process control chapter 4
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CHEE 3550
Process Dynamics and Control
Lecture Notes #4: Transfer Function Models
of Dynamical Processes
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Lecture Notes Content
Transfer Functions [Seborg 3.1]
Poles & Zeros [Seborg 6.1]
Transfer Functions for Nonlinear Systems [Seborg 4.4]
Simplifying Multiple Transfer Functions [Seborg 4.3]
First Order Systems [Seborg 5.2,5.3]
Second Order Systems [Seborg 5.4]
Higher Order Systems [Seborg 6.3]
Systems with Time Delays [Seborg 6.2]
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Linear SISO Control Systems
General form of a linear SISO control system:
this is a underdetermined higher order differential
equation
the function must be specified for this ODE to
admit a well defined solution
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Transfer Functions
The expression
describes the dynamic behavior of the process explicitly
The Laplace domain function is called thetransfer function
between and
Transfer functions are usually represented inblock diagram form
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Transfer Function
Heated stirred-tank model (constant )
Taking the Laplace transform yields:
or letting
T(s) = s+1
T(0) 1s+1
Tin(s) +1
CpF
s+1Q(s)
Transfer functions
T, F
Tin, F
Q
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Transfer Function
Heated stirred tank example
e.g. The block is called thetransfer function relating Q(s) to T(s)
+
++
T(s) = s+1
T(0) + 1s+1
Tin(s) +1
CpF
s+1Q(s)
s+1
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Process Control
Time Domain
Transfer functionModeling, Controller
Design and Analysis
Process Modeling,Experimentation and
Implementation
Laplace Domain
Ability to understand dynamics in Laplace andtime domains is extremely important in the
study of process control
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Transfer Functions
Transfer functions are generally expressed as a ratio ofpolynomials
Where
The polynomial is called thecharacteristic polynomialof
Roots of are thezeros of
Roots of are thepoles of
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Transfer Function
Order of underlying ODE is given by degree ofcharacteristic polynomial
Q: What is the order of the following transfer function?
Q: What is the order of the following transfer functions?
Order of the process is the degree of the characteristic
(denominator) polynomial Therelative order is the difference between the degree of the
denominator polynomial and the degree of the numerator
polynomial
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Transfer Function
Steady state behavior of the process obtained from thefinal value theorem
e.g. First order process
For a unit-step input,
From the final value theorem, the ultimate value of is
This implies that the limit exists, i.e. that the system is stable.
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Transfer Function
Transfer function is the unit impulse response
e.g. First order process,
Unit impulse response is given by
In the time domain,
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Transfer Function
Unit impulse response of a 1st order process
dt = 0.1;
Tfinal = 6;
Time = 0:dt:Tfinal;num = [1]; den = [1 1];
sys = tf(num,den);
Output = impulse(sys,Time);
plot(Time,Output);
xlabel('Time');
ylabel('Output');
MATLAB
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Deviation Variables
To remove dependence on initial condition
e.g.
Compute equilibrium condition for a given and
Definedeviation variables
Rewrite linear ODE
or
0
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Deviation Variables
Assume that we start at equilibrium
Transfer functions express extent of deviation from a givensteady-state
Q: Why do we do all this?
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
T0(s) = 1s+1
T0in
(s) + Ks+1
Q0(s)
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Deviation Variables
In mechanical systems, the equilibrium is usually selectedas the initial rest position
Cruise control example
Suspension system example
Satellite system
DC motor
Using initial condition such that the output is at zero,avoids the need for deviation variables
Initial conditions must be an equilibrium of the system
v0(t) = v(t) (0), u0(t) = u(t) (0)
x0(t) = x(t), y0(t) = y(t), r0(t) = r(t)
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Deviation Variables
Example (the ball and beam example)
This is anonlinear system of ordinary differential equations
Must be linearized about an equilibrium to obtain a transfer
function model
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Deviation Variables
Pendulum example System equations are nonlinear in
For small perturbation about the vertical position , the
nonlinearity can approximated (1st order Taylor series expansion)
Linearized model
Starting at rest, , taking the Laplace transform
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Process Modeling
Gravity tank
Modeling Objectives: Control height of liquid in tank
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
h
L
F
Fo
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Transfer Functions
From mass balance and Newtons law,
A system ofsimultaneous ordinary differential equations results
Q: Is it a linear or nonlinear system of ODEs?
