

e.g. Landing on Mars

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 y
P 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
Process
Output variables
System Physical
Boundary
Transfer of fundamental quantities
Mass, Energy and Momentum
Physical
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
Controller
Observation monitor
The controller is designed to provide regulation of process outputs in the presence of disturbances

What is required for the development of a control system ?
1. The Plant (e.g. SPP of Nylon)
Nylon
Reheater
Dehumidifier
Gas Make-up
Relief
Pot
Vent
Steam
Heater
Water
Blower

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
Controller u
Process y
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 r
+
e
Computer
C
Actuator
A d
P
Process
M
Sensor y

Examples
 Driving an automobile r
+
e
Driver
C
Steering
A P
Automobile
M
Visual and tactile measurement y
Desired trajectory r
Actual trajectory y

Examples
 Stirred-Tank Heater
T in
, w
Heater
T
R
+
e
Controller
C
Q
TC
Thermocouple
T in
, w
Heater
A P
Tank
M
Thermocouple
T, w y

Examples
 Measure T, adjust Q
T
R
+
e
Controller
C
Heater
A
M
Thermocouple
Feedback control
T in
, w
P
Tank
Controller: where
Q=K(T
R
-T)+Q nominal
Q nominal
=wC(T-T in
)
Q: Is this positive or negative feedback?
T

Examples
 Measure T i
, adjust Q
T i
M
Q i
C
+
+
D
Q
Q
A
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
Process
Other
Control
Controller

Example w i
, T i
P c
L w c
, T ci h T
P o w c
, T co w o
, T o
Variables
T
• w i
, w o
: Tank inlet and outlet mass flows
• T i
, T o
: Tank inlet and outlet temperatures
• w c
:
• P c
:
Cooling jacket mass flow
Position of cooling jacket inlet valve
• P o
: Position of tank outlet valve
• T ci
, T co
: Cooling jacket inlet and outlet temperatures
• h: Tank liquid level

Example
Variables Inputs Outputs
Disturbances Manipulated Measured Unmeasured Control w o
T o
P c
P o w i
T i
T ci w c h
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 r e
Controller u d
Process
Model
Design
Implementation y

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
55
Steady-State 2
50
45
40
0 50 100 150
Time
200 250
 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
300

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
1.2
1.1
1
0.9
0.8
0.7
0.6
0.5
0.4
0
 Empirical vs Mechanistic models
 Empirical models
 do not rely on underlying mechanisms

Fit specific function to match process

Mathematical French curve
1.3
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)

 Modeling procedure
Process Modeling
 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
F in h
F
 Modeling objective : Control of tank level
 Fundamental quantity: Mass
 Assumptions : Incompressible flow

Process Modeling
Total mass in system = r
V = r
Ah
Flow in = r
F in
Flow out = r
F
Total mass at time t = r
Ah(t)
Total mass at time t+
D t = r
Ah(t
+D t
)
Accumulation r
Ah(t
+D t
) r
Ah(t) =
D t( r
F in
r
F ), r
Ah t
+ t
D
)
-
D t r
D lim t

0 r
(
+ t
D
)
-
D t r
 r
( F in
-
F ),
 r
( F in
-
F ), r
A dh dt
 r
( F in
-
F ).

Process Modeling
Model consistency
“Can we solve this equation?”
Variables: h, r
, F in
, F, A
Constants: r
, A
5
2
Inputs: F in
, F
Unknowns: h
Equations
2
1
1
Degrees of freedom 0
There exists a solution for each value of the inputs F in
, F

Process Modeling
Solve equation
 Specify initial conditions h(0)=h
0 and integrate h t
 h 0
+ 
0 t

 F in
 -
F

A

 d

2
1.5
1
0.5
0
0
1.3
1.2
1.1
1
0.9
0
Fin
10 20 30 40 50 60 70 80 90 100
10 20 30 40 50 60 70 80 90 100
F

 Energy balance
Process Modeling
M
T in
, 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

H in
-
H out
Q W s dt
 Typically work done on system by external forces is negligible dH

H in
-
H out
+
Q dt
 Assume that the heat capacities are constant such that
H in

 r r
C V T
P
(
-
T ref
( in
-
T ref
) out
 r
P
(
-
T ref
)
)

Process Modeling
After substitution, d ( r
(
-
T ref
))
 r dt
Since T ref
,C p
( in
-
T ref
)
r
(
-
T ref
)
+
Q is fixed and we assume constant r
(
-
T ref
) r dt
Divide by r
C p
V
 r
( in
-
T ref
)
r
(
-
T ref
)
+
Q dT dt
 w
( T in
-
T )
+
V
Q r
C V
P

Process Modeling
Resulting equation: dT dt

F
( T in
-
T )
+
V
Q r
VC
P
Model Consistency
Variables: T, F, V, T in
, Q, C p
, r
Constants: V, C p
, r
Inputs: F, T in
, Q
Unknown: T
7
3
3
1
1 Equations
There exists a unique solution

Process Modeling
Assume F is fixed
T t

T 0 e
t /

+  t e
(
 t )/

(
T in

+ Q
 r
0 where
 
V/F is the tank residence time (or time constant)
) d

If F changes with time then the differential equation does not have a closed form solution.
dT t

F t dt V
( T in t
-
T t
+ r
VC
P
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 dt

M dt
 u t
bv t
Model consistency
Variables: M, v, b, u
Constants: M, b
Inputs: u
Unknowns v
4
2
1
1

Process Modeling
 Gravity tank
F o 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 dt

F
A o -
A dv dt
 hg
-
L r
A
P
2
A system of simultaneous ordinary differential equations results
Linear or nonlinear?

Process Modeling
Model consistency
Variables
Constants
F o
, A, A p
, v, h, g, L, K
F
, r 9
A, A p
, g, L, K
F
, r 6
1 Inputs F o
Unknowns h, v 2
2 Equations
Model is consistent

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

 dt specify r,
C
P,
F
( T in
V and V
-
T )
+
 specify T(0)
 specify Q(t) and F(t)
Q r
VC
P

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
1
0.5
0
0 1 2 3 4 5
Time
6 7 8 9 10
1.4
1.2
1
0.8
0.6
0.4
0.2
0
0 1 2 3 4 5 6 7 8 9 10

Common Input Signals
2. Ramp Input: A sustained constant rate of change e.g.
9
8
7
6
3
2
5
4
1
0
0 1 2 3 4 5
Time
6 7 8 9 10
8
7
6
5
2
1
4
3
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
60
50
40
30
20
10
0
0 1 2 3 4 5
Time
6 7 8 9 10
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.05
0 1 2 3 4 5
Time
6 7 8 9 10

Common Input Signals
3. Pulses: e.g. Rectangular Pulse
1.5
1
0.5
0
0 1 2 3 4 5
Time
6 7 8 9 10
1.2
1
0.8
0.6
0.4
0.2
0
-0.2
0 1 2 3 4 5
Time
6 7 8 9 10

Common Input Signals
4. Sinusoidal input
1.5
1
0.5
0
-0.5
-1
-1.5
0 5 10 15
Time
20 25 30
0
-0.2
-0.4
-0.6
-0.8
0
0.8
0.6
0.4
0.2
5 10 15
Time
20 25 30

Common Input Signals
5. Random Input
1.5
1
0.5
0
-0.5
-1
-1.5
0 5 10 15
Time
20 25 30
0.6
0.4
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

F s
  f t
  t


0
( )
st f t e dt
Integrate out time and transform to
Laplace domain
 dt y 0
 c ay t
+ bu t
Integration
Multiplication

Common Transforms
Useful Laplace Transforms
1. Exponential f t
 e
bt

[ e
bt
]

[ e
bt
]


 e
bt e
st dt

  e
0 0
 e


 

