Process Controll

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Lec 1: Introduction to Process Control

Learning objectives: To introduce the student to process control, the need for it and itsapplications.

Learning outcomes: On completion of this topic, students should be able to describe asimple control system, types of controllers and parameters in a control system.

Process Control is the study of automatic control principles applied to chemical

processes. It applies principles of mathematics and engineering science to the regulation

of the dynamic operation of process systems. To be successful, you need strong appliedmathematics skills and process understanding (most of which is just common sense).

The skills and tasks you've been exploring in the first three years of ChE classes are

predominantly analytical. They are used for diagnosis and understanding of processesand problems. This year, your design classes will work on synthesis skills for devising

new processes. In Process Control, we will use some analytical tools (old and new) andsynthetic skills to understand the dynamic (time dependent) behavior of processes andways to regulate plant operation.

Since the primary function of control systems is to compensate for dynamic changes in

process systems, we need to understand the dynamics of processes -- how their behavior

changes with time -- if we are to develop workable solutions. We address this need

through dynamic modeling of the chemical processes. Mathematically, this means wewill be dealing with differential equations.

Note on course organization

Control courses can be difficult for an instructor to organize. There are often multiple

ways of approaching concepts, each with its own "dialect" of terminology and equation.

Topics often wrap back around, so books and instructors sometimes have a tendency touse terms and ideas before they are fully defined. So let me know if this is happening in

this course.


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Putting all this together, the two main functions of control systems are:

1. setpoint tracking -- the ability to shift from one desired operating point to another(like you driving your car)2. disturbance rejection -- the ability to maintain an operating point despite

fluctuating conditions and external forces (like your thermostat)

Disturbances can never be completely eliminated; however, a good control system can

greatly attenuate their consequences and reduce the variability of process parameters. Ifwe can reduce variability, we need smaller margins of error and contingency allowances,

and so can operate much closer to optimum conditions, reducing waste and saving

money.

Safety Systems

No feedback control loop, no matter how well-designed and tuned, can guarantee safeoperation. Consequently, a regulatory process control system cannot be trusted as the

primary safety system. Almost all chemical plants have a second, parallel control system

to handle safety alarms and shutdown. While we will always consider the safety aspectsof control systems, we will not study the design of these alarm/shutdown systems.

Control Objectives

The objectives of a control system fit into a hierarchy -- that is, some objectives are given

priority over others. One way of ordering the hierarchy is by the purpose of the controlsystem components:

1. Safety2. Environmental Protection3. Equipment Protection4. Smooth Plant Operation5. Product Quality


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6. Profit Optimization7. Monitoring and Diagnosis

According to this structure, control loops responsible for safety-related tasks will always

have priority over all other tasks; loops for product quality will have priority over loops

whose primary task is optimization; and so forth. Most of the techniques we study in thiscourse will apply directly to the operating and quality objectives.

Be aware that a loop can serve more than one purpose and that its place in the hierarchyis not always cut-and-dried.

One of the themes of our study this semester will be the tradeoffs between plant design

and plant operation. Control systems are part of the day to day operation of a plant. This

suggests another way of ordering the hierarchy of control objectives: "achievability".After all, until your plant is operating, controls aren't needed at all. This sort of hierarchy

tends to group loops by function as much as it does by objective:

1. Production Rate & Inventory Controls2. Safety/Environmental Controls3. Equipment and Operating Constraint Controls4. Product Quality Controls5. Optimization

Feedback Control Loops

Imagine taking a shower from a two-knob faucet. You want to set the rate of water flow

and its temperature so that the shower is effective and comfortable. You can control the

hot water flow and the cold water flow separately. Throughout your shower, there may be"disturbances" in the form of changes in water supply temperature and pressure.

So how do you regulate your shower?

1. You test the flow and temperature2. Decide if it is OK3. If not OK, decide what adjustment to make4. Adjust the knobs5. Repeat from the first step


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This is the structure of a basicfeedback control loop. Information from the process (water

flow and temperature) is "fed back" by a measurement device (you) through a controller(you again) to a control element (the knobs) to change the process input.

All control systems have these same basic components:

measurements of variables affecting the system a specified desired value or range of values for the controlled variable (the

setpoint).

a control calculation or algorithm

a way of adjusting the system to reflect the results of the control calculation (thecontrol element).

The system can be represented by a block diagram where lines are used to representvariables or signals and boxes for actions.

Block diagrams represent the logic and mathematical model of a control loop. We might

also choose to represent the equipment used to construct the loop in a piping and

instrument diagram or P&ID. Complete P&IDs show every piece of equipment, wiring,

etc., that need to be installed, and so can be very complex. For our purposes, we just want

to see the main pieces, so a simplified drawing is used.

The loop shown consists of:


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a sensoror transducerelement which measures some output process variable.

a transmitterwhich converts the sensor output to a signal suitable for transmissionthrough the plant

a controllerwhich compares the transmitted measurement to a setpoint value and

then determines what control action is required.

a final control element (usually a control valve) that transforms the controlleroutput signal into a change in a manipulated variable.

These are the instruments that are used to control the process.

Table 1.1 of your textbook has a handy list of commonly used symbols for thesediagrams.

Control System Design

When an engineer sets out to design a control system, the steps are:

1. determine control objectives2. identify measurable variables, available manipulators3. pair variables (choose controller structure)4. select controller algorithms5. tune controller (adjust sensitivity)

In this class, we will deal mainly with the tools and concepts needed to model the

process, examine the stability, and select and tune controllers.

Terminology

Steady State: A steady state system does not change with time. Mathematically, this

means the time derivatives in the balance equations (the accumulation terms) are zero.Often, systems will reach steady state if given a long time to settle -- usually, real systems

don't get the time. This leads to another mathematical approximation -- steady state is the

behavior of the system as time approaches infinity. Some people use the words static or

stationary as synonyms for steady state.

Dynamic (or transient) systems are time dependent. All real systems are dynamic; thismakes process control necessary. Dynamic systems must be modeled using differential

equations, unlike steady state systems where algebraic systems will suffice.

Inputs and Outputs are not necessarily material flows. An input is a variable that causesan output to change. Both inputs and outputs may be measurable or they may not.

Disturbances are inputs that cannot be adjusted, and often they are not measurable.

Erroris the difference between the measured behavior of a process output and its desired

behavior or setpoint. Never forget that the measured values of the outputs are onlyrepresentations of the real values, and may be limited in accuracy.


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Feedback Control: information from an output of a system is used to adjust a manipulator

to change an input to the system to try and compensate for disturbances after they havechanged the system.

Feedforward Control: information from measured disturbances is used to adjust a

manipulator to try and compensate for disturbances as they occur. Feedforward allows forthe possibility of "perfect control", but only if all disturbances are measured and the

adjustments are fully understood. This means you must have a complete and veryaccurate model of the process -- not an easy achievement. Feedback control adjusts for all

disturbances and does not require an exact process model.

Negative feedbackreduces the difference between the actual and desired values, so it is

beneficial. Positive feedback increases the difference, so it is undesired.

When a system is operating without control, we say it is operating Open Loop. A ClosedLoop system has controllers on-line.

One of the most important things we will be watching is the stability of the system. The

error of an unstable system becomes larger and larger (unbounded) with time, often

leading to undesirable consequences.

Control Loop Hardware

A control loop is built from mechanical and electrical devices. These usually include

a sensor a transmitter a controller

an actuator, and

afinal control element

Riggs (2001) lumps some of these into subsystems. He calls the sensor and transmitter

the "sensor system" and the actuator and control element the "actuator system".

The controller will usually be located in a control room; typically, it exists as software

within a Distributed Control System (DCS) computer. The other parts are physical

equipment built into or adjacent to the process equipment.

Information is passed around the control loop in the form of signals. These may beanalog or digital, electrical or pneumatic. Converters or transducers transform signals

from one type to another.