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Nonlinear ODEs
Q: If the model of the process is nonlinear, how do weexpress 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(x0)
f(x)
f
xx( )0
xx0
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li
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Nonlinear Systems
First order Taylor series expansion
1. Function of one variable
2. Function of two variables
3. ODEs
f x u f xs usf x u
xx xs
f x u
uu us
s s s s( , ) ( , )( , )
( )( , )
( ) + +
f x f xs f xx x xss( ) ( ) ( ) ( ) +
& ( ) ( )( )
( )x f x f xsf xs
xx xs= +
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T f F i
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Transfer Function
Procedure to obtain transfer function from nonlinearprocess models
Find an equilibrium point of the system
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
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T f F i
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Transfer Function
Ball and beam example Linearize the system of equations about equilibrium
The nonlinear model is given by
Linearize (1st order Taylor series expansion about equilibrium)
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T f F i
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Transfer Function
Linearization gives the linear system
Taking Laplace transform
Transfer function
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Bl k Di
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Block Diagrams
Transfer functions of complex systems can be representedin block diagram form.
3 basic arrangements of transfer functions:
1. Transfer functions in series
2. Transfer functions in parallel
3. Transfer functions in feedback form
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Bl k Di
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Block Diagrams
Transfer functions in series
Overall operation is the multiplication of transfer functions
Resulting overall transfer function
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T f F ti
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Transfer Functions
DC Motor example: In terms of angular velocity
In terms of the angle
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T f F ti
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Transfer Functions
Transfer function in parallel
Overall transfer function is the addition of TFs in parallel
+
+
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T f F ti
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Transfer Functions
Transfer function in parallel
Overall transfer function is the addition of TFs in parallel
+
+
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T f F ti
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Transfer Functions
Transfer functions innegative feedback form
Overall transfer function
>> s = tf(s);
>> G1 = 2/(3*s+1);
>> G2 = 6.5/(4.2*s+3);
>>feedback(G1,G2,-1)
Transfer function:8.4 s + 6
----------------------
12.6 s^2 + 13.2 s + 16
MATLAB
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Transfer Function
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Transfer Function
Example
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Transfer Function
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Transfer Function
Example
A positive feedback loop
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Transfer Function
Example
Two systems in parallel
Replace by
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Transfer Function
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Transfer Function
Example
Two systems in parallel
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Transfer Function
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Transfer Function
Example
A negative feedback loop
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Transfer Function
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Transfer Function
Example
Two process in series
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First Order Systems
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First Order Systems
First order systems are systems whose dynamics aredescribed by the transfer function
where
is the systems (steady-state) gain
is thetime constant
First order systems are the most common behaviourencountered in practice
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First Order Systems
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First Order Systems
Examples, Liquid storage
Assume:
Incompressible flow
Outlet flow due to gravity
Balance equation: Total
Flow In
Flow Out
Fin
h
F
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First Order Systems
Balance equation:
Deviation variables about the equilibrium
Laplace transform
First order system with
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First Order Systems
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First Order Systems
Liquid Storage Tank
Speed of a car
DC Motor
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
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First Order Systems
Step response of first order process
Step input signal of magnitudeM
The ultimate change in is given by
Q: Could we get the same result in another way?
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First Order Systems
Step responseclc;clear;
dt = 0.1;
Tfinal = 6;
Time = 0:dt:Tfinal;
num = [1]; den = [1 1];
sys = tf(num,den);Output = step(sys,Time);
plot(Time,Output);
xlabel('Time');
ylabel('Output');
MATLAB
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First Order Systems
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First Order Systems
What do we look for? Systems Gain: Steady-State Response
Process Time Constant:
What do we need?
System initially at equilibrium
Step input of magnitudeM
Measure process gain from new steady-state
Measure time constant
Time Required to Reach
63.2% of final value
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First Order Systems
First order systems are also called systems with finite
settling time
Thesettling time is the time required for the system to come
within 5% of the total change and remain there for all times
Consider the step response
The overall change is
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First Order Systems
Settling time
num = [1]; den = [1 1];
sys = tf(num,den);
ltiview('step',sys);
% set settling time to 5%
% (Edit => Viewer Pref.