0

1 dt
2. Cosine

[cos(
 t f t
)]
 cos(
 t )
 e
-
+
2

1
2





0 e
- 
) t dt

+  e
0 e

1
2 s
-
1 j

+ s
+
1 j

 s s
2 +  2 s j

) t dt




Common Transforms
Useful Laplace Transforms
3. Sine f t
 sin(
 t )
 e
e
-
2 j


[sin(
2
1
 j s t
-
)]
1 j


1
2 j




 e
0
- 
) t dt

-  e
0
s
+
1 j

 s
2

+  2 s j

) t dt




Common Transforms
Operators
1. Derivative of a function f(t) dt du
 df v
 e
st
 df
[ ] dt
 uv
 
0

[ df dt
]

 
0

-  udv

( )
st
0
( )
st dt
f
 
0

( sf t e
st
) dt
0
 sF s
f ( )
2. Integral of a function f(t)


0 t f
 


 

 e
st
0
( t
 f
 
) dt
0
 s

Common Transforms
Operators
3. Delayed function f(t-

)
   


 e
st
0
0 f t

) t t
 
  dt
+


 e
st
(

) dt
 
( )
  e
s

( )

Common Transforms
Input Signals
1. Constant f t
 a

[ ]


 ae
st dt
0 ae
st s
)

0
 a s
2. Step
 f t
 f t


 ae
st dt
0


0 a t t


0
0 ae
st s
)

0
 a s
3. Ramp function f t

0 t

0 at t

0
    

0 ate
st dt
 e
st at s



 +

0 0
 ae
st s dt
 a s
2

Common Transforms
Input Signals
4. Rectangular Pulse
0 f t
  a
 0 t

0
0 t t w t
 t w
   t
  w
0 ae
st dt
 a s
( 1
e
-
5. Unit impulse
)
  
( )
  t
1 w lim (

0 t s
1
e
-
  
( )
  t w lim

0 se
s

1
)

Laplace Transforms
Final Value Theorem
Limitations: t lim


( )
  lim s

0
 sY s
 y t

C
1
, lim s

0
 sY s

exists
 s s

0
Initial Value Theorem y 0
 lim s

 sY 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
Result dy
5 4 2 dt
+ y

, y ( )

1 y t

.
+
.
e
-
.
t

Solution of Linear ODEs
Stirred-tank heater (with constant F) dT dt
T

F
( T in
-
T )
+
V
Q r
VC
P

T
0 taking Laplace

V
F

 dT dt


  
T t in
( )
 -    +
1 r
FC
P
T s
-
T 0

T s in
-
T s
+
P
( )

 s

+
1
T
+
 s
1
+
1
T s in
( )
+
 s
K
+
P
1
  
To get back to time domain, we must

Specify Laplace domain functions Q(s), T in
(s)

Take Inverse Laplace

Linear ODEs
Notes:
 The expression

 s

+
1
T
+
 s
1
+
1
T s in
( )
+
 s
K
+
P
1 describes the dynamic behavior of the process explicitly

The Laplace domain functions multiplying T(0),
T in
(s) and Q(s) are transfer functions
T in
(s)
Q(s)
T(0)
 s
1
+
1
 s
K
+
P

 s
+
1
1
+
+
+
T(s)

Laplace Transform
Assume T in
(t) = sin(
 t) then the transfer function gives directly
 s
1
+
1
T in
( )

( s
2

+  2
)(
 s
+
1 )
Cannot invert explicitly, but if we can find A and B such that s
2 +
A
 2
+
 s
B
+
1


( s
2 +  2
)(
 s
+
1 ) we can invert using tables.
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,
(x 2 +bx+c) m

Partial fraction Expansions
1. q(s) has real and distinct factors q s
 i n


1
(
+ i
) expand as r s
 i n


1
 i
+ i
2. q(s) has real but repeated factor q s
 +
) n expanded


+
1 +
(

+
2
)
2
 n
(
+
) n

Partial Fraction Expansion
Heaviside expansion
For a rational function of the form
  i n


1
(
Constants are given by
+ i
)
 i n


1
(
 i
+ i
)
 i s b i
)

 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

( s
2 + +
0
) n where d
2
 d
0
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

 s

+
1
T
T in
(s)
Q(s)
T(0)
+
 s
1
+
1
T s in
( )
+
 s
K
+
P
1
 s
1
+
1
 s
K
+
P

 s
+
1
1
+
+
+
T(s) e.g. The block
 s
K
+
P
1 is called the transfer function relating Q(s) to T(s)

Time Domain
Process Control
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


K s
+
P
1
Second order processes

 2 2 s
K
P
+
2
 s
+
1
 Steady-state value obtained directly e.g. First order response to unit step function

K p
 +
1 )
Final value theorem lim s

0

( )
  lim s

0

( )
 
K
P
 Transfer functions are additive and multiplicative

Transfer function
 Effect of many transfer functions on a variable is additive

 s

+
1
T
+
 s
1
+
1
T s in
( )
+
 s
K
+
P
1
T in
(s)
Q(s)
T(0)
 s
1
+
1
 s
K
+
P

 s
+
1
1
+
+
+
T(s)

Transfer Function
 Effect of consecutive processes in series in multiplicative
U(s) K
P
 s
+
1
 Transfer Function
Y
1
(s)
( )


K s
+
P
1
( )
Y
2
( )
( )



K s
+
P
1
( )



K s
+
P
1



 
K s
+
P
1


K
P
 s
+
1
Y
2
(s)

Deviation Variables
 e.g.
To remove dependence on initial condition

 s

+
1
T
+
 s
1
+
1
T s in
( )
+
 s
K
+
P
1
Remove dependency on T(0)

( )

 s
1
+
1 in

( )
+
 s
K
+
P
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, T in
F c
, T cin h
F c
, T c
F, T
Assumptions:
 Constant hold-up in tank and jacket
 Constant heat capacities and densities

Incompressible flow
Model dT dt

F
( T in
-
T )
+
V r dT c dt

F c
( T cin
-
T c
)
-
V c r
( T c
-
T )
( T c
-
T )

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(x
0
)
 f
 x
( x
0
) x
0 x

Nonlinear systems
 First order Taylor series expansion
1. Function of one variable f x
 f xs )
+


( s
) x
( x
xs )
2. Function of two variables f x u
 f xs us )
+

( , s
)
 x
( x
xs )
+

( , s
)
 u
(
3. ODEs x 

( )

( )
+

( )
 x
( x
xs )
)

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

1
U
2 s

1

First order Processes
Examples, Liquid storage
F i h
F r
A dh dt
 r
F r
A dh
 dt i

r r

F
 r
F i
h
F i
 h

 dh
+ 
K F dt dh dt
 +  
K F


First Order Processes
Examples: Speed of a Car
M dv dt u bv
M b dv 1 dt b u v
 dv dt
 
K u
 - 
Stirred-tank heater r
V
F dT
  dt dT
  r
1 dt C F r
C FT
 + 
Q T
 dT dt
 
K Q
 - 
Note:
T

( )

0

K p
 s
+
1

( )

K p
 s
+
1

Liquid Storage Tank
First Order Processes
K p r
/

 r
A/

Speed of a car M/b 1/b
Stirred-tank heater 1/ r
C p
F 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 : r
A

Resistance to flow : 1/

Car:

Capacity to store momentum: M
 Resistance to momentum transfer : 1/ b
Stirred-tank heater
 Capacity to store energy: r
C p
V

Resistance to energy transfer : 1/ r
C p
F
Time Constant =

= (Storage capacitance)*
(Resistance to flow)

First order process
 Step response of first order process
Y s

K p
 s
+
1
M s
Step input signal of magnitude M
0.5
0.4
0.3
0.2
0.1
0
0
1
0.9
0.8
0.7
0.632
0.6
1 2 t/

