Process Control Slides

8/17/2019 Process Control Slides

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Development of Dynamic Models

I l lustrative Example: A Blending Process

An unsteady-state mass balance for the blending system:

rate of accumulation rate of rate of

(2-1)of mass in the tank mass in mass out

= −



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or

where w1, w2, and w are mass flow rates.

( )

1 2

ρ(2-2)

d V w w w

dt = + −

The unsteady-state component balance is:

( )1 1 2 2

ρ

(2-3)

d V x

w x w x wxdt = + −

The corresponding steady-state model was derived in Ch. 1 (cf.

Eqs. 1-1 and 1-2).

1 2

1 1 2 2

0 (2-4)

0 (2-5)

w w w

w x w x wx

= + −

= + −



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General Modeling Principles

• The model equations are at best an approximation to the real process.

• Adage: “All models are wrong, but some are useful.”

• Modeling inherently involves a compromise between model

accuracy and complexity on one hand, and the cost and effort

required to develop the model, on the other hand.

• Process modeling is both an art and a science. Creativity is

required to make simplifying assumptions that result in an

appropriate model.

• Dynamic models of chemical processes consist of ordinary

differential equations (ODE) and/or partial differential equations

(PDE), plus related algebraic equations.



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Table 2.1. A Systematic Approach for

Developing Dynamic Models

1. State the modeling objectives and the end use of the model.

They determine the required levels of model detail and model

accuracy.2. Draw a schematic diagram of the process and label all process

variables.

3. List all of the assumptions that are involved in developing themodel. Try for parsimony; the model should be no more

complicated than necessary to meet the modeling objectives.

4. Determine whether spatial variations of process variables are

important. If so, a partial differential equation model will be

required.

5. Write appropriate conservation equations (mass, component,

energy, and so forth).



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Table 2.1. (continued)

6. Introduce equilibrium relations and other algebraic

equations (from thermodynamics, transport phenomena,

chemical kinetics, equipment geometry, etc.).

7. Perform a degrees of freedom analysis (Section 2.3) to

ensure that the model equations can be solved.

8. Simplify the model. It is often possible to arrange the

equations so that the dependent variables (outputs) appear

on the left side and the independent variables (inputs)

appear on the right side. This model form is convenient

for computer simulation and subsequent analysis.

9. Classify inputs as disturbance variables or as manipulated

variables.



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Table 2.2. Degrees of Freedom Analysis

1. List all quantities in the model that are known constants (or

parameters that can be specified) on the basis of equipment

dimensions, known physical properties, etc.

2. Determine the number of equations N E and the number of

process variables, N V . Note that time t is not considered to be a

process variable because it is neither a process input nor a

process output.

3. Calculate the number of degrees of freedom, N F = N V - N E .

4. Identify the N E output variables that will be obtained by solving

the process model.

5. Identify the N F input variables that must be specified as either

disturbance variables or manipulated variables, in order to

utilize the N F degrees of freedom.



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Conservation LawsTheoretical models of chemical processes are based on

conservation laws.

Conservation of Mass

rate of mass rate of mass rate of mass(2-6)

accumulation in out

= −

Conservation of Component i

rate of component i rate of component i

accumulation in

rate of component i rate of component i (2-7)out produced

=

− +



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Conservation of Energy

The general law of energy conservation is also called the First

Law of Thermodynamics. It can be expressed as:

= −

+ +

rate of energy rate of energy in rate of energy out

accumulation by convection by convection

net rate of heat addition net rate of work

to the system from performed on the sys

the surroundings

tem (2-8)

by the surroundings

The total energy of a thermodynamic system, U tot , is the sum of itsinternal energy, kinetic energy, and potential energy:

int(2-9)

tot KE PE U U U U = + +



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For the processes and examples considered in this book, it

is appropriate to make two assumptions:

1. Changes in potential energy and kinetic energy can beneglected because they are small in comparison with changes

in internal energy.

2. The net rate of work can be neglected because it is smallcompared to the rates of heat transfer and convection.

For these reasonable assumptions, the energy balance in

Eq. 2-8 can be written as

( )int (2-10)dU

wH Qdt

= −∆ +

int the internal energy of

the system

enthalpy per unit mass

mass flow rate

rate of heat transfer to the system

U

H

w

Q

=

=

==

( )

denotes the difference between outlet and inlet

conditions of the flowing

streams; therefore

-∆ wH = rate of enthalpy of the inletstream(s) - the enthalpy

of the outlet stream(s)

∆ =

h l i f l i i i



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The analogous equation for molar quantities is,

( )

int (2-11)dU

wH Q

dt

= −∆ +

where is the enthalpy per mole and is the molar flow rate. w

In order to derive dynamic models of processes from the general

energy balances in Eqs. 2-10 and 2-11, expressions for Uint and

or are required, which can be derived from thermodynamics.

ˆ

The Blending Process Revisited

For constant , Eqs. 2-2 and 2-3 become: ρ

1 2 (2-12)dV

w w wdt

ρ = + −

( )1 1 2 2 (2-13)

d Vxw x w x wx

dt

ρ = + −

E i 2 13 b i lifi d b di h l i



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Equation 2-13 can be simplified by expanding the accumulation

term using the “chain rule” for differentiation of a product:

( ) (2-14)d Vx dx dV V xdt dt dt

ρ ρ ρ = +

Substitution of (2-14) into (2-13) gives:

1 1 2 2 (2-15)dx dV

V x w x w x wxdt dt

ρ ρ + = + −

Substitution of the mass balance in (2-12) for in (2-15)

gives:

/dV dt ρ

dx( )1 2 1 1 2 2 (2-16)V x w w w w x w x wx

dt ρ + + − = + −

After canceling common terms and rearranging (2-12) and (2-16),

a more convenient model form is obtained:

( )

( ) ( )

1 2

1 21 2

1(2-17)

(2-18)

dV w w w

dt

w wdx x x x x

dt V V

ρ

ρ ρ

= + −

= − + −



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Stirred-Tank Heating Process

Figure 2.3 Stirred-tank heating process with constant holdup, V .



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Stirred-Tank Heating Process (cont’d.)

Assumptions:

1. Perfect mixing; thus, the exit temperature T is also thetemperature of the tank contents.

2. The liquid holdup V is constant because the inlet and outlet

flow rates are equal.3. The density and heat capacity C of the liquid are assumed to

be constant. Thus, their temperature dependence is neglected.

4. Heat losses are negligible.

ρ



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Model Development - I

For a pure liquid at low or moderate pressures, the internal energyis approximately equal to the enthalpy, U int , and H depends

only on temperature. Consequently, in the subsequent

development, we assume that U int = H and where thecaret (^) means per unit mass. As shown in Appendix B, a

differential change in temperature, dT , produces a corresponding

change in the internal energy per unit mass,

≈

ˆ ˆint U H =

ˆ,int dU

intˆ ˆ (2-29)dU dH CdT = =

where C is the constant pressure heat capacity (assumed to be

constant). The total internal energy of the liquid in the tank is:

int intˆ (2-30)U VU ρ =



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Model Development - II

An expression for the rate of internal energy accumulation can bederived from Eqs. (2-29) and (2-30):

int (2-31)dU dT

VC

dt dt

ρ =

Note that this term appears in the general energy balance of Eq. 2-

10.

Suppose that the liquid in the tank is at a temperature T and has anenthalpy, . Integrating Eq. 2-29 from a reference temperature

T ref to T gives,

ˆ

( )ˆ ˆ

(2-32)ref ref H H C T T − = −where is the value of at T ref . Without loss of generality, we

assume that (see Appendix B). Thus, (2-32) can be

written as:

ˆref

ˆ

ˆ 0ref H =

( )ˆ (2-33)ref H C T T = −



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Model Development - III

For the inlet stream

( )ˆ (2-34)i i ref H C T T = −

Substituting (2-33) and (2-34) into the convection term of (2-10)

gives:

( ) ( ) ( )ˆ (2-35)i ref ref wH w C T T w C T T −∆ = − − −

Finally, substitution of (2-31) and (2-35) into (2-10)

( ) (2-36)i

dT V C wC T T Q

dt

ρ = − +

D f F d A l i f th Sti d T k



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Degrees of Freedom Analysis for the Stirred-Tank

Model:

, ,V C ρ 3 parameters:

4 variables:

1 equation: Eq. 2-36

, , ,iT T w Q

Thus the degrees of freedom are N F = 4 – 1 = 3. The process

variables are classified as:

1 output variable: T

3 input variables: T i , w, Q

For temperature control purposes, it is reasonable to classify the

three inputs as:

2 disturbance variables: T i , w1 manipulated variable: Q



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Biological Reactions

• Biological reactions that involve micro-organisms and enzyme

catalysts are pervasive and play a crucial role in the natural

world.• Without such bioreactions, plant and animal life, as we know

it, simply could not exist.

• Bioreactions also provide the basis for production of a widevariety of pharmaceuticals and healthcare and food products.

• Important industrial processes that involve bioreactions include

fermentation and wastewater treatment.

• Chemical engineers are heavily involved with biochemical and

biomedical processes.

Bi i



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Bioreactions

• Are typically performed in a batch or fed-batch reactor.

• Fed-batch is a synonym for semi-batch .

• Fed-batch reactors are widely used in the pharmaceuticaland other process industries.

• Bioreactions:

• Yield Coefficients:

substrate more cells + products (2-90)cells→

/ (2-92) P S mass of product formed Y mass of substrated consumed to form product =

/ (2-91) X S mass of new cells formed Y

mass of substrated consumed to form new cells=

F d B t h Bi t



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Fed-Batch Bioreactor

Figure 2.11. Fed-batch reactor

for a bioreaction.

Monod Equation

Specific Growth Rate

(2-93) g r X µ =

max (2-94) s

S

K S µ µ =

+



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Modeling Assumptions

1. The exponential cell growth stage is of interest.2. The fed-batch reactor is perfectly mixed.

3. Heat effects are small so that isothermal reactor operation can

be assumed.4. The liquid density is constant.

5. The broth in the bioreactor consists of liquid plus solid

material, the mass of cells. This heterogenous mixture can beapproximated as a homogenous liquid.

6. The rate of cell growth r g is given by the Monod equation in (2-

93) and (2-94).



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Modeling Assumptions (continued)

7. The rate of product formation per unit volume r p can beexpressed as

/ (2-95) p P X g r Y r =

where the product yield coefficient Y P/X is defined as:

/ (2-96) P X mass of product formed Y mass of new cells formed =

8. The feed stream is sterile and thus contains no cells.

General Form of Each Balance

{ } { } { } (2-97) Rate of accumulation rate in rate of formation= +



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Individual Component Balances

Cells:

Product:

Substrate:

Overall Mass Balance

Mass:

( ) (2-98) g d XV V r

dt =

1 1(2-100) f g P

X / S P / S

d( SV ) F S V r V r

dt Y Y −= −

( )d V F =

( ) (2-99) pd PV Vr

dt =

(2-101)dt

L l T f



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Laplace Transforms

• Important analytical method for solving linear ordinary

differential equations.

- Application to nonlinear ODEs? Must linearize first.

• Laplace transforms play a key role in important process

control concepts and techniques.

- Examples:

• Transfer functions

• Frequency response

• Control system design

• Stability analysis

Definition



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Definition

The Laplace transform of a function, f (t ), is defined as

[ ] ( )0( ) ( ) (3-1) st F s f t f t e dt

∞ −= =

∫L

where F ( s) is the symbol for the Laplace transform, L is the

Laplace transform operator, and f (t ) is some function of time, t .

Note: The L operator transforms a time domain function f (t )

into an s domain function, F ( s). s is a complex variable:

s = a + bj, 1 j −

Inverse Laplace Transform L-1:



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Inverse Laplace Transform, L

By definition, the inverse Laplace transform operator, L-1,

converts an s-domain function back to the corresponding time

domain function:

( ) ( )1 f t F s− = L

Important Properties:

Both L and L-1 are linear operators. Thus,

( ) ( ) ( ) ( )

( ) ( ) (3-3)

ax t by t a x t b y t

aX s bY s

+ = +

= +

L L L



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where:

- x(t ) and y(t ) are arbitrary functions

- a and b are constants

- ( ) ( ) ( ) ( ) X s x t Y s y t

L Land

Similarly,

( ) ( ) ( ) ( )

1 aX s bY s ax t b y t − + = +

L

Laplace Transforms of Common



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Laplace Transforms of Common

Functions

1. Constant Function

Let f (t ) = a (a constant). Then from the definition of theLaplace transform in (3-1),

( )0

0

0 (3-4) st st a a aa ae dt e

s s s

∞

∞ − − = = − = − − =

∫L

2 St F ti



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2. Step Function

The unit step function is widely used in the analysis of process

control problems. It is defined as:

( ) 0 for 0 (3-5)1 for 0

t S t t

< ≥

Because the step function is a special case of a “constant”, it

follows from (3-4) that

( ) 1 (3-6)S t s

= L

3 D i ti



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3. Derivatives

This is a very important transform because derivatives appear

in the ODEs we wish to solve. In the text (p.53), it is shown

that

( ) ( )0 (3-9)df sF s f dt

= − L

initial condition at t = 0

Similarly, for higher order derivatives:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

11 2

2 1

0 0

... 0 0 (3-14)

nn n n

n

n n

d f s F s s f s f

dt

sf f

− −

− −

= − − −

− − −

L

where:



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- n is an arbitrary positive integer

- ( ) ( )0

0k

k

k t

d f f

dt =

Special Case: All Initial Conditions are Zero

Suppose Then( )

In process control problems, we usually assume zero initial

conditions. Reason: This corresponds to the nominal steady statewhen “deviation variables” are used, as shown in Ch. 4.

( ) ( )0 0 ... 0 . f f f = = =( ) ( )1 1n−

( )n

n

n

d f s F s

dt

=

L

4. Exponential Functions



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Consider where b > 0. Then,( ) bt f t e−=

( )

( )

0 0

0

1 1(3-16)

b s t bt bt st

b s t

e e e dt e dt

eb s s b

∞ ∞ − +− − −

∞− +

= =

= − = + +

∫ ∫L

5. Rectangular Pulse Function

It is defined by:

( )0 for 0

for 0 (3-20)

0 for

w

w

t f t h t t

t t

<= ≤ <

≥



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h

( ) f t

wt

Time, t

The Laplace transform of the rectangular pulse is given by

( ) ( )1 (3-22)wt sh

F s e s

−= −

6. Impulse Function (or Dirac Delta Function)



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The impulse function is obtained by taking the limit of the

rectangular pulse as its width, t w, goes to zero but holdingthe area under the pulse constant at one. (i.e., let )

Let,

Then,

1

w

ht

=

( )t δ impulse function

( ) 1t δ = L

Solution of ODEs by Laplace Transforms

Procedure:

1. Take the L of both sides of the ODE.

2. Rearrange the resulting algebraic equation in the s domain tosolve for the L of the output variable, e.g., Y ( s).

3. Perform a partial fraction expansion.

4. Use the L-1 to find y(t ) from the expression for Y ( s).

Table 3.1. Laplace Transforms



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Table 3.1. Laplace Transforms

See page 54 of the text.

Example 3.1



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Solve the ODE,

( )5 4 2 0 1 (3-26)dy

y ydt

+ = =

First, take L of both sides of (3-26),

( )( ) ( )2

5 1 4 sY s Y s s

− + =

Rearrange,

( )( )5 2

(3-34)5 4

sY s

s s

+=+

Take L-1,

( )( )

1 5 25 4 s y t

s s− +=

+ L

From Table 3.1,

( ) 0.80.5 0.5 (3-37)t y t e−= +

Partial Fraction Expansions



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Basic idea: Expand a complex expression for Y ( s) into

simpler terms, each of which appears in the LaplaceTransform table. Then you can take the L-1 of both sides of

the equation to obtain y(t ).

Example:

( )

( )( )

5(3-41)

1 4

sY s

s s

+=

+ +Perform a partial fraction expansion (PFE)

( )( )1 25

(3-42)1 4 1 4

s

s s s s

α α +

= ++ + + +

where coefficients and have to be determined.1α

2α

To find : Multiply both sides by s + 1 and let s = -11α



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1

11

5 4

4 3 s

s

sα =−

+

∴ = =+

To find : Multiply both sides by s + 4 and let s = -42α

24

5 1

1 3 s

s

sα

=−

+∴ = = −

+

A General PFE

Consider a general expression,

( ) ( )

( )( )

( )1

(3-46a)n

ii

N s N sY s

D s s bπ

=

= =

+

Here D( s) is an n-th order polynomial with the roots

all being real numbers which are distinct so there are no repeated( )i s b= −



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all being real numbers which are distinct so there are no repeated

roots.

The PFE is:

( ) ( )

( ) 11

(3-46b)n

in

iii

i

N sY s

s b

s b

α

π =

=

= =+

+

∑

Note: D( s) is called the “characteristic polynomial”.

Special Situations:Two other types of situations commonly occur when D( s) has:

i) Complex roots: e.g.,

ii) Repeated roots (e.g., )

For these situations, the PFE has a different form. See SEM

text (pp. 61-64) for details.

3b b= = − ( )3 4 1

ib j j= ± −

1 2

Example 3.2 (continued)



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Recall that the ODE, , with zero initial

conditions resulted in the expression6 11 6 1 y y y y+ + + + =

( )

( )3 2

1(3-40)

6 11 6

Y s

s s s s

=

+ + +

The denominator can be factored as

( ) ( ) ( ) ( )3 2

6 11 6 1 2 3 (3-50) s s s s s s s s+ + + = + + + Note: Normally, numerical techniques are required in order to

calculate the roots.

The PFE for (3-40) is

( )( )( )( )

31 2 41(3-51)

1 2 3 1 2 3

Y s

s s s s s s s s

α α α α = = + + +

+ + + + + +

Solve for coefficients to get



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1 2 3 4

1 1 1 1, , ,

6 2 2 6

α α α α = = − = = −

(For example, find , by multiplying both sides by s and then

setting s = 0.)

α

Substitute numerical values into (3-51):

1/ 6 1/ 2 1/ 2 1/ 6( )

1 2 3

Y s

s s s s

= − + ++ + +

Take L-1 of both sides:

( )1 1 1 1 11/ 6 1/ 2 1/ 2 1/ 6

1 2 3Y s s s s s

− − − − −

= − + + + + + L L L L L

From Table 3.1,

( ) 2 31 1 1 1 (3-52)6 2 2 6

t t t y t e e e− − −= − + −

Important Properties of Laplace Transforms



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1. Final Value Theorem

It can be used to find the steady-state value of a closed loop

system (providing that a steady-state value exists.

Statement of FVT:

( ) ( )0

limlimt s

sY s y t →∞ →

=

providing that the limit exists (is finite) for all

where Re ( s) denotes the real part of complexvariable, s.

( )Re 0, s ≥

Example:



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Suppose,

( )( )5 2

(3-34)5 4

sY s

s s

+=

+

Then,

( ) ( )0

5 2lim 0.5lim

5 4t s

s y y t

s→∞ →

+ ∞ = = = +

2. Time Delay

Time delays occur due to fluid flow, time required to do an

analysis (e.g., gas chromatograph). The delayed signal can berepresented as

( )θ θ time delay y t − =

Also,( ) ( )θ

θ s y t e Y s− − = L

Transfer Functions



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• Convenient representation of a linear , dynamic model.

• A transfer function (TF) relates one input and one output:

( )( )

( )( )

system x t y t

X s Y s→ →

The following terminology is used:

y

output

response

“effect”

x

input

forcing function

“cause”

Definition of the transfer function:



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Let G( s) denote the transfer function between an input, x, and an

output, y. Then, by definition

( ) ( )

( )

Y sG s

X s

where:

( ) ( )

( ) ( )

Y s y t

X s x t

L

L

Development of Transfer Functions

Example: Stirred Tank Heating System



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Figure 2.3 Stirred-tank heating process with constant holdup, V .

Recall the previous dynamic model, assuming constant liquid

holdup and flow rates:



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holdup and flow rates:

( ) (1)idT V C wC T T Qdt

ρ = − +

Suppose the process is initially at steady state:

( ) ( ) ( ) ( )0 , 0 , 0 2i iT T T T Q Q= = =

where steady-state value of T, etc. For steady-stateconditions:

T

)0 (3)iwC T T Q= − +

Subtract (3) from (1):

( ) ( ) ( ) (4)i i

dT

V C wC T T T T Q Qdt ρ = − − − + −

But,

)d T TdT −



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) because is a constant (5)

d T T dT T

dt dt

=

Thus we can substitute into (4-2) to get,

( ) (6)idT V C wC T T Qdt

ρ ′ ′ ′ ′= − +

where we have introduced the following “deviation variables”,

also called “perturbation variables”:

, , (7)i i iT T T T T T Q Q Q′ ′ ′− − −

Take L of (6):

( ) ( ) ( ) ( ) ( )0 (8)iV C sT s T t wC T s T s Q s ρ ′ ′ ′ ′ ′ − = = − −

( )0 .T t ′ =Evaluate



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By definition, Thus at time, t = 0,.T T T ′ −

( ) ( )0 0 (9)T T T ′ = −

But since our assumed initial condition was that the process

was initially at steady state, i.e., it follows from (9)

that

Note: The advantage of using deviation variables is that the

initial condition term becomes zero. This simplifies the later

analysis.

( )0T T =( )0 0.T ′ =

Rearrange (8) to solve for T s( ) :′

( ) ( ) ( )1

(10)

1 1

i

K T s Q s T s

s sτ τ

′ ′ ′= +

+ +

where two new symbols are defined:

1 V



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( )1

and 11V

K

wC w

τ

Transfer Function Between andQ′ T ′

Suppose is constant at the steady-state value. Then,iT

Then we can substitute into

(10) and rearrange to get the desired TF:

( ) ( ) ( )0 0.i i i iT t T T t T s′ ′= ⇒ = ⇒ =

( )

( )(12)

1

T s K

Q s sτ

′=

′ +

Transfer Function Between andT ′ :iT ′



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Suppose that Q is constant at its steady-state value:

( ) ( ) ( )0 0Q t Q Q t Q s′ ′= ⇒ = ⇒ =

Thus, rearranging

( )

( )

1(13)

1i

T s

T s sτ

′=

′ +

Comments:

1. The TFs in (12) and (13) show the individual effects of Q and

on T . What about simultaneous changes in both Q and ?iT iT

• Answer: See (10). The same TFs are valid for simultaneous

changes



9

C h a

p t e r 4

changes.