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Lec 2:Process Diagrams

Block diagrams

At an early stage or to provide an overview of a complex process or plant, a drawing is

made with rectangular blocks to represent individual processes or groups of operations,together with quantities and other pertinent properties of key streams between the blocksand into and from the process as a whole. Such block flowsheets are made at the

beginning of a process design for orientation purposes or later as a summary of the

material balance of the process.

Process flow diagrams (and to some extent PID)

Process flowsheets embody the material and energy balances between and the sizing of

the major equipment of the plant. They

include all vessels such as reactors, separators, and drums; special processing equipment,heat exchangers, pumps, and so on. Numerical data include flow quantities,

compositions, pressures, temperatures, and so on. Inclusion of major instrumentation that

is essential to process control and to complete understanding of the flowsheet without

reference to other information is required particularly during the early stages of a job,since the process flowsheet is drawn first and is for some time the only diagram

representing the process. As the design develops and a mechanical Boarsheet gets

underway, instrumentation may be taken off the process diagram to reduce the clutter. Achecklist of the information that usually is included on a process flowsheet is giventogether with an example of a PFD.


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Essential components

Process lines, but including only those

bypasses essential to an understanding of

the process

All process equipment. Spares are indicatedby letter symbols or notes

Major instrumentation essential to processcontrol and to understanding of thediagram

Valves essential to an understanding of thediagram

Design basis, including stream factor

Essential information

Stream Number

Temperature (C)

Pressure (bar)

Vapor Fraction

Total Mass Flow Rate (kg/h)

Total Mole Flow Rate (kmol/h)

Individual Component Flow Rates

(kmol/h)

Figure: Example of a Process flow diagram


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Figures: Examples of PID


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Drawing of diagrams

Flowsheets are intended to represent and explain processes. To make them easy tounderstand, they are constructed with a consistent set of symbols for equipment, piping,

and operating conditions. At present there is no generally accepted industrywide body of

drafting standards, although every large engineering office does have its internal

standards.Equipment symbols are a compromise between a schematic representation of the

equipment and simplicity and ease of drawing.Common symbols:


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Since a symbol does not usually speak entirely for itself but also carries a name and a

letter-number identification, the flowsheet can be made clear even with the roughest ofequipment symbols. The letter-number designation consists of a letter or combination to

designate the class of the equipment and a number to distinguish it from others of the

same class, as two heat exchangers by E-112 and E-215. Example of letters used:


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Chapter 3: Mathematical models

Learning objectives: To introduce the student to the use of balance equations in processcontrol and how it renders transfer functions.

Learning outcomes: Students should be able model simple systems and express them in

the form of block diagram.

IntroductionVery often engineers use "models" of a process to aid understanding. A model can be a

description, a picture or physical model, or a mathematical or statistical construct that

emulates the behavior of the real, physical system, although often in an idealized way.

Or as Adage put it: All models are wrong, but some are useful

The degree of complexity of a model is linked to decisions made in the modeling process.

Sometimes it is desirable to start with afundamental orfirst principles model -- modelingequations are developed starting from the material and energy balances, chemical and

physical laws.

The steps in developing a fundamental model are: Preparation

o Decide what kind of model is needed. What scale? How detailed? Howaccurate? What features cannot be neglected?

o Define and sketch the system.

o Select variables Model Development

o Write balance equations (mass, component, energy) to describe the

system.

o Write (descriptive) constitutive equations (transport, equilibrium, kinetic)needed to implement the balance equations.

o Check for consistency of units, independence of equations, degrees of

freedom Solution (Simulation)

o Solve the equations (analytically or numerically)

o Check and verify the solution

This type of model will emerge as a system of differential balance equations (ordinary or

partial) accompanied by a set of algebraic constitutive equations. Depending on the

intended use, the model can be adapted in several ways made steady-state (time derivatives approach zero)

linearized

o differential equation form

o state-space form

o transformed to transfer function form


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When fundamental models are not possible.

In other cases, the fundamental behavior of a process is poorly understood or

prohibitively complex to model based on first principles. In these cases, models may bedeveloped from experimental dynamic data. Developing a model from experimental data

is often called process identification. Identification techniques can be very simple(process reaction curve analysis of step inputs) or complex (ARIMA modeling of PRBSinputs). Essentially, these approaches "curve fit" the data to produce what are sometimes

called "Input/Output Models" or "Black Box" models.

The choice of a model type depends on the scale of the problem. You probably don't want

to model molecular dynamics unless they make a difference!

Basic Modeling Equations

The dynamic models used for process modeling and control can be mathematicallyrepresented by a set of balance equations (conservation equations). These may be

supplemented by one or more constitutive equations that further define terms in the

balance equations.

Every dynamic model will include at least one balance equation. The balance equations

will have the same general form you've been using for several years now:

Accumulation = Transport Generation

= [InOut] [ProductionConsumption]

The accumulation term will supply the time derivative and produce a differential

equation. If the model is distributed, the equation will also include position derivatives.Transport terms will typically be initially written in terms of the transport fluxes.

Material Balance

A mass balance is needed whenever you are interested in the "holdup" of a system.Holdup is typically measured using level for liquid systems or pressure for gas/vapor

systems. You should expect your mass balance equation to have either or both of these

variables inside the accumulation term. Unless you are dealing with a nuclear reaction,

the mass balance will not have a generation term.


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A mass balance may be written over each system or subsystem that you can define within

your process.

Constitutive equations may be needed to define system properties such as density in

terms of composition, temperature, pressure, etc.

Exercise: Write the mass balance for a mixing tank, using the V, the volume of the tank,qi and qo, the volume flowrate in and out respectively.

________________________________________

Exercise: Get together in groups and state common assumptions used in writing andsimplifying mass balances.

____________________

____________________

____________________

Component BalanceA component balance must be written whenever composition changes are to be

examined.

Almost all _________ or _________ problems will involve a component balance.Compositions are usually expressed in terms of mole fractions.

You can write one balance for each component over each subsystem, but remember that

the sum of all component balances is the total material balance, so normally will use onetotal mass balance and (Ncomponent-1) component balances.

Initially, transport terms in a component balance will be expressed in terms of thetransport fluxes (both molecular and convective). Constitutive equations defining the

fluxes will thus be needed. When reactors are modeled, generation terms will be required

and will be written in terms of reaction rate expressions. These will also need to bedefined by a constitutive equation.

The generation term is commonly expresses through the rate of reaction; rA

kmoles/(m3s).

As we can see from the unit to determine the generation we have to multiply the rate ofreaction with the __________________.

Exercise: Write the component balances for a CSTR with an A-->B reaction:

___________________________________________

___________________________________________


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Common assumptions include:

___________________________________

___________________________________

Energy Balance

You will need to write an energy balance whenever the __________ within your system

changes; which will almost always be inside the derivative. Reference temperatures for

enthalpies can complicate things, so be careful.

One energy balance can be written for each separable system or subsystem.

Energy transport fluxes and thermodynamic property relations will require constitutive

equations for fuller definition.

The enthalpy for a flow system contains the properties; flowrate, ________,_________

and ___________.

Exercise: Write the energy balance for a heated tank, which has a heater suppliant a heattransfer q.

________________________________________________________________

Exercise: What simplifications would commonly be helpful?

___________________________

___________________________

___________________________

Constitutive Equations


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All models will include one or more balance equations. Most will also use a set of

constitutive equations to better define specific terms in the balance equations. Commonconstitutive relationships include:

property relations and equations of state

transport flux relations reaction rate expressions

equilibrium expressions fluid flow relations

Property Relations / Equations of State

Physical and thermodynamic properties (density, heat capacity, enthalpy, etc.) vary withtemperature, pressure, and composition. These relationships usually must be incorporated

into dynamic models.

Enthalpy is typically expressed as a function of temperature ____________________

Equations of state are typically used to express vapor densities in terms of systemtemperature and pressure. Often, the ideal gas equation is adequate.