3 4 5 6

First order process
 What do we look for?
 Process Gain: Steady-State Response lim s

0



 s
K
+ p
1


  K p

Overall Change in y

Overall Change in u
D y
D u

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:
Y s

K p a
 s
+
1 s
2
Ramp input of slope a
2.5
2
1.5
1
0.5
0
0
5
4.5
4
3.5
3
0.5
a

1 1.5
2 t/

2.5
3 3.5
4 4.5
5

First order Process
Sinusoidal response
Y s


K s
+
P
1 s
2
A
+

 2 t lim
 

 -
1


 
1
+
Sinusoidal input Asin(
 t)
  2 sin(
 f
)
2
0.5
0
1.5
1
AR
-0.5
-1
-1.5
0 2 4 f
6 8 t/

10 12 14 16 18 20

First order Processes
Bode Plots
10
0
10
-1
Corner Frequency
High Frequency
Asymptote
10
-2
10
-2
0
-20
-40
-60
-80
-100
10
-2
10
-1
 p

0
10
-1
Amplitude Ratio
AR

K
1
+   2
 p
10
0

10
1
10
1
Phase Shift f  tan (

)
10
2
10
2

Integrating Processes
Example: Liquid storage tank
F i h
F r
A dh dt
 r
F i
r
F
A dh dt

F i
-
F
F i
F F is
, F
  -
F s
A dh dt


 -
1 / s
A
F i


( )

1 / s
A
Process acts as a pure integrator

Process Modeling
 Step input of magnitude M
Y s

K M s s

KM s
2
Slope = KM
Time y t

0 t

0
KMt t

0
Time

Integrating processes
 Unit impulse response
Y s

K
M s

KM s
Time y t

0 t

0
KM t

0
Time
KM

Integrating Processes
 Rectangular pulse response

K s
M
( 1
e
s
)

KM
( 1
e
s
2
)
Time y t

KMt t
 t w
KMt w t
 t w
Time

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)
K
P 1
 s
+
1
1
P 2

+
2
1
Y(s)
U(s)
(

1 s
K K
P 1 P
+
1 )(

2
2 s
+
1 )
Y(s)

By multiplicative property of transfer functions

(

1 s
+
K K
P 1
1 )(

P
2
2 s
+
1 )

Second Order Processes
 Inherently second order process: e.g. Pneumatic Valve p x
Momentum Balance d
M dt dx dt
 pA
-
Kx
-
C dx dt
M
K dt
( )
( )

+
C
K dx dt
A p
K
M
K s
2
A
+
K
C
K s
+
1

Second order Processes
 Second order process:
 Assume the general form


2 2 s
K
+
2
P
 s
+
1 where
K
P
= 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
 
1
4
  -
4
 2
2
 2
 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
Y S


2 2 s
+
K
2
P
 s
+
1
M s
1
0.8
0.6
0.4
0.2
0
0
2
1.8
1.6
1.4
1.2
1 2
0
0.2
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
T in
, w
Response of T
2
T in to is an example of an overdamped second order process
Q
T
1
, w
M
T
2
, w
Q

Second order Processes
Characteristics of underdamped second order process
1. Rise time, t r
2. Time to first peak, t p
3. Settling time, t s
4. Overshoot:
OS b
5. Decay ratio: exp



-

1
 2
 


DR b exp



-
2

1
 2




Second order Processes
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
b a
0.2
0
0 t r t p
5
P c
10 15 t s
20 25 30 35 40
+5%
-5%
45 50

Second Order Process
 Sinusoidal Response
Y s


K p A

+
2
 s
+
1 s
2 +  2

1

2
2

+ ( ) 2 sin( where f  tan
-
1


 
1
-
2

(

)
2


 
AR n

1

2
1
2

+ (
2
 ) 2
+
)

10
-1
10
-1
0
-50
-100
-150
10
-1
 Bode Plots
10
1
Second Order Processes
0.1
10
0

=1
10
0
0.1
10
0

=1
10
1
10
1

More Complicated processes
Transfer function typically written as rational function of polynomials
  a
0
+ a s  b
0
+ b s  b s
 r where r(s) and q(s) can be factored as q s
 b
0
(

1 s r s
 a
0
(
 a 1 s
+
+
1 )(

2
1 )(
 s a
+
2 s
1 )  (
+


1 )  (
 s
+ a

1 ) s
+
1 ) s.t.
G s

K
(
 a
1 s
(

1 s
+
+
1 )  (
 a r
1 )  (

 s s
+
1 )
+
1 )

Poles and zeroes
Definitions:
 the roots of r(s) are called the zeros of G(s) z
1
 -
1
 a
1
,  , z r
 -
1
 a r
 the roots of q(s) are called the poles of G(s) p
1
 -
1

1
,  , p 
 -
1


Poles: Directly related to the underlying differential equation
If Re( p i
)<0, then there are terms of the form e -p i t in y(t) y(t) vanishes to a unique point
If any Re( p i
)>0 then there is at least one term of the form e p i t - y(t) does not vanish

Poles e.g. A transfer function of the form with 0 s (

1 s
+
1 )(

K
+
2

2 s
+
1 )
1 can factored to a sum of

A constant term from s

A e -t/
 from the term (

1 s+1)
 A function that includes terms of the form e
-
 t

2 sin( 1
 2 t

2
) e
-
 t

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

 Calculation performed easily in MATLAB
 Function ROOTS e.g.
q s
 s
3 + s
2 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
0.4
0.2
0
-0.2
-0.4
-0.6
1
0.8
0.6
-0.8
-1
-1.2
-1 -0.8
-0.6
-0.4
Real axis q s
 s
3 + s
2 s
-0.2
1
Roots: -1.0, 1.0
j , -1.0
j
0 0.2

Poles
Process Behavior with purely complex poles
Unit Step Response
1.8
1.6
1.4
1.2
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
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
Real axis
0.1
0.2
0.3
0.4
0.5
2 s
3 +
.
s
2 + +
1
Roots: -0.4368, -0.4066+0.9897
j ,
-0.4066-0.9897
j

Poles
Process behavior with mixed real and complex poles
Unit Step Response
1
0.9
0.8
0.7
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
0.5
0
-0.5
-1
-1.5
-0.8
-0.6
-0.4
-0.2
Real axis
0 0.2
0.4
0.6
2 s
4 +
.
s
3 +
3 s
2 + -
Roots: -0.7441, -0.3805+1.0830
j ,
-0.3805-1.0830
j, 0.2550

Poles
Process behavior with unstable pole
Unit Step Response
160
140
120
100
80
60
40
20
0
-20
0 2 4 6 8 10 t
12 14 16 18 20

Zeros
Transfer function:

K
(

1 s p
(
 a
+
1 )(
 s
+
2 s
1 )
+
1 )
( )

K M 



1
+

 a
-


1
1
-
2 e
t

1 +

 a
-


2
1
-
2 e
t

2




Let
3

1
> 
2
2.5
16

1 is the dominant time constant
2
1.5
1
0.5
0
8
4
2
1
0
-1
-2
-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
 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

K
1
1 s
+
1
U(s)
K
2

2 s
+
1
G s

K
(

1 s
(

+ a s
+
1 )(

2
1 ) s
+
1 )
K

K
1
+
K
2
 a

K
 +
K
K
1
+
K
2

Y(s)
If gains are of opposite signs and time constants are different then a right half plane zero occurs

Dead Time
F i
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
( )
 e
 d 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
 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
0.6
0.5
0.4
0.3
0.2
0.1
0
0

D 0.5
1 1.5
2 2.5
3 3.5
4 4.5
5 5.5
t/tau
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 e
 s  e
 s 
( )