• Note that (10) shows that the effects of changes in both Q

and are additive. This always occurs for linear, dynamic

models (like TFs) because the Principle of Superposition isvalid.

iT

2. The TF model enables us to determine the output response to

any change in an input.

3. Use deviation variables to eliminate initial conditions for TF

models.

Properties of Transfer Function Models



10

C h a

p t e r 4

1. Steady-State Gain

The steady-state of a TF can be used to calculate the steady-

state change in an output due to a steady-state change in theinput. For example, suppose we know two steady states for an

input, u, and an output, y. Then we can calculate the steady-

state gain, K , from:

2 1

2 1

(4-38) y y

K u u

−=

−

For a linear system, K is a constant. But for a nonlinear

system, K will depend on the operating condition ( ), .u y

Calculation of K from the TF Model:



11

C h a

p t e r 4

If a TF model has a steady-state gain, then:

( )0

lim (14) s

K G s→

=

• This important result is a consequence of the Final Value

Theorem

• Note: Some TF models do not have a steady-state gain (e.g.,

integrating process in Ch. 5)

2. Order of a TF Model

Consider a general n th order linear ODE:



12

C h a

p t e r 4

Consider a general n-th order, linear ODE:

1

1 1 01

1

1 1 01(4-39)

n n m

n n mn n m

m

m m

d y dy dy d ua a a a y b

dt dt dt dt

d u dub b b udt dt

−

− −

−

− −

+ + + = +

+ + +

…

…

TakeL

, assuming the initial conditions are all zero. Rearranginggives the TF:

( ) ( )( )

0

0

(4-40)

mi

iin

ii

i

b sY sG s

U sa s

=

=

= = ∑∑

Definition:

Th d f th TF i d fi d t b th d f th d i t



13

C h a p t e r 4

The order of the TF is defined to be the order of the denominator

polynomial.

Note: The order of the TF is equal to the order of the ODE.

Physical Realizability:

For any physical system, in (4-38). Otherwise, the system

response to a step input will be an impulse. This can’t happen.

Example:

n m≥

0 1 0 and step change in (4-41)du

a y b b u udt

= +

3. Additive Property

Suppose that an output is influenced by two inputs and that



14

C h a p t e r 4

Suppose that an output is influenced by two inputs and that

the transfer functions are known:

( )

( )

( ) ( )

( )

( )1 2

1 2

andY s Y s

G s G sU s U s

= =

Then the response to changes in both and can be written

as:1U 2U

( ) ( ) ( ) ( ) ( )1 1 2 2Y s G s U s G s U s= +

The graphical representation (or block diagram) is:

G1( s)

G2( s)

Y ( s)

U 1( s)

U 2( s)

4. Multiplicative Property

Suppose that



15

C h a p t e r 4

Suppose that,

( )( )

( ) ( )

( ) ( )2

2 32 3

andY s U s

G s G sU s U s

= =

Then,

( ) ( ) ( ) ( ) ( ) ( )2 2 2 3 3Y s G s U s and U s G s U s= =

Substitute,

( ) ( ) ( ) ( )2 3 3Y s G s G s U s=Or,

( )( )

( ) ( ) ( ) ( ) ( ) ( )2 3 3 2 33

Y sG s G s U s G s G s Y s

U s=

Linearization of Nonlinear Models



16

C h a p t e r 4

• So far, we have emphasized linear models which can be

transformed into TF models.

• But most physical processes and physical models are nonlinear.

- But over a small range of operating conditions, the behaviormay be approximately linear.

- Conclude: Linear approximations can be useful, especially

for purpose of analysis.

• Approximate linear models can be obtained analytically by a

method called “linearization”. It is based on a Taylor Series

Expansion of a nonlinear function about a specified operating

point.

• Consider a nonlinear, dynamic model relating two process

variables, u and y:



17

C h a p t e r 4

( ), (4-60)dy f y udt

=

Perform a Taylor Series Expansion about and andtruncate after the first order terms, u u=

y y=

( ) ( ), , (4-61) y y

f f

f u y f u y u yu y

∂ ∂′ ′= + +

∂ ∂where and . Note that the partial derivative

terms are actually constants because they have been evaluated at

the nominal operating point,

Substitute (4-61) into (4-60) gives:

u u u′ = − y y y′ = −

( ), .u y

( ), y y

dy f f f u y u ydt u y

∂ ∂′ ′= + +∂ ∂

The steady-state version of (4-60) is:

( )0 ,f u y=



18

C h a p t e r 4

( )0 , f u y

,dy dy

dt dt ′=Substitute into (7) and recall that

(4-62) y y

dy f f u ydt u y

′ ∂ ∂′ ′= +∂ ∂

Linearized

model

Example: L iquid Storage System

Mass balance:

Valve relation:

A = area, C v = constant

(1)i

dh A q q

dt

= −

(2)vq C h=

qi

h

q

Combine (1) and (2),

dh



19

C h a p t e r 4

(3)i v

dh A q C h

dt = −

Linearize term,

( )1

(4)2

h h h hh

≈ − −

Or

1(5)h h h′≈ −

where:

2 h

h h h′ −

Substitute linearized expression (5) into (3):

1dh ′



20

C h a p t e r 4

1(6)

i v

dh A q C h h

dt R

′= − −

The steady-state version of (3) is:

0 (7)i vq C h= −

Subtract (7) from (6) and let , noting thatgives the linearized model:

i i iq q q′ − dh dhdt dt

′=

1 (8)idh A q hdt R

′ ′ ′= −

Summary:

In order to linearize a nonlinear dynamic model:



21

C h a p t e r 4

In order to linearize a nonlinear, dynamic model:

1. Perform a Taylor Series Expansion of each nonlinear term

and truncate after the first-order terms.

2. Subtract the steady-state version of the equation.

3. Introduce deviation variables.

State-Space Models



22

C h a p t e r 4

• Dynamic models derived from physical principles typically

consist of one or more ordinary differential equations (ODEs).

• In this section, we consider a general class of ODE modelsreferred to as state-space models.

Consider standard form for a linear state-space model ,

(4-90)

(4-91)

x = Ax + Bu + Ed

y = C x

where:

x = the state vector



23

C h a p t e r 4

u = the control vector of manipulated variables (also calledcontrol variables)

d = the disturbance vector

y = the output vector of measured variables. (We use

boldface symbols to denote vector and matrices, and

plain text to represent scalars.)

• The elements of x are referred to as state variables.

• The elements of y are typically a subset of x , namely, the state

variables that are measured. In general, x , u , d , and y arefunctions of time.

• The time derivative of x is denoted by

• Matrices A, B , C , and E are constant matrices.

( )d / d .t =x x

Example 4.9

Show that the linearized CSTR model of Example 4.8 can

be written in the state space form of Eqs 4 90 and 4 91



24

C h a p t e r 4

be written in the state-space form of Eqs. 4-90 and 4-91.

Derive state-space models for two cases:

(a) Both c A and T are measured.

(b) Only T is measured.

Solution

The linearized CSTR model in Eqs. 4-84 and 4-85 can be written

in vector-matrix form:

11 12

21 22 2

0

(4-92)

A A

s

dc a a cdt

T dT

a a T b

dt

′ ′ ′= + ′ ′

Let and , and denote their time derivatives by

and . Suppose that the steam temperature T s can be1 x c′ 2 x T ′ 1 x

2 x



25

C h a p t e r 4

manipulated. For this situation, there is a scalar control variable,, and no modeled disturbance. Substituting these

definitions into (4-92) gives, su T ′

1 11 12 1

2 21 22 2 2

0(4-93)

x a a xu

x a a x b

= +

B A

which is in the form of Eq. 4-90 with x = col [ x1, x2]. (The symbol

“col ” denotes a column vector.)

a) If both T and c A are measured, then y = x , and C = I in

Eq. 4-91, where I denotes the 2x2 identity matrix. A and B are

defined in (4 93)



26

C h a p t e r 4

defined in (4-93).

b) When only T is measured, output vector y is a scalar,

and C is a row vector, C = [0,1]. y T ′=

Note that the state-space model for Example 4.9 has d = 0

because disturbance variables were not included in (4-92). Bycontrast, suppose that the feed composition and feed temperature

are considered to be disturbance variables in the original

nonlinear CSTR model in Eqs. 2-60 and 2-64. Then the linearized

model would include two additional deviation variables,

and .ic′

iT ′

Dynamic Behavior



1

C h a p t e r 5

In analyzing process dynamic and process control systems, it is

important to know how the process responds to changes in the

process inputs.

A number of standard types of input changes are widely used for

two reasons:

1. They are representative of the types of changes that occur

in plants.

2. They are easy to analyze mathematically.

1. Step Input

A sudden change in a process variable can be approximated by



2

A sudden change in a process variable can be approximated by

a step change of magnitude, M :

C h a p t e r 5

0 0(5-4)

0

s

t U

M t

<

≥

• Special Case: If M = 1, we have a “unit step change”. We

give it the symbol, S (t ).

• Example of a step change: A reactor feedstock is suddenly

switched from one supply to another, causing sudden

changes in feed concentration, flow, etc.

The step change occurs at an arbitrary time denoted as t = 0.

Example:

The heat input to the stirred-tank heating system in Chapter 2 is



3

( ) ( ) ( )

( ) ( )

8000 2000 , unit step

2000 , 8000 kcal/hr

Q t S t S t

Q t Q Q S t Q

= +

′ = − = =

2. Ramp Input

• Industrial processes often experience “drifting

disturbances”, that is, relatively slow changes up or downfor some period of time.

• The rate of change is approximately constant.

suddenly changed from 8000 to 10,000 kcal/hr by changing theelectrical signal to the heater. Thus,

and

C h a p t e r 5

0 0t <

We can approximate a drifting disturbance by a ramp input:



4

C h a p t e r 5

( )0 0

(5-7)at 0

R

t U t

t

< ≥

Examples of ramp changes:

1. Ramp a setpoint to a new value. (Why not make a step

change?)

2. Feed composition, heat exchanger fouling, catalystactivity, ambient temperature.

3. Rectangular Pulse

It represents a brief, sudden change in a process variable:

( )

0 for 0

for 0 (5-9) RP w

t

U t h t t

<

≤ <



5

0 for wt t ≥

0

h

X RP

T w Time, t

C h a p t e r 5

Examples:

1. Reactor feed is shut off for one hour.

2. The fuel gas supply to a furnace is briefly interrupted.



6

C h a p t e r 5

4. Sinusoidal Input

Processes are also subject to periodic, or cyclic, disturbances.



7

They can be approximated by a sinusoidal disturbance:

( ) ( )sin 0 for 0 (5-14)sin for 0

t U t A t t ω

< ≥

where: A = amplitude, ω = angular frequency C h a p t e r 5

Examples:

1. 24 hour variations in cooling water temperature.

2. 60-Hz electrical noise (in the USA)

5. Impulse Input

• Here, ( ) ( ). I U t t δ =



8

C h a p t e r 5

Examples:

1. Electrical noise spike in a thermo-couple reading.2. Injection of a tracer dye.

• It represents a short, transient disturbance.

• Useful for analysis since the response to an impulse input

is the inverse of the TF. Thus,

( ) ( )

( )

( )

( ) ( )

( )

u t y t G s

U s Y sHere,

( ) ( ) ( )(1)Y s G s U s=

The corresponding time domain express is:



9

( ) ( ) ( )0

τ τ τ (2)t

y t g t u d = −∫where:

C h a p t e r 5

( ) ( )1 (3) g t G s− L

Suppose . Then it can be shown that:( ) ( )u t t δ =

( ) ( ) (4) y t g t =

Consequently, g (t ) is called the “impulse response function”.

First-Order System

The standard form for a first-order TF is:



10

C h a p t e r 5

where:

Consider the response of this system to a step of magnitude, M :

Substitute into (5-16) and rearrange,

( )

( )(5-16)

τ 1

Y s K

U s s=

+

steady-state gain

τ time constant

K

( ) ( )for 0U t M t U s s

= ≥ ⇒ =

( )( )

(5-17)

τ 1

KM Y s

s s

=

+

Take L-1 (cf. Table 3.1),

/ τ1 (5-18)t y t KM e−= −



11

C h a p t e r 5

( ) ( )( )y

Let steady-state value of y(t ). From (5-18),∞ . y KM ∞ =

t ___

0 0

0.632

0.865

0.9500.982

0.993

y y∞

τ

2τ

3τ4τ

5τ

0

0

21 543

1.0

0.5

y

y∞

τ

t

Note: Large means a slow response.τ

Integrating Process

Not all processes have a steady-state gain. For example, an



12

C h a p t e r 5

“integrating process” or “integrator” has the transfer function:

Consider a step change of magnitude M . Then U ( s) = M / s and,

( )

( ) ( )constant

Y s K

K U s s= =

( ) ( )2

KM Y s y t KMt

s

= ⇒ =

Thus, y(t ) is unbounded and a new steady-state value does not

exist.

L-1

Common Physical Example:

Consider a liquid storage tank with a pump on the exit line:



13

C h

a p t e r 5

- Assume:

1. Constant cross-sectional area, A.

2.

- Mass balance:

- Eq. (1) – Eq. (2), take L, assume steady state initially,

- For (constant q),

( )q f h≠

(1) 0 (2)i i

dh A q q q q

dt

= − ⇒ = −

( ) ( ) ( )1

i s Q s Q s s

′ ′ ′ = −

( ) 0Q s′ =

( )

( )

1

i

H s

Q s As

′

=′

h

qi

q

• Standard form:

Second-Order Systems



14

( )

( ) 2 2(5-40)

τ 2ζτ 1

Y s K

U s s s=

+ +

which has three model parameters:

steady-state gain

τ "time constant" [=] time

ζ damping coefficient (dimensionless)

K

• Equivalent form:1

natural frequency τnω

=

( )

( )

2

2 22ζ

n

n n

Y s K

U s s s

ω

ω ω

=

+ +

C h

a p t e r 5

• The type of behavior that occurs depends on the numerical

value of damping coefficient, :ζ



15

It is convenient to consider three types of behavior:

Complex conjugatesUnderdamped

Real and =Critically damped

Real and ≠Overdamped

Roots of Charact.

Polynomial

Type of ResponseDamping

Coefficient

ζ 1>

ζ 1=

0 ζ 1≤ <

• Note: The characteristic polynomial is the denominator of thetransfer function:

2 2τ 2ζτ 1 s s+ +

• What about ? It results in an unstable systemζ 0<

C h

a p t e r 5



16

C h

a p t e r 5



17

C h

a p t e r 5

Several general remarks can be made concerning the

responses show in Figs. 5.8 and 5.9:



18

C h

a p t e r 5

1. Responses exhibiting oscillation and overshoot ( y/ KM > 1) are

obtained only for values of less than one.

2. Large values of yield a sluggish (slow) response.

3. The fastest response without overshoot is obtained for the

critically damped case

ζ

ζ

( )ζ 1 .=



19

C h

a p t e r 5

1. Rise Time: is the time the process output takes to first

reach the new steady-state value.

2 Ti t Fi t P k i th ti i d f th t t t

r t

t



20

C h

a p t e r 5

2. Time to First Peak: is the time required for the output toreach its first maximum value.

3. Settling Time: is defined as the time required for the

process output to reach and remain inside a band whose widthis equal to ±5% of the total change in y. The term 95%

response time sometimes is used to refer to this case. Also,

values of±

1% sometimes are used.4. Overshoot: OS = a/b (% overshoot is 100a/b).

5. Decay Ratio: DR = c/a (where c is the height of the second

peak).

6. Period of Oscillation: P is the time between two successive

peaks or two successive valleys of the response.

pt

st

More General Transfer Function Models

• Poles and Zeros:



1

C h

a p t e r 6

• The dynamic behavior of a transfer function model can be

characterized by the numerical value of its poles and zeros.

• General Representation of ATF:

There are two equivalent representations:

( ) 0

0

(4-40)

mi

i

in

ii

i

b s

G s

a s

=

=

=

∑

∑

( ) ( )( ) ( )

( )( ) ( )1 2

1 2

(6-7)m m

n n

b s z s z s z G s

a s p s p s p

− − −=

− − −

…

…



2

C h

a p t e r 6

1 2n n

where { z i} are the “zeros” and { pi} are the “poles”.

• We will assume that there are no “pole-zero” calculations. That

is, that no pole has the same numerical value as a zero.

• Review: in order to have a physically realizable system.n m≥

Example 6.2

For the case of a single zero in an overdamped second-order

transfer function,



3

C h

a p t e r 6

( ) ( )

( )( )1 2

τ 1(6-14)

τ 1 τ 1

a K sG s

s s

+=

+ +

calculate the response to the step input of magnitude M and plot

the results qualitatively.

Solution

The response of this system to a step change in input is

( ) 1 2τ τ τ τ/ τ / τ1 21 (6-15)τ τ τ τ1 2 2 1

t t a a y t KM e e − −− −= + + − −

Note that as expected; hence, the effect of

including the single zero does not change the final value nor does

it change the number or location of the response modes. But the

ero does affect ho the response modes (e ponential terms) are

( ) y t KM → ∞ =



4

C h

a p t e r 6

zero does affect how the response modes (exponential terms) are

weighted in the solution, Eq. 6-15.

A certain amount of mathematical analysis (see Exercises 6.4, 6.5,

and 6.6) will show that there are three types of responses involved

here:

Case a:

Case b:

Case c:

1τ τa >

10 τ τa< ≤

τ 0a <



5

C h

a p t e r 6

1. Poles

Summary: Effects of Pole and Zero Locations



6

• Pole in “right half plane (RHP)”: results in unstable system

(i.e., unstable step responses)

( )1

p a bj

j

= +

= −

x

x

x

Real axis

Imaginary axis

x = unstable pole

• Complex pole: results in oscillatory responses

Real axis

Imaginary axis

x

xx = complex poles

C h

a p t e r 6

2 Zeros

• Pole at the origin (1/s term in TF model): results in an“integrating process”



7

2. Zeros Note: Zeros have no effect on system stability.

• Zero in RHP: results in an inverse response to a step change in

the input

• Zero in left half plane: may result in “overshoot” during a step

response (see Fig. 6.3).

x ⇒ y 0

t

inverseresponse

Real

axis

Imaginary axis

C h

a p t e r 6

Inverse Response Due to Two Competing Effects



8

C h

a p t e r 6

An inverse response occurs if:

2 2

1 1

τ(6-22)

τ

K

K

− >

Time DelaysTime delays occur due to:

1 Fluid flow in a pipe



9

C h

a p t e r 6

1. Fluid flow in a pipe

2. Transport of solid material (e.g., conveyor belt)

3. Chemical analysis- Sampling line delay

- Time required to do the analysis (e.g., on-line gas

chromatograph)

Mathematical description:

A time delay, , between an input u and an output y results in thefollowing expression:

θ

( ) ( )

0 for θ

(6-27)θ for θ

t

y t u t t

<

= − ≥

Example: Turbulent f low in a pipe

Let, fluid property (e.g., temperature or composition) at

point 1

u



10

C h

a p t e r 6

point 1

fluid property at point 2 y

Assume that the velocity profile is “flat”, that is, the velocity

is uniform over the cross-sectional area. This situation is

analyzed in Example 6.5 and Fig. 6.6.

Fluid In

Point 1

Fluid Out

Figure 6.5

Point 2



11

C h

a p t e r 6

Example 6.5

For the pipe section illustrated in Fig. 6.5, find the transfer

functions:

(a) relating the mass flow rate of liquid at 2 w to the mass flow



12

C h

a p t e r 6

(a) relating the mass flow rate of liquid at 2, w2, to the mass flow

rate of liquid at 1, wt,

(b) relating the concentration of a chemical species at 2 to theconcentration at 1. Assume that the liquid is incompressible.

Solution

(a) First we make an overall material balance on the pipe

segment in question. Since there can be no accumulation

(incompressible fluid),

material in = material out ⇒ =( ) ( )1 2w t w t

Putting (6-30) in deviation form and taking Laplace transformsyields the transfer function,

( )

( )2

1

W s

W s

′

=′



13

C h

a p t e r 6

( )1W s =

(b) Observing a very small cell of material passing point 1 at time

t , we note that in contains Vc1(t ) units of the chemical species of

interest where V is the total volume of material in the cell. If, at

time t + , the cell passes point 2, it contains units of

the species. If the material moves in plug flow, not mixing at allwith adjacent material, then the amount of species in the cell is

constant:

θ ( )2 θVc t +

( ) ( )2 1θ (6-30)Vc t Vc t + =or

( ) ( )2 1θ (6-31)c t c t + =

An equivalent way of writing (6-31) is

( ) ( )2 1 θ (6-32)c t c t = −

if th fl t i t t P tti (6 32) i d i ti f d



14

if the flow rate is constant. Putting (6-32) in deviation form and

taking Laplace transforms yields

C h

a p t e r 6

( )

( )2 θ

1

(6-33) sC se

C s

−′=

′

Time Delays (continued)

Transfer Function Representation:

( )

( )θ (6-28) sY s

eU s

−=

Note that has units of time (e.g., minutes, hours)θ

Polynomial Approximations to θ : se−

For purposes of analysis using analytical solutions to transfer

functions, polynomial approximations for are commonlyused Example: simulation software such as MATLAB and

θ s

e

−



15

C h

a p t e r 6

, p y pp yused. Example: simulation software such as MATLAB and

MatrixX.

Two widely used approximations are:

1. Taylor Series Expansion:2 2 3 3 4 4

θ θ θ θ1 θ (6-34)

2! 3! 4!

s s s se s− = − + − + +…

The approximation is obtained by truncating after only a few

terms.