Transport Flux Expressions

Transport flux expressions are usually used to quantify heat and mass transfer. When

transport is purely molecular, these are nothing more than statements of Fick's Law,

Fourier's Law, and Newton's Law of Viscosity. They are expressed as:

12

1,2,

TThAQ

CCkN AAcA

Reaction Rate Expressions

The reaction rate expressions used in dynamic modeling are typically based on the

principles of mass action. The Arrhenius expression must be incorporated directly when

rate constants depend on temperature; otherwise, the energy balance won't adequately

describe temperature changes.

n

ART

E

A CekrA

0

or for constant temperature _____________________________


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Equilibrium Expressions

Phase equilibrium expressions are often needed when modeling separation systems.

Raoult's law, ______________, equilibrium K-values,_____________, and relativevolatility,_____________, all are used. The choice is the modelers.

Chemical equilibrium expressions are needed less often. If they are needed, they areusually incorporated as part of the reaction rate expression.

Fluid Flow Relations

Fluid flow relationships are typically used when it is necessary to relate pressure drop to

flow rate. These usually take the form of a momentum balance (equation of motion) or

mechanical energy balance. Momentum balances are typically required for gravity flow

problems (where the balance may reduce to Torricelli's Law).

For systems involving flow through a weir or across a valve, a reduced form of themechanical energy balance is often used. Rather than deriving these from the balance, itis usually reasonable to select an appropriate eqaution, for instance

Though it is not uncommon to have to use the full Bernoullis law and the continuity

relation.

Example 3.1:

Flow rates in and out, no reaction Simple material balance

tpktq

sticscharecteriValve

thktq

lawsBernoulli

'


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Material balance: dm/dt = min-mout

or in variables in the flowchart: d(V)/dt=qinin- qoutoutAssume constant density for all streams: d(V)/dt=qin- qout

Bernoullis equation:

Atmospheric pressure in both ends; p1=p2

Substitute: h=h1-h2The liquid surface is moving very slowly: v1=0

Hence, we get just two terms: _______________

with the relation between flow rate and velocity we get a description of qout: __________And finally we got equation describing the system:

Example 3.2:

Component balance: outin qcqcdt

dVc

There will take a certain amount of time for the flow to get from the tank to the pointwhere the fluid exits the pipe, which we have to take into account. So the concentration

of the outlet is the concentration in the tank a certain amount of time earlier.

From physics we knows=vtTd=l/v

We also know that: v=q/aHence: Td=la/q

And we finally get: qlatqcqcdt

dcV in /

From this we can see that the goal is to describe at least one of the variables on left handside in the variable that is present in the differentiation. So in example 1 described qoutas

a function of h and coutis described as c.

22

2

222

2

111

vghp

vghp

qA

hA

gah

dt

d 12


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Exercise:


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Lec 4: Laplace transform

Learning objectives: In this topic, Laplace transforms are introduced and itsuse in process control and solving differential equations are discussed.

Learning outcomes: At the end of the lesson, students must be able to carryout Laplace and inverse Laplace transforms using tables, use Laplace

transform to solve simple differential equations and change a control

equation into one utilizing deviation variables.

IntroductionA mathematical transform takes an expression in one mathematical "language" andconverts it to another. If done correctly, a transform doesn't change the meaning of the

expression, but may make it easier to interpret, as with a unit conversion. We usually

choose to use transforms to gain insight or simplify a problem.

The Laplace transform is commonly used in process modeling and control. The

transform is given by

The transform produces several changes in the equation:

Variable is t(time) Variable is s (dimensions of inverse time)

tis Real number s is Complex number

Solutions from "Time Domain" Solutions from "Laplace Domain"

Differential equation Algebraic equation

The last point is the biggest single reason the Laplace transform is valued -- it transforms

linear differential equations into algebraic equations, which many people find easier to

solve.

As an example, we'll apply the definition of the Laplace transform to the unit step

function. The step function is very important in control modeling. It is given by:

at

atat

0

1


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Most of the time, the "trigger" time (the value a) is set to zero, and the unit step function

defines what we typically think of as a constant (that exists from "the beginning oftime"!):

f(t)=K(t)

Applying the definition of Laplace transform to this function gives:

sK

sK

seKdteKdtetKtKLtfLsF

ststst

100 00

You can always use the definition to find the Laplace transform of a function, but thatusually is more trouble than it is worth. First of all, you can probably work the vast

majority of control problems with a set of about 4 transforms -- so you'll probably end upmemorizing those.

Properties of the Laplace Transform

The Laplace transform is a "linear operator", so we have that

L{f(t)+g(t)}=L{f(t)}+ L{g(t)}=F(s)+ G(s)

L{cf(t)}= cL{f(t)}= cF(s)

Derivatives

Derivatives also produce a very nice result:

00 fssFdt

tdfLorxssX

dt

dxL

This is the key result that causes time domain differential equations to transform to

Laplace domain algebraic equations. In the Laplace domain, multiplication by s isequivalent to differentiation in the time domain. Similarly, Laplace domain division by s

can be shown to be equivalent to integration in the time domain.

This can be extended to second order derivative:

0)0(0)0(0 22

2

xsxsXsxxssXsx

dt

dxsL

dt

dx

dt

dL

dt

xdL

And for third order

0)0()0(233

2

xxsxssXsdt

xdL

And so on


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Initial Value and Final Value Theorems

There are two very useful theorems involving Laplace transforms, which basically

enables you to find out where your system is heading (the new steady state for aparticular system) or where the system started from (the original steady state), without

out actually having to determine the functionf(t) from the ODE.

Final Value Theorem

which means that the final value off(t) is the same as initial value ofsF(s).

Initial Value Theorem

which means that the initial value off(t) is the same as final value ofsF(s).


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Exercise: You have a flow system as in Example 3.1, but we assume that qout can bemodeled as kh, which gives the following balance equation Adh/dt=q-kh. The values ofAand kcan be taken to be 5m and 1 m/s respectively.

1. Carry out the Laplace transform of the balance equation to determine the function H(s).

2. For q=0.5 m3/s (which gives Q(s)=0.5/s) use the final value theorem to determine the

final steady state levels of the tank.

Solving ODEs with Laplace Transforms

The major reason people bother with Laplace transforms is that they can make it easier to

obtain analytical solutions of many linear ordinary differential equations. The procedure

can be described in diagram form as:

Example 4.1:

Consider the ODE and initial condition:

2003 xxdt

dx

We start with transforming the equation

0/00

333

20

sL

XxLxL

sXxsXdt

dxL

Then we rearrange the equation to findX.

3

2

23032

sX

XsXsX


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Now, all that is necessary is to invert the Laplace transform to find the time domainsolution. Look in the table for a useful transform:

as

eL at

1

and then use this to invert the solution:

tes

Ls

Ls

Ltx 3111 23

12

3

12

3

2

Using this approach, many linear ODEs can be solved with a little algebra and a table ofLaplace transforms.

Partial Fractions Expansion

Not every solution is likely to be found in a table like;

But using the properties of the transform you see that you need to break problems up intosmaller, more familiar ones were the first step is to expand the equation:

Followed by partial fractioning

Any fraction with a polynomial denominator can be expressed as the sum of terms withfirst order denominators and then Laplace transform every single term to obtain:

We can note that it the roots are real and ri>0 we get exponential growth of that term and

ifri


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111 21

s

c

s

c

ssX

Once expanded, the terms look pretty simple to work with. All that is needed is to find

values for the constants c1 and c2. The steps are:

1. multiply through by one of the denominator factors2. set s=r1. All the constants but one will vanish.3. repeat for the other factors

110

0

10

10.2

11

1

11

1.1

12

1

21

21

cc

cs

s

csc

ss

c

s

c

sss

(All but one of the constants vanishes each time, so it isn't really even necessary to write

those terms out.):

1/s=c2 for s=-1 c2 = -1And now the inversion is easy:

te

sL

sL

ssL

ssLtf

1

1

11

1

11

1

1)( 1111

Exercise 4.1:The method becomes more complicated when roots are repeated. You must always have

one term, and one unknown constant, for each root. So for repeated roots you get:

1111 3

2

21

2

s

c

s

c

s

c

ss

For c1 and c2 we got on as before, multiply by the denominator of the constant and insertthe root in place of s to get: c1=___________________ and c2=___________________.However if you try to do the same for you will end up with infinities. What you will have

to do is to multiply by the highest order present of the root connected to the c3, in this

case (s+1) which gives:

1

1132

2

1

sccs

sc

s

and then differentiate with regard to s to get:

32

2

11

2

1121c

s

sc

s

sc

s

After which we can insert the s=-1 and continue and get c3=___________________.Which means we have got it properly fractioned and can carry out the inverse Laplace

transform to getf(t)=____________________________________.