1
-
1
+

2

2 s s
( )

1
1
  2
s
+
2 12
  2
+ s
+
2 12 s
2 s
2

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
G s
 e
 s
K p
 s
+
1

1
-
1
+
 s
2
 
K s
+ p
1
2
 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
0.5
0.4
0.3
0.2
0.1
0
0 1 2 3
- First Order plus dead time o Second Order
4 5 6 7 8

Process Approximation
 Second order overdamped or first order plus dead time?
0.6
0.4
0.2
0
-0.2
0
1.2
1
0.8
1
-- First order plus dead time
- Second order overdamped o Actual process
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


K e
P
 s
+
1
D s
 Second order processes


2 2 s

K e
P

D s
+
2 s
+
1
Y(s)
U(s)
 e D s
G(s)

Process Approximation
 More complicated processes
 Higher order processes (e.g. N tanks in series)
U(s)
(

1 s
+
K
P 1
K
P 2
1 )(

2 s
+
 K
1 )  (
PN

N s
+
1 )
Y(s)

For two dominant time constants

1 process well approximated by and

2

(

1 s e
 s
K
+ p
1 )(

2 s
+
1 )
  i
N


3
 i
 For one dominant time constant

1
, process well approximated by
 e
 s
K p
(

1 s
+
1 )
  i
N


2
 i

Process Approximation
 Example

1
( 10 s
+
)( s
+
1 )( s
+
1 )
2
1.2
1
0.8
0.6
0.4
0.2
0
G
2
( )

( ) e
-
12 s

25 s
+
1 e
-
2 s
( 10 s
+
)( s
+
1 )
-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., K p
,
 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
+ 
1 x
Process Description y
 
0
+ 
1 x
+  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)
SSR
 i n


1
( y i

0

1 x i
)
2

Empirical Modeling
Solution
Differentiate SSR with respect to parameters

SSR

 
0
SSR
 
1
 -
2 i n


1
( y i
 -
2 i n


1
-
( i
 
0
 
1 x i
)
 
0
 
1 x i

0
)

0
These are called the normal equations .
Solving for parameters gives:
 
0
  
1 x
 
1
 i n


1 x y
nxy i n


1 x i
2 nx
2 where x
 i n


1 x i n
, y
 i n


1 y i n

Empirical Modeling
Compact form
Define
Y






 y
1 y
2
 y n






, X







1
1

1 x x x

1
2 n






,
 




0

1



Then
E






 y y y n
1
2
-
-


0
0

-
-

 x x

0

1 x n






Y X
Problem find value of
 that minimize SSR

SSR

E E

Empirical Modeling
Solution in Compact Form
Normal Equations can be written as

 

0 which can be shown to give
X X
  
X Y or
  
( ) -
1
X Y
In practice
 Manipulations are VERY easy to perform in
MATLAB

Extends to general linear model (GLM)
E y
 
0
+  x 

 Polynomial model
E y
  +  x
+  x
2
0 1 1 11 1

Empirical Modeling
Control Implementation:
 previous technique applicable to process model that are linear in the parameters (GLM, polynomials in x, etc…) i.e. such that, for all i, the derivatives a function of

 e i
  are not
 typical process step responses
 first order
[ ( )]

K M ( 1
e
t /

)
Nonlinear in K p and

 overdamped second order
[ ( )]

K M



1
-

1 e
t /

1



2
1
-
2 e
t /

2



Nonlinear in K p
,

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 t
1 y t
2
 y t n or, simply y t

[ y y
2
,  , y n
]
First Order process (step response)
[ i
]

K M ( 1
e
t i
/

)
Least squares problem becomes the minimization of
SSR
 n
 i

1
( y i
-
K M ( 1
e
t i
/

))
2
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
2.5
2
1.5
1
0.5
0
-0.5
0
4
3.5
3
4.5
Model
E y t

3 0 K p
( 1
e
t /

)
Data
Step Response
10 20 30 40 t
50 60 70 80

Empirical Modeling
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
-0.5
0
Results:
Using MATLAB function “leastsq” obtained
K p

.
,
 
.
Resulting Fit
Step Response
10 20 30 40 t
50 60 70 80

Empirical Modeling
Approximation using delayed transfer functions

For first order plus delay processes
[ i
]



 K M ( 1
e
0
( t i
 
)
0 t
 

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

1
( 10 s
+
)( s
+
1 )( s
+
1 )
2
Data
3.5
3
2.5
2
1.5
1
0.5
0
-0.5
0 20 40 60 t
80 100 120 140

Empirical Modeling
2
1.5
1
0.5
0
3.5
3
2.5
-0.5
0
Fit of a first order plus dead time
( )

.
e
-
11 s
( .
s
+
1 )
Second order plus dead time
G
2
( )

.
e
-
2 s
( .
s
+
)( .
s
+
1 )
20 40 60 t
80 100 120 140

Empirical Modeling
Process reaction curve method:
 based on approximation of process using first order plus delay model
M/s
D(s)
Y * (s)
G c
G p
U(s) Y(s)
G s
Y m
(s)
Manual Control
1. Step in U is introduced
2. Observe behavior y m
(t)
3. Fit a first order plus dead time model
Y m
( )

KMe
 s s (
 s
+
1 )

First order plus dead-time approximations
1.2
1
0.8
0.6
0.4
KM
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
0.9
0.8
0.7
Estimation of time constant and dead-time from process reaction curves
 find times at which process reaches 35.3% and
85.3%

1
0.6
0.5
0.4
0.3
0.2
0.1
0
0 20 t
1
40
 Estimate
60 t
2
80 t
100 120 140 160




.
t
1
-
.
t
2
0 67 ( t
2
t
1
)

Empirical Process
Example
For third order process
Estimates:

1
( 10 s
+
)( s
+
1 )( s
+
1 )
2 t
1

23 , t
2

.
,
 
.
Compare:
Least Squares Fit Reaction Curve
( )

.
e
-
11 s
( .
s
+
1 )
( )

.
e
-
11 78 s
( .
s
+
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
F in
,T in
TT TC
IP
LT
P s
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
T in
, F
T
R
+
e
Controller
C
Valve
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
T

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) Y(s)
G p
R(s)
+

-
In process control, we are concerned with the dynamic behavior of a controlled or Closed-loop process controller actuator process
D(s)
+ Y(s)
G c
G v
G p
+ sensor
G m

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
R(s)
+
controller
G c actuator
G v
D(s) process
G p
+
+ Y(s) sensor
G m
G p
(s)
G c
(s)
- Process Transfer Function
- Controller Transfer Function
G m
(s) - Sensor Transfer Function
G v
(s) - Actuator Transfer Function

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
( )

G p
( ) ( )
( )

G p
( ) v
( ) ( )
( )

G p
( ) v
( ) ( ) ( )
( )

G p
( ) v
( )

( )
-
Y m
( )

( )

G p
( ) v
( ) ( )

( )
-
G m
( ) ( )


Isolate Y(s)

G p
( ) v
( ) ( )
1
+
G p s G s G s G m s

Closed-Loop Transfer Function
2. Closed-loop Regulatory Response
 Transfer Function relating D(s) to Y(s) at
R(s)=0
( )

( )
+
G p
( ) ( )
( )

( )
+
G p
( ) v
( ) ( )
( )

( )
+
G p
( ) v
( ) ( ) ( )
( )

( )
+
G p
( ) v
( ) ( )

0
-
Y m
( )

( )

( )
+
G p
( ) v
( ) ( )

0
-
G m
( ) ( )

 Isolating Y(s)

1
1
+
G p s G s G s G m s

Closed-loop Transfer Function
2. Regulatory Response with Disturbance
Dynamics

G d
( )
1
+
G p s G s G s G m s
G d
(s) Disturbance (or load) transfer function
3. Overall Closed-Loop Transfer Function
Servo