% => Options)% Right-click and select
% settling time
MATLAB
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First Order Systems
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First Order Systems
Process initially at equilibrium subject to a step of
magnitude 1
num = [3.2]; den = [7 1];sys = tf(num,den);
ltiview('step',sys);
MATLAB
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First Order Systems
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First Order Systems
Ramp response:
Ramp input of slope a
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 50
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
a = 1 , = 1
TimeStep = 0.01;
Time = 0:TimeStep:5;
Input = Time;
G = tf([1],[1 1]);
[Output] = lsim(G,Input,Time);
plot(Time,Input); hold on;plot(Time,Output); hold off;
MATLAB
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First Order Systems
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First Order Systems
10 10 10 10 10
Bode Plots
10 10 10 10 10
-2 -1 0 1 210-2
10-1
100
High Frequency
AsymptoteCorner Frequency
Amplitude Ratio Phase Shift
-2 -1 0 1 2-100
-80
-60
-40
-20
0
num = [1]; den = [1 1];
sys = tf(num,den);
ltiview(bode',sys);
MATLAB
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Integrating Systems
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teg at g Syste s
Example: Liquid storage tank
Laplace domain dynamics
If there is no outlet flow,
H
F
Fi
H(s) = 1As
Fin(s)1As
F(s)
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Integrating Systems
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g g y
Example
Capacitor
Dynamics of both systems is equivalent
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Integrating Systems
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g g y
Rectangular pulse response
TimeTime
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Second Order Systems
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y
Second order process:
Assume the general form
where = Process steady-state gain
= Process time constant
= Damping Coefficient
Three families of processes
UnderdampedCritically Damped
Overdamped
Q:What is a physical example of an overdamped system?
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y
Three types of second order process:
1. Two First Order Systems in series or in parallel
e.g. Two holding tanks in series
2. Inherently second order processes: Mechanical systemspossessing inertia and subjected to some external force
e.g. A pneumatic valve
3. Processing system with a controller: Presence of a
controller induces oscillatory behaviore.g. Feedback control system
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y
Multicapacity Second Order Processes
Naturally arise from two first order processes in series
Q: What is an example of such a system?
By multiplicative property of transfer functions
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Transfer Functions
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First order systems in parallel
Overall transfer function a second order process (with one zero)
+
+
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y
Inherently second order process:
e.g. Pneumatic Valve
x
pBy Newtons law
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Feedback Control Systems
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Second order process:
Assume the general form
where = Process steady-state gain
= Process time constant
= Damping Coefficient
Three families of processes
UnderdampedCritically Damped
Overdamped
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Second Order Systems
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Step response of magnitudeM
0 1 2 3 4 5 6 7 8 9 100
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
=2
=0
=0.2
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Second Order Systems
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Characteristics of underdamped second order process
1. Rise time,
2. Time to first peak,
3. Settling time,
4. Overshoot:
5. Decay ratio:
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Step response
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Sinusoidal Response
where
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More Complicated Systems
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Transfer function typically written as rational function of polynomials
where and can be factored following the discussion on partialfraction expansion
s.t.
where and appear as real numbers or complex conjugate pairs
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Poles and Zeroes
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Definitions:
the roots of are called thezeros ofG(s)
the roots of are called thepoles ofG(s)
Poles: Directly related to the underlying differential equation
If , then there are terms of the form in
vanishes to a unique point
If any then there is at least one term of the form
does not vanish
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Poles
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e.g. A transfer function of the form
with can factored to a sum of
A constant term from
A from the term
A function that includes terms of the form
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
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Poles
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Calculation performed easily in MATLAB
Function POLE, PZMAP
e.g.
s=tf('s');
sys=1/s/(s+1)/(4*s^2+2*s+1);
Transfer function:
1
-------------------------4 s^4 + 6 s^3 + 3 s^2 + s
pole(sys)
ans =
0
-1.0000
-0.2500 + 0.4330i
-0.2500 - 0.4330i
MATLAB
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Poles
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Function PZMAP ( pzmap(sys))
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One constant pole
integrating feature
One real negative pole
decaying exponential
A pair of complex roots with negative real part
decaying sinusoidal
Q:What is the dominant feature?
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Step Response
Integrating factor dominates the dynamic behaviour
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Example
Poles
-1.0000
-0.0000 + 1.0000i
-0.0000 - 1.0000i
One negative real pole
Two purely complex poles
Q: What is the dominant feature?
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Step Response
Purely complex poles dominate the dynamics
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Example
Poles
-8.1569
-0.6564
-0.1868
Three negative real poles
Q: What is the dominant dynamic feature?