Exercise 4.2:What would be the procedure if there was a root of third order instead?

Two last comments:


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partial fractions expansion works even if the denominator factors are complex

numbers, it just makes the algebra a little trickier if you only care about the "shape" of the solution, not the relative weightings, you

may not need to evaluate the constants.

Deviation VariablesMost real process variables are functions of time. Typically when if the system is affectedby a disturbance, the values will fluctuate around a value, sometimes slightly higher,

sometimes lower, but in the end it will commonly settle in on that value, which would be

a new steady state. When we are controlling a system, we want to make small

compensatory changes to flows, etc., to try and pull the process back to set point. Itmakes more sense to calculate the change needed rather than calculating the new valve

position from scratch every time. This is made easier if we keep track of how a variable

differs from its steady-state value instead of tracking its total value.

We can define any variable x as the sum of two parts: the average or steady-state value(xss) and the deviation orperturbation from that value (x, xor Y).

txtxtx ss or rearranged txtxtx ss In control system analysis, we are typically more interested in the deviation variable

(a.k.a. perturbation variable), so it is common practice to rewrite systems in terms of

deviation variables.

Many analysis techniques (such as Laplace transforms) are limited to linear systems.Linear systems have certain big advantages when using perturbation variables:

constant terms in many ODEs vanish

if we use perturbation variables and linearize around the steady-state, initialconditions are zero

These may be best illustrated by examples.

Example:Take the equation

tmcxdt

dx

into deviation variables. Begin by noticing that at steady state we get:

00 mcxormcx ssss What we do next is to subtract the steady state equation from the ODE;

ssss mtmccxxdt

dx

And then substitute the deviation variable for any difference between actual variable

and steady state variable;

tmxdt

dx

The problem here is that the differentiation doesnt contain the deviation variable, but ifdifferentiate the deviation variable we get:

dt

dx

dt

dx

dt

dx

dt

xxd

dt

xd ssss


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because the steady state value is a constant and hence have a derivative that is zero. So

we have that we can write the ODE with all deviation variables:

tmxdt

xd

The additive constant "biasing" terms have vanished. Also notice that using deviationvariables means that we do not need a value for the steady-state positions ofx and m tosolve the equations, or rather they naturally become zero.

Exercise:

Show that the last statement is correct by taking the Laplace transform ofdt

xd.


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Lec 5: Transfer functions and block diagrams

Learning objectives: To introduce the students to the transfer function andblock diagrams and its use in process control.

Learning outcomes: On completion of this topic, students must be able tocarry out manipulation of transfer functions to block diagrams and vice versa

and carry out block diagram manipulations.

A block diagram is a common way to represent a dynamic system. In these, signals

(variables) are represented by lines and functional relationships (transfer functions) by

blocks.

So if we return to the example with deviation variables from the previous chapter:

tmxdtxd

.

Carrying out the Laplace transform we get;

__________________________

This might be translated into a transfer function (G(s)) by collecting all variables on one

side we get:

1

1

ssM

sXsG

These ODE and the transfer function are two different representations of the sameequation. A third approach is to represent the system with a block diagram:

This diagram shows that X(s) (the output) is produced by the transfer function in theblock acting on the inputM(s), the equation given by:

sMssX

1

1

When you read an equation from a block diagram, the easiest way is to start at the output

(often on the right) and work backwards adding in the elements as you see them.Block diagrams make it easy to represent the connection patterns between processes, and

provide a way of visualizing that connection and converting it to math.

Let us revisit the heated tank:


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For which we have the differential equation:

The output variable of a system is the variable that commonly appears in the

differentiation, which means that the output variable is:________________________

And all the other variables are therefore inputs (whether they are disturbances or

controlled or manipulated variable). In this case it means that the disturbance is :_____and the manipulated variable is:____________.

Write the steady state material balance:____________________________________

Subtract it from the original ODE:________________________________________

Specify the deviation variables and insert it into the ODE:_____________________

As we can see there is not really much change of the ODE, but as we remember we do

this because_______________________.

We saw initially that we had two different inputs, which means we must have two

transfer functions as well qVsc 1

and qVs

q

.

Most transfer function can be separated into two different parts depending on whetherthey will determine how they affect the final change (difference between initial and final

steady state) and how they affect the behavior between the steady states. The constants

that affect the size of the change are the gains, commonly denoted by Ks, and theconstants regarding the behavior are the time constants, s.The standard form of representing the gains and the time constants in the transfer

function is:

1

s

KG

or if we have two time constants

11 21

ss

KG

To use the standard definition we have to rearrange the transfer function to the standardform, which for the previous examples means:

inqTc

uqT

dt

dTV


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qV

qcK

sqV

qc

sqVqcqVscG

111 ,

1

1

1

1

11

and for the second part we get

G2=___________ which gives K2=___________ and 2=_____________

To have a quick look at how the time constant and gain affects the behavior of the system

we study the part regarding heating elements effect on the temperature in the heatingtank, we have:

us

KT

11

1

Its common to assume that the heat changes according to a step change, which meansthat U(s)=a/s where a is the magnitude of the change, for simplicity assume a=1.

ss

KT

1

11

1

partial fractioning givess

K

s

KT 1

1

11

1

and if we carry out the inverse

Laplace transform we get: 111 1)( KeKtTt

One part is exponential (assuming that 1 >0), which will disappear as the system

progresses and the final change will just be K1.

Exercise:Make a simple plot ofT(t) from t= 0 s to t= 20 s for the four possible combinations ofvalues for 1= 0.5 and 1 and K1= 1 and 2.

Block diagram

Block diagram is a very useful tool to graphically represent the transfer functions,especially when it comes to system with interconnected transfer functions. Block

diagrams can be used to show an array of mathematical operations; summing, subtractionand multiplication, which are then combined to more advanced transfer functions.

Summing junctions are used to show addition


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221121 GuGuyyy

or subtraction

221121 GuGuyyy

Which, can, with the addition of a splitting point, can be used to represent parallel

systems;


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If we just study the input-output system we could represent the system with just a singleblock that describes the system.

We must have that the block is representing;

A case that commonly turns up when we have to deal with summing points is distributive

properties. This can be seen as doing a block expansion, getting more block than westarted with:

sUsGsG

sUsGsUsGsYsYsY

21

2121

sUsGsGsUsGsY 21

sGsGsG 21


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Exercise: Draw the distributive equivalent of the system.

Before we go onto more complex systems, we introduce the case with transfer functions

in series as well;

This diagram is written with a help function y1, which might be a helpful idea to use

when you come to block diagram reduction. This means that we can use what we learnt

so far by writing this system as two separate systems:

But if we use the definition of the help functiony1=u2 we get:

This means that blocks in parallel are the same thing as multiplication and just as

multiplication they are associative and commutative.

The summing, subtraction, parallel system, series systems and the distributive properties

are basically all we need to be able fro block scheme reduction or to determine the overall

transfer function of a complex system. But we commonly include the feedback system asstandard procedures as well as they turn up so often. The simplest feedback system looks

like this (this is not totally true as we could make it even simpler by putting H(s)=1):

222111 uGyanduGy

1122

222112

uGGy

uGyanduGu


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The note that should be made of how to simplify this diagram is that what is fed back isthe output itself, hence we could draw diagram as:

This becomes a very straight forward case giving us the following algebraic equation

describing the system_________________________.