G p
( ) v
( ) ( )
1
+
G p s G s G s G m s
+
G d
( )
1
+
G p s G s G s G m s
Regulatory

PID Controllers
The acronym PID stands for:
 P - Proportional

I - Integral

D - 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:
( )

K c


 ( )
+

1
I 0
 t e
  
D de dt


 + u
R

PID Control
PID Control Equation
Proportional
Action
Derivative
Action
( )

K c


 ( )
+

1
I
 t
0 e
  
D de dt


 + u
R
Integral
Action
Controller
Bias
PID Controller Parameters
K c

I

D u
R
Proportional gain
Integral Time Constant
Derivative Time Constant
Controller Bias

PID Control
PID Controller Transfer Function
 
( )
u
R
  
( )

K c



1
+

I
1 s
+ 
D
 s E s or:

( )



P
I 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:
( )

K c



1
+

I
1 s
+ 
D s


 or,
G s



P
I s
Ds


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
R

( )
Transfer function: or,
'( )

( )
Closed-loop form:
( )

K c

G p
( ) v
( ) c
1
+
G p
( ) v
( ) ( )
+
1
1
+
G p
( ) v
( ) ( )

Proportional Feedback
Example:
Given first order process:
G p
( )

K p
 s
+
1
, ( )

1 , G m
( )

1 for P-only feedback closed-loop dynamics:



 1
+
KpKc
1
+
KpKc
 
KpKc 
1
+


 1
+
 
KpKc 


 1
+

KpKc
1


1
KpKc
1
Closed-Loop
Time Constant

Proportional Feedback
Final response: t lim


1
KpKc
+
, lim
KpKc t

Note:
 for “zero offset response” we require

1
1
+
KpKc t lim
 y servo t

1 , t lim
 y reg t

0
Tracking Error 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
0.9
0.8
0.7
0.6
0.5
10.0
5.0
1.0
0.4
0.3
0.5
0.2
0.1
K c
0
0 1 2 3 4 t/

5 6 7 8 9
- increasing controller gain eliminates off-set
0.01
10

Proportional Feedback
High-order process e.g. second order underdamped process
1.5
1
0.5
5.0
2.5
1.0
0.5
0
0 5 10 15
0.01
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
( )

K c


 ( )
+

1
I
 t
0 e
 


 + u
R
Transfer function model

( )

K c



1
+

I
1 s




PI Feedback
Closed-loop response

1
+
G p
( ) v
( ) c
G p s G s K c 
 



I
 s
I
I
 s
I
+ s
+ s
1


1


G m
( )
1
+
1
G p
( ) v
( ) c



I
 s
I
+ s
1


G m
( )
+
 more complex expression
 degree of denominator is increased by one

PI Feedback
Example
PI control of a first order process
G p
( )

K p
 s
+
1
, ( )

1 , G m
( )

1
Closed-loop response



 K K p


 s
2 +




I
1 s
+
+
1
K K
K K p p


 
I s
+
1






K K p


 s
2 + 


K K p


 s
2 +

I
K K p



1
+
K K
K K p p


 s


 
I s
+
1
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

 damping coefficient
K K p
 
1
2
K p


K
 


K K p
K K p
+
1



 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
0.8
0.6
0.4
0.2
0
0
1.8
1.6
1.4
1.2
0.01
1
0.1
2
0.5
1.0
3 4
K c
=1
5 6 7 8 9 10

PI Feedback
Effect of controller gain
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.9
1
10.0
5.0
0.8
0.7
1 2
1.0
3
0.5
4 5
0.1
6 7 8
 affects speed of response
 increasing gain eliminates offset quicker

I
=1
9 10

PI Feedback
Effect of integral action of regulatory response
0.4
0.35
0.3
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
K c
 
D de dt
 if error is increasing, decrease control action
 if error is decreasing, decrease control action

Always implemented in PID form
( )

K c


 ( )
+

1
I
 t
0 e
  
D de dt


 + u
R

PID Feedback
Transfer Function

( )

K c



1
+

I
1 s
+ 
D
 s E s
Closed-loop Transfer Function

G p
( ) v
( ) c



1
+
G p
( ) v
( ) c



1
+
G p
( ) v
( ) c



1 s
2

I
+ 
I s
+ s
1 

 s
2

I
+ 
I s
+ s
1 


G m
( ) s
2

I
+ 
I s
+ s
1 


G m
( )
+

Slightly more complicated than PI form

PID Feedback
Example:
PID Control of a first order process
G p
( )

K p
 s
+
1
, ( )

1 , G m
( )

1
Closed-loop transfer function



 K K p
+
 




 K K p



+
K K p




2
2 s
2
2
+




1
I s
+
+
1
K K p
K K p


 
I s
+
1




I
K K p


 s



1
+
K K
K K p p


 
I s
+
1
+

PID Feedback
Effect of derivative action on servo dynamics
1
0.8
0.6
0.4
0.2
0
0
1.6
1.4
1.2
1
0.1
0.5
1.0
2.0
2 3 4 5 6 7 8 9 10

PID Feedback
Effect of derivative action on regulatory response
0.05
0
-0.05
0.25
0.2
0.15
0.1
0.1
0.5
1.0
2.0
-0.1
0 1 2 3 4 5 6 7 8 9
 increasing derivative action reduces impact of disturbances on control variable
 slows down servo response and affects oscillation of process
10

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

1
+
G G G p
Y
* s
+
1
+
1
G
OL
 if G
OL is a rational function then the closed-loop transfer functions are rational functions and take the form
 and factor as
 a
0
+ a s  b
0
+ b s  b s
 r
G s

K
(
 a
1 s
(

1 s
+
+
1 )  (
 a r
1 )  (

 s s
+
1 )
+
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 a s n + a n
-
1 s n
-
1
 a s a
0

0 is
1
2
3
4
 n
+
1 a a n a n
-
2 n b
1 c

-
1
1 a n
-
3 b
2 c
2 z
1 a n
-
4 a n
-
5 b
3



 where b
1
 c
1
 a n b a
n
1 a
n
3
b
1
2
a n
-
1
b a a n n
-
-
1
3
, a n c
2
, b
2

 a n
-
1 a b
1 n
-
4
a n
-
1 b a n
-
5
b a n
-
1 a n
-
5 a n
, 
, 
Elements of left column must be positive to have roots with negative real parts

Example: Routh Array
Characteristic polynomial
.
s
5 +
.
s
4 -
.
s
3 +
.
s
2 +
.
s
+
.

0
Polynomial Coefficients a
5

2 36 , a
4

149 , a
3
 -
. , a
2
 a
1

0 42 , a
0

Routh Array a
5
2 36 ) a
3
(
-
0 58 ) a
1
0 42 ) a
4
149 ) a
2 b
1
(
-
2 50 ) b
2
(
-
0 82 ) a
0 b
3
0 78 ) c
1 d
1
0 72
189 ) e
1
0 78 )
) c
2 d
2
0 78 )
 Closed-loop system is unstable

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 K c poles that yields purely complex
 Strategy:

Start with characteristic polynomial
1
+
( ) p
( ) m
( )

Write characteristic equation:
( )
+
( )

0

Substitute for complex pole ( s=j

)
(

)
+
(

)

0

Solve for K c and

K c

Example: Direct Substitution
Characteristic equation
1
+
K c s
3 + s
3 +
.
s
2 s
+
-
1
.
s
-
.

0
.
s
2 -
0 5 s
-
.
+
K s
+
K c

0 s
3 +
.
s
2 +
( K c
-
0 5 s
+
( K c
-
.
)

0
Substitution for s=j

( j

)
3 +
0 5 j

)
2 +
( K c
-
j
 3  2 +
( K c
j
 +
( K c
j
 +
( K c
-
.
)

0
.
)

0
Real Part
-
.
 2 +
K c
-
.

0
Complex Part
( K c
  3 
0

K c

 -

 2 +
2 +
.