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Step Response
Slowest exponential ( ) dominates the dynamicfeature (yields a time constant of 1/0.1868= 5.35 s
num = [1]; den = [1 9 7 1];
sys = tf(num,den);
ltiview('step',sys);
MATLAB
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Poles
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Example
Poles
-4.6858
0.3429 + 0.3096i
0.3429 - 0.3096i
One negative real root
Complex conjuguate roots withpositive real part
The system is unstable (grows without bound)
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Step Response
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Two types of poles
Slow (or dominant) poles
Poles that are closer to the imaginary axis
Smaller real part means smaller exponential term and
slower decay
Fast poles Poles that are further from the imaginary axis
Larger real part means larger exponential term and faster
decay
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Poles dictate the stability of the process
Stable poles
Poles with negative real parts
Decaying exponential
Unstable poles
Poles with positive real parts Increasing exponential
Marginally stable poles
Purely complex poles
Pure sinusoidal
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Stable Unstable
Marginally
Stable
Pure Exponential
Oscillatory
Behaviour
Integrating
Behaviour
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Poles and Zeros
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Definitions:
the roots of are called thezeros ofG(s)
the roots of are called thepoles ofG(s)
Zeros:
Do not affect the stability of the system but can modify the dynamicalbehaviour
G(s) =B(s)A(s)
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Types of zeros:
Slow zeros are closer to the imaginary axis than the dominant poles of thetransfer function
Affect the behaviour
Results in overshoot (or undershoot)
Fast zeros are further away to the imaginary axis
have a negligible impact on dynamics
Zeros with negative real parts, , arestable zeros
Slow stable zeros lead to overshoot
Zeros with positive real parts, , are unstable zeros Slow unstable zeros lead to undershoot (or inverse response)
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Zeros
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Zeros can result from two processes in parallel
If gains are of opposite signs and time constants are different then a right halfplane zero occurs
+
+
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Example
Poles
-8.1569
-0.6564
-0.1868
Zeros
-10.000
Q:What is the effect of the zero on the dynamic behaviour?
num = [0.1 1]; den = [1 9 7 1];
sys = tf(num,den);
pole(sys)
zero(sys)
pzmap(sys)
ltiview('step',sys);
MATLAB
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Poles and zeros
Dominant
PoleFast, Stable
Zero
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Unit step response:
Effect of zero is negligible
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Example
Poles
-8.1569
-0.6564
-0.1868
Zeros
-0.1000
Q: What is the effect of the zero on the dynamic behaviour?
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Poles and zeros:
Slow (dominant)
Stable zero
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Unit step response:
Slow dominant zero yields an overshoot.
num1 = [0.1 1]; num2 = [10 1];
den = [1 9 7 1];
sys1 = tf(num1,den);
sys2 = tf(num2,den);
ltiview('step',sys1,sys2);
MATLAB
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Example
Poles
-8.1569
-0.6564
-0.1868
Zeros
10.000
Q: What is the effect of the zero on the dynamic behaviour?
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Poles and Zeros
Fast UnstableZero
DominantPole
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Unit Step Response:
Effect of unstable zero is negligible
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Example
Poles
-8.1569
-0.6564
-0.1868
Zeros
0.1000
Q: What is the effect of the zero on the dynamic behaviour?
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Poles and zeros:
SlowUnstable Zero
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Unit step response:
Slow unstable zero causes undershoot (inverse response)
Q: Can the inverse response exist in reality? If yes, give examples?
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Observations:
Adding a stable zero to an overdamped system yields overshoot
Adding an unstable zero to an overdamped system yieldsundershoot (inverse response)
Inverse response is observed when the zeros lie in right half
complex plane,
Overshoot or undershoot are observed when the zero is dominant(closer to the imaginary axis than dominant poles)
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Example: System with complex zeros
Dominant
Pole
SlowStable Zeros
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Poles and zeros
Poles have negative real parts: The system is stable
Dominant poles are real: yields an overdamped behaviour
A pair of slow complex stable zeros: yields overshoot
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Unit step response:
Effect of slow zero is significant and yields oscillatory behaviour
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Example: System with zero at the origin
Zero at
Origin
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Unit step response
Zero at the origin: eliminates the effect of the step Q: Why?
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Observations:
Complex (stable/unstable) zero that is dominant yields an
(overshoot/undershoot)
Complex slow zero can introduce oscillatory behaviour
Zero at the origin eliminates (or zeroes out) the system response
to a step
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Time required for the
fluid to reach the valve
usually approximated as
dead time
h
Fi
Control loop
Manipulation of valve does not lead to immediate
change in level
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l d f f i
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Delayed transfer functions
e.g. First order plus dead-time
Second order plus dead-time
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Dead time (delay)
Most processes will display some type of lag time
Dead time is the moment that lapses between input changes and
process response
Step response of a first order plus dead time processCHEE 3550 - Winter 2012 - S.Blouin
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Delayed step response:
Q: What is the transfer function?
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Problem
use of the dead time approximation makes analysis (poles and
zeros) more difficult
Approximate dead-time by a rational (polynomial)
function Most common is Pade approximation
es s+2ks+2k
k
, k {1, 2, . . .}
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