Now the only thing we have to do is to rearrange the equation to get all the outputs on the

left hand side of the equation and the input on the right hand side of the equation to get:

sR

sHsG

sGsY

1


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We could also have so called positive feedback, where both signs at the summing point

are positive, which essentially means that the transfer functionH(s) is multiplied by -1. Itwill have the same effect on the final transfer function, hence getting a negative sign in

the denominator.

When dealing with feedback system we occasionally come across open loop transferfunction, which is the transfer function going from the y(t) that is fed back to the y(t) of

the output. For this transfer function the variable L(s) is used and hence for the simplecase above we have:L(s)= G(s) H(s).

Block diagram reduction

Block diagram reduction is a very important tool in process control to enable to simplify

the system (in the sense of the number of blocks) and enables us the get one single

transfer function that is easier to handle in the continuing applications of process control.

When we are dealing with simple straight forward systems and/or systems with a singlefeedback loop, we can just start applying the rules of summing, subtraction, parallel,series, distributive and feedback in a direct block diagram manipulation.

The alternative is to carry algebraic manipulation instead, which basically means that we

keep going from left to write adding in the different inputs as we go along and using

addition, subtraction and multiplication as it comes and in the end getting an algebraic

equation that need some manipulation to collect all inputs and outputs on opposite sides.

Exercise/Example:

So let us study the two different techniques for the following feedback system withdisturbance where the different blocks represent different functions in a normal controlsystem:

Start with the easy things, we see three block in series that we can change into one:


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For simplicity we put d=0 and get:

For which is just have to apply the feedback law to get the transfer function:_________Than we repeat the procedure to get the transfer function from d to y, starting withputting ysp=0 and rearranging the block diagram.

For which we again apply the feedback law to get the transfer function:____________.We should note that we have got the same denominator in both transfer functions, which

is due to the two systems having the same open loop transfer functionL.

Now lets try doing with algebraic manipulation.Start from left with y

spfollowed by a negative summing of yGM that gives: __________

Next we go through Gc , followed by Gv , and finally Gp that we multiply with the

previous transfer function to get:______________________We come to another summing point where we add dGMto get___________________.And then we finally reach y so we set our previous function equal to y to get

y=(ysp

-yGM) Gc Gv Gp + dGMFinally we rearrange it to get y=________ ysp +_____________ d.

Occasionally we will come across even more complicated systems for which we willneed another tool or trick, which is moving of summing places, which is fairly often


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complemented by introducing help functions. This is done to simplify the block diagram

and enable us to use the standard techniques of block diagram reduction or algebraic

manipulation.

Example/Exercise:

The following diagram is actually a standard diagram for a feedback/feedforward controlsystem. The algebraic manipulation can be carried out directly starting from the left at

r(t) going straight toy(t) and carrying out the algebraic operators along the way and thecarrying out the algebraic manipulation:

Y(s)=______________________________________________

If we, on the other hand, wish to carry out block diagram reduction it would be

troublesome when we come to finding the transfer function G2=Y(s)/V(s) as it have two

different entries into the loop, while we should be able to write down the transfer function

G1=Y(s)/R(s) after just a quick look at the diagram and using the formula for feedback

loops.

G1=_____________________________________

If we just had one summing point for G2 it would be just as straightforward. When we

remove summing points we have to be sure to compensate for whatever changes we areincurring so we dont change the system outlined.Lets take a look at the part of the diagram that we want to change:


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The goal is to remove one of the summing points and carry out compensation for the

changes so that overall transfer function remains unchanged.It doesnt matter which summing point we remove but the tendency is to keep the onefurthest to the right (ii in this case), which means connect signal (3) to summing point

(ii). We see that if we draw the line (3) straight into (ii) we dont have the signal passingthrough the Gu block and we would have altered the system. So as soon as we bypass a

block we have to compensate by multiplying by that transfer function:

As we end up with a system of parallel block we simplify to get a single block:Gv2=FfGu+Gv

The opposite will be the case if we move the other way around, i.e. move signal (4) to

summing point (i) as it makes the signal pass by an extra block. So as it passes an extrablock, we compensate by dividing the original function by the extra transfer function.


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This is also a case of two parallel blocks that can be simplified in the same way:

Gv2=____________________________________

After that its just a matter of drawing the modified block diagram and carry out theapproach from the previous example, i.e. making the v be the only input and imagine or

redraw the system on the standard feedback form and end up with a single transfer

function, Gvv, for the relation between Y(s) and V(s).

Gvv=_______________________________________

In the previous examples the block diagrams could be solved directly with the algebraicmanipulationnot needing to do any manipulations, which might imply that its superiorto the reduction method. However, this is not always the case especially if we start

having internal loops in the system, like the diagram below. When that is the case wehave to simplify the loop to a single block.

We study the part by itself:

which obviously is a feedback loop and can be simplified by the normal reduction

method to get Gfb,internal=_________________


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When that is done, we just substitute it into the block diagram in the place of the loop that

we studied.

References:

Smith and Corripio: Chapter 3

Seborg, Edgar, Mellichamp: Chapter 4, Chapter 11

Marlin: Chapter 4


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Lec 6: Linearization

Learning objectives: To introduce the student to response chart ofprocesses and the method to linearise non-linear transfer functions.

Learning outcomes: At the end of the lesson, students should be able todraw a response diagram, relate the look to transfer function parameters and

to linearise non-linear transfer functions.

Many of an engineer's tools for analyzing dynamic systems apply only to linear systems.

The Laplace transform, for instance, only works if the equations to be transformed are

linear, except for the time variable t.

What makes an equation "linear"?

all variables present only to the first power

no product terms where variables are multiplied (constants are ok) no square roots, exponentials, products, etc. involving variables

These can be understood by looking at some examples.

tmdt

dxa or tma

dt

dx are linear as long as a is a constant and m(t) is linear.

tmdt

dxa is nonlinear because of the square root.

tmdt

dxaxxa

dt

dxa 23212

11 is nonlinear because of the cross-product x1x2 term, while

tmxdtdxa

dtdxa 3

22

11 is linear when m(t) is linear.

2t

dt

dxa is linear when a is a constant.

Linearity is useful, because if f(x) is a linear differential equation the following

statements are true:1. Ifx1 is a solution to the equation and c1 a constant, then c1x1 is also a solution2. Ifx1 andx2 are solutions to the equation, thenx1+x2 is also a solution.

The latter means that for a linear process, the result of two input changes is the sum of the

results of the individual changes.

Making a Model LinearMany chemical engineering systems are highly nonlinear and general methods for

working with nonlinear models are few, so it is important to know how to approximatenonlinear equations with linear ones.


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The approach is really pretty straightforward:

First, expand all nonlinear terms in a Tayalor series, usually around the steady state value(a):

32!3!2!1

axaf

axaf

axaf

afxf

Second, truncate the expansion after the 1st

order terms (i.e. remove the non linear parts)This gives a general result for linearizing equations:

axafafxf

Notice that when you linearize, you do so around a specific point. Choice of this point is

important. If the linear version of your model is to work, you must be operating close to

the chosen point, so that you remain within the region where the linear approximation is

valid. The steady state value is the usual choice since control systems are most often used

to reject disturbances moving the plant away from steady state.

Example:Linearize: around x=1.

The nonlinear terms are on the right hand side and its usually easier to linearize themseparately:

Linearization in higher dimensionsThe principle is the same, but just as mentioned earlier we have nonlinearity as soon as

we have any multiplication or division between variables as well.We have to linearize regarding all variables that are present in the nonlinear function but

only those that are nonlinear, so if z, u, and y are variables and f(z,u,y)=zu+y the

nonlinearity is justzu and hence the only part we have to linearize.To do this we would have to carry out Taylor expansion in all dimension that are

nonlinear:

xxxy sin2 23

xy

xxxxfxxxf

xxxxxfxxf

45.776.4

11cos21sin41sin2sin2_________________sin2)(

3211313)(

22

23323


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and keep adding terms as the dimensions increase.