2. Padé Approximations:

Many are available. For example, the 1/1 approximation is,

θ12 s−



16

C h

a p t e r 6

θ 2 (6-35)θ

1

2

se

s

− ≈

+

Implications for Control:

Time delays are very bad for control because they involve a

delay of information.

Interacting vs. Noninteracting Systems

• Consider a process with several invariables and several output

variables. The process is said to be interacting if:



17

C h

a p t e r 6

o Each input affects more than one output.

or

o A change in one output affects the other outputs.

Otherwise, the process is called noninteracting .

• As an example, we will consider the two liquid-level storage

systems shown in Figs. 4.3 and 6.13.

• In general, transfer functions for interacting processes are morecomplicated than those for noninteracting processes.

Figure 4.3. A noninteracting system:

two surge tanks in series.



18

C h

a p t e r 6

Figure 6.13. Two tanks in series whose liquid levels interact.

Figure 4.3. A noninteracting system:

two surge tanks in series.



19

C h

a p t e r 6

1

1 1

(4-48)i

dh A q q

dt = −Mass Balance:

Valve Relation:1 1

1

1(4-49)q h

R=

Substituting (4-49) into (4-48) eliminates q1:

11 11

1 (4-50)idh A q hdt R= −

Putting (4-49) and (4-50) into deviation variable form gives

11 1

1

1(4-51)i

dh A q h

dt R

′′ ′= −

1



20

C h

a p t e r 6

1 11

1(4-52)q h

R′ ′=

The transfer function relating to is found by

transforming (4-51) and rearranging to obtain

( )1 s′ ( )1iQ s′

( )

( )1 1 1

1 1 1

(4-53)1 τ 1i

H s R K

Q s A R s s

′= =

′ + +

where and Similarly, the transfer function

relating to is obtained by transforming (4-52).1 1 K R 1 1 1τ . R

( )1Q s′ ( )1 s′

( )( )

1

1 1 1

1 1(4-54)

Q s

H s R K

′ = =′

The same procedure leads to the corresponding transfer functionsf T k 2



21

for Tank 2,

( )

( )2 2 2

2 2 2 2 (4-55)1 τ 1

H s R K

Q s A R s s

′

= =′ + +

C h

a p t e r 6

( )

( )2

2 2 2

1 1(4-56)

Q s

H s R K

′

= =′

where and Note that the desired transfer

function relating the outflow from Tank 2 to the inflow to Tank 1

can be derived by forming the product of (4-53) through (4-56).

2 2 K R

2 2 2.τ R

( )( )

( )( )

( )( )

( )( )

( )( )

2 2 2 1 1

2 1 1

(4-57)i i

Q s Q s H s Q s H s

Q s H s Q s H s Q s

′ ′ ′ ′ ′=

′ ′ ′ ′ ′

or



22

C h

a p t e r 6

or

( )

( )2 2 1

2 2 1 1

1 1

(4-58)τ 1 τ 1i

Q s K K

Q s K s K s

′

=′ + +

which can be simplified to yield

( )

( ) ( )( )2

1 2

1(4-59)

τ 1 τ 1i

Q s

Q s s s

′=

′ + +

a second-order transfer function (does unity gain make sense on

physical grounds?). Figure 4.4 is a block diagram showing

information flow for this system.

Block Diagram for Noninteracting

Surge Tank System



23

g y

Figure 4.4. Input-output model for two liquid surge tanks in

series.

Dynamic Model of An Interacting Process



24

C h

a p t e r

6

Figure 6.13. Two tanks in series whose liquid levels interact.

( )1 1 21

1 (6-70)q h h R

= −

The transfer functions for the interacting system are:

( )( )

( )( )

2 22 2

22 2

(6-74)τ 2ζτ 1

12ζ 1

i

H s RQ s s s

Q sQ s

′ =′ + +

′ =′



25

( )

( )

( )

( )

( )

2 2

1 1

2 2

1 2 2 11 2 1 2 2 1 2

1 2

τ 2ζτ 1

τ 1(6-72)

τ 2ζτ 1

whereτ τ

τ= τ τ , ζ , and τ /2 τ τ

i

a

i

a

Q s s s

H s K s

Q s s s

R A R A R R

+ +

′ ′ +=

′ + +

+ ++

C h

a p t e r

6

In Exercise 6.15, the reader can show that ζ>1 by analyzing the

denominator of (6-71); hence, the transfer function is

overdamped, second order, and has a negative zero.

Model Comparison• Noninteracting system

( )

( ) ( ) ( )

2

1 2

1(4-59)

τ 1 τ 1i

Q s

Q s s s

′=

′ + +



26

1 1 1 2 2 2where τ and τ . A R A R

• Interacting system( )

( )

1 2

2

2 2

where ζ 1 and τ τ τ

1

τ 2ζτ 1i

Q s

Q s s s

>

′=

′ + +

• General Conclusions

1. The interacting system has a slower response.(Example: consider the special case where τ = τ1= τ2.)

2. Which two-tank system provides the best damping

of inlet flow disturbances?

Approximation of Higher-OrderTransfer Functions

In this section, we present a general approach for

i ti hi h d t f f ti d l ith



27

C h

a p t e r

6

approximating high-order transfer function models with

lower-order models that have similar dynamic and steady-state

characteristics.

In Eq. 6-4 we showed that the transfer function for a timedelay can be expressed as a Taylor series expansion. For small

values of s,

0θ

01 θ (6-57) s

e s− ≈ −

• An alternative first-order approximation consists of the transfer

function,



28

0

0

θ

θ

0

1 1(6-58)

1 θ

s

se

se

− = ≈

+

where the time constant has a value of

• Equations 6-57 and 6-58 were derived to approximate time-

delay terms.

• However, these expressions can also be used to approximatethe pole or zero term on the right-hand side of the equation by

the time-delay term on the left side.

0θ .

C h

a p t e r

6

Skogestad’s “half rule”

• Skogestad (2002) has proposed a related approximation method

for higher-order models that contain multiple time constants.



29

• He approximates the largest neglected time constant in the

following manner.

• One half of its value is added to the existing time delay (if any)

and the other half is added to the smallest retained time

constant.• Time constants that are smaller than the “largest neglected time

constant” are approximated as time delays using (6-58).

C h

a p t e r

6

( ) ( )( )( )( )

0.1 1(6-59)

5 1 3 1 0 5 1

K sG s − +=

Example 6.4

Consider a transfer function:



30

( )( )( )( )

( )5 1 3 1 0.5 1 s s s+ + +

( )θ

(6-60)

τ 1

s KeG s

s

−

=

+

using two methods:

(a) The Taylor series expansions of Eqs. 6-57 and 6-58.

(b) Skogestad’s half rule

C h

a p t e r

6

Compare the normalized responses of G( s) and the approximate

models for a unit step input.

Derive an approximate first-order-plus-time-delay model,

Solution

(a) The dominant time constant (5) is retained. Applying

the approximations in (6-57) and (6-58) gives:0 1 ( )s



31

0.10.1 1 (6-61) s s e−− + ≈

and3 0.51 1

(6-62)3 1 0.5 1

s se e s s

− −≈ ≈+ +

Substitution into (6-59) gives the Taylor series

approximation, ( ) :TS G s

( )0.1 3 0.5 3.6

(6-63)5 1 5 1

s s s s

TS

Ke e e KeG s

s s

− − − −= =

+ +

C h

a p t e r

6

(b) To use Skogestad’s method, we note that the largest neglectedtime constant in (6-59) has a value of three.

• According to his “half rule”, half of this value is added to the

next largest time constant to generate a new time constant

τ 5 0 5(3) 6 5= + =



32

θ 1.5 0.1 0.5 2.1= + + =

C h

a p t e r

6 • The other half provides a new time delay of 0.5(3) = 1.5.

• The approximation of the RHP zero in (6-61) provides anadditional time delay of 0.1.

• Approximating the smallest time constant of 0.5 in (6-59) by

(6-58) produces an additional time delay of 0.5.• Thus the total time delay in (6-60) is,

τ 5 0.5(3) 6.5.= + =

and G(s) can be approximated as:

( )2.1

(6-64)

6.5 1

s

Sk

KeG s

s

−=

+

Th li d t f G( ) d th t i t



33

C h

a p t e r

6

The normalized step responses for G(s) and the two approximate

models are shown in Fig. 6.10. Skogestad’s method provides

better agreement with the actual response.

Figure 6.10

Comparison of theactual and

approximate models

for Example 6.4.

Example 6.5

Consider the following transfer function:

( ) ( )( )( )( )( )

1 (6-65)12 1 3 1 0.2 1 0.05 1

s

K s eG s s s s s

−

−=+ + + +



34

( )( )( )( )

Use Skogestad’s method to derive two approximate models:(a) A first-order-plus-time-delay model in the form of (6-60)

(b) A second-order-plus-time-delay model in the form:

( )( )( )

θ

1 2

(6-66)τ 1 τ 1

s KeG s

s s

−=

+ +

C h

a p t e r

6

Compare the normalized output responses for G(s) and the

approximate models to a unit step input.



(b) An analogous derivation for the second-order-plus-time-delay

model gives:

0.2θ 1 0.05 1 2.15

2= + + + =



36

C h

a p t e r

6 1 2

2

τ 12, τ 3 0.1 3.1= = + =

In this case, the half rule is applied to the third largest time

constant (0.2). The normalized step responses of the original and

approximate transfer functions are shown in Fig. 6.11.

Multiple-Input, Multiple Output(MIMO) Processes

• Most industrial process control applications involved a number

of input (manipulated) and output (controlled) variables



37

C h

a p t e r

6

of input (manipulated) and output (controlled) variables.

• These applications often are referred to as multiple-input/multiple-output (MIMO) systems to distinguish them from the

simpler single-input/single-output (SISO) systems that have

been emphasized so far.

• Modeling MIMO processes is no different conceptually than

modeling SISO processes.

• For example, consider the system illustrated in Fig. 6.14.

• Here the level h in the stirred tank and the temperature T are to

be controlled by adjusting the flow rates of the hot and cold

streams wh and wc, respectively.



38

C h

a p t e r

6• The temperatures of the inlet streams T h and T c represent

potential disturbance variables.• Note that the outlet flow rate w is maintained constant and the

liquid properties are assumed to be constant in the following

derivation.

(6-88)



39

C h

a p t e r

6

Figure 6.14. A multi-input, multi-output thermal mixing process.



40

C h

a p t e r

6

Development of Empirical Models

From Process Data• In some situations it is not feasible to develop a theoretical

(physically-based model) due to:



1

C h

a p t e r

7

(physically based model) due to:

1. Lack of information2. Model complexity

3. Engineering effort required.

• An attractive alternative: Develop an empirical dynamic

model from input-output data.

• Advantage: less effort is required

• Disadvantage: the model is only valid (at best) for therange of data used in its development.

i.e., empirical models usually don’t extrapolate very

well.

Simple Linear Regression: Steady-State Model• As an illustrative example, consider a simple linear model

between an output variable y and input variable u,

1 2β β ε y u= + +



2

where and are the unknown model parameters to be

estimated and ε is a random error.

• Predictions of y can be made from the regression model,

C h

a p t e r

7

ˆ

1β 2β

1 2ˆ ˆˆ β β (7-3) y u= +where and denote the estimated values of β1 and β2,

and denotes the predicted value of y.1β̂ 2β̂

1 2β β ε (7-1)i i iY u= + +

• Let Y denote the measured value of y. Each pair of (ui , Y i)

observations satisfies:

• The least squares method is widely used to calculate the

values of β1 and β2 that minimize the sum of the squares of

the errors S for an arbitrary number of data points, N :

The Least Squares Approach



3

( )221 1 2

1 1

ε β β (7-2)

N N

i i

i i

S Y u= =

= = − −∑ ∑

y p

• Replace the unknown values of β1 and β2 in (7-2) by theirestimates. Then using (7-3), S can be written as:

2

1

where the -th residual, , is defined as,

ˆ (7 4)

N

ii

i

i i i

S e

i e

e Y y

=

=

− −

∑

C h

a p t e r

7

• The least squares solution that minimizes the sum ofsquared errors, S , is given by:

The Least Squares Approach (continued)



4

( )1 2

ˆβ (7-5)

uu y uy u

uu u

S S S S

NS S

−

= −

( )2 2β̂ (7-6)

uy u y

uu u

NS S S

NS S

−

= −

where:

2

1 1

N N

u i uu ii i

S u S u= =

∆ ∆∑ ∑1 1

N N

y i uy i ii i

S Y S u Y = =

∆ ∆∑ ∑

C h

a p t e r

7

• Least squares estimation can be extended to more general

models with:1. More than one input or output variable.

Extensions of the Least Squares Approach



5

C h

a p t e r

7 2. Functionals of the input variables u, such as poly-

nomials and exponentials, as long as the unknown parameters appear linearly.

• A general nonlinear steady-state model which is linear in the parameters has the form,

1

β ε (7-7) p

j j

j

y X

=

= +∑

where each X j is a nonlinear function of u.

The sum of the squares function analogous to (7-2) is

2

1 1

β (7-8) p N

i j iji j

S Y X = =

= −

∑ ∑which can be written as,



6

C h

a p t e r

7

which can be written as,

( ) ( ) (7-9)T S = −β βY - X Y X

where the superscript T denotes the matrix transpose and:

1 1β

βn p

Y

Y

= =

β Y

11 12 1

21 22 2

1 2

p

p

n n np

X X X

X X X

X X X

=

X



7

C h

a p t e r

7

The least squares estimates is given by,β̂

( )1

ˆ (7-10)−

=β T T

X X X Y

providing that matrix X T

X is nonsingular so that its inverse exists.

Note that the matrix X is comprised of functions of u j; for

example, if:2

1 2 3β β β ε y u u= + + +

This model is in the form of (7-7) if X 1 = 1, X 2 = u, and

X 3 = u2.

• Simple transfer function models can be obtained graphicallyfrom step response data.

Fitting First and Second-Order ModelsUsing Step Tests



8

C h

a p t e r

7 • A plot of the output response of a process to a step change in

input is sometimes referred to as a process reaction curve.

• If the process of interest can be approximated by a first- or

second-order linear model, the model parameters can beobtained by inspection of the process reaction curve.

• The response of a first-order model, Y ( s)/U ( s)= K /(τ s+1), to

a step change of magnitude M is:

( ) /(1 ) (5-18)t y t KM e τ −= −

1(7-15)

d y

d KM

=

• The initial slope is given by:



9

C h

a p t e r

7 0

( )τt dt KM =

• The gain can be calculated from the steady-state changes

in u and y:

where ∆ is the steady-state change in .

y y K u M

y y

∆ ∆= =∆



10

C h a p t e r

7

Figure 7.3 Step response of a first-order system and

graphical constructions used to estimate the time constant, τ.

First-Order Plus Time Delay Model-θ

( )

τ 1

Ke sG s

s

=

+For this FOPTD model, we note the following charac-

teristics of its step response:



11

teristics of its step response:

C h a p t e r

7

1. The response attains 63.2% of its final response

at time, t = τ+θ.

2. The line drawn tangent to the response atmaximum slope (t = θ) intersects the y/KM =1

line at (t = τ + θ ).

3. The step response is essentially complete at t =5τ.

In other words, the settling time is t s=5τ.



12

C h a p t e r

7

Figure 7.5 Graphical analysis of the process reaction curve

to obtain parameters of a first-order plus time delay model.

There are two generally accepted graphical techniques for

determining model parameters τ, θ, and K .

Method 1: Slope-intercept method

First, a slope is drawn through the inflection point of the



13

First, a slope is drawn through the inflection point of the

process reaction curve in Fig. 7.5. Then τ and θ aredetermined by inspection.

Alternatively, τ can be found from the time that the

normalized response is 63.2% complete or from

determination of the settling time, t s. Then set τ=t s/5. C h a p t e r

7

Method 2. Sundaresan and Krishnaswamy’s Method

This method avoids use of the point of inflection

construction entirely to estimate the time delay.

• They proposed that two times, t 1 and t 2, be estimated from astep response curve, corresponding to the 35.3% and 85.3%

response times, respectively.

Sundaresan and Krishnaswamy’s Method



14

C h a p t e r

7p , p y

• The time delay and time constant are then estimated from thefollowing equations:

( )1 2

2 1

θ 1.3 0.29

(7-19)τ 0.67

t t

t t

= −

= −

• These values of θ and τ approximately minimize the

difference between the measured response and the model, based on a correlation for many data sets.

• In general, a better approximation to an experimental step

response can be obtained by fitting a second-order model to

the data.

Estimating Second-order Model ParametersUsing Graphical Analysis



15

C h a p t e r

7

• Figure 7.6 shows the range of shapes that can occur for thestep response model,

( )

( ) ( )1 2

(5-39)

τ 1 τ 1

K G s

s s

=

+ +• Figure 7.6 includes two limiting cases: , where the

system becomes first order, and , the critically

damped case.

• The larger of the two time constants, , is called the

dominant time constant.

2 1τ / τ 0=

2 1τ / τ 1=

1τ

7



16

C h a p t e r

7

Figure 7.6 Step response for several overdamped second-order systems.

7

• Assumed model:

( )θ

2 2τ 2ζτ 1

s KeG s

s s

−

=+ +

Smith’s Method



17

1. Determine t 20 and t 60 from the step response.2. Find ζ and t 60/τ from Fig. 7.7.

3. Find t 60/τ from Fig. 7.7 and then

t 60 is known).

C h a p t e r

7ζ

• Procedure:

calculate τ (since

7



18

C h a p t e r

7

7

Fitting an Integrator Modelto Step Response Data

In Chapter 5 we considered the response of a first-order process

to a step change in input of magnitude M :



19

C h a p t e r

7

( ) ( )/ τ

1 M 1 (5-18)t

y t K e−

= −For short times, t < τ, the exponential term can be approximated

by

/ τ 1τ

t t e− ≈ −

so that the approximate response is:

( )1

MM 1 1 (7-22)

τ τ

t K y t K t

≈ − − =

7

is virtually indistinguishable from the step response of theintegrating element

( )2

2 (7-23)

K

G s s=In the time domain, the step response of an integrator is



20

C h a p t e r

7

( )2 2 (7-24) y t K Mt =

Hence an approximate way of modeling a first-order process is

to find the single parameter

2 (7-25)τ

K K =

that matches the early ramp-like response to a step change in

input.

7

If the original process transfer function contains a time delay(cf. Eq. 7-16), the approximate short-term response to a step

input of magnitude M would be

( ) ( ) ( )θ θ KM

y t t S t t

= − −



21

C h a p t e r

7

where S (t-θ) denotes a delayed unit step function that starts at

t =θ.

7



22

C h a p t e r

7

Figure 7.10. Comparison of step responses for a FOPTD

model (solid line) and the approximate integrator plus time

delay model (dashed line).

7

Development of Discrete-TimeDynamic Models

• A digital computer by its very nature deals internally with

discrete-time data or numerical values of functions at equally

spaced intervals determined by the sampling period.

Th di i d l h diff i



23

C h a p t e r

7 • Thus, discrete-time models such as difference equations are

widely used in computer control applications.

• One way a continuous-time dynamic model can be converted to

discrete-time form is by employing a finite differenceapproximation.

• Consider a nonlinear differential equation,

( )( ), (7-26)

dy t f y u

dt =

where y is the output variable and u is the input variable.

7

• This equation can be numerically integrated (though with someerror) by introducing a finite difference approximation for the

derivative.

• For example, the first-order, backward differenceapproximation to the derivative at is

( ) ( )1y k y kdy

t k t = ∆



24

C h a p t e r

7

where is the integration interval specified by the user and

y(k ) denotes the value of y(t ) at . Substituting Eq. 7-26into (7-27) and evaluating f ( y, u) at the previous values of y and

u (i.e., y(k – 1) and u(k – 1)) gives:

( ) ( )1

(7-27)

y k y k dy

dt t

− −≅ ∆

t ∆

t k t = ∆

( ) ( )( ) ( )( )

( ) ( ) ( ) ( )( )

11 , 1 (7-28)

1 1 , 1 (7-29)

y k y k f y k u k

t

y k y k tf y k u k

− − ≅ − −∆

= − + ∆ − −

7

Second-Order DifferenceEquation Models

• Parameters in a discrete-time model can be estimated directlyfrom input-output data based on linear regression.

• This approach is an example of system identification (Ljung,



25

C h a p t e r

7

( ) ( ) ( ) ( ) ( )1 2 1 21 2 1 2 (7-36) y k a y k a y k b u k b u k = − + − + − + −

pp p y f ( j g

1999).

• As a specific example, consider the second-order difference

equation in (7-36). It can be used to predict y(k ) from data

available at time (k – 1) and (k – 2) .

• In developing a discrete-time model, model parameters a1, a2,

b1, and b2 are considered to be unknown.

t ∆ t ∆

β β β βa a b b

• This model can be expressed in the standard form of Eq. 7-7,

1

β ε (7-7) p

j j

j

y X =

= +∑

by defining:

7



26

( ) ( )

( ) ( )

1 1 2 2 3 1 4 2

1 2

3 4

β , β , β , β

1 , 2 ,

1 , 2

a a b b

X y k X y k

X u k X u k

− −

− −

2

1 1

β (7-8) p N

i j ij

i j

S Y X = =

= −

∑ ∑

• The parameters are estimated by minimizing a least squares

error criterion:

C h a p t e r

7

Equivalently, S can be expressed as,

( ) ( ) (7-9)T

S = −β βY - X Y X

where the superscript T denotes the matrix transpose and:

7



27

C h a p t e r

1 1β

βn p

Y

Y

= =

β Y

The least squares solution of (7-9) is:

( )1

ˆ (7-10)−

=β T T

X X X Y

Feedback Controllers

8



1

C h a p t e r

Figure 8.1 Schematic diagram for a stirred-tank blending

system.

8

Basic Control Modes Next we consider the three basic control modes starting with the

simplest mode, proportional control .

Proportional Control



2

C h a p t e r In feedback control, the objective is to reduce the error signal to

zero where

( ) ( ) ( ) (8-1) sp me t y t y t = −

and( )

( )

( )

error signal

set point

measured value of the controlled variable

(or equivalent signal from the sensor/transmitter)

sp

m

e t

y t

y t

=

=

=

8

Although Eq. 8-1 indicates that the set point can be time-varying,in many process control problems it is kept constant for long

periods of time.