( .
.

0
K c

1
 2 +
.
-
. )
  3 
0
 System is unstable if K c
>
1

1.5
0
-0.5
1
0.5
-1
-1.5
-1.5
Root Locus Diagram
Old method that consists in plotting poles of characteristic polynomial as controller gain is changed e.g. s
3 + s
2 +
( K c
-
0 5 s
+
( K c
-
. )

0
-1
K c
-0
K c
-0
1
1 1.5
-0.5
0
Real Axis
0.5

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
C
R

G G p
1
+
G G p
Isolate G c
G c

1
G p




1
C
-
R
C
R




For a desired trajectory ( C/R ) d and plant model G pm
, controller is given by
G c

G
1 pm
1
-
( )
( ) d 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
R

 d

1
 cannot be achieved, requires infinite gain
2. Closed-loop process with finite settling time

 C
R

 d

 c s
1
+
1
 For 1st order G p
, it leads to PI control
 For 2nd order, get PID control
3. Processes with delay


C
R


   e c
s
 c
+ s
1
 requires
 c
 
 again, 1st order leads to PI control

2nd order leads to PID control

IMC Controller Tuning
R
+
-
G c
*
G p
G pm
+
+
D
C
-
+
Closed-loop transfer function
C

G G p
1
+
G G p
-
G pm
)
R
+
1
-
G G p
1
+
G G p
-
G pm
)
D
In terms of implemented controller, G c
G c

G
* c
1
-
G G pm

IMC Controller Tuning
1. Process model factored into two parts
G pm

G
+
G
pm pm where contains dead-time and RHP zeros, steady-state gain scaled to 1.
2. Controller
G
* c

G
1
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
G p
( )
 e
-
9 s
30 s
+
1
Process parameters: K=0.3,
30, 9
1.
IMC Design: K c
=6.97,

I
=34.5,
 d
=3.93

Filter f

1
12 s
+
1
2. Direct Synthesis: K c
=4.76,

I
=30

Servo Transfer function

 C
R

 d
 e
-
9 s
12 s
+
1

Example
Result: Servo Response
 IMC and direct synthesis give roughly same results
25
20
IMC
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 y(t)
30
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
Y * (s)
G c
1/s
U(s)
G p
G s
Y m
(s)
Manuel Control
1. Step in U is introduced
2. Observe behavior y m
(t)
3. Fit a first order plus dead time model
Y m
( )

Ke
 s
 s
+
1
D(s)
Y(s)

Tuning Relations
Process response
1.2
1
0.8
0.6
0.4
KM
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 K c
P-only
PI
PID
( 1 / K p
)(

/
( 0 .
9 /
( 1 .
2 /
K p
)(

K p
)(


/
/
)


)
)
T i
3 .
3

2 .
0

T d
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:
G p
( )
 e
-
9 s
30 s
+
1
K c
= 10
PID: K c
= 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
30
Direct Synthesis
25
20
0 50 100 t
150 200 250 300

Example
Regulatory Response
25
20
15
10
0
40
35
30
Direct Synthesis
Z-N PI
Z-N PID
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
PI
K c
( 1 /
( 1 /
K p
K
)(
 p
)(

/

)[ 1
+ 
/

)[ 0 .
9
+
/ 3

]

/ 12

]
PID
( 1 / K p
)(

/

)[
3
 +
16
12


]
T i
T d

[ 30
9
+
+
3 (

20 (

/
/


)
)]

[ 32
13
+
+
6 (

8 (

/

/

)
)]
11
+
4

2 (

/

)
Example:
PI: K c
=10.27
K c
=15.64
 d
=3.10

I
=18.54

I
=19.75

Tuning relations
Cohen-coon: Servo
55
50
45
40
C-C PID
35
30
25
C-C PI
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
15
10
5
0 50 100

Highly oscillatory

Very aggressive t
150
C-C PID
200 250 300

Integral Error Relations
1. Integral of absolute error (IAE)
IAE

 
( )
0
2. Integral of squared error (ISE)
ISE

 
0
( )
 penalizes large errors
2 dt
3. Integral of time-weighted absolute error
(ITAE)
ITAE

 
( )
0
 penalizes errors that persist

ITAE is most conservative

ITAE is preferred

ITAE Relations
Choose K c
,

I and
 d that minimize the ITAE:
 For a first order plus dead time model, solve for:


ITAE
K c

0 ,

ITAE

I

0 ,

ITAE
 d

0
 Design for Load and Setpoint changes yield different ITAE optimum
Type of
Input
Load
Load
Set point
Set point
Type of
Controller
PI
PID
PI
PID
Mode A
I
P
I
P
D
P
I
I
P
D
0.859
0.674
1.357
0.842
0.381
0.586
1.03
0.965
0.796
0.308
B
-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

KK c



I

 d

Setpoint Settings:
Y

A
( )
B

KK c

 d
 ,


I
A B
Example
G s

.
e
-
9 s
30 s
+
1
, G
L

1

ITAE Relations
Example (contd)
Setpoint Settings
KK c

( ) -


K c

I
I




 d
 d


.
K
 .
.
.
 30

( )
.
 
.
Load Settings:
KK c

K c

I



I



.
K
 .
-
 30
 d
 d



( )
.
 
.
-











ITAE Relations
Servo Response
60
55
50
ITAE(Load)
30
25
20
0
45
40
35
ITAE(Setpoint)
50 100 150 200 250 300
 design for load changes yields large overshoots for set-point changes

ITAE Relations
Regulatory response
25
20
15
10
5
0
0
40
35
30
ITAE(Setpoint)
100
ITAE(Load)
150 200 50 250

Tuning relations are based G
L
=G p

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


 In general,
 d

I

0 25
 Ziegler-Nichols and Cohen-Coon tuning relations yield aggressive control with oscillatory response
(requires detuning)
 ITAE provides conservative performance (not aggressive)

Frequency Response of
Linear Control Systems

First order Process
Response to a sinusoidal input signal t lim
 

 -
1
 


1
+   2 sin(
 f
)
2
1.5
1
AR
0.5
0
-0.5
-1 f
-1.5
0 2 4 6 12 14 16 18 20 8 t/

10
Recall: Sinusoidal input Asin(
 t) yields sinusoidal output caharacterized by AR and f

First order Processes
Bode Plots
10
0
10
-1
Corner Frequency
High Frequency
Asymptote
10
-2
10
-2
0
-20
-40
-60
-80
-100
10
-2
10
-1
 p

0
10
-1
Amplitude Ratio
AR

K
1
+   2
 p
10
0

10
1
10
1
Phase Shift f  tan (

)
10
2
10
2

Second Order Process
 Sinusoidal Response
Y s


K p A

+
2
 s
+
1 s
2 +  2

1

2
2

+ ( ) 2 sin( where f  tan
-
1


 
1
-
2

(

)
2


 
AR n

1

2
1
2

+ (
2
 ) 2
+
)

10
1
10
0
10
-1
10
-1
0
-50
-100
-150
10
-1
Second Order Processes
Bode Plot
0.1
Amplitude reaches a maximum at resonance frequency

=1
10
0
10
1
0.1
10
0


=1
10
1

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
 w b

Im b
 a w
Re
It follows that a
 w w
 such that w
2 + w
2 w
 w e j

  where b
 w w
 tan
-
1

 


Frequency Response
Some facts: ii) Let z=a-bj and w= a+bj then w
 z z
 w iii) For a first order process
G s

K p
 s
+
1
Let s=j

G j

)

K p
+
( 1
-
1 ( 1
-
)
)