Example:

Linearization and deviation variablesLinearization in combination with deviation variables has particular advantages. Recall

that a deviation variable is defined as

0xtxx so if we linearize aound the steady state, the value of the deviation variable will be zero.All of the "pure constant" terms will then vanish from the linearized deviation variables.Moreover, when we switch to perturbation variables and then linearize around the steady-

state, the initial conditions are zero. This means we can drop the x(0) terms as we take

Laplace transforms.

Example: Take the two variable case above into deviation variables.

wherez0u0will disappear as well due to the introduction of the deviation variable in thefirst place.

The last example showed that by combining linearization with perturbation variables, you

effectively change your nonlinear side of the differential equation:

Dynamic behavior

The dynamic behavior in this section is mostly dealing with how an uncontrolled systemchanges from one steady state to another caused by the change of one or more of the

inputs. In its simplest form its just a matter of plotting the output for the input change,but we want to do a little bit more than that. We also study what is the common behaviorof a system and what can be deduced if we are just given the response diagram (the plot

of the output).

But before we can look at the behavior of the system we take a look at the most commonways to describe the input.

)()(),(),( 0,

0

,

00

0000

uuu

fzz

z

fuzfuzf

uzuz

)()(),(

)()(),(),(

),(

000000

0

,

0

,

00

0000

uuzzzuuzuzf

uuu

fzz

z

fuzfuzf

zuuzf

uzuz

uzzuuzuzf

uuzzzuuzuzf

0000

000000

),(

)()(),(

uu

fz

z

fuzf

uzuz

0000 ,,

),(


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Process inputs

Step changeA step change is a sudden change of a function from one value (commonly zero) to

another value at time t=a (also commonly at zero)

The mathematical description of the function (), its Laplace transform (, note that theexponential function will disappear ifa=0) and diagram of the function:

Ramp changeA ramp change means that we have a change occurring where the value keep increasing

linearly as time progresses. The ramp could be delayed as well so that it change appear at

a time after t=0.

ImpulseAn impulse is, by definition, a change that occurs during an infinitesimal (basically 0)

amount of time and has a magnitude of the change being infinite. This is simulated by

making a rectangular pulse occurring during a time a and with a magnitude 1/a while weare trying to get the limit a0 . This is basically impossible to achieve (just imagine to

having to bring a Temperature or a flowrate to infinity in zero time), but the dynamic

behavior of the system due to a impulse change have many handy properties, which is

why its included here aswell as why its used in industry.

s

es

atz

atat

as

0

2

0

00

s

asX

tat

ttx

R

R

1

0

01

00

sX

awhen

ta

t

x

I

I


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Sinusoidal inputPeriodic variables also turn up quite frequently, for example if we were trying to controlthe temperature in a swimming pool, the ambient temperature would vary periodically,

with a 24 hour rhythm.

Rectangular pulseA rectangular pulse is basically a step change that just occurs for a limited amount of

time. The modeling of the pulse is basically done by taking the difference between to step

changes:

And in mathematical form:

To obtain Laplace transform we carry out the same steps as for the graphical derivation

above, giving thatxRP= (t)-(t-a) which can easily transformed using tables.

22

0sin

00

s

AsX

ttA

t

tx

asRP

RP

es

zsX

atatz

t

x

1

00

00


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1st Order SystemsA first order system is described by

In this model,y represents the measured and controlled output variable and x(t) the inputfunction. (Referred to as the forcing function of the ODE.)To simplify the problem (for the Laplace transformed case) we subtract the steady state to

get it expressed using deviation variables instead:

The equation is often rearranged to the standard form

This model is linear as long asx(t) is not a function ofy thus it can be transformed into atransfer function

This type of transfer function is known as afirst order lag with a steady state gain ofK.

Step ResponseIf we let the forcing function (X(s) or x(t) depending on what framework we are dealing

with) we get the output response that is known as a step response of a 1st order system.The shape should over the rest of the course become very familiar, as any 1

storder

system forced by a step function will response of this shape. The unit step response of asystem with time constant 1.0 is shown in the figure. "Unit step response" means that the

forcing function (the step) has magnitude 1.0 (i.e.x(t)=1 orX(s)=1/s.

ctbxtyadt

tdya )()(

)(01

)(')(')('

01 tbxtyadt

tdya

)(')(')('

tKxtydt

tdy

)()()( sKXsYssY

)(1

)( sXs

KsY


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As the system approaches steady state, the response approaches a constant value. In the

plot, this value is 1.0 (which is the value of the gain of the process). If we looked at ageneral case (with a forcing step function of magnitude z) we get:

and then carried out the inverse Laplace transform we have:

/exp1)( tzKty which indicates that the final value will be given by Kz.

Initial ResponseThe initial response (time close to zero) has a slope of 1.0. This is true of all first order

systems.

Time ConstantNext, consider what happens to the function when the elapsed time is equal to one time

constant (i.e. when t=)632.0)1( 1 ey

Thus, when one time constant has elapsed, the process output will have achieved 63.2%

of its final value (in the plot, 0.632).

The response will keep getting increasingly closer to the new steady state value, but

technically not reaching it in a finite time space. So we need an approximation of when

the steady states have been reached, which is commonly specified by getting to within

1% final value. As it turns out we can get a close approximation of that time by using 5as that would give:

1%0067.0)1( 5 ey

Ramp and Impulse ResponseBriefly, let's take a look at the response of the first order process to two additional typesof inputs.

First, consider a ramp function, CA0=Rt. Then

Partional fractioning and inversed Laplace transform yields;

Which is represented graphically as:

s

z

s

KsY

s

zsX

1

22 1

s

z

s

KsY

s

zsXtztx

ttzKezKtys

C

s

B

s

AsY t/

1)(


49/134

The straight line represents the input and the initially curved is the output. From this wecan see that the response is having an initial transient period, which is analogous to the

transient behavior of the step change and when the response settle in at a constant slope it

happens with a shift equal to (this is assuming that Kz is unitary, else we would get adifferent slope of the response compared to the input)

Notice that the impulse response is the derivative of the step response. In some cases, it iseasier to find the impulse response function by taking the derivative of the step response

than by integrating the impulse forms.


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Response of 2nd

Order SystemsThe first-order system considered in the previous section yields well-behaved

exponential responses. Second-order systems can be much more exciting since they

can give an oscillatory or under damped response.

We will consider only linear second order function ODEs that have constant

coefficients:

ctbxtyadt

tdya

dt

tyda )()(

)()(012

2

2

To simplify we can subtract the steady state:

cbxya )0()0(0

and apply deviation variables to get

020

1

0

2

2

22

,

2

,

)()()(

2)(

a

bK

aa

a

a

awhere

tKXtYdt

tdY

dt

tYd

This yields the following transfer function:

sXss

KsY

1222

Which have the following poles:

11112

2

2

2

1

1

sands

The two roots 1/1and 1/2 is of course a simples description of the case which we canuse when roots are easily found and specified to get the second order system described on

the form:

sXss

KsY11 21

As we save in chapter 4, if the pole (i.e. the root of a denominator) is real and negative or

imaginary with a negative real part we got an exponential decay which would mean that

the response would reach a steady state. If that were not the case we got exponential

growth towards, positive or negative infinity.The behavior can hence be very easily deduced for the transfer function described by 1and 2., as both of them have to be negative (or negative real parts) for stability.For the case when the second order system is described using and , we can see fromthe equations why they were chosen the way they were as it has made the stability of only

one variable; .In both of the square roots inside the roots we have -1, which means that the square rootgives rise to a imaginary root only when -1


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The last case of real poles are when -1, which of course means that the roots arepositive and hence exhibit exponential growth and is called a runaway system

When we have a case where the poles are imaginary, for stability we need to be in theinterval between 0 and 1 as directly determines the whether the roots real part ispositive or not. The system is then called an underdamped system. The common response

diagrams for these systems are shown in the figure on the following page. Before we goon and study underdamped systems we not that if is in the interval between -1 and 0 weget a positive real part and hence exponential growth, but with oscillations as its aimaginary pole.