For proportional control, the controller output is proportional tothe error signal,

+



3

C h a p t e r ( ) ( ) (8-2)c p t p K e t

= +

where:

( ) controller output bias (steady-state) value

controller gain (usually dimensionless)c

p t p

K

==

=

8



4

C h a p t e r

8

1. The controller gain can be adjusted to make the controller

output changes as sensitive as desired to deviations between

set point and controlled variable;

2. the sign of K c can be chosed to make the controller output

increase (or decrease) as the error signal increases.

The key concepts behind proportional control are the following:



5

C h a p t e r

( ) g

For proportional controllers, bias can be adjusted, a procedure

referred to as manual reset .

Some controllers have a proportional band setting instead of a

controller gain. The proportional band PB (in %) is defined as

p

100% (8-3)c

PB K

8

In order to derive the transfer function for an ideal proportional

controller (without saturation limits), define a deviation variable

as( ) p t ′

( ) ( ) (8-4) p t p t p′ −

Then Eq. 8-2 can be written as

′ =



6

C h a p t e r ( ) ( ) (8-5)c p t K e t

=

The transfer function for proportional-only control:

( )( )

(8-6)c P s K E s

′ =

An inherent disadvantage of proportional-only control is that a

steady-state error occurs after a set-point change or a sustained

disturbance.

8

Integral ControlFor integral control action, the controller output depends on the

integral of the error signal over time,

( ) ( )0

1* * (8-7)

τ

t

I

p t p e t dt = + ∫



7

C

h a p t e r where , an adjustable parameter referred to as the integral time

or reset time, has units of time.τ I

Integral control action is widely used because it provides an

important practical advantage, the elimination of offset.

Consequently, integral control action is normally used in

conjunction with proportional control as the proportional-integral

(PI) controller:

( ) ( ) ( )0

1* * (8-8)

τ

t

c

I

p t p K e t e t dt

= + +

∫

8

( )

( )τ 11

1 (8-9)τ τ

I c c

I I

P s s

K K E s s s

′ +

= + =

Some commercial controllers are calibrated in terms of 1/ τ

The corresponding transfer function for the PI controller inEq. 8-8 is given by



8

C

h a p t e r

(repeats per minute) rather than (minutes, or minutes per

repeat).

I τ I

Reset Windup

• An inherent disadvantage of integral control action is a

phenomenon known as reset windup or integral windup.

• Recall that the integral mode causes the controller output to

change as long as e(t *) ≠ 0 in Eq. 8-8.

8

• When a sustained error occurs, the integral term becomesquite large and the controller output eventually saturates.

• Further buildup of the integral term while the controller is

saturated is referred to as reset windup or integral windup.

Derivative Control



9

C

h a p t e r

The function of derivative control action is to anticipate the future behavior of the error signal by considering its rate of change.

• The anticipatory strategy used by the experienced operator can

be incorporated in automatic controllers by making the

controller output proportional to the rate of change of the error

signal or the controlled variable.

• Thus, for ideal derivative action,

( ) ( )τ (8-10) D

de t p t p

dt = +

where , the derivative time, has units of time.

For example, an ideal PD controller has the transfer function:

τ D

8



10

( )

( ) ( )1 τ (8-11)c D

P s K s

E s

′= +

• By providing anticipatory control action, the derivative mode

tends to stabilize the controlled process.

• Unfortunately, the ideal proportional-derivative control

algorithm in Eq. 8-10 is physically unrealizable because it

cannot be implemented exactly.

C

h a p t e r

8

• For analog controllers, the transfer function in (8-11) can beapproximated by

( )

( )τ

1 (8-12)ατ 1

Dc

D

P s s

K E s s

′

= + +

α



11

C

h a p t e r

where the constant typically has a value between 0.05 and0.2, with 0.1 being a common choice.

• In Eq. 8-12 the derivative term includes a derivative mode

filter (also called a derivative filter ) that reduces the sensitivityof the control calculations to high-frequency noise in the

measurement.

8

• Many variations of PID control are used in practice.

• Next, we consider the three most common forms.

Proportional-Integral-Derivative (PID) Control Now we consider the combination of the proportional, integral,

and derivative control modes as a PID controller.



12

C

h a p t e r Paral lel Form of PID Control

The parallel form of the PID control algorithm (without a

derivative filter) is given by

( ) ( ) ( )

( )0

1

* * τ

(8-13)τ

t

c D I

de t

p t p K e t e t dt dt

= + + +

∫

8

The corresponding transfer function is:

( )

( )1

1 τ (8-14)τ

c D

I

P s K s

E s s

′ = + +

Series Form of PID Control



13

C

h a p t e r

Historically, it was convenient to construct early analogcontrollers (both electronic and pneumatic) so that a PI element

and a PD element operated in series.

Commercial versions of the series-form controller have aderivative filter that is applied to either the derivative term, as in

Eq. 8-12, or to the PD term, as in Eq. 8-15:

( )

( )τ 1 τ 1

(8-15)τ ατ 1

I Dc

I D

P s s s K

E s s s

′ + +=

+

8

( ) ( ) ( ) ( )

0* * (8-16)

t

c I D

de t p t p K e t K e t dt K

dt = + + +∫

Expanded Form of PID Control

In addition to the well-known series and parallel forms, the

expanded form of PID control in Eq. 8-16 is sometimes used:



14

C

h a p t e r Features of PID Controllers

Elimination of Derivative and Proportional Kick

• One disadvantage of the previous PID controllers is that a

sudden change in set point (and hence the error, e) will cause the

derivative term momentarily to become very large and thus provide a derivative kick to the final control element.

• This sudden change is undesirable and can be avoided by basingthe derivative action on the measurement, ym, rather than on the

error signal, e.

• We illustrate the elimination of derivative kick by consideringthe parallel form of PID control in Eq. 8-13.

• Replacing de/dt by – dym/dt gives 8



15

( ) ( ) ( ) ( )

0

1* * τ (8-17)

τ

t mc D

I

dy t p t p K e t e t dt

dt

= + + −

∫

Reverse or Direct Action

• The controller gain can be made either negative or positive.

C

h a p t e r

8

• For proportional control, when K c > 0, the controller output p(t )increases as its input signal ym(t ) decreases, as can be seen by

combining Eqs. 8-2 and 8-1:

( ) ( ) ( ) (8-22)c sp m p t p K y t y t − = −

• This controller is an example of a reverse-acting controller.

Wh K 0 h ll i id b di i b



16

C

h a p t e r

• When K c < 0, the controller is said to be direct acting becausethe controller output increases as the input increases.

• Equations 8-2 through 8-16 describe how controllers perform

during the automatic mode of operation.

• However, in certain situations the plant operator may decide to

override the automatic mode and adjust the controller outputmanually.

Figure 8.11 Reverse

d di t ti

8



17

and direct-acting proportional

controllers. (a) reverse

acting ( K c > 0. (b)direct acting ( K c < 0) C

h a p t e r

• Example: Example: Flow Control Loop

8



18

Assume FT is direct-acting.

1. Air-to-open (fail close) valve ==> ?

2. Air-to-close (fail open) valve ==> ?

C

h a p t e r

e

8

Automatic and Manual Control Modes

• Automatic Mode

Controller output, p(t), depends on e(t), controllerconstants, and type of controller used.

( PI vs. PID etc.)

M l M d



19

C

h a p t e r • Manual Mode

Controller output, p(t), is adjusted manually.

• Manual Mode is very useful when unusual

conditions exist:

plant start-up

plant shut-down

emergencies

• Percentage of controllers "on manual” ??(30% in 2001, Honeywell survey)

Example: Example: Liquid Level Control• Control valves are air-to-open

• Level transmitters are direct acting

e

8



20

C

h a p t e r

Questions: Questions: 1. Type of controller action?

2. What type of fish?

e

8

On-Off Controllers

• Simple

• Cheap

• Used In residential heating and domestic refrigerators



21

C

h a p t e r

• Used In residential heating and domestic refrigerators• Limited use in process control due to continuous

cycling of controlled variable ⇒ excessive wear

on control valve.



Practical case (dead band)

e r 8



23

C

h a p t e

e r 8



24

C

h a p t e



e r 8

• Integral action eliminates steady-state error

(i.e., offset) Why??? e ≠ 0 ⇒ p is changing with

time until e = 0, where p reaches steady state.

τ+=′

s11K

E(s)(s)P

I

c• Transfer function for PI control



26

y sp

C

h a p t e

Derivative Control Action

• Ideal derivative action

• Some controllers are calibrated in 1/τI

("repeats per minute") instead of τI .

p

e r 8

• For PI controllers, is not adjustable.



27

• Ideal derivative action

• Used to improve dynamic response of thecontrolled variable

• Derivative kick (use db/dt )

• Use alone?

dt

de p)t( p Dτ+=

C

h a p t e

PID Controller

• Ideal controller

τ+′′

τ++=

∫

t

0D

Ic dt

detd)t(e

1)t(eK p)t( p

e r 8

• Transfer function (ideal)



28

τ ∫0I dt

τ+

τ+=

′s

s

11K

E(s)

(s)PD

I

c

C

h a p t e

• Transfer function (actual)

α = small number (0.05 to 0.20)

+ατ

+τ

τ

+τ=

′

1s

1s

s

1sK

E(s)

(s)P

D

D

I

Ic

lead / lag units

e r 8

PI - More complicated to tune (K τ )

P - Simplest controller to tune (K c).- Offset with sustained disturbance or setpoint

change.

Controller Comparison



29

C

h a p t e

PID - Most complicated to tune (K c, τI, τD) .

- Better performance than PI

- No offset

- Derivative action may be affected by noise

PI - More complicated to tune (K c, τI) .

- Better performance than P

- No offset

- Most popular FB controller

Typical Response of Feedback Control Systems

Consider response of a controlled system after a

sustained disturbance occurs (e.g., step change in

the disturbance variable)

yte r 8



30

y

C

h a p t e

Figure 8.12. Typical process responses with feedback control.

y

te r 8

Figure 8.13.Proportional control:

effect of controller

gain.



31

C

h a p t e

Figure 8.15. PID

control: effect of

derivative time.

y y

te r 8



32

C

h a p t e

Figure 8.14. PI control: (a) effect of reset time (b) effect of

controller gain.



There are two alternative forms of the digital PID control

equation, the position form and the velocity form. Substituting (8-

24) and (8-25) into (8-13), gives the position form,

( )11 1

(8-26)

k D

k c k j k k j

t

p p K e e e et

τ

τ −=

∆

= + + + − ∆ ∑t e r 8



34

( )1 1 j= + + + ∆ ∑

C

h a p t

Where pk is the controller output at the k th sampling instant. The

other symbols in Eq. 8-26 have the same meaning as in Eq. 8-13.

Equation 8-26 is referred to as the position form of the PIDcontrol algorithm because the actual value of the controller output

is calculated.

t e r 8

In the velocity form, the change in controller output is

calculated. The velocity form can be derived by writing the

position form of (8-26) for the (k -1) sampling instant:

( )11 1

(8-26)k

Dk c k j k k

j

t p p K e e e e

t

τ

τ −

=

∆= + + + −

∆ ∑

Note that the summation still begins at j = 1 because it is assumed



35

C

h a p t g j

that the process is at the desired steady state for

and thus e j = 0 for . Subtracting (8-27) from (8-26)

gives the velocity form of the digital PID algorithm:

0 j ≤ 0 j ≤

( ) ( )1 1 1 22

(8-28)

Dk k k c k k k k k k

I

t

p p p K e e e e e et

τ

τ − − − −

∆

∆ = − = − + + − + ∆

t e r 8

The velocity form has three advantages over the position form:

1. It inherently contains anti-reset windup because the

summation of errors is not explicitly calculated.

2. This output is expressed in a form, , that can be utilized

directly by some final control elements, such as a control

valve driven by a pulsed stepping motor.

k p∆



36

C

h a p t y p pp g

3. For the velocity algorithm, transferring the controller from

manual to automatic mode does not require any initializationof the output ( in Eq. 8-26). However, the control valve (or

other final control element) should be placed in the

appropriate position prior to the transfer.

p

t e r 9

Control System Instrumentation



1

C

h a p

Figure 9.3 A typical process transducer.

Transducers and Transmitters

• Figure 9.3 illustrates the general configuration of a measurementtransducer; it typically consists of a sensing element combined

with a driving element (transmitter).

p t e r 9

• Transducers for process measurements convert the magnitude of

a process variable (e.g., flow rate, pressure, temperature, level,or concentration) into a signal that can be sent directly to the

controller.

• The sensing element is required to convert the measured

quantity, that is, the process variable, into some quantity more

appropriate for mechanical or electrical processing within the

transducer.



2

C

h a p

Standard Instrumentation Signal Levels

• Before 1960, instrumentation in the process industries utilized

pneumatic (air pressure) signals to transmit measurement and

control information almost exclusively.

• These devices make use of mechanical force-balance elements

to generate signals in the range of 3 to 15 psig, an industry

standard.

p t e r 9

• Since about 1960, electronic instrumentation has come into

widespread use.

Sensors

The book briefly discusses commonly used sensors for the most

important process variables. (See text.)

TransmittersA i ll h i l l l



3

C

h a p • A transmitter usually converts the sensor output to a signal level

appropriate for input to a controller, such as 4 to 20 mA.

• Transmitters are generally designed to be direct acting .

• In addition, most commercial transmitters have an adjustable

input range (or span).• For example, a temperature transmitter might be adjusted so that

the input range of a platinum resistance element (the sensor) is

50 to 150 °C.

p t e r 9

• In this case, the following correspondence is obtained:

20 mA150 °C

4 mA50 °C

OutputInput

• This instrument (transducer) has a lower limit or zero of 50 °C



4

C

h a p

( )

and a range or span of 100 °C.

• For the temperature transmitter discussed above, the relation between transducer output and input is

( ) ( )( )

20 mA 4 mA

mA 50 C 4 mA150 C 50 C

mA0.16 C 4 mA

C

mT T

T

−

= − + −

= −

p t e r 9

The gain of the measurement element K m is 0.16 mA/°C. For any

linear instrument:

range of instrument output(9-1)

range of instrument input

m K =

Final Control Elements

• Every process control loop contains a final control element



5

C

h a p (actuator), the device that enables a process variable to be

manipulated.

• For most chemical and petroleum processes, the final control

elements (usually control valves) adjust the flow rates of

materials, and indirectly, the rates of energy transfer to and

from the process.

p t e r 9



6

C

h a p

Figure 9.4 A linear instrument calibration showing its zero

and span.

p t e r 9

Control Valves

• There are many different ways to manipulate the flows of

material and energy into and out of a process; for example, the

speed of a pump drive, screw conveyer, or blower can beadjusted.

• However, a simple and widely used method of accomplishing

this result with fluids is to use a control valve, also called anautomatic control valve.



7

C

h a p

• The control valve components include the valve body, trim,

seat, and actuator.

Air-to-Open vs. Air-to-Close Control Valves

• Normally, the choice of A-O or A-C valve is based on safety

considerations.

p t e r 9



8

C

h a p

Figure 9.7 A pneumatic control valve (air-to-open).

p t e

r 9

• We choose the way the valve should operate (full flow or no

flow) in case of a transmitter failure.

• Hence, A-C and A-O valves often are referred to as fail-open

and fail-closed , respectively.

Example 9.1

Pneumatic control valves are to be specified for the applications

listed below. State whether an A-O or A-C valve should be used

f h f ll i i l d i bl d i ( )



9

C

h a p for the following manipulated variables and give reason(s).

a) Steam pressure in a reactor heating coil.

b) Flow rate of reactants into a polymerization reactor.

c) Flow of effluent from a wastewater treatment holding tank into

a river.

d) Flow of cooling water to a distillation condenser.

p t e

r 9

Valve Positioners

Pneumatic control valves can be equipped with a valve

positioner , a type of mechanical or digital feedback controller

that senses the actual stem position, compares it to the desired

position, and adjusts the air pressure to the valve accordingly.

Specifying and Sizing Control Valves

A design equation used for sizing control valves relates valve



10

C

h a p g q g

lift to the actual flow rate q by means of the valve coefficient

C v, the proportionality factor that depends predominantly onvalve size or capacity:

( ) (9-2)vv s

P q C f g

∆=

p t e

r 9

• Here q is the flow rate, is the flow characteristic, is the pressure drop across the valve, and g s is the specific gravity of

the fluid.

• This relation is valid for nonflashing fluids.

• Specification of the valve size is dependent on the so-called

valve characteristic f .

• Three control valve characteristics are mainly used.

( ) f

v P ∆



11

C

h a p

• For a fixed pressure drop across the valve, the flow

characteristic is related to the lift , thatis, the extent of valve opening, by one of the following relations:( )0 1 f f ≤ ≤ ( )0 1≤ ≤

1

Linear:

Quick opening: (9-3)

Equal percentage:

f

f

f R −

=

=

=

p t e

r 9



12

C

h a p

Figure 9.8 Control valve characteristics.

p t e

r 9

where R is a valve design parameter that is usually in the range

of 20 to 50.

Rangeability

The rangeability of a control valve is defined as the ratio ofmaximum to minimum input signal level. For control valves,

rangeability translates to the need to operate the valve within the

range 0.05 ≤ f ≤ 0.95 or a rangeability of 0.95/0.05 = 19.



13

C

h a p

To Select an Equal Percentage Valve:

a) Plot the pump characteristic curve and , the system pressure drop curve without the valve, as shown in Fig. 9.10.

The difference between these two curves is . The pump

should be sized to obtain the desired value of , forexample, 25 to 33%, at the design flow rate qd .

s P ∆

v P ∆

/v s P ∆ ∆

p t e

r 9



14

C

h a

Figure 9.10 Calculation of the valve pressure drop

from the pump characteristic curve and the system pressure

drop without the valve

( )v P ∆

( ).

s P ∆

p t e

r 9

b) Calculate the valve’s rated C v, the value that yields at least

100% of qd with the available pressure drop at that higherflow rate.

c) Compute q as a function of using Eq. 9-2, the rated C v,

and from (a). A plot of the valve characteristic (q vs. )

should be reasonably linear in the operating region of

interest (at least around the design flow rate). If it is not

suitably linear, adjust the rated C v and repeat.

v∆

Example 9 2



15

C

h a Example 9.2

A pump furnishes a constant head of 40 psi over the entire flowrate range of interest. The heat exchanger pressure drop is 30 psig

at 200 gal/min (qd ) and can be assumed to be proportional to q2.

Select the rated C v of the valve and plot the installed characteristicfor the following case:

a) A linear valve that is half open at the design flow rate.

a p t e

r 9



16

C

h a

Figure 9.9 A control valve placed in series with a pump and

a heat exchanger. Pump discharge pressure is constant.

a p t e

r 9

Solution

First we write an expression for the pressure drop across the heat

exchanger

2

(9-5)30 200

he P q∆ =

2

30 (9-6)200

s he q P P ∆ = ∆ =



17

C

h a

Because the pump head is constant at 40 psi, the pressure drop

available for the valve is2

40 40 30 (9-7)

200

v he

q P P

∆ = − ∆ = −

Figure 9.11 illustrates these relations. Note that in all four design

cases at qd ./ 10 / 30 33%v s P P ∆ ∆ = =

a p t e

r 9



18

C

h a

Figure 9.11 Pump characteristic and system pressure drop

for Example 9.2.

a p t e

r 9

a) First calculate the rated C v.

200126.5 (9-8)

0.5 10vC = =

We will use C v = 125. For a linear characteristic valve, use the

relation between and q from Eq. 9-2:

(9-9)v v

q

C P =

∆



19

C

h a v v

Using Eq. 9-9 and values of from Eq. 9-7, the installedvalve characteristic curve can be plotted.

v P ∆

a p t e

r 9



20

C

h a

Figure 9.12 Installed valve characteristics for Example 9.2.

a p t e

r 9



21

C

h a

Figure 9.16 Schematic diagram of a thermowell/thermocouple.



a p t e

r 9



23

C

h a

Figure 9.13 Analysis of types of error for a flow instrument

whose range is 0 to 4 flow units.

a p t e

r 9

Figure 9 14 Analysis of



24

C

h a Figure 9.14 Analysis of

instrument error showing theincreased error at low readings

(from Lipták (1971)).

a p t e

r 9



25

C

h a

Figure 9.15 Nonideal instrument behavior: (a) hysteresis,

(b) deadband.

a p t e

r 1 1

Dynamic Behavior and Stability ofClosed-Loop Control Systems

• In this chapter we consider the dynamic behavior of processes that are operated using feedback control.

• This combination of the process, the feedback controller,

and the instrumentation is referred to as a feedback control

loop or a closed-loop system.



1

C

h a

Block Diagram Representation

To illustrate the development of a block diagram, we return to a

previous example, the stirred-tank blending process considered inearlier chapters.

a p t e

r 1 1



2

C

h a

Figure 11.1 Composition control system for a stirred-tank

blending process.

a p t e

r 1 1

Next, we develop a transfer function for each of the five elements

in the feedback control loop. For the sake of simplicity, flow ratew1 is assumed to be constant, and the system is initially operating

at the nominal steady rate.

Process

In section 4.3 the approximate dynamic model of a stirred-tank

blending system was developed:

K K



3

C

h ( ) ( ) ( )1 21 2 (11-1)

τ 1 τ 1

K K X s X s W s

s s

′ ′ ′= + + +

where

11 2

ρ 1, , and (11-2)wV x K K w w w

τ −= = =

a p t e

r 1 1



4

C

h

Figure 11.2 Block diagram of the process.

h a p t e

r 1 1

Composition Sensor-Transmitter (Analyzer)

We assume that the dynamic behavior of the composition sensor-

transmitter can be approximated by a first-order transfer function:

( )( )(11-3)

τ 1m m

m

X s K

X s s

′=′ +

Controller

Suppose that an electronic proportional plus integral controller is

used. From Chapter 8, the controller transfer function is



5

C

hp

( )( )

11 (11-4)τ

c I

P s K E s s

′ = +

where and E ( s) are the Laplace transforms of the controlleroutput and the error signal e(t ). Note that and e are

electrical signals that have units of mA, while K c is dimensionless.