K p
1
+   2
-
1
+
K p

  2 such that
G j

)

1
+
K p
(

AR )
  2
G j

))
 tan
-
1
(

)(

Phase Lag) j

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

).
i.e.
The Phase Shift is given by the argument of the transfer function model in the frequency domain.
AR

G j

)

G j

))
2 +
G j

))
2
Phase Angle tan
-
1


G j

))

G j

))


Frequency Response
For a general transfer function
  e

( s
( s
s
p
1
) z
1

) 
(
( s
z m s
p n
)
)
Frequency Response summarized by
G j

)

G j

) e j
 where
 is the modulus of G(j
( )
G(j

)

) and

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 function G(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
G j

)

K j
 
 -
j
 j
 

 -
K
 j
AR

K

, f  tan
-
1


-
K
0
/
 

 -

2
2. Dead Time G(s)=e -
 s
G j

)
 e
j

AR

1, f  

Frequency Response
Examples:
3. n process in series
( )

( )  ( )
Frequency response of G(s)
G j

)

G
1
( j

)  G n
( j

)
G
1
( j

) e j f
1  G n
( j

) e j f n therefore
 f
AR

(

) n
  i

1

(

)
G j

))
 i n


1 arg( (

))
 i n


1 f i

Frequency Response
Examples.
4. n first order processes in series
G s


K
1 s
1
+
1

 n
K n s
+
1
AR

K
1
1
+   2
 f  tan
-
1
1
+
K n
  2 tan
-
1 (  n
)
5. First order plus delay
 s
G s

 s
+
1
AR

K p
1
+   2
, 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

10
1
10
0
10
-2
-89
-89.5
-90
-90.5
-91
10
-2
Pure Capacitive Process
AR

K

  -

2
10
2
10
-1
10
-1
Frequency (rad/sec)
Bode Plot
10
0
10
0

Bode Plot
10
0
G s

G s G s G s
( )

1
10 s
+
1
, ( )

1
5 s
+
1
, ( )

1 s
+
1
10
-2
G
1
G
3
G
2
10
-4
10
-4
0
10
-3
10
-2
10
-1
10
0
10
1
G
-100
-200
-300
10
-4
10
-3
10
-2
10
-1
10
0
10
1
G j

)

1

(
+ 2  2
)(
+ 2  2
)(
+ 2  2
)
 tan
-
1
( 10

)
tan
-
1
( 5

)
tan
-
1 

Bode Plot
10
-2
10
-4
10
-4
0
-100
-200
-300
10
-4
G s
 e
 s
G j

)

1 ,
   
Example: Effect of dead-time
G d
( )
 e
-
2 s
( )
2
( )
3
( )
10
0
10
-3
10
-3
10
-2
10
-2
10
-1
10
0
G=G d
10
1
10
-1
G d
10
0
10
1
G

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
G s

1 s
3 +
3 s
2 +
3 s
+
1
Effect of dead-time (second order process)

1 s
2 +
3 s
+
1
, Gd s
 2 s

Che 446: Process Dynamics and
Control
Frequency Domain
Controller Design

PI Controller
10
3
10
2
AR
10
1
10
0
10
-3
-20

-40
-60
-80
0
-100
10
-3
10
-2
10
-2
AR

K c
 
1
 
I
2
+
1 tan
-
1
(
-
1 /

I
)
10
-1
10
-1

10
0
10
0
10
1
10
1

PID Controller
10
3
AR
10
2
10
1
10
0
10
-3
100

50
0
-50
-100
10
-3
AR

K c





D
-
1

I


2
 + 1
 tan
-
1 



D
-
1

I



10
-2
10
-2
10
-1
10
-1

10
0
10
0
10
1
10
1

Bode Stability Criterion
Consider open-loop control system
R(s)
+
-
G c
U(s)
G p
+
D(s)
+
Y(s)
G s
Y m
(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?

Bode Stability Criterion
“A closed-loop system is unstable if the frequency of the response of the open-loop
G
OL has an amplitude ratio greater than one at the critical frequency. Otherwise it is stable. “
Strategy:
1. Solve for
 in arg( G
OL
( j

))
 
2. Calculate AR
AR

G
OL
( 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
K cu

1
AR

Bode Criterion
Consider the transfer function and controller

5 e
-
.
s
( s
+
)( .
s
+
1 )
G s

0 4 1
+
1
.
s


- Open-loop transfer function
G
OL
( )

5 e
-
.
s
( s
+
)( .
s
+
1 )


+
1
.
s


- Amplitude ratio and phase shift
AR

5
1
+  2
1
.
 2
+
.
1
 2 f  -
0 1
 tan
-
1  tan
-
1
0 5

)
tan
-
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, K u
, is 1/M where M is the amplitude of the open-loop system

Ultimate Period
P u

2

 c
Ziegler-Nichols Tunings
P K u
/2
PI K u
/2.2
PID K u
/1.7
P u
/1.2
P u
/2 P u
/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 G
OL 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 G
2. Plot G
OL
(j
OL
(s)

) in the complex plane
3. Count encirclements of (-1,0) in the clockwise direction

Nyquist Criterion
Consider the transfer function

5 e
-
.
s
( s
+
)( .
s
+
1 ) and the PI controller
G s

0 4 1
+
1
.
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.

Advanced Control Techniques:
1. Feedforward Control

Feedforward Control
Feedback control systems have the general form:
D(s)
U
R
(s)
R(s)
+
G c
+ +
U(s)
G v
G p
G
D
+
+
Y(s)
G s
Y m
(s) where U
R
(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 G
D
(s)
Feedforward Control is possible.
R(s)
+
G c
+ +
U(s)
G v
G f
Feedforward
Controller
D(s)
G p
G
D
+
+
Y(s)
G s
Y m
(s)
Feedback/Feedforward Controller
Structure

Feedforward Control
Heated Stirred Tank
F,T in
TT
TT
TC1
P s
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?

A suggestion:
Feedforward Control
F,T in
TT
TT
+
+
TC2
P s
TC1
Steam
Condensate
How do we design TC2?
F,T

Feedforward Control
The feedforward controller:
D(s)
G f
U
R
(s)
+ +
U(s)
G v
G p
G
D
+
+
Y(s)
Transfer Function
( )

G
D
( ) ( )
+
G
P
( ) v
( ) ( )
( )

G
D
( ) ( )
+
G
P
( ) v
( )(
R
( )
+
G f
( ) ( ))
( )

( G
D
( )
+
G p
( ) v
( ) f
( )) ( )
+
G p
( ) v
( )
R
( )
( )

( G
D
( )
+
G p
( ) v
( ) f
( )) ( )
+
Y
R
( )
Tracking of Y
R requires that
G
D
( )
+
G p
( ) v
( ) f
( )

0

G f
( )
 -
G
D
( )
G p s G s

Feedforward Control
Ideal feedforward controller:
G f
( )
 -
G
D
( )
G p s G s

Exact cancellation requires perfect plant and perfect disturbance models.
G
D
( )
+
G p
( ) v
( ) f
( )

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
R(s)
+
G c
+ +
U(s)
G v
G f
D(s)
G p
G
D
+
+
Y(s)
Y m
(s)
G s
What is the impact of G f on the closed-loop performance of the feedback control system?