Underdamped systemsThe most commonly studied case is the underdamped system as they are the mostcommonly occurring systems in process control. Hence a number of terms are used to

describe the underdamped response quantitatively. Equations for some of these terms are

listed below for future reference. In general, the terms depend of and/or . They were allderived from the step response formula, which is obtained the same way as for the 1

st

order system, step is applied as a forcing function followed by partial fractioning andinverse Laplace transform:

tttzKty sinexp

1

1

2

Where

21

21

tan,1

; however, the mathematical derivations are left

out.


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The characteristics of the underdamped system is quite obvious from the graph above.

The response is quite slow to start with and then picks up with an increasing slope, itovershoots the final value and start oscillating around what is the final value (the new

steady state) with decreasing amplitude to finally settle in at a new steady state.The following five terms are what we use to describe second order system, where we can

both use the function itself or read off the graph to determine the different terms:

1. Overshoot. Overshoot (or actually the relative overshoot) is a measure of how muchthe response exceeds the ultimate value following a step change and is expressed as the

ratioA/B in the figure.

2. Decay ratio. The decay ratio is defined as the ratio of the sizes of successive

peaks and is given by C/B.

3.Rise time. This is the time required for the response to first reach its ultimate

value and is labeled tr , in figure.


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4. Settling time or Response time. This is the time required for the response to come

within 5 percent of its ultimate value and remain there. The response time is indicated inthe figure by ts. The limits of 5 percent are arbitrary, and other limits have been used in

other texts for defining a settling time.

5Period of oscillation, time between consecutive peaks or troughs.First we have to determine the angular or radian frequency

The relation between angular frequency and frequency is the same as for SHM hence the

frequency (i.e. the inverse period) is :

The dampening ratio will affect all of the terms mentioned above as could seen analyzing

the transfer function, but also by a simple plot for various dampening ratios, .

Figure of the step response diagrams for different dampening ratios.


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Lec 7 Dynamic response of more advanced processesIn the previous chapter we discussed the behavior of processes that could be modeled

using a firstor second-order model. There are of course processes that are considerablymore complicated than those examples. These more advanced processes have the effect

on the transfer function that it either increases the order of the denominator, but it would

also be possible to have a function ofs in the numerator.

Poles and zeros and their effect on process responseOne feature of the first- and second-order processes described in the previous chapter isthat their responses are very easily deduced by studying the factors ( for 1storder and and for 2nd order) in the denominator of the transfer function. Things wont be asstraightforward, when we have additional poles or zeros. These things are highlighted

through examples:

Example; a system with a third poleWe have a system with the transfer function:

14.01

12

ssTs

sG

where we have added an additional pole, located at s=-1/Tto a second order system with

= _____ and = _____. If the system is subjected to a step change we get the following

behavior for the system:

From the diagram we can see that the extra pole creates a slower system compared to the

second order system by itself (T=0). As we keep increasing Twe ultimately get a root at

the same real value as the complex conjugate poles of the first order function ( T=5) for

which the system is dominated by the real pole, and the system behaves in a similarfashion to a first order system.

Example; a system with a zeroWe have a system with the transfer function:

14.0

1

ss

TssG

where we have added a zero at s=-1/T with the same values for and as in the previousexample. Again we study the step respone behavior:

From the diagram we can see that an additional zero is increasing the overshootespecially when the zero is close to the imaginary axis (i.e. when T is increasing).

The reason for the overshoot can be illustrated by assuming that y0(t) is the output for a

system without a zero (i.e. T=0), which means that the output including the zerobecomes:

tyTtytYTsLty 0001 1

The derivative of the signaly0(t) would initially be of the same magnitude as y0(t) itself,but the multiplication with Tmeans that the system with zero will get a greater addition


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the greater the value ofTis. This explains why we get an increased overshoot if we add a

zero close to the imaginary axis. From the figure we can see that y0(t) is approaching aconstant value as time progresses, which means that the derivative ofy0(t) approaching

zero. With the derivative disappearing for large times we can see that the final value ofthe system with a zero is the same as the one without the zero.

Zeros placed to the left of the imaginary axis is not very common in physical systems, but

the will turn up in most standard feedback controllers, as we want as fast behavior as

possible, which is obviously achieved with a zero as seen in the figure. If we instead adda zero to the right of the imaginary axis (i.e. a positive zero) we get a case that is more

common for physical systems.

Example: a system with a positive zeroThe transfer function we will use is made up of two additive systems, one has a slow

dynamics (( ) ( ) ) which will force the system in the positive direction andanother system that is fast (and with an amplification of T) which is force the system in

the opposite direction.

sT

s

TsG

2.01

1

1

1

The overall effect is that we get a positive zero; s=T. And we can see from the stepresponse figure below that the system initially becomes negative as the zero is positive,

before the slower dynamics of the other part makes the system positive. The magnitude

of the negative overshoot is increasing with the value ofT. IfT is chosen negative we get

the same behavior as we had for negative root.If we, again, assume thaty0(t) is the output for a system without a zero we get:

tyTtyty 00 This again means that we get the big effect of the derivative and that its amplified by the

size of T as well as the disappearance of the effect of the zero as the system converges.

Another way the system complexity can be increased is when the system has time delayalso referred to as a dead time, which commonly occurs in process where we have a

transportation over a distance, like fluid flow along a long pipe or if you have chemical

analysis for which you can get a fast reading, like a gas chromatographer.

Example: a system with time delayA time delay system would be a system of the form:

atatx

atty

0

The reason for the function being specified as followingx exactly but with that there is a

delay of the value, so if we were dealing with a first order system we would get thestandard first order response diagram, but with starting from a instead of at 0.


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Approximation of higher order systemsWhen an extra pole was added to the system we saw that for particular values we could

make the system behave very similar to a first order system. This is a fairly commonoccurrence and will happen as soon as we a dominating pole, which will be the case assoon as we have a pole that is closer to the imaginary axis then all other poles.

The addition of a positive zero was also described with the system then experiencing a

negative overshoot, which, if we neglect the overshoot itself, bears a resemblance to thebehavior of a system with time delay.

From this we could get the idea that we can simplify systems with positive zeros by

describing the zero as a time delay instead.

The simplification approximation of a higher order system can be highlighted by studyinga system that a multiple roots. The following figure has a pole repeated 5, 20, 100 time

and as the multiplicity increases, the system start to resemble a system with a time dealy:

This is the basis of the practice of approximating higher order systems by the use of time

delays. The only thing needed is an approximation relation between time constants and


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time delays, which can be achieved by studying the Taylor expansion of a time delay (i.e.

an exponential function):

!3!2

133

0

22

00

0stst

stest

which means if we just keep the two first of the expansion we get:

ste st 010

and with opposite signs:

stest

010

This is the simplest form to approximate a zero. To approximate a pole we use that

which gives:

stee

st

st

01

110

0

Those are the formulas that are commonly used to simplify a high order function down to

a system of first- or second-order with time delay. This is done by substituting the least

dominant time delays (poles farthest away from the imaginary axis) by using the Taylor

expansion above backwards (i.e. 1+5s becomes e5s or 1/(1+5s) becomes e-5s ).

Simple Taylor expansion approximation procedureFirst decide on if you want a first order or second orders model as the final result. Then

you find the dominant pole (for first order) or the two dominating poles (for a second

order model) and move them to your model. After that you go ahead and change the otherpoles and zeros into time delays using the Taylor expansions formulas.