The error signal is expressed as

( ) P s′( ) p t ′ p′

h a p t e

r 1 1

( ) ( ) ( )(11-5)

sp me t x t x t ′ ′= −

or after taking Laplace transforms,

( ) ( ) ( ) (11-6) sp m E s X s X s′ ′= −

The symbol denotes the internal set-point composition

expressed as an equivalent electrical current signal. This signalis used internally by the controller. is related to the actual

composition set point by the composition sensor-

( ) sp x t ′

( ) sp x t ′( )sp x t ′



6

C

hp p y p

transmitter gain K m

:( ) sp

( ) ( ) (11-7) sp m sp x t K x t ′ ′=

Thus

( )

( )(11-8)

spm

sp

X s K

X s

′=

′

h a p t e

r 1 1



7

C

h

Figure 11.3 Block diagram for the composition sensor-

transmitter (analyzer).



h a p t e

r 1 1

Figure 11.5 Block diagram for the I/P transducer.



9

C

h

Figure 11.6 Block diagram for the control valve.

h a p t e

r 1 1



10

C

h

Figure 11.7 Block diagram for the entire blending processcomposition control system.

h a p t e

r 1 1

Closed-Loop Transfer Functions

The block diagrams considered so far have been specifically

developed for the stirred-tank blending system. The more general

block diagram in Fig. 11.8 contains the standard notation:

disturbance variable (also referred to as load

variable)

D =

manipulated variableU =

controlled variableY =



11

C

h

internal set point (used by the controller)

set pointY sp =

measured value of Y Y m =

error signal E =controller output P =

)

spY =

h a p t e

r 1 1



12

C

h

Figure 11.8 Standard block diagram of a feedbackcontrol system.

h a p t e

r 1 1

disturbance transfer functionGd =

process transfer functionG p =

transfer function for final control element

(including K IP , if required)

Gv =controller transfer functionGc =

change inY d = Y due to D

change inY u = Y due to U



13

C

h

steady-state gain for Gm K m =

transfer function for measuring element andtransmitter Gm =

h a p t e

r 1 1

Block Diagram Reduction

In deriving closed-loop transfer functions, it is often convenient to

combine several blocks into a single block. For example, consider

the three blocks in series in Fig. 11.10. The block diagram

indicates the following relations:

1 1

2 2 1

3 3 2

(11-11)

X G U

X G X X G X

=

==

By successive substitution



14

C

h By successive substitution,

3 3 2 1 (11-12) X G G G U =

or

3 (11-13) X GU =

where 3 2 1.G G G G

h a p t e

r 1 1 Figure 11.10 Three blocks in series.



15

C

h

Figure 11.11 Equivalent block diagram.

h a p t e

r 1 1

Set-Point Changes

Next we derive the closed-loop transfer function for set-point

changes. The closed-loop system behavior for set-point changes is

also referred to as the servomechanism ( servo) problem in the

control literature.

(11-14)

0 (because 0) (11-15)(11-16)

d u

d d

u p

Y Y Y

Y G D DY G U

= +

= = ==



16

C

Combining gives

(11-17) pY G U =

h a p t e

r 1 1

Figure 11.8 also indicates the following input/output relations for

the individual blocks:

(11-18)

(11-19)

(11-20)

(11-21)

(11-22)

v

c

sp m

sp m sp

m m

U G P

P G E

E Y Y

Y K Y

Y G Y

=

=

= −

=

=

Combining the above equations gives



17

C

g q g

( )

( )

(11-23)

(11-24)

(11-25)

p v p v c

p v c sp m

p v c m sp m

Y G G P G G G E

G G G Y Y

G G G K Y G Y

= =

= −

= −

h a p t e

r 1 1

Rearranging gives the desired closed-loop transfer function,

(11-26)1

m c v p

sp c v p m

K G G GY

Y G G G G=

+

Disturbance Changes

Now consider the case of disturbance changes, which is also

referred to as the regulator problem since the process is to beregulated at a constant set point. From Fig. 11.8,

(11 27)Y Y Y G D G U



18

C (11-27)d u d p

Y Y Y G D G U = + = +

Substituting (11-18) through (11-22) gives

( ) (11-28)d p p v c m sp mY G D G U G G G K Y G Y = + = −

C h a p t e

r 1 1

Because Y sp = 0 we can arrange (11-28) to give the closed-loop

transfer function for disturbance changes:

(11-29)1

d

c v p m

GY

D G G G G=

+

A comparison of Eqs. 11-26 and 11-29 indicates that both

closed-loop transfer functions have the same denominator,

1 + GcGvG pGm. The denominator is often written as 1 + GOL

where GOL is the open-loop transfer function, .OL c v p mG G G G G



19

CAt different points in the above derivations, we assumed that D = 0 or Y sp = 0, that is, that one of the two inputs was constant.

But suppose that D ≠ 0 and Y sp ≠ 0, as would be the case if a

disturbance occurs during a set-point change. To analyze thissituation, we rearrange Eq. 11-28 and substitute the definition of

GOL to obtain

C h a p t e

r 1 1

(11-30)1 1

m c v pd spOL OL

K G G GGY D Y G G= ++ +

Thus, the response to simultaneous disturbance variable and set-

point changes is merely the sum of the individual responses, as

can be seen by comparing Eqs. 11-26, 11-29, and 11-30.

This result is a consequence of the Superposition Principle forlinear systems.



20

C

C h a p t e

r 1 1

General Expression for Feedback Control Systems

Closed-loop transfer functions for more complicated block

diagrams can be written in the general form:

(11-31)1

f

i e

Z

Z

Π=

+ Π

where:

is the output variable or any internal variable within the

t l l

Z



21

C

= product of every transfer function in the feedback loop

= product of the transfer functions in the forward path from

Z i to Z

is an input variable (e.g., Y sp or D) Z i

control loop

f Π

eΠ

C h a p t e

r 1 1

Example 11.1

Find the closed-loop transfer function Y /Y sp for the complex

control system in Figure 11.12. Notice that this block diagram has

two feedback loops and two disturbance variables. Thisconfiguration arises when the cascade control scheme of Chapter

16 is employed.



22

C

Figure 11.12 Complex control system.

C h a p t e

r 1 1



23

C

Figure 11.13 Block diagram for reduced system.

C h a p t e

r 1 1



24

C

Figure 11.14 Final block diagrams for Example 11.1.

C h a p t e

r 1 1

Solution

Using the general rule in (11-31), we first reduce the inner loop to

a single block as shown in Fig. 11.13. To solve the servo problem,

set D1 = D2 = 0. Because Fig. 11.13 contains a single feedback

loop, use (11-31) to obtain Fig. 11.14a. The final block diagram isshown in Fig. 11.14b with Y /Y sp = K m1G5. Substitution for G4 and

G5 gives the desired closed-loop transfer function:

1 1 2 1 2 3

2 1 2 1 2 3 1 2 11

m c c

sp c m c m c

K G G G G GY

Y G G G G G G G G G=

+ +



25

C

Closed-Loop Responses of Simple

Control Systems

In this section we consider the dynamic behavior of several

elementary control problems for disturbance variable and set-

point changes.

C h a p t e

r 1 1

The transient responses can be determined in a straightforward

manner if the closed-loop transfer functions are available.

Consider the liquid-level control system shown in Fig. 11.15. The

liquid level is measured and the level transmitter (LT) output issent to a feedback controller (LC) that controls liquid level by

adjusting volumetric flow rate q2. A second inlet flow rate q1 is

the disturbance variable. Assume:

1. The liquid density ρ and the cross-sectional area of the tank A



26

Care constant.

2. The flow-head relation is linear, q3 = h/ R.

3. The level transmitter, I/P transducer, and control valve have

negligible dynamics.

4. An electronic controller with input and output in % is used (full

scale = 100%).

C h a p t e

r 1 1



27

C

Figure 11.15 Liquid-level control system.

C h a p t e

r 1 1

Derivation of the process and disturbance transfer functions

directly follows Example 4.4. Consider the unsteady-state mass balance for the tank contents:

1 2 3ρ ρ ρ ρ (11-32)dh

A q q qdt = + −

Substituting the flow-head relation, q3 = h/ R, and introducing

deviation variables gives

1 2 (11-33)dh h

A q qdt R

′ ′′ ′= + −



28

C

Thus, we obtain the transfer functions

( )( )

( )2

(11-34)τ 1

p p K H s G s

Q s s′ = =′ +

C h a p t e

r 1 1

( )( ) ( )1

(11-35)τ 1

pd

K H sG s

Q s s

′= =′ +

where K p = R and = RA. Note that G p( s) and Gd ( s) are identical

because q1 and q2 are both inlet flow rates and thus have the sameeffect on h.

τ

Proportional Control and Set-Point Changes

If a proportional controller is used, then Gc( s) = K c. From Fig.

11.6 and the material in the previous section, it follows that the



29

C p ,

closed-loop transfer function for set-point changes is given by

( )

( )

( )

( )

/ τ 1

(11-36)1 / τ 1

c v p m

sp c v p m

K K K K s H s

H s K K K K s

+′

=′ + +

C h a p t e

r 1 1



30

C

Figure 11.16 Block diagram for level control system.



C h a p t e

r 1 1

From Eq. 11-37 it follows that the closed-loop response to a unit

step change of magnitude M in set point is given by

( ) ( )1/ τ

1 1 (11-41)t

h t K M e−′ = −

This response is shown in Fig. 11.17. Note that a steady-state

error or offset exists because the new steady-state value is K 1 M

rather than the desired value of M . The offset is defined as

( ) ( )offset (11-42) sph h′ ′∞ − ∞



32

CFor a step change of magnitude M in set point, .From (11-41), it is clear that . Substituting these

values and (11-38) into (11-42) gives

( ) sph M ′ ∞ =( ) 1h K M ′ ∞ =

1offset (11-43)1 OL

M K M K

= − =+

C h a p t e

r 1 1



33

C

Figure 11.17 Step response for proportional control (set-

point change).

C h a p t e r 1 1

Proportional Control and Disturbance Changes

From Fig. 11.16 and Eq. 11-29 the closed-loop transfer function

for disturbance changes with proportional control is

( )( )

( )( )1

/ τ 1 (11-53)1 / τ 1

p

OL

K s H sQ s K s

+′ =′ + +

Rearranging gives

( )

( )2

1 1

(11-54)τ 1

H s K

Q s s

′=

′ +



34

where is defined in (11-39) and K 2 is given by1τ

2 (11-55)1

p

OL

K K K

= +

C h a p t e r 1 1

• A comparison of (11-54) and (11-37) indicates that both closed-

loop transfer functions are first-order and have the same timeconstant.

• However, the steady-state gains, K 1 and K 2, are different.

• From Eq. 11-54 it follows that the closed-loop response to a

step change in disturbance of magnitude M is given by

( ) ( )1/ τ2 1 (11-56)t h t K M e−′ = −

The offset can be determined from Eq. 11-56. Now ( ) 0h′ ∞ =



35

since we are considering disturbance changes and

for a step change of magnitude M .

Thus,

( ) sp( ) 2h K M ′ ∞ =

( ) 2offset 0 (11-57)1

p

OL

K M h K M

K ′= − ∞ = − = −

+

C h a p t e r 1 1



36

Figure 11.18 Set-point responses for Example 11.2.

C h a p t e r 1 1



37

Figure 11.19 Load responses for Example 11.3.

C h a p t e r 1 1

PI Control and Disturbance Changes

For PI control, . The closed-loop transfer

function for disturbance changes can then be derived from Fig.

11.16:

( ) ( )1 1/ τc c I G s K s= +

( )( )

( )( ) ( )1

/ τ 1(11-58)

1 1 1/ τ / τ 1

p

OL I

K s H s

Q s K s s

+′=

′ + + +

Clearing terms in the denominator gives

( )

( ) ( )

τ

1

(11-59)

1

I p s

I OL I

K H s

Q K

′=



38

( ) ( )1 τ τ 1 τ I OL I Q s s s K s′ + +Further rearrangement allows the denominator to be placed in the

standard form for a second-order transfer function:

( )

( )3

2 21 3 3 3

(11-60)τ 2ζ τ 1

H s K s

Q s s s

′=

′ + +

C h a p t e r 1 1

where

3

3

3

τ / (11-61)

1 τ1ζ (11-62)

2 τ

τ ττ / (11-63)

I c v m

OL I

OL

I OL

K K K K

K

K K

= +

=

=

For a unit step change in disturbance, , and (11-59) becomes ( )1 1/Q s s′ =

( ) 32 2

3 3 3

(11-64)τ 2ζ τ 1

K H s

s s′ =

+ +



39

3 3 3

For , the response is a damped oscillation that can be

described by30 ζ 1< <

( ) 3 3ζ / τ 23

3 32

3 3

sin 1 ζ / τ (11-65)τ 1 ζ

t K h t e t

− ′ = − −



C h a p t e r 1 1



41

Figure 11.22 Liquid-level control system with pump in exit line.

C h a p t e r 1 1

If the level transmitter and control valve in Eq. 11.22 have

negligible dynamics, the Gm( s) = K

mand G

v( s) = K

v. For PI

control, . Substituting these expressions

into the closed-loop transfer function for disturbance changes

( ) ( )1 1/ τc c I G s K s= +

( )( )1

(11-68)1

d

c v p m

H s GQ s G G G G

′ =′ +

and rearranging gives

( )

( )4

2 21 4 4 4

(11-69)τ 2ζ τ 1

H s K s

Q s s s

′=

′ + +

h



42

where

4

4

4

τ / (11-70)

τ τ / (11-71)

ζ 0.5 τ (11-72)

c v m

I OL

OL I

K K K K

K

K

= −

=

=

And K OL = K c K v K p K m with K p = - 1/ A.

C h a p t e r 1 1

Stability of Closed-Loop Control Systems

Example 11.4

Consider the feedback control system shown in Fig. 11.8 with

the following transfer functions:

1(11-73)

2 1c c vG K G

s= =

+

1 1(11-74)

5 1 1

p d mG G G

s s

= = =



43

5 1 1 s s+ +

Show that the closed-loop system produces unstable responses if

controller gain K c is too large.

C h a p t e r 1 1



44

Figure 11.23. Effect of controller gains on closed-loopresponse to a unit step change in set point (example 11.1).



C h a p t e r 1 1

Character istic Equation

As a starting point for the stability analysis, consider the block

diagram in Fig. 11.8. Using block diagram algebra that was

developed earlier in this chapter, we obtain

(11-80)1 1

m c v p d sp

OL OL

K G G G GY Y D

G G= +

+ +

where GOL is the open-loop transfer function,

GOL = GcGvG pGm. For the moment consider set-point changes

only, in which case Eq. 11-80 reduces to the closed-loop

transfer function



46

transfer function,

(11-81)1

m c v p

sp OL

K G G GY

Y G=

+

C h a p t e r 1 1

Comparing Eqs. 11-81 and 11-82 indicates that the poles are also

the roots of the following equation, which is referred to as thecharacteristic equation of the closed-loop system:

1 0 (11-83)OLG+ =

General Stabi l i ty Criter ion. The feedback control system in Fig.

11.8 is stable if and only if all roots of the characteristic

equation are negative or have negative real parts. Otherwise, the

system is unstable.

Example 11.8



47

Consider a process, G p = 0.2/- s + 1), and thus is open-loop

unstable. If Gv = Gm = 1, determine whether a proportional

controller can stabilize the closed-loop system.

C h a p t e r 1 1

Figure 11.25Stability regions

in the complex

plane for roots

of the charact-

eristic equation.



48

C h a p t e r 1 1

Figure 11.26

Contributions of

characteristic

equation roots to

closed-loopresponse.



49

C h a p t e r 1 1

Solution

The characteristic equation for this system is

0.2 1 0 (11-92)c s K + − =

Which has the single root, s = -1 + 0.2 K c. Thus, the stabilityrequirement is that K c < 5. This example illustrates the important

fact that feedback control can be used to stabilize a process that

is not stable without control.

Routh Stability Criterion



50

The Routh stability criterion is based on a characteristic equation

that has the form

11 1 0 0 (11-93)n n

n na s a s a s a−−+ + + + =…

C h a p t e r 1 1

Routh array:

z 1n + 1

c2c14

b3b2b13

an-5an-3an-12

an-4an-2an1Row

where:



51

1 2 31

1

1 4 521

(11-94)

(11-95)

n n n n

n

n n n nn

a a a ab

a

a a a ab a

− − −

−

− − −−

−=

−=

C h a p t e r 1 1

and:

1 3 1 21

1

1 5 1 32

1

(11-96)

(11-97)

n n

n n

b a a bc

bb a a b

cb

− −

− −

−=

−=

Routh Stabil i ty Cri ter ion:

A necessary and sufficient condition for all roots of the



52

A necessary and sufficient condition for all roots of the

characteristic equation in Eq. 11-93 to have negative real parts

is that all of the elements in the left column of the Routh array

are positive.

C h a p t e r 1 1

Example 11.9

Determine the stability of a system that has the characteristic

equation4 3 25 3 1 0 (11-98) s s s+ + + =

Solution

Because the s term is missing, its coefficient is zero. Thus,

the system is unstable. Recall that a necessary condition for

stability is that all of the coefficients in the characteristic

equation must be positive.



53

C h a p t e r 1 1

Example 11.10

Find the values of controller gain K c that make the feedback

control system of Eq. 11.4 stable.

Solution

From Eq. 11-76, the characteristic equation is

3 2

10 17 8 1 0 (11-99)c s s s K + + + + =All coefficients are positive provided that 1 + K c > 0 or K c < -1.

The Routh array is



54

c1

b2b1

1 + K c17

810

C h a p t e r 1 1

To have a stable system, each element in the left column of the

Routh array must be positive. Element b1 will be positive if K c < 7.41/0.588 = 12.6. Similarly, c1 will be positive if K c > -1.

Thus, we conclude that the system will be stable if

1 12.6 (11-100)c K − < <

Direct Substitution Method

• The imaginary axis divides the complex plane into stable andunstable regions for the roots of characteristic equation, as

indicated in Fig. 11.26.

• On the imaginary axis, the real part of s is zero, and thus we can



55

g y , p ,write s=jω. Substituting s=jω into the characteristic equation

allows us to find a stability limit such as the maximum value of

K c.

• As the gain K c is increased, the roots of the characteristic

equation cross the imaginary axis when K c = K cm.

C h a p t e r 1 1

Example 11.12

Use the direct substitution method to determine K cm for the system

with the characteristic equation given by Eq. 11-99.

Solution

Substitute and K c = K cm into Eq. 11-99:ω s j=

3 2

10 ω 17ω 8 ω 1 0cm j j K − − + + + =or (11-105)

( ) ( )2 31 17ω 8ω 10ω 0

cm K j+ − + − =



56

( ) ( )

C h a p t e r 1 1

Equation 11-105 is satisfied if both the real and imaginary parts

are identically zero:21 17ω 0 (11-106a)cm K + − =

( )3 28ω 10ω ω 8 10ω 0 (11-106b)− = − =

Therefore,

2

ω 0.8 ω 0.894 (11-107)= ⇒ = ±and from (11-106a),

12.6cm K =



57

C h a p t e r 1 1

Root Locus Diagrams

Example 11.13

Consider a feedback control system that has the open-loop

transfer function,

( )

( )( )( )

4(11-108)

1 2 3

cOL

K G s

s s s=

+ + +

Plot the root locus diagram for 0 20.c K ≤ ≤

Solution



58

The characteristic equation is 1 + GOL = 0 or

( )( )( )1 2 3 4 0 (11-109)c s s s K + + + + =

C h a p t

e r 1 1

• The root locus diagram in Fig. 11.27 shows how the three rootsof this characteristic equation vary with K c.

• When K c = 0, the roots are merely the poles of the open-loop

transfer function, -1, -2, and -3.



59

C h a p t

e r 1 1



60

Figure 11.27 Root locus diagram for third-order system. Xdenotes an open-loop pole. Dots denote locations of the closed-

loop poles for different values of K c. Arrows indicate change of

pole locations as K c increases.

C h a p t

e r 1 1 Figure 11.29. Flowchart

for performing a stability

analysis.



61

C h a p t

e r 1 2

Controller Tuning: A Motivational Example



1

Fig. 12.1. Unit-step disturbance responses for the candidate controllers(FOPTD Model: K = 1, θ 4, τ 20).= =

C h a p t

e r 1 2

PID Controller Design, Tuning, and

Troubleshooting

Performance Criteria For Closed-Loop Systems• The function of a feedback control system is to ensure that

the closed loop system has desirable dynamic and steady-

state response characteristics.• Ideally, we would like the closed-loop system to satisfy the

following performance criteria:

1. The closed-loop system must be stable.



2

2. The effects of disturbances are minimized, providing

good disturbance rejection.3. Rapid, smooth responses to set-point changes are

obtained, that is, good set-point tracking .

C h a p t

e r 1 2

4. Steady-state error (offset) is eliminated.

5. Excessive control action is avoided.

6. The control system is robust, that is, insensitive to

changes in process conditions and to inaccuracies in the process model.

PID controller settings can be determined by a number

of alternative techniques:1. Direct Synthesis (DS) method

2. Internal Model Control (IMC) method

3 Controller tuning relations



3

3. Controller tuning relations

4. Frequency response techniques

5. Computer simulation

6. On-line tuning after the control system is installed.

C h a p t

e r 1 2

Direct Synthesis Method

• In the Direct Synthesis (DS) method, the controller design is based on a process model and a desired closed-loop transferfunction.

• The latter is usually specified for set-point changes, butresponses to disturbances can also be utilized (Chen andSeborg, 2002).

• Although these feedback controllers do not always have a PIDstructure, the DS method does produce PI or PID controllersfor common process models.

• As a starting point for the analysis, consider the block diagramof a feedback control system in Figure 12 2 The closed loop



4

of a feedback control system in Figure 12.2. The closed-looptransfer function for set-point changes was derived in Section

11.2:(12-1)

1

m c v p

sp c v p m

K G G GY

Y G G G G=

+

C h a p t

e r 1 2

Fig. 12.2. Block diagram for a standard feedback control system.