Feedforward Control
Regulatory transfer function of feedforward/feedback loop

G
D
( )
+
G f
( ) v
( ) p
( )
1
+
G s G s G p s G m s
Perfect control requires that (as above)
G f
( )
 -
G
D
( )
G 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
( )

K f
(

1 s
(

2 s
+
+
1 )
1 )
G f
( )
 -
K
D
K K p

Feedforward Control
Tuning: In absence of disturbance model lead-lag approximation may be good
G f
( )

K f
(

1 s
(

2 s
+
+
1 )
1 )
 K f obtained from open-loop data
K f
 -
K
D
K K p

1 and

2
 from open-loop data

1
  p
,

2
 
D
 from heuristics



1 
Trial-and-error


1
2


1

2
 c

Feedforward Control
Example:
Plant:
G p
( )

10
( 10 s
+
)( s
+
1 )( s
+
1 )
G
D
( )

1
( .
s
+
1 )( s
+
1 )
Plant Model:
G pm
( )

10 e
-
6 s
10 s
+
1
G
Dm
( )
 e
s
.
s
+
1
Feedback Design from plant model: IMC PID tunings
K c

0 26 ,

I

13 ,

D


Feedforward Control
Possible Feedforward controllers:
1. From plant models:
G f
( )
 -

Not realizable e
5 s
10
( 10 s
+
1 )
( .
s
+
1 )
2. Lead-lag unit

1

10 ,

2

K f
 -
1
K f
 -
1
10

Feedforward Control
For Controller 2 and 3
Disturbance Controller with Feedforward
0.6
0.4
0.2
0
-0.2
-0.4
1.2
1
0.8
.. - Gain Controller
-- - Lead-Lag Controller
- - No FF Controller
-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, H
4
-controllers, etc…)
 Use process knowledge (and intuition)

Advanced Control Techniques
2. Cascade Control

Jacketed Reactor:
F,T in TT
Cascade Control
TT
TC1
P s
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,T in TT
TC1 FC
P s
Steam
FT
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

Cascade Control

Cascade Control
Closed-loop transfer function
1. Inner loop
C
2
R
2

G p 2
G G c 2
1
+
G p 2
G G G m 2
2. Outer loop

G cl 2
C
1
R
1

G G cl 2
G c 1
1
+
G G cl 2
G G m 1
Characteristic equation
1
+
G G cl 2
G G m 1

0
1
+
G p 1
1
+
G p 2
G G c 2
G p 2
G G G m 2
1
+
G p 2
G G G m 2
+
G G
G G m 1 p 2

0
G G G G m 1

0

Cascade Control
Stability of closed-loop process is governed by
1
+
G p 2
G G G m 2
+
G G p 2
G G G G m 1

0
Example
G p 1

K

1 s p 1
+
1
, G c 1

K c 1
, G v 1

G m 1

1
G p 2


K p 2
2 s
+
1
, G c 2

K c 2
, G m 2

1
(

1 s
+
1
+
1 )(

K c 2
2 s
K

2 s p 2
+
1
+
K c 1
K

2 s p 2
+
K
1

1 s p 1
+
1

K c 2
K p 2
(

1 s
0
K K p 2
K p 1

0 s
2 +
(

1
+ 
2
+
K c 2
K p

) s
K K p 2
K p 1

0
K c 2
K p 2
+

Cascade Control
Design a cascade controller for the following system:
1. Primary:
G p 1 s
 e
-
.
s
0 5 s
+
1 )( s
+
1 )
, G m 1

1 ,
G

Kc
1
 c 1 
2. Secondary:
1
+
1

I s

 p 2

1
G
.
s
+
1
G c 2

K c 2
, G v 2

G m 2

1

Cascade Control
1. PI controller only
G
OL 1

K c 1 

1
+
1  1 s

0 1 s
+
( .
s e
+
-
.
s
1 )( s
+
1 )
AR

K c 1
1
+
1
 2
1
.
 2 +
1
1
 2 +
1
1
 2 +
1
  tan
-
1


1
 

tan
-
1
( .

)
tan
-
1
0 5

)
tan
-
1  -
0 1

Critical frequency
 c

2 99 ,
Maximum gain
AR

K c 1

5 61

AR
Bode Plots
Cascade Control

 ln(

)

Cascade Control
2. Cascade Control
 Secondary loop
G
Ol 2

K c 2
01 s
1
+
1
 no critical frequency gain can be large
 Let K c2
=10.

Primary loop
G

OL 1

K c 1 

1
+
1 s


10
.
s
+
1
1
.
s
+
1 e
-
.
s
( .
s
+
1 )( s
+
1 )
K c 1
10
11 e
-
.
s s ( .
s
+
1 )(
11 s
+
1 )

Cascade Control
Closed-loop stability: f
AR
K c 1

10 1
11


2
1 1
1
+ 

.
11

 2
 2
.
 2
0 1
 tan
-
1


11



tan
-
1
0 5

)
 Bode
 c

AR


Maximum gain K c1
=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

Advanced Control Techniques
3. Dead-time Compensation

R
Dead-time Compensation
Consider feedback loop:
D
C
G c
G p 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

Dead-time Compensation
Motivation e
 s
 s
2 +
3 s
+
2


+
1 s


0.75
0.5 0.25
0.1

R
Dead-time Compensation
Use plant model to predict deviation from setpoint
D
C
G c
G p e -
 s
G pm
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

Dead-time Compensation
Closed-loop transfer function

1 ,
 s

1
+
G G pm
Characteristic Equation becomes
1
+
G G pm

0

Effect of dead-time on closed-loop stability is removed
 Controller is tuned to stabilize undelayed process model
 No disturbance rejection

R
Dead-time Compensation
D
C
+
1  s

1 s
2 +
3 s
+
2
1 s
2 +
3 s
+
2 e -0.5s
Servo Response
1.5
1
0.5
0
0
1
0.8
0.6
0.4
0.2
0
0
1
1
2 3 4 5 6
Regulatory Response
7 8 9 10
2 3 4 5 6 7 8 9 10

Dead-time Compensation
R
+
Include effect of disturbances using model predictions

( )

( )

( )

( )

G e
 s
( )
-
G pm e
  s
( )
Adding this to previous loop gives
-
G c
G
G p pm e -
 s e -
 s
D
+
+
-
+
C
+
+
G pm

Dead-time Compensation
Closed-loop transfer function

1
+
1
G G
+ pm
( e
+
 s c
e
  s
) G G pm p
 s -
G pm e
  s
)
 s

1
+
G G pm
+ c p
 s -
G pm e
  s
)
Characteristic Equation
1
+
G G pm
+
G G e
 s -
G pm e
  s
)

0
Fast Slow
Dynamics Dynamics

Effect of dead-time on stability is removed

Disturbance rejection is achieved

Controller tuned for undelayed dynamics

Dead-time Compensation
R
+
+
-
+
1  s

+ 1 s
2 +
3 s
+
2
1 s
2 +
3 s
+
2
1 s
2 +
3 s
+
2 e e -0.5s
-0.5s
+
D
C
+
-
( )
+
Servo Response
1.5
1
0.5
0
0
1
1
0.5
0
-0.5
0 1
2 3 4 5 6
Regulatory Response
7 8 9 10
2 3 4 5 6 7 8 9 10

Dead-time Compensation
Alternative form
R
+ -
G c
G p e -
 s
D
+
+
C
+
+
G pm
(1-e -
 s )
Reduces to classical feedback control system with c
* 
1
+
G pm
( 1
e
 m s
) called a Smith-Predictor

Dead-time compensation
Smith-Predictor Design
1. Determine delayed process model

( )

G pm
( )
 m s
2. Tune controller G c for the undelayed transfer function model G pm
3. Implement Smith-Predictor as
* c

1
+
G pm
( 1
e
 m s
)
4. Perform simulation studies to tune controller and estimate closed-loop performance over a range of modeling errors ( G pm and
 m
)

Dead-time Compensation
R
+
+
Effect of dead-time estimation errors:
-
+
1  s

+ 1 s
2 +
3 s
+
2
1 s
2 +
3 s
+
2
1 s
2 +
3 s
+
2 e -0.5s
e -
 s
+
D
C
+
-
( )
+