Example/ExerciseReduce the transfer function given below to a first order model with time delay:

As we want a first order we just keep the pole closest to the imaginary axis, which is the

pole with the highest time constant, which in this case is:_____________________

Than we go through the other poles and change them into time delays;1/(3s+1)=__________ and 1/(0.5s+1)=_____________ and finally we do the same thing

for the zero(s) (1-0.1s)=____________.

This should give us a transfer function looking like this, which only needs the final touchto combine the time delays into one.

15

sG5.031.0

s

eeKesss

If a second order model was desired we would keep the two most dominating poles

instead ending up with:

1315

sG6.0

ss

Kes

15.01315

1.01

sss

sKsG


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Skogestads half ruleThere is a tendancy, when using the Taylor expansion straight as in the previous case, to

over emphasize the time delays in the system. This would essentially mean that the

approximation wouldnt fit very well initially. To get a better fit for the initial behaviorSkogestad proposed that we take the largest neglected (i.e. the largest of the time

constants that we didnt keep) and add half of it to the smallest kept times constant, whilethe other half is translated into a time delay using Taylor expansion. The other poles andzeros are(is) treated in the similar way as for the Taylor expansion method.

ExampleReduce the transfer function given below to a first order model with time delay using

Skogestads half rule.:

Just as before the time 1/(5s+1) is kept.

That would leave us to neglect (3s+1) and (0.5s+1), but we are supposed to take half ofthe biggest neglected time constant and add it to the smallest (in this case the only, thisbecomes important when producing second order models).

So we get ({5+3/2}s+1)=6.5s+1.

After that its just a matter of carrying on like in the previous example with addition ofmaking sure not to forget the other half of the biggest neglected time constant:

1/(3/2s+1)=____________, 1/(0.5s+1)=___________ and (1-0.1s)=__________

and we get:

We can compare the step response behavior for the two methods and the actual case. Inthe figure we can clearly see that Skogestads half rule creates a better approximation in

the initial stages creating a better approximation over all.

If we wanted a second order model, we would have taken out the two biggest time

constant to start with, (in this case 5 and 3) leaving 0.5 as the biggest neglected one. We

would take half of that and add to the smallest kept one. This would modify our kept timeconstants to 5 and 3.25, while we would have to translate the time constant left over

(0.25) as a time delay together with the zero to get:

15.01315

1.01

sss

sKsG

15.6

sG1.2

s

Ke s

125.315

sG35.0

ss

Kes


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Exercise:Use Skogestads approach to derive a first order with time delay and a second order withtime delay model for the following transfer function:

First order case: Find the biggest time constant:________.

Find the biggest neglected time constant:_______.

Add half of it to the biggest time constant to get the time constant used for themodel__________.

Change all the other poles and zeros into time delays getting:1.5+___+___+___+1=______

and we get the final function:

Second order case:(You should get a time delay=2.15)

105.012.013112

1

ssss

esKsG

s

15.13

75.3

s

KesGs


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Pad approximationsThe major advantage of the Laplace transform is that it translates the differential equation

models into transfer function models. However as a part of transformation we commonly

end up with exponential functions (due to time delays or an impulse). This is occasionally

causing some problem to describe the system in a simple form, which means we can

factor the transfer function into a form with just zeros and poles.One solution would be to use the Taylor expansion we used in the previous case, that

however would generally not be considered to be exact enough. To increase accuracy weuse the simplest pole-zero approximation (i.e. approximating the exponential part with a

combination of poles and zeros) which is the 1/1 Pad approximation;

st

stsG

e

ee

st

stst

0

01/15.0

5.0

5.01

5.010

0

0

Its called 1/1 Pad approximation because its first order in both numerator anddenominator.Performing long division of the approximation we get:

42133

0

22

001/1 stststsG

which if we compare with the Taylor expansion is correct in the three first terms. There

are higher order Pad approximations, for example the 2/2 Pad approximation:

1221

1221

22

00

22

00

2/20

stst

stst

sGest

Interacting and noninteracting processesMost of the systems that we have studied so far have been simple processes with a single

input and a single output, which commonly could be isolated and treated separately.

However many processes are not like that. Those processes usually have one or morevariables that interact with at least another variable, which creates a case of internal

feedback within the system, which is referred to as an interacting system. Most

interacting systems have more complicated transfer functions compared to thenoninteracting processes.

Example: noninteracting processThere is a dual tank system arranged as follows as seen on the following figure. The

flows through the valves are assumed to be linear relations on the following form:

qi=1/Rihi which means that we consider the valve acting as a resistance to the flow.In Chapter 3 and 4 a single tank was studied and we got the following balance equation:

outin qqdt

dhA which for our cases will be:

i

i

iini h

Rq

dt

dhA

1, where i is the number of

the tank with the following Laplace transfer function:1,

AsR

R

Q

H

i

i

iin

i


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The flow equation for the valves can also be transformed: ii

iout

RHQ 1,

This gives four equations in total, one material balance and one valve equation for each

tank. As everything is happening in after each other, we could draw a block diagram

showing the different transfer functions connected in series from Qin,1 to Qout,2 or H2,

whichever output is of interest. For example if the level in the second tank was the

controlled variable we would get:

111111

1 21

2

2211

2

22

2

111

1

2,

2

1

1,

1,

1

1,

2

ss

K

sARsAR

R

sAR

R

RsAR

R

Q

H

H

Q

Q

H

Q

H

in

out

inin

This could be generalized for any number of tanks in series as it would just be a matter of

continuing multiplying the additional transfer function together:

n

i

i

n

n

i

n

in

n

s

K

sAR

R

Q

H

11

111, 11

That is very easy to deal with if we wanted to find the behavior we could go straight forpartial factoring followed by the inversed Laplace transform.

Example: interacting process

Next consider an example of an interacting process that is similar to the previousexample. The tanks are still in series but now the level of tank 2 will affect the level of

tank 1 as the flow through the valve between the tanks will be given by:

)(1

21

1

1, hhR

Qout which you could derived using Bernoullis equation.


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The balance equations will essentially be the same as before until we substitute the

definition of the outflow from tank 1.

21

21

1

222

2

21

1

22

21

1

1,1121

1

1,1

1

1111

11

H

R

HH

R

sHAh

R

hh

Rdt

dhA

HHR

QsHAhhR

qdt

dhA

Laplace

in

Laplace

in

Substituting the H2 we get from the tank 2 equation into the tank 1 equation gives:

12

)1('

1)(

)1)((

)(

)(22

1

121122

2

2121

21

22121

1,

1

ss

sK

sARARARsAARR

sRR

ARRRR

sQ

sH a

in

This means we have gone from a first order description to a second order model with anadditional zero as well. The process can be reversed as well substituting the tank 1

balance into the tank 2 balance:

12

'

)(

)(22

2

1,

2

ss

K

sQ

sH

in

The analysis has become more complicated for the interacting system compared to the

non-interacting system. The denominator is not directly factored into two time constantsbut instead is in the form a quadratic equation, and it appears for both tanks and not as in

the previous case just for the overall system.

Multiple-Input, Multiple-Output (MIMO) processesMost industrial processes will contain multiple inputs (manipulated variables) and

multiple outputs (controlled variables). These are referred to as MIMO systemsdistinguishing them from the single-input, single-output (SISO) case. The modeling of a

MIMO system is basically the same as for the SISO system and follows the same steps,there are just more variables to deal with.

Example: MIMO systemLets return to trying to control the temperature in a tank by heating, but without thevolume being a constant.The balances we get are:


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qqqdt

dVMB

TTqTTqTTqdt

TTVdEB

ch

refrefccrefhh

ref

:

:

with Tref=0

qqqdt

dVMB

qTTqTqdt

VTdEB

ch

cchh

:

:

First we expand the derivative in the energy balance using the chain rule: dt

dTV

dt

dVT

dt

VTd

The second term in the expansion contains dV/dt which is present in the MB as well andhence

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