5

C h a p t

e r 1 2

For simplicity, let and assume that Gm = K m. ThenEq. 12-1 reduces to

v p mG G G G

(12-2)1

c

sp c

G GY

Y G G=

+

Rearranging and solving for Gc gives an expression for thefeedback controller:

/1(12-3a)

1 /

sp

c sp

Y Y G

G Y Y

=

− • Equation 12-3a cannot be used for controller design because the

closed-loop transfer function Y /Y sp is not known a priori.

• Also, it is useful to distinguish between the actual process G



6

and the model, , that provides an approximation of the process behavior.

• A practical design equation can be derived by replacing theunknown G by , and Y /Y sp by a desired closed-loop transfer

function, (Y /Y s )d :

G

G

C h a p t

e r 1 2

( )( )

/1(12-3b)1 /

sp

d c

sp d

Y Y

G G Y Y

= −

• The specification of (Y /Y sp)d is the key design decision and will

be considered later in this section.

• Note that the controller transfer function in (12-3b) contains

the inverse of the process model owing to the term.• This feature is a distinguishing characteristic of model-based

control.

1/ G

Desired Closed-Loop Transfer Function



7

p

For processes without time delays, the first-order model in

Eq. 12-4 is a reasonable choice,1

(12-4)1 sp cd

Y

Y sτ

= +

C h a p t

e r 1 2

• The model has a settling time of ~ 4 , as shown in

Section 5. 2.• Because the steady-state gain is one, no offset occurs for set-

point changes.

• By substituting (12-4) into (12-3b) and solving for Gc, thecontroller design equation becomes:

τc

1 1 (12-5)τ

cc

G sG

=

• The term provides integral control action and thuseliminates offset.

1/ τc s



8

• Design parameter provides a convenient controller tuning

parameter that can be used to make the controller moreaggressive (small ) or less aggressive (large ).

τc

τcτc

C h a p t

e r 1 2

• If the process transfer function contains a known time delay ,

a reasonable choice for the desired closed-loop transferfunction is:

θ

θ

(12-6)τ 1

s

sp cd

Y e

Y s

−

= +

• The time-delay term in (12-6) is essential because it is physically impossible for the controlled variable to respond toa set-point change at t = 0, before t = .

• If the time delay is unknown, must be replaced by anestimate. θ

θ



9

• Combining Eqs. 12-6 and 12-3b gives:

θ

θ

1(12-7)

τ 1

s

c sc

eG

G s e

−

−=

+ −

C h a p t

e r 1 2

• Although this controller is not in a standard PID form, it is

physically realizable.• Next, we show that the design equation in Eq. 12-7 can be used

to derive PID controllers for simple process models.

• The following derivation is based on approximating the time-delay term in the denominator of (12-7) with a truncated Taylorseries expansion:

θ 1 θ (12-8) se s− ≈ −

Substituting (12-8) into the denominator of Eq. 12-7 and

rearranging givesθ



10

( )

θ1(12-9)

τ θ

−

=

+

s

c s

c

eG

G

Note that this controller also contains integral control action.

C h a p t

e r 1 2

First-Order-plus-Time-Delay (FOPTD) Model

Consider the standard FOPTD model,

( )θ

(12-10)

τ 1

s KeG s

s

−

=

+

Substituting Eq. 12-10 into Eq. 12-9 and rearranging gives a PIcontroller, with the following controller

settings:

( )1 1/ τ ,c c I G K s= +

1 τ, τ τ (12-11)

θ τc I

c

K K

= =+

Second-Order-plus-Time-Delay (SOPTD) Model



11

Consider a SOPTD model,

( )( )( )

θ

1 2

(12-12)τ 1 τ 1

s KeG s

s s

−

=+ +

C h a p t

e r 1 2

Substitution into Eq. 12-9 and rearrangement gives a PIDcontroller in parallel form,

11 τ (12-13)

τc c D

I

G K s s

= + +

where:

1 2 1 21 2

1 2

τ τ τ τ1, τ τ τ , τ (12-14)

τ τ τc I D

c

K K θ

+= = + =

+ +

Example 12.1

Use the DS design method to calculate PID controller settings forthe process:



12

p

( )( )

2

10 1 5 1

seG

s s

−

=+ +

C h a p t

e r 1 2

Consider three values of the desired closed-loop time constant:. Evaluate the controllers for unit step changes in

both the set point and the disturbance, assuming that Gd = G.Repeat the evaluation for two cases:

1, 3, and 10c

τ =

a. The process model is perfect ( = G).

b. The model gain is = 0.9, instead of the actual value, K = 2.Thus,

G

K

0.9 se−

( )( )10 1 5 1G

s s=

+ +

The controller settings for this example are:

0 6821 883 75τ 1c = τ 3c = 10cτ =

( )2K K =



13

3.333.333.33

151515

1.514.178.33

0.6821.883.75( )2c K K =

( )0.9c K K =

τ I

τ D

C h a p t

e r 1 2

The values of K c decrease as increases, but the values of

and do not change, as indicated by Eq. 12-14.

τc τ I

τ D



14

Figure 12.3 Simulation results for Example 12.1 (a): correctmodel gain.

C h a p t

e r 1 2



15

Fig. 12.4 Simulation results for Example 12.1 (b): incorrectmodel gain.

C h a p t

e r 1 2

Internal Model Control (IMC)

• A more comprehensive model-based design method, Internal

Model Control ( IMC ), was developed by Morari andcoworkers (Garcia and Morari, 1982; Rivera et al., 1986).

• The IMC method, like the DS method, is based on an assumed process model and leads to analytical expressions for the

controller settings.• These two design methods are closely related and produce

identical controllers if the design parameters are specified in a

consistent manner.• The IMC method is based on the simplified block diagram



16

The IMC method is based on the simplified block diagramshown in Fig. 12.6b. A process model and the controller

output P are used to calculate the model response, .

G

Y

C h a p t

e r 1 2

• The model response is subtracted from the actual response Y ,and the difference, is used as the input signal to the IMC

controller, .

• In general d e to modeling errors and nkno n

Y Y −

*cG

Y Y≠ ( )G G≠

Figure 12.6.Feedback control

strategies



17

• In general, due to modeling errors and unknowndisturbances that are not accounted for in the model.

• The block diagrams for conventional feedback control andIMC are compared in Fig. 12.6.

Y Y ≠ ( )G G≠( )0 D ≠

C h a p t e r 1 2

*

cG

• It can be shown that the two block diagrams are identical if

controllers Gc and satisfy the relation

*

*(12-16)

1

cc

c

GG

G G

=

−

• Thus, any IMC controller is equivalent to a standard

feedback controller Gc, and vice versa.• The following closed-loop relation for IMC can be derived from

Fig. 12.6b using the block diagram algebra of Chapter 11:

*cG

* *1(12 17)c cG G G G

Y Y D−



18

( ) ( )* *(12-17)

1 1c c

sp

c c

Y Y DG G G G G G

= ++ − + −

C h a p t e r 1 2

For the special case of a perfect model, , (12-17) reduces toG G=

( )* *1 (12-18)c sp cY G GY G G D= + −

The IMC controller is designed in two steps:

Step 1. The process model is factored as

(12-19)G G G+ −=

where contains any time delays and right-half planezeros.

• In addition, is required to have a steady-state gain equalto one in order to ensure that the two factors in Eq. 12-19

G+

G+



19

are unique.

C h a p t e r 1 2

Step 2. The controller is specified as

* 1 (12-20)cG f G−

=

where f is a low-pass filter with a steady-state gain of one. Ittypically has the form:

( )

1

(12-21)τ 1 r c

f s=

+

In analogy with the DS method, is the desired closed-loop timeconstant. Parameter r is a positive integer. The usual choice isr = 1

τc



20

r 1.

C h a p t e r 1 2

For the ideal situation where the process model is perfect ,

substituting Eq. 12-20 into (12-18) gives the closed-loopexpression

( )G G=

( )1 (12-22) spY G fY fG D+ += + −

Thus, the closed-loop transfer function for set-point changes is

(12-23) sp

Y G f

Y +=

Selection of τc

• The choice of design parameter is a key decision in both theDS and IMC design methods

τc



21

DS and IMC design methods.

• In general, increasing produces a more conservativecontroller because K c decreases while increases.

τc

τ I

C h a p t e r 1 2

• Several IMC guidelines for have been published for themodel in Eq. 12-10:

τc

1. > 0.8 and (Rivera et al., 1986)

2. (Chien and Fruehauf, 1990)

3. (Skogestad, 2003)

τ / θc τ 0.1τc >

τ τ θc

> >

τ θc =

Controller Tuning RelationsIn the last section, we have seen that model-based designmethods such as DS and IMC produce PI or PID controllers for

certain classes of process models.



22

IMC Tuning Relations

The IMC method can be used to derive PID controller settingsfor a variety of transfer function models.

C h a p t e r 1 2

Table 12.1 IMC-Based PID Controller Settings for Gc( s)

(Chien and Fruehauf, 1990). See the text for the rest of thistable.



23

C h a p t e r 1 2



24

C h a p t e r 1 2

Tuning for Lag-Dominant Models

• First- or second-order models with relatively small time delaysare referred to as lag-dominant models.

• The IMC and DS methods provide satisfactory set-pointresponses, but very slow disturbance responses, because thevalue of is very large.

• Fortunately, this problem can be solved in three different ways.

Method 1: Integrator Approximation

τ I

( )θ / τ 1

*Approximate ( ) by ( )1

s s

Ke K eG s G s s s

−θ −θ

= =τ +



25

where * / . K K τ

• Then can use the IMC tuning rules (Rule M or N)to specify the controller settings.

C h a p t e r 1 2

Method 2. Limit the Value of I

• For lag-dominant models, the standard IMC controllers for first-order and second-order models provide sluggish disturbanceresponses because is very large.

• For example, controller G in Table 12.1 has where isvery large.

• As a remedy, Skogestad (2003) has proposed limiting the valueof :

τ I

τ τ I = τ

τ I

( ){ }1τ min τ , 4 τ θ (12-34) I c= +

where τ1 is the largest time constant (if there are two).

Method 3. Design the Controller for Disturbances, Rather



26

Method 3. Design the Controller for Disturbances, Rather

Set-point Changes

• The desired CLTF is expressed in terms of (Y/D)des, rather than (Y/Y sp)des

• Reference: Chen & Seborg (2002)

C h a p t e r 1 2

Example 12.4

Consider a lag-dominant model with θ / τ 0.01:=

( )100

100 1

sG s e s

−=

+

Design four PI controllers:

a) IMC b) IMC based on the integrator approximation

c) IMC with Skogestad’s modification (Eq. 12-34)

d) Direct Synthesis method for disturbance rejection (Chen andSeborg 2002): The controller settings are K = 0 551 and

( )τ 1c =( )τ 2c =

( )τ 1c

=



27

Seborg, 2002): The controller settings are K c 0.551 andτ 4.91. I

=

C h a p t e r 1 2

Evaluate the four controllers by comparing their performance for unit step changes in both set point and disturbance. Assume thatthe model is perfect and that Gd ( s) = G( s).

Solution

The PI controller settings are:

4.910.551(d) DS-d80.5(c) Skogestad

50.556(b) Integrator approximation

1000.5(a) IMC

K c

Controller I

τ



28

C h a p t e r 1 2

Figure 12.8. Comparison

of set-point responses(top) and disturbanceresponses (bottom) for

Example 12.4. Theresponses for the Chenand Seborg and integratorapproximation methodsare essentially identical.



29

C h a p t e r 1 2

Tuning Relations Based on Integral

Error Criteria

• Controller tuning relations have been developed that optimize

the closed-loop response for a simple process model and aspecified disturbance or set-point change.

• The optimum settings minimize an integral error criterion.

• Three popular integral error criteria are:

1. I ntegral of the absolute value of the error (I AE)

( )IAE (12-35)e t dt

∞

= ∫



30

0∫

where the error signal e(t ) is the difference between the set point and the measurement.

C h a p t e r 1 2

C h a p t e r 1 2 a

Figure 12.9. Graphical

interpretation of IAE.The shaded area is theIAE value.



31

C h a p t e r 1 2

2. I ntegral of the squared error (I SE)

( ) 2

0

ISE (12-36)e t dt

∞

= ∫3. I ntegral of the time-weighted absolute error (I TAE)

( )0

ITAE (12-37)t e t dt ∞

= ∫

See text for ITAE controller tuning relations.

Comparison of Controller Design and

Tuning Relations

Alth h th d i d t i l ti f th i ti



32

Although the design and tuning relations of the previous sections

are based on different performance criteria, several generalconclusions can be drawn:

C h a p t e r 1 2

1. The controller gain K c should be inversely proportional to the

product of the other gains in the feedback loop (i.e., K c 1/ K where K = K v K p K m).

2. K c should decrease as , the ratio of the time delay to the

dominant time constant, increases. In general, the quality ofcontrol decreases as increases owing to longer settlingtimes and larger maximum deviations from the set point.

3. Both and should increase as increases. For manycontroller tuning relations, the ratio, , is between 0.1 and0.3. As a rule of thumb, use = 0.25 as a first guess.

4. When integral control action is added to a proportional-onlycontroller, K c should be reduced. The further addition ofderi ati e action allo s K to be increased to a al e greater

θ / τ

θ / τ

τ I τ D θ / ττ / τ D I

τ / τ D I

∝



33

derivative action allows K c to be increased to a value greater

than that for proportional-only control.

C h a p t e r 1 2

Controllers With Two Degrees

of Freedom• The specification of controller settings for a standard PID

controller typically requires a tradeoff between set-pointtracking and disturbance rejection.

• These strategies are referred to as controllers with two-

degrees-of-freedom.• The first strategy is very simple. Set-point changes are

introduced gradually rather than as abrupt step changes.

• For example, the set point can be ramped as shown in Fig.12.10 or “filtered” by passing it through a first-order transferf ti



34

function,* 1

(12-38)τ 1

sp

sp f

Y

Y s=

+

C h a p t e r 1 2

where denotes the filtered set point that is used in the control

calculations.

• The filter time constant, determines how quickly the filteredset point will attain the new value, as shown in Fig. 12.10.

* spY

τ f



35

Figure 12.10 Implementation of set-point changes.

C h a p t e r 1 2

• A second strategy for independently adjusting the set-pointresponse is based on a simple modification of the PID control

law in Chapter 8,

( ) ( ) ( )* *

0

1(8-7)

t m

c D

I

dy p t p K e t e t dt

dt τ

τ

= + + −

∫

where ym is the measured value of y and e is the error signal..

• The control law modification consists of multiplying the set point in the proportional term by a set-point weighting factor, :

sp me y y−

β

( ) ( ) ( )

( )* *

β

1τ (12-39)

c sp m

t m

c D

p t p K y t y t

dy K e t dt

= + −

+ − ∫



36

( )0

( )τ

c D

I

dt

∫

The set-point weighting factor is bounded, 0 < ß < 1, and serves asa convenient tuning factor.

C h a p t e r 1 2

Fi 12 11 I fl f i i h i l d l



37

Figure 12.11 Influence of set-point weighting on closed-loop

responses for Example 12.6.

C h a p t e r 1 2

On-Line Controller Tuning

1. Controller tuning inevitably involves a tradeoff between

performance and robustness.

2. Controller settings do not have to be precisely determined. Ingeneral, a small change in a controller setting from its bestvalue (for example, ±10%) has little effect on closed-loopresponses.

3. For most plants, it is not feasible to manually tune each

controller. Tuning is usually done by a control specialist(engineer or technician) or by a plant operator. Because each person is typically responsible for 300 to 1000 control loops, itis not feasible to tune every controller.



38

4. Diagnostic techniques for moni toring control systemperformance are available.

C h a p t e r 1 2

Continuous Cycling Method

Over 60 years ago, Ziegler and Nichols (1942) published aclassic paper that introduced the continuous cycling method forcontroller tuning. It is based on the following trial-and-error

procedure:

Step 1. After the process has reached steady state (at leastapproximately), eliminate the integral and derivative control

action by setting to zero and to the largest possible value.

Step 2. Set K c equal to a small value (e.g., 0.5) and place thecontroller in the automatic mode.

Step 3. Introduce a small, momentary set-point change so that thecontrolled variable moves away from the set point. Gradually

τ D τ I



39

y p y

increase K c in small increments until continuous cycling occurs.The term continuous cycling refers to a sustained oscillation witha constant amplitude. The numerical value of K c that produces

C h a p t e r 1 2

continuous cycling (for proportional-only control) is called theultimate gain, K cu. The period of the corresponding sustained

oscillation is referred to as the ultimate period , P u.

Step 4. Calculate the PID controller settings using the Ziegler- Nichols (Z-N) tuning relations in Table 12.6.

Step 5. Evaluate the Z-N controller settings by introducing asmall set-point change and observing the closed-loop response.

Fine-tune the settings, if necessary.

The continuous cycling method, or a modified version of it, isfrequently recommended by control system vendors. Even so, thecontinuous cycling method has several major disadvantages:

1. It can be quite time-consuming if several trials are required and



40

q g q

the process dynamics are slow. The long experimental testsmay result in reduced production or poor product quality.

C h a p t e r 1 2

P u



41

Figure 12.12 Experimental determination of the ultimate gain K cu.

C h a p

t e r 1 2



42

C h a p

t e r 1 2

2. In many applications, continuous cycling is objectionable

because the process is pushed to the stability limits.3. This tuning procedure is not applicable to integrating or

open-loop unstable processes because their control loops

typically are unstable at both high and low values of K c,while being stable for intermediate values.

4. For first-order and second-order models without time delays,

the ultimate gain does not exist because the closed-loopsystem is stable for all values of K c, providing that its sign iscorrect. However, in practice, it is unusual for a control loop

not to have an ultimate gain.



43

C h a p

t e r 1 2

Relay Auto-Tuning

• Åström and Hägglund (1984) have developed an attractivealternative to the continuous cycling method.

• In the relay auto-tuning method, a simple experimental test is

used to determine K cu and P u.

• For this test, the feedback controller is temporarily replaced byan on-off controller (or relay).

• After the control loop is closed, the controlled variable exhibitsa sustained oscillation that is characteristic of on-off control

(cf. Section 8.4). The operation of the relay auto-tuner includesa dead band as shown in Fig. 12.14.

• The dead band is used to avoid frequent switching caused by



44

measurement noise.

C h a p

t e r 1 2



45

Figure 12.14 Auto-tuning using a relay controller.

C h a p

t e r 1 2

• The relay auto-tuning method has several important advantages

compared to the continuous cycling method:

1. Only a single experiment test is required instead of a

trial-and-error procedure.2. The amplitude of the process output a can be restricted

by adjusting relay amplitude d .

3. The process is not forced to a stability limit.

4. The experimental test is easily automated usingcommercial products.



46

C h a p

t e r 1 2

Step Test Method

• In their classic paper, Ziegler and Nichols (1942) proposed asecond on-line tuning technique based on a single step test.The experimental procedure is quite simple.

• After the process has reached steady state (at leastapproximately), the controller is placed in the manual mode.

• Then a small step change in the controller output (e.g., 3 to5%) is introduced.

• The controller settings are based on the process reaction curve

(Section 7.2), the open-loop step response.• Consequently, this on-line tuning technique is referred to as the

step test method or the process reaction curve method.



47

C h a p

t e r 1 2



48

Figure 12.15 Typical process reaction curves: (a) non-self-regulating process, (b) self-regulating process.

C h a p

t e r 1 2

An appropriate transfer function model can be obtained from thestep response by using the parameter estimation methods of

Chapter 7.

The chief advantage of the step test method is that only a singleexperimental test is necessary. But the method does have fourdisadvantages:1. The experimental test is performed under open-loop conditions.

Thus, if a significant disturbance occurs during the test, no

corrective action is taken. Consequently, the process can beupset, and the test results may be misleading.

2. For a nonlinear process, the test results can be sensitive to themagnitude and direction of the step change. If the magnitude ofthe step change is too large, process nonlinearities caninfluence the result. But if the step magnitude is too small, the



49

step response may be difficult to distinguish from the usualfluctuations due to noise and disturbances. The direction of thestep change (positive or negative) should be chosen so that

C h a p

t e r 1 2

the controlled variable will not violate a constraint.

3. The method is not applicable to open-loop unstable processes.

4. For analog controllers, the method tends to be sensitive tocontroller calibration errors. By contrast, the continuous

cycling method is less sensitive to calibration errors in K c because it is adjusted during the experimental test.

Example 12.8

Consider the feedback control system for the stirred-tank blending process shown in Fig. 11.1 and the following step test. Thecontroller was placed in manual, and then its output was suddenly

changed from 30% to 43%. The resulting process reaction curve isshown in Fig. 12.16. Thus, after the step change occurred at t = 0,the measured exit composition changed from 35% to 55%



50

(expressed as a percentage of the measurement span), which isequivalent to the mole fraction changing from 0.10 to 0.30.Determine an appropriate process model for . IP v p mG G G G G

C h a p

t e r 1 2

Figure 11.1 Composition control system for a stirred-tank



51

g p y

blending process.

C h a p

t e r 1 2



52

Figure 12.16 Process reaction curve for Example 12.8.

C h a p

t e r 1 2

Figure 12.17 Block diagram for Example 12.8.



53

C h a p

t e r 1 2

Solution

A block diagram for the closed-loop system is shown in Fig.12.17. This block diagram is similar to Fig. 11.7, but the feedbackloop has been broken between the controller and the current-to- pressure (I/P) transducer. A first-order-plus-time-delay model can

be developed from the process reaction curve in Fig. 12.16 usingthe graphical method of Section 7.2. The tangent line through theinflection point intersects the horizontal lines for the initial and

final composition values at 1.07 min and 7.00 min, respectively.The slope of the line is

55 35%3.37%/min7.00 1.07 minS

−= = −

and the normalized slope is



54

13.37%/min0.259min

43% 30%

S R

p

−= = =∆ −

C h a p

t e r 1 2

The model parameters can be calculated as

( )55% 35% 1.54 dimensionless43% 30%

θ 1.07 min

τ 7.00 1.07 min 5.93 min

m x K p

∆ −= = =∆ −

=

= − =

The apparent time delay of 1.07 min is subtracted from the

intercept value of 7.00 min for the calculation.The resulting empirical process model can be expressed as

τ

( )( ) ( )1.07

1.545.93 1

s

m X s eG s P s s

−′= =′ +

Example 12.5 in Section 12.3 provided a comparison of PI



55

p p pcontroller settings for this model that were calculated usingdifferent tuning relations.

C h a p

t e r 1 2

Guidelines For Common Control Loops

(see text)

Troubleshooting Control Loops• If a control loop is not performing satisfactorily, then

troubleshooting is necessary to identify the source of the problem.

• Based on experience in the chemical industry, he has observedthat a control loop that once operated satisfactorily can become

either unstable or excessively sluggish for a variety of reasonsthat include:

a. Changing process conditions, usually changes in



56

throughput rate. b. Sticking control valve stem.

C h a p

t e r 1 2

c. Plugged line in a pressure or differential pressuretransmitter.

d. Fouled heat exchangers, especially reboilers fordistillation columns.

e. Cavitating pumps (usually caused by a suction pressurethat is too low).

The starting point for troubleshooting is to obtain enough background information to clearly define the problem. Manyquestions need to be answered:

1. What is the process being controlled?

2. What is the controlled variable?3. What are the control objectives?

4. Are closed-loop response data available?



57

5. Is the controller in the manual or automatic mode? Is itreverse or direct acting?

C h a p

t e r 1 2

6. If the process is cycling, what is the cycling frequency?

7. What control algorithm is used? What are the controllersettings?

8. Is the process open-loop stable?

9. What additional documentation is available, such ascontrol loop summary sheets, piping and instrumentationdiagrams, etc.?



58

C h a p

t e r 1 3

Frequency Response Analysis

Sinusoidal Forcing of a First-Order Process

For a first-order transfer function with gain K and time constant ,

the response to a general sinusoidal input, is:

τ

( ) sin ω= x t A t

( ) ( )/ τ

2 2 ωτ ωτ cosω sin ω (5-25)

ω τ 1

−= − +

+

t KA y t e t t

Note that y(t) and x(t) are in deviation form. The long-time

response, yl (t), can be written as:

( ) ( )2 2

sin ω φ for (13-1)ω τ 1

= + → ∞+

KA y t t t

where:



1

( )1φ tan ωτ−= −

C h a p

t e r 1 3

Figure 13.1 Attenuation and time shift between input and output

sine waves ( K = 1). The phase angle of the output signal is given

by , where is the (period) shift and P

φ

φ Time shift / 360= − × P t ∆



2

is the period of oscillation.

C h a p

t e r 1 3

Frequency Response Characteristics of

a First-Order Process( ) ( )ˆFor ( ) sin , sin ω φ as where :ω = = + → ∞

x t A t y t A t t

( )12 2

ˆ and φ tan ωτω τ 1

−= = −+

KA A

1. The output signal is a sinusoid that has the same frequency, ω,as the input.signal, x(t) = Asinωt .

2. The amplitude of the output signal, , is a function of the

frequency ω and the input amplitude, A:

ˆ

2 2

ˆ (13-2)ω τ 1

=+

KA A



3

3. The output has a phase shift, φ, relative to the input. The

amount of phase shift depends on ω.

C h a p

t e r 1 3

Dividing both sides of (13-2) by the input signal amplitude A

yields the amplitude ratio (AR)

2 2

ˆ

AR (13-3a)ω τ 1= = +

A K

A

which can, in turn, be divided by the process gain to yield the

normalized amplitude ratio (AR N)

N2 2

1AR (13-3b)

ω τ 1=

+



4

C h a p

t e r 1 3

Shortcut Method for Finding

the Frequency ResponseThe shortcut method consists of the following steps:

Step 1. Set s=jω in G( s) to obtain .

Step 2. Rationalize G( jω); We want to express it in the form.

G( jω)= R + jI where R and I are functions of ω. Simplify G( jω) by

multiplying the numerator and denominator by the

complex conjugate of the denominator.

Step 3. The amplitude ratio and phase angle of G(s) are given

by:

( )ωG j

2 2AR R I



5

2 2

1

AR

tan ( / )

R I

I ϕ −

= +=

Memorize

C h a p

t e r 1 3

Example 13.1

Find the frequency response of a first-order system, with

( )1

(13-16)τ 1

G s s

=+

Solution

First, substitute in the transfer functionω s j=

( )1 1

ω (13-17)τ ω 1 ωτ 1

G j j j= =+ +

Then multiply both numerator and denominator by the complex

conjugate of the denominator, that is, ωτ 1 j− +

( )( )( ) 2 2

ωτ 1 ωτ 1ω

ωτ 1 ωτ 1 ω τ 1

j jG j

j j

− + − += =

+ − + +



6

( )2 2 2 2

ωτ1(13-18)

ω τ 1 ω τ 1 j R jI

−= + = +

+ +

C h a p

t e r 1 3

2 2

1(13-19a)

ω τ 1 R =

+where:

2 2

ωτ(13-19b)

ω τ 1 I

−=

+

From Step 3 of the Shortcut Method,

2 22 2

2 2 2 2

1 ωτAR

ω τ 1 ω τ 1

− = + = +

+ + R I

( )( )

2 2

2 2 22 2

1 ω τ

1AR (13-20a)ω τ 1ω τ 1

+

= = ++

Also,

( ) ( )1 1 1t t t (13 20b) I

or



7

( ) ( )1 1 1φ tan tan ωτ tan ωτ (13-20b)− − − = = − = − I R

C h a p

t e r 1 3

Complex Transfer Functions

Consider a complex transfer G( s),

( ) ( ) ( ) ( )

( ) ( ) ( )1 2 3

(13-22)=

a b cG s G s G sG s

G s G s G s

( ) ( ) ( ) ( )

( ) ( ) ( )1 2 3

ω ω ωω (13-23)

ω ω ω=

a b cG j G j G jG j

G j G j G j

Substitute s=jω,

From complex variable theory, we can express the magnitude and

angle of as follows:( )ωG j

( ) ( ) ( ) ( )( ) ( ) ( )1 2 3

ω ω ωω (13-24a)

ω ω ω=

a b cG j G j G jG j

G j G j G j

( ) ( ) ( ) ( )ω ω ω ωbG j G j G j G j∠ ∠ ∠ ∠



8

( ) ( ) ( ) ( )( ) ( ) ( )1 2 3

ω ω ω ω

[ ω ω ω ] (13-24b)

a b cG j G j G j G j

G j G j G j

∠ = ∠ + ∠ + ∠ +− ∠ + ∠ + ∠ +

C h a p

t e r 1 3

Bode Diagrams

• A special graph, called the Bode diagram or Bode plot , provides a convenient display of the frequency response

characteristics of a transfer function model. It consists of

plots of AR and as a function of ω.• Ordinarily, ω is expressed in units of radians/time.

φ

Bode Plot of A First-order System

Recall:

( )1 N

2 2

1AR and φ tan ωτ

ω τ 1

−= = −

+

N

ω 0 and ω 1) :

AR 1 and

ω 0 and ω 1) :

• → τ

= ϕ = 0

At low frequencies (

At high frequencies (



9

N

ω 0 and ω 1) :

AR 1/ ωτ and

• → τ

= ϕ = −90

At high frequencies (

C h a p

t e r 1 3



10

Figure 13.2 Bode diagram for a first-order process.

C h a p

t e r 1 3

• Note that the asymptotes intersect at , known as the

break frequency or corner frequency. Here the value of AR Nfrom (13-21) is:

ω ω 1/ τb= =

( ) N

1AR ω ω 0.707 (13-30)

1 1b= = =

+

• Some books and software defined AR differently, in terms of

decibels. The amplitude ratio in decibels AR d is defined as

dAR 20 log AR (13-33)=



11

C h a p

t e r 1 3

Integrating Elements

The transfer function for an integrating element was given inChapter 5:

( ) ( )

( )(5-34)

Y s K G s

U s s= =

( )AR ω (13-34)ω ω

K K G j

j= = =

( ) ( )φ ω 90 (13-35)G j K = ∠ = ∠ − ∠ ∞ = −

Second-Order ProcessA general transfer function that describes any underdamped,

critically damped, or overdamped second-order system is

K



12

( )2 2

(13-40)τ 2ζτ 1

K G s

s s=

+ +

C h a p

t e r 1 3

Substituting and rearranging yields:ω s j=

( ) ( )2 22 2

AR (13-41a)

1 ω τ 2ωτζ

K =

− +

1

2 2

2ωτζφ tan (13-41b)

1 ω τ

− − = −



13

Figure 13.3 Bode diagrams for second-order processes.

C h a p

t e r 1 3

Time Delay

Its frequency response characteristics can be obtained bysubstituting ,ω s j=

( ) ωθω (13-53) jG j e−=

which can be written in rational form by substitution of the

Euler identity,

( ) ωθ

ω cos ωθ sin ωθ (13-54)

−

= = − j

G j e jFrom (13-54)

( )( )

2 2

1

AR ω

cos ωθ

sin ωθ

1 (13-55)sin ωθ

φ ω tancosωθ

G j

G j −

= = + =

= ∠ = −

or θ (13 56)



14

or φ ωθ (13-56)= −

C h a p

t e r 1 3



15

Figure 13.6 Bode diagram for a time delay, .θ se−

C h a p

t e r 1 3

Figure 13.7 Phase angle plots for and for the 1/1 and 2/2

Padé approximations (G1 is 1/1; G2 is 2/2).

θ se−



16

Padé approximations (G is 1/1; G is 2/2).

C h a p

t e r 1 3

Process Zeros

Consider a process zero term,

( ) ( τ 1)= +G s K s

Substituting s=jω gives

( )ω ( ωτ 1)= +G j K j

Thus:

( )

( ) ( )

2 2

1

AR ω ω τ 1

φ ω tan ωτ−

= = +

= ∠ = +

G j K

G j

Note: In general, a multiplicative constant (e.g., K ) changes

the AR by a factor of K without affecting .φ



17

the AR by a factor of K without affecting .φ

C h a p

t e r 1 3

Frequency Response Characteristics of

Feedback Controllers

Proportional Control ler . Consider a proportional controller with positive gain

( ) (13-57)c cG s K =

In this case , which is independent of ω.

Therefore,( )ωc cG j K =

AR (13-58)c c K =and

φ 0 (13-59)c =



18

C h a p

t e r 1 3

Proportional-I ntegral Control ler . A proportional-integral (PI)

controller has the transfer function (cf. Eq. 8-9),

( ) τ 11

1 (13-60)

τ τ

I c c c

I I

sG s K K

s s

+= + =

Substitute s=jω:

( ) τ 11 11 1τ τ τ

ω +ω = + = = − ω ω ω

I c c c c I I I

jG j K K K j j j

Thus, the amplitude ratio and phase angle are:

( )( )

( )2

2

ωτ 11AR ω 1 (13-62)

ωτωτ

I c c c c

I I

G j K K +

= = + =

( ) ( ) ( )1 11/ 90 (13 63)G − −∠



19

( ) ( ) ( )1 1φ ω tan 1/ ωτ tan ωτ 90 (13-63)c c I I G j − −= ∠ = − = −

C h a p

t e r 1 3

Figure 13.9 Bode plot of a PI controller, ( ) 10 1210

c sG s + =



20

g p , ( ) 10 s+

C h a p

t e r 1 3

I deal Proportional-Der ivative Control ler . For the ideal

proportional-derivative (PD) controller (cf. Eq. 8-11)

( ) ( )1 τ (13-64)c c DG s K s= +

The frequency response characteristics are similar to those of a

LHP zero:

( )2

AR ωτ 1 (13-65)c c D K = +

( )1φ tan ωτ (13-66) D−=

Proportional-Der ivative Control ler with F ilter . The PD

controller is most often realized by the transfer function

( )

τ 1

(13-67)ατ 1

D

c cD

s

G s K s

+

= +



21

( ) ( ) D +

C h a p

t e r 1 3

Figure 13.10 Bode

plots of an ideal PD

controller and a PDcontroller with

derivative filter.

Idea:

With Derivative

Filter:

G s( ) ( )2 4 1c s= +

( ) 4 12

0.4 1c

sG s

s+ = +



22

C h a p

t e r 1 3

PID Controller Forms

Parallel PID Controller . The simplest form in Ch. 8 is

( )1

11 τ

τc c DG s K s

s

= + +

Series PID Control ler . The simplest version of the series PID

controller is

( ) ( )1

1

τ 1τ 1 (13-73)

τc c D

sG s K s

s

+= +

Ser ies PID Control ler with a Derivative F il ter .

( ) 1τ 1 τ 1τ τ 1

Dc c s sG s K s sα

+ += +



23

( )1τ τ 1 D s sα

+

C h a p

t e r 1 3

Figure 13.11 Bode plots of ideal parallel

PID controller and

series PID controllerwith derivative filter

(α = 1).

Idea parallel:

Series with

Derivative Filter:

( )1

2 1 410

cG s s s

= + +

( )10 1 4 1

2 s s

G s + + =



24

( ) 210 0.4 1

cG s s s

= +

C h a p

t e r 1 3

Nyquist Diagrams

Consider the transfer function

( )

1(13-76)

2 1G s

s=

+with

( ) ( )2

1AR ω (13-77a)

2ω 1

G j= =+

and

( ) ( )1φ ω tan 2ω (13-77b)G j −= ∠ = −



25

C h a p

t e r 1 3

Figure 13.12 The Nyquist diagram for G( s) = 1/(2 s + 1)plotting and( )( )Re ωG j ( )( )Im ωG j



26

plotting and( )( )Re ωG j ( )( )Im ω .G j

C h a p

t e r 1 3

Figure 13.13 The Nyquist diagram for the transfer

function in Example 13.5:

65(8 1)( ) (20 1)(4 1)

s s eG s s s

−+= + +



27

( )( )

C h a p

t e r 1 4

Control System Design Based on

Frequency Response AnalysisFrequency response concepts and techniques play an importantrole in control system design and analysis.

Closed-Loop Behavior

In general, a feedback control system should satisfy the following

design objectives:1. Closed-loop stability

2. Good disturbance rejection (without excessive control action)

3. Fast set-point tracking (without excessive control action)

4. A satisfactory degree of robustness to process variations and

model uncertainty



15. Low sensitivity to measurement noise

C h a p

t e r 1 4

• The block diagram of a general feedback control system isshown in Fig. 14.1.

• It contains three external input signals: set point Y sp, disturbance D, and additive measurement noise, N .

(14-1)1 1 1

m c v pd c sp

c c c

K G G GG G GY D N Y G G G G G G

= − ++ + +

(14-2)1 1 1

d m m m sp

c c c

G G G K E D N Y

G G G G G G= − − +

+ + +

(14-3)1 1 1d m c v m c v m c v

spc c c

G G G G G G G K G GU D N Y

G G G G G G= − − +

+ + +

where G G G G



2

where G GvG pGm.

C h a p

t e r 1 4

Figure 14.1 Block diagram with a disturbance D andmeasurement noise N .



3

C h a p

t e r 1 4

Example 14.1

Consider the feedback system in Fig. 14.1 and the followingtransfer functions:

0.5, 1

1 2 p d v mG G G G

s

= = = =

−Suppose that controller Gc is designed to cancel the unstable pole in G p:

3 (1 2 )1c s

G s

−= −+

Evaluate closed-loop stability and characterize the outputresponse for a sustained disturbance.



4

C h a p

t e r 1 4

Solution

The characteristic equation, 1 + GcG = 0, becomes:

3 (1 2 ) 0.51 0

1 1 2

s

s s

−+ =

+ −

or 2.5 0 s + =

In view of the single root at s = -2.5, it appears that the closed-loop system is stable. However, if we consider Eq. 14-1 for N = Y sp = 0,

( )0.5 11 (1 2 )( 2.5)

d

c

sGY D DG G s s

− += =+ − +



5

C h a p

t e r 1 4

• This transfer function has an unstable pole at s = +0.5. Thus,the output response to a disturbance is unstable.

• Furthermore, other transfer functions in (14-1) to (14-3) alsohave unstable poles.

• This apparent contradiction occurs because the characteristicequation does not include all of the information, namely, theunstable pole-zero cancellation.

Example 14.2

Suppose that Gd = G p, Gm = K m and that Gc is designed so that the

closed-loop system is stable and |GGc | >> 1 over the frequencyrange of interest. Evaluate this control system design strategy forset-point changes, disturbances, and measurement noise. Also

consider the behavior of the manipulated variable, U .



6

C h a p

t e r 1 4

Solution

Because |GGc | >> 1,

10 and 1

1 1c

c c

G G

G G G G≈ ≈

+ +

The first expression and (14-1) suggest that the output responseto disturbances will be very good because Y/D ≈ 0. Next, weconsider set-point responses. From Eq. 14-1,

1

m c v p

sp c

K G G GY

Y G G=

+

Because Gm = K m, G = GvG p K m and the above equation can bewritten as,

1c

sp c

G GY Y G G= +



7

sp c

C h a p

t e r 1 4

For |GGc | >> 1,

1 sp

Y

Y ≈

Thus, ideal (instantaneous) set-point tracking would occur.

Choosing Gc so that |GGc| >> 1 also has an undesirableconsequence. The output Y becomes sensitive to noise becauseY ≈ - N (see the noise term in Eq. 14-1). Thus, a design tradeoffis required.

Bode Stability Criterion

The Bode stability criterion has two important advantages incomparison with the Routh stability criterion of Chapter 11:

1. It provides exact results for processes with time delays, while

the Routh stability criterion provides only approximate resultsdue to the polynomial approximation that must be substituted



8for the time delay.

C h a p

t e r 1 4

2. The Bode stability criterion provides a measure of the relative

stability rather than merely a yes or no answer to the question,

“Is the closed-loop system stable?”Before considering the basis for the Bode stability criterion, it isuseful to review the General Stability Criterion of Section 11.1:

A feedback control system is stable if and only if all roots of the

characteristic equation lie to the left of the imaginary axis in the

complex plane.

Before stating the Bode stability criterion, we need to introducetwo important definitions:

1. A critical frequency is defined to be a value of forwhich . This frequency is also referred to asa phase crossover frequency.

2. A gain crossover frequency is defined to be a value of

ωc ω

( )φ ω 180OL = −

ω g ω

( )



9for which .( )ω 1OL AR =

C h a p

t e r 1 4

For many control problems, there is only a single and asingle . But multiple values can occur, as shown in Fig. 14.3

for .

ωcω g

ωc

Fi 14 3 B d l t hibiti lti l iti l f i



10

Figure 14.3 Bode plot exhibiting multiple critical frequencies.

C h a p

t e r 1 4

Bode Stabil i ty Criter ion . Consider an open-loop transfer function

GOL=GcGvG pGm that is strictly proper (more poles than zeros)

and has no poles located on or to the right of the imaginary axis,with the possible exception of a single pole at the origin. Assume

that the open-loop frequency response has only a single critical

frequency and a single gain crossover frequency . Then theclosed-loop system is stable if AROL( ) < 1. Otherwise it is

unstable.

ωc ω g ωc

Some of the important properties of the Bode stability criterionare:

1. It provides a necessary and sufficient condition for closed-loop stability based on the properties of the open-loop transferfunction.

2. Unlike the Routh stability criterion of Chapter 11, the Bodestability criterion is applicable to systems that contain time



11delays.

C h a p

t e r 1 4

3. The Bode stability criterion is very useful for a wide range of process control problems. However, for any GOL( s) that does

not satisfy the required conditions, the Nyquist stabilitycriterion of Section 14.3 can be applied.

4. For systems with multiple or , the Bode stability

criterion has been modified by Hahn et al. (2001) to provide asufficient condition for stability.

ωc ω g

• In order to gain physical insight into why a sustained oscillationoccurs at the stability limit, consider the analogy of an adult pushing a child on a swing.

• The child swings in the same arc as long as the adult pushes atthe right time, and with the right amount of force.

• Thus the desired “sustained oscillation” places requirements on

both timing (that is, phase) and applied force (that is,amplitude)



12

amplitude).

C h a p

t e r 1 4

• By contrast, if either the force or the timing is not correct, thedesired swinging motion ceases, as the child will quickly

exclaim.

• A similar requirement occurs when a person bounces a ball.

• To further illustrate why feedback control can producesustained oscillations, consider the following “thoughtexperiment” for the feedback control system in Figure 14.4.

Assume that the open-loop system is stable and that nodisturbances occur ( D = 0).

• Suppose that the set point is varied sinusoidally at the critical

frequency, y sp(t) = A sin(ωct), for a long period of time.• Assume that during this period the measured output, ym, is

disconnected so that the feedback loop is broken before the

comparator.



13

C h a p

t e r 1 4

Figure 14.4 Sustained oscillation in a feedback control system.



14

C h a p t e r 1 4

• After the initial transient dies out, ym will oscillate at theexcitation frequency ωc because the response of a linear system

to a sinusoidal input is a sinusoidal output at the same frequency(see Section 13.2).

• Suppose that two events occur simultaneously: (i) the set point

is set to zero and, (ii) ym is reconnected. If the feedback controlsystem is marginally stable, the controlled variable y will thenexhibit a sustained sinusoidal oscillation with amplitude A and

frequency ωc.• To analyze why this special type of oscillation occurs only when

ω = ωc, note that the sinusoidal signal E in Fig. 14.4 passes

through transfer functions Gc , Gv , G p , and Gm before returning tothe comparator.

• In order to have a sustained oscillation after the feedback loop is

reconnected, signal Y m must have the same amplitude as E and a-180° phase shift relative to E .



15

C h a p t e r 1 4

• Note that the comparator also provides a -180° phase shift dueto its negative sign.

• Consequently, after Y m passes through the comparator, it is in phase with E and has the same amplitude, A.

• Thus, the closed-loop system oscillates indefinitely after thefeedback loop is closed because the conditions in Eqs. 14-7and 14-8 are satisfied.

• But what happens if K c is increased by a small amount?• Then, AROL(ωc) is greater than one and the closed-loop system

becomes unstable.

• In contrast, if K c is reduced by a small amount, the oscillationis “damped” and eventually dies out.



16

C h a p t e r 1 4

Example 14.3

A process has the third-order transfer function (time constant inminutes),

3

2( )

(0.5 1)

p sG

s

=

+Also, Gv = 0.1 and Gm = 10. For a proportional controller,evaluate the stability of the closed-loop control system using the

Bode stability criterion and three values of K c: 1, 4, and 20.

Solution

For this example,

3 3

2 2( )(0.1) (10)

(0.5 1) (0.5 1)

ccOL c v p m

K G G G G G K

s s

= = =

+ +



17

C h a p t e r 1 4

Figure 14.5 shows a Bode plot of GOL for three values of K c.

Note that all three cases have the same phase angle plot because

the phase lag of a proportional controller is zero for K c > 0. Next, we consider the amplitude ratio AROL for each value of K c.

Based on Fig. 14.5, we make the following classifications:

Unstable520Marginally stable14

Stable0.251

Classification K c ( )for ω ωOL c AR =



18

C h a p t e r 1 4

Figure 14.5 Bode plots for GOL = 2 K c/(0.5 s+1)3.



19

C h a p t e r 1 4

In Section 12.5.1 the concept of the ultimate gain was introduced.For proportional-only control, the ultimate gain K cu was defined to

be the largest value of K c that results in a stable closed-loopsystem. The value of K cu can be determined graphically from aBode plot for transfer function G = GvG pGm. For proportional-

only control, GOL= K cG. Because a proportional controller haszero phase lag if K c > 0, ωc is determined solely by G. Also,

AROL(ω)=K c ARG(ω) (14-9)

where ARG denotes the amplitude ratio of G. At the stability limit,ω = ωc, AROL(ωc) = 1 and K c= K cu. Substituting these expressionsinto (14-9) and solving for K

cu

gives an important result:

1(14-10)

(ω )cuG c

K AR

=

The stability limit for K c can also be calculated for PI and PIDcontrollers, as demonstrated by Example 14.4.



20

, y p

C h a p t e r 1 4

Nyquist Stability Criterion

• The Nyquist stability criterion is similar to the Bode criterionin that it determines closed-loop stability from the open-loopfrequency response characteristics.

• The Nyquist stability criterion is based on two concepts fromcomplex variable theory, contour mapping and the Principle

of the Argument .

Nyquist Stabi l i ty Cr iter ion . Consider an open-loop transfer function GOL( s) that is proper and has no unstable pole-zero

cancellations. Let N be the number of times that the Nyquist plot

for GOL(s) encircles the -1 point in the clockwise direction. Alsolet P denote the number of poles of GOL(s) that lie to the right of

the imaginary axis. Then, Z = N + P where Z is the number of

roots of the characteristic equation that lie to the right of theimaginary axis (that is, its number of “zeros”). The closed-loop

system is stable if and only if Z = 0



21 system is stable if and only if Z 0.

C h a p t e r 1 4

Some important properties of the Nyquist stability criterion are:

1. It provides a necessary and sufficient condition for closed-

loop stability based on the open-loop transfer function.

2. The reason the -1 point is so important can be deduced fromthe characteristic equation, 1 + G

OL( s) = 0. This equation can

also be written as GOL( s) = -1, which implies that AROL = 1and , as noted earlier. The -1 point is referred toas the critical point .

3. Most process control problems are open-loop stable. Forthese situations, P = 0 and thus Z = N . Consequently, theclosed-loop system is unstable if the Nyquist plot for G

OL

( s)encircles the -1 point, one or more times.

4. A negative value of N indicates that the -1 point is encircled

in the opposite direction (counter-clockwise). This situationimplies that each countercurrent encirclement can stabilizeone unstable pole of the open-loop system

φ 180OL = −



22one unstable pole of the open loop system.

23

C h a p t e r 1 4

5. Unlike the Bode stability criterion, the Nyquist stabilitycriterion is applicable to open-loop unstable processes.

6. Unlike the Bode stability criterion, the Nyquist stabilitycriterion can be applied when multiple values of oroccur (cf. Fig. 14.3).

ωc ω g

Example 14.6

Evaluate the stability of the closed-loop system in Fig. 14.1 for:

4( )5 1

s

pe

sG s

−=

+

(the time constants and delay have units of minutes)

Gv = 2, Gm = 0.25, Gc = K c

Obtain ωc and K cu from a Bode plot. Let K c =1.5 K cu and drawthe Nyquist plot for the resulting open-loop system.



23

24

C h a p t e r 1 4

Solution

The Bode plot for GOL and K c = 1 is shown in Figure 14.7. For

ωc = 1.69 rad/min, φOL = -180° and AROL = 0.235. For K c = 1, AROL = ARG and K cu can be calculated from Eq. 14-10. Thus, K cu = 1/0.235 = 4.25. Setting K c = 1.5 K cu gives K c = 6.38.

Figure 14.7

Bode plot forExample 14.6, K c = 1.



24

25

C h a p t e r 1 4

Figure 14.8 Nyquist plot for Example 14.6,

K c = 1.5 K cu = 6.38.



25

26

C h a p t e r 1 4

Gain and Phase Margins

Let ARc be the value of the open-loop amplitude ratio at thecritical frequency . Gain margin GM is defined as:ωc

1

(14-11)cGM AR

Phase margin PM is defined as

180 φ (14-12) g PM +

• The phase margin also provides a measure of relative stability.

• In particular, it indicates how much additional time delay can beincluded in the feedback loop before instability will occur.

• Denote the additional time delay as .• For a time delay of , the phase angle is .

maxθ∆

maxθ∆ maxθ ω−∆



26max max

27

C h a p t e r 1 4

Figure 14.9 Gainand phase margins

in Bode plot.



27

28

C h a p t e r 1 4

max c180

= θ ω (14-13) PM π

∆

or

maxc

PMθ = (14-14)

ω 180

π ∆

where the factor converts PM from degrees to radians.( )/180π

• The specification of phase and gain margins requires a

compromise between performance and robustness.

• In general, large values of GM and PM correspond to sluggishclosed-loop responses, while smaller values result in lesssluggish, more oscillatory responses.

Guideline. In general, a well-tuned controller should have a gain

margin between 1.7 and 4.0 and a phase margin between 30° and45°.



28

29

C h a p t e r 1 4

Figure 14.10 Gain and phase margins on a Nyquist plot.



29

30

C h a p t e r 1 4

Recognize that these ranges are approximate and that it may not be possible to choose PI or PID controller settings that result in

specified GM and PM values.

Example 14.7

For the FOPTD model of Example 14.6, calculate the PIDcontroller settings for the two tuning relations in Table 12.6:

1. Ziegler-Nichols

2. Tyreus-Luyben

Assume that the two PID controllers are implemented in the

parallel form with a derivative filter (α = 0.1). Plot the open-loopBode diagram and determine the gain and phase margins for eachcontroller.



30

31

C h a p t e r 1 4

Figure 14.11Comparison of GOL

Bode plots forExample 14.7.



31



33

C h a p t e r 1 4

The open-loop transfer function is:

2

5 1

se

G G G G G GOL c v p m c s

−

= = +

Figure 14.11 shows the frequency response of GOL for the two

controllers. The gain and phase margins can be determined byinspection of the Bode diagram or by using the MATLABcommand, margin.

0.7976°1.8Tyreus-Luyben

1.0240°1.6Ziegler- Nichols

ωc (rad/min) PM GM Controller



34

C h a p t e r 1 4

The Tyreus-Luyben controller settings are more conservativeowing to the larger gain and phase margins. The value of

is calculated from Eq. (14-14) and the information in the abovetable:

maxθ∆

max

(76°)(π rad)θ = = 1.7 min

(0.79 rad/min)(180°)∆

Thus, time delay can increase by as much as 70% and stillmaintain closed-loop stability.

θ



35

C h a p t e r 1 4

Figure 14.12 Nyquist plot where the gain and phase margins aremisleading.



36

C h a p t e r 1 4

Closed-Loop Frequency Response and

Sensitivity Functions

Sensitivity Functions

The following analysis is based on the block diagram in Fig.14.1. We define G as and assume that Gm=K m andGd = 1. Two important concepts are now defined:

v p mG G G G

1 sensitivity function (14-15a)1

complementary sensitivity function (14-15b)1

c

c

c

S G G

G G

T G G

+

+



37

C h a p t e r 1 4

Comparing Fig. 14.1 and Eq. 14-15 indicates that S is theclosed-loop transfer function for disturbances (Y/D), while T is

the closed-loop transfer function for set-point changes (Y/Y sp). Itis easy to show that:

1 (14-16)S T + =

As will be shown in Section 14.6, S and T provide measures ofhow sensitive the closed-loop system is to changes in the process.

• Let |S ( j )| and |T ( j )| denote the amplitude ratios of S and T ,respectively.

• The maximum values of the amplitude ratios provide usefulmeasures of robustness.

• They also serve as control system design criteria, as discussed below.

ω ω



C h a p t e r 1 4

• Define M S to be the maximum value of |S ( j )| for allfrequencies:

ω

ωmax | ( ω) | (14-17)S M S j

The second robustness measure is M T , the maximum value of

|T ( j )|:ω

ωmax | ( ω) | (14-18)T M T j

M T is also referred to as the resonant peak. Typical amplituderatio plots for S and T are shown in Fig. 14.13.

It is easy to prove that M S and M T are related to the gain and

phase margins of Section 14.4 (Morari and Zafiriou, 1989):

1 1GM , PM 2sin (14-19)

1 2

S

S S

M

M M

− ≥ ≥

−



C h a

p t e r 1 4

Figure 14.13 Typical S and T magnitude plots. (Modified fromMaciejowski (1998)).

Guideline. For a satisfactory control system, M T should be in therange 1.0 – 1.5 and M S should be in the range of 1.2 – 2.0.



C h a

p t e r 1 4

It is easy to prove that M S and M T are related to the gain and phase margins of Section 14.4 (Morari and Zafiriou, 1989):

1 1GM , PM 2sin (14-19)

1 2S

S S

M

M M

− ≥ ≥ −

1GM 1 , PM 2sin (14-20)2

T T M M

− ≥ + ≥

1 1



C h a

p t e r 1 4

Bandwidth

• In this section we introduce an important concept, the

bandwidth. A typical amplitude ratio plot for T and thecorresponding set-point response are shown in Fig. 14.14.

• The definition, the bandwidth ω BW is defined as the frequency atwhich |T ( jω)| = 0.707.

• The bandwidth indicates the frequency range for which

satisfactory set-point tracking occurs. In particular, ω BW is themaximum frequency for a sinusoidal set point to be attenuated by no more than a factor of 0.707.

• The bandwidth is also related to speed of response.• In general, the bandwidth is (approximately) inversely

proportional to the closed-loop settling time.



C h a

p t e r 1 4

Figure 14.14 Typical closed-loop amplitude ratio |T ( jω)| andset-point response.



C h a

p t e r 1 4

Closed-loop Performance Criteria

Ideally, a feedback controller should satisfy the following

criteria.1. In order to eliminate offset, |T ( jω)|→ 1 as ω → 0.

2. |T ( jω)| should be maintained at unity up to as high asfrequency as possible. This condition ensures a rapidapproach to the new steady state during a set-point change.

3. As indicated in the Guideline, M T should be selected so that1.0 < M T < 1.5.

4. The bandwidth ω BW and the frequency ωT at which M T

occurs, should be as large as possible. Large values result inthe fast closed-loop responses.

Nichols Chart

The closed-loop frequency response can be calculated analyticallyfrom the open-loop frequency response.



C h a

p t e r 1 4

Figure 14.15 A Nichols chart. [The closed-loop amplitude ratio ARCL ( ) and phase angle are shown in familiesof curves.]

( )φCL − − −



C h a

p t e r 1 4

Example 14.8

Consider a fourth-order process with a wide range of timeconstants that have units of minutes (Åström et al., 1998):

1(14-22)

( 1)(0.2 1)(0.04 1)(0.008 1)

v p mG G G G

s s s s

= =

+ + + +Calculate PID controller settings based on following tuningrelations in Chapter 12

a. Ziegler-Nichols tuning (Table 12.6)

b. Tyreus-Luyben tuning (Table 12.6)

c. IMC Tuning with (Table 12.1)τ 0.25 minc

d. Simplified IMC (SIMC) tuning (Table 12.5) and a second-

order plus time-delay model derived using Skogestad’smodel approximation method (Section 6.3).

=



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Determine sensitivity peaks M S and M T for each controller.Compare the closed-loop responses to step changes in the set-

point and the disturbance using the parallel form of the PIDcontroller without a derivative filter:

( ) 11 τ (14-23)( ) τc D

I

P s

K s E s s

′

= + +

Assume that Gd ( s) = G( s).



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Controller Settings for Example 14.8

1.161.580.1801.2221.8SimplifiedIMC

1.001.120.1671.204.3IMC

0.089

0.070

1.45

2.38

M S

1.231.2513.6Tyreus-Luyben

2.410.2818.1Ziegler- Nichols

M T K cController τ (min) I τ (min) D



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Figure 14.16 Closed-loop responses for Example 14.8. (A set-

point change occurs at t = 0 and a step disturbance at t = 4 min.)



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Robustness Analysis

• In order for a control system to function properly, it shouldnot be unduly sensitive to small changes in the process or toinaccuracies in the process model, if a model is used to designthe control system.

• A control system that satisfies this requirement is said to berobust or insensitive.

• It is very important to consider robustness as well as performance in control system design.

• First, we explain why the S and T transfer functions inEq. 14-15 are referred to as “sensitivity functions”.



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Sensitivity Analysis

• In general, the term sensitivity refers to the effect that a

change in one transfer function (or variable) has on anothertransfer function (or variable).

• Suppose that G changes from a nominal value G p0 to an

arbitrary new value, G p0 + dG.

• This differential change dG causes T to change from itsnominal value T

0to a new value, T

0+ dT .

• Thus, we are interested in the ratio of these changes, dT /dG,and also the ratio of the relative changes:

/sensitivity (14-25)

/

dT T

dG G

W i h l i i i i i i l f



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We can write the relative sensitivity in an equivalent form:

/(14-26)/

dT T dT G

dG G dG T

=

The derivative in (14-26) can be evaluated after substituting the

definition of T in (14-15b):

2 (14-27)cdT

G S dG

=

Substitute (14-27) into (14-26). Then substituting the definition ofS in (14-15a) and rearranging gives the desired result:

/ 1(14-28)

/ 1 c

dT T S

dG G G G= =

+

• Equation 14 28 indicates that the relative sensitivity is equal to



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• Equation 14-28 indicates that the relative sensitivity is equal toS .

• For this reason, S is referred to as the sensitivity function.• In view of the important relationship in (14-16), T is called the

complementary sensitivity function.

Effect of Feedback Control on Relative Sensitivity

• Next, we show that feedback reduces sensitivity by comparing

the relative sensitivities for open-loop control and closed-loopcontrol.

• By definition, open-loop control occurs when the feedbackcontrol loop in Fig. 14.1 is disconnected from the comparator.

• For this condition:

(14-29)OL c sp OL

Y T G GY

=

S b tit ti T f T i E 14 25 d ti th t dT /dG G



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Substituting T OL for T in Eq. 14-25 and noting that dT OL/dG = Gc

gives:

/ 1 (14-30)/

OL OL OLc

OL c

dT T dT G GGdG G dG T G G

= = =

• Thus, the relative sensitivity is unity for open-loop control andis equal to S for closed-loop control, as indicated by (14-28).

• Equation 14-15a indicates that |S| <1 if |GcG p| > 1, which

usually occurs over the frequency range of interest.• Thus, we have identified one of the most important properties

of feedback control:

• Feedback control makes process performance less sensitive to

changes in the process.



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Feedforward and Ratio Control

In Chapter 8 is was emphasized that feedback control is animportant technique that is widely used in the process industries.

Its main advantages are as follows.

1. Corrective action occurs as soon as the controlled variable

deviates from the set point, regardless of the source and type

of disturbance.

2. Feedback control requires minimal knowledge about the

process to be controlled; it particular, a mathematical model

of the process is not required, although it can be very useful

for control system design.

3. The ubiquitous PID controller is both versatile and robust. If

process conditions change, retuning the controller usually

produces satisfactory control.

However feedback control also has certain inherent



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However, feedback control also has certain inherent

disadvantages:

1. No corrective action is taken until after a deviation in thecontrolled variable occurs. Thus, perfect control , where the

controlled variable does not deviate from the set point during

disturbance or set-point changes, is theoretically impossible.

2. Feedback control does not provide predictive control action

to compensate for the effects of known or measurable

disturbances.

3. It may not be satisfactory for processes with large time

constants and/or long time delays. If large and frequent

disturbances occur, the process may operate continuously in atransient state and never attain the desired steady state.

4. In some situations, the controlled variable cannot be

measured on-line, and, consequently, feedback control is not

feasible.

Introduction to Feedforward Control



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Introduction to Feedforward Control

The basic concept of feedforward control is to measure important

disturbance variables and take corrective action before they upsetthe process. Feedforward control has several disadvantages:

1. The disturbance variables must be measured on-line. In many

applications, this is not feasible.

2. To make effective use of feedforward control, at least a crude

process model should be available. In particular, we need to

know how the controlled variable responds to changes in both

the disturbance and manipulated variables. The quality of

feedforward control depends on the accuracy of the process

model.

3. Ideal feedforward controllers that are theoretically capable of

achieving perfect control may not be physically realizable.

Fortunately, practical approximations of these ideal controllersoften provide very effective control.



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Figure 15.2 The feedback control of the liquid level in a boiler

drum.

• A boiler drum with a conventional feedback control system is



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• A boiler drum with a conventional feedback control system is

shown in Fig. 15.2. The level of the boiling liquid is measured

and used to adjust the feedwater flow rate.• This control system tends to be quite sensitive to rapid changes

in the disturbance variable, steam flow rate, as a result of the

small liquid capacity of the boiler drum.

• Rapid disturbance changes can occur as a result of steam

demands made by downstream processing units.

The feedforward control scheme in Fig. 15.3 can provide better

control of the liquid level. Here the steam flow rate is

measured, and the feedforward controller adjusts the feedwater flow rate.



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Figure 15.3 The feedforward control of the liquid level in a

boiler drum.

• In practical applications feedforward control is normally used



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• In practical applications, feedforward control is normally used

in combination with feedback control.

• Feedforward control is used to reduce the effects of measurabledisturbances, while feedback trim compensates for inaccuracies

in the process model, measurement error, and unmeasured

disturbances.



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Figure 15.4 The feedfoward-feedback control of the boiler

drum level.

Ratio Control



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Ratio Control

• Ratio control is a special type of feedforward control that has

had widespread application in the process industries.

• The objective is to maintain the ratio of two process variables

as a specified value.• The two variables are usually flow rates, a manipulated

variable u, and a disturbance variable d .

• Thus, the ratio

(15-1)u

Rd

=

is controlled rather than the individual variables. In Eq. 15-1,

u and d are physical variables, not deviation variables.

Typical applications of ratio control:



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Typical applications of ratio control:

1. Setting the relative amounts of components in blending

operations

2. Maintaining a stoichiometric ratio of reactants to a reactor

3. Keeping a specified reflux ratio for a distillation column

4. Holding the fuel-air ratio to a furnace at the optimum value.



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Figure 15.5 Ratio control, Method I.

• The main advantage of Method I is that the actual ratio R is



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calculated.

• A key disadvantage is that a divider element must be includedin the loop, and this element makes the process gain vary in a

nonlinear fashion. From Eq. 15-1, the process gain

1 (15-2) pd

R K u d

∂ = = ∂

is inversely related to the disturbance flow rate d .

• Because of this significant disadvantage, the preferred scheme

for implementing ratio control is Method II, which is shown in

Fig. 15.6.



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Figure 15.6 Ratio control, Method II

• Regardless of how ratio control is implemented, the process



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variables must be scaled appropriately.

• For example, in Method II the gain setting for the ratio station K d must take into account the spans of the two flow

transmitters.

• Thus, the correct gain for the ratio station is

(15-3)d R d

u

S K R

S =

where Rd is the desired ratio, S u and S d are the spans of the

flow transmitters for the manipulated and disturbance streams,

respectively.

Example 15.1



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A ratio control scheme is to be used to maintain a stoichoimetric

ratio of H2 and N2 as the feed to an ammonia synthesis reactor.Individual flow controllers will be used for both the H2 and N2

streams. Using the information given below, do the following:

a) Draw a schematic diagram for the ratio control scheme.

b) Specify the appropriate gain for the ratio station, K R.

Available Information



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i. The electronic flow transmitters have built-in square root

extractors. The spans of the flow transmitters are 30 L/min forH2 and 15 L/min for N2.

ii. The control valves have pneumatic actuators.

iii. Each required current-to-pressure ( I / P ) transducer has a gain

of 0.75 psi/mA.

iv. The ratio station is an electronic instrument with 4-20 mAinput and output signals.

Solution

The stoichiometric equation for the ammonia synthesis reaction is

2 2 33H N 2NH+

In order to introduce the feed mixture in stoichiometric



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proportions, the ratio of the molar flow rates (H2/N2) should be

3:1. For the sake of simplicity, we assume that the ratio of the

molar flow rates is equal to the ratio of the volumetric flow rates.

But in general, the volumetric flow rates also depend on the

temperature and pressure of each stream (cf., the ideal gas law).

a) The schematic diagram for the ammonia synthesis reaction is

shown in Fig. 15.7. The H2 flow rate is considered to be the

disturbance variable, although this choice is arbitary because both the H2 and N2 flow rates are controlled. Note that the ratio

station is merely a device with an adjustable gain. The input

signal to the ratio station is d m, the measured H

2flow rate. Its

output signal u sp serves as the set point for the N2 flow control

loop. It is calculated as u sp = K Rd m.



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5

Figure 15.7 Ratio control scheme for an ammonia synthