Process Control Chapter 4

8/3/2019 Process Control Chapter 4

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CHEE 3550

Process Dynamics and Control

Lecture Notes #4: Transfer Function Models

of Dynamical Processes

CHEE 3550 - Winter 2012 - S.Blouin


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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]

2CHEE 3550 - Winter 2012 - S.Blouin


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3

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


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= +


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


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)


T f F i


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Transfer Function

Linearization gives the linear system

Taking Laplace transform

Transfer function


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


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


T f F ti


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Transfer Functions

Transfer function in parallel

Overall transfer function is the addition of TFs in parallel

+

+


T f F ti


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Transfer Functions

Transfer function in parallel

Overall transfer function is the addition of TFs in parallel

+

+


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


Transfer Function


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Transfer Function

Example

A positive feedback loop


Transfer Function


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Transfer Function

Example

Two systems in parallel

Replace by


Transfer Function


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Transfer Function

Example

Two systems in parallel


Transfer Function


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Transfer Function

Example

A negative feedback loop


Transfer Function


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Transfer Function

Example

Two process in series


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


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


First Order Systems


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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


First Order Systems


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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?


First Order Systems


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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


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


First Order Systems


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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


First Order Systems


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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


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);


MATLAB


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


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)


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


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


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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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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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


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


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


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


Poles


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Function PZMAP ( pzmap(sys))


Poles


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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?


Poles


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Step Response

Integrating factor dominates the dynamic behaviour


Poles


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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?


Poles


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Step Response

Purely complex poles dominate the dynamics


Poles


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Example

Poles

-8.1569

-0.6564

-0.1868

Three negative real poles

Q: What is the dominant dynamic feature?


Poles


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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);


MATLAB


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)


Poles


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Step Response


Poles


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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


Poles


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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


Poles


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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)


Zeros


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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)


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

+

+


Zeros


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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)


MATLAB


Zeros


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Poles and zeros

Dominant

PoleFast, Stable

Zero


Zeros


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Unit step response:

Effect of zero is negligible


Zeros


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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?


Zeros


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Poles and zeros:

Slow (dominant)

Stable zero


Zeros


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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


Zeros


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Example

Poles

-8.1569

-0.6564

-0.1868

Zeros

10.000



Zeros


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Poles and Zeros

Fast UnstableZero

DominantPole


Zeros


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Unit Step Response:

Effect of unstable zero is negligible


Zeros


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Example

Poles

-8.1569

-0.6564

-0.1868

Zeros

0.1000



Zeros


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Poles and zeros:

SlowUnstable Zero


Zeros


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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?


Zeros


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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)


Zeros


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Example: System with complex zeros

Dominant

Pole

SlowStable Zeros


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


Zeros


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Unit step response:

Effect of slow zero is significant and yields oscillatory behaviour


Zeros


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Example: System with zero at the origin

Zero at

Origin


Zeros


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Unit step response

Zero at the origin: eliminates the effect of the step Q: Why?


Zeros


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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


Delay


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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


Delay

l d f f i


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Delayed transfer functions

e.g. First order plus dead-time

Second order plus dead-time


Delay


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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

Delay


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Delayed step response:

Q: What is the transfer function?


Delay


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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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Process Control Chapter 4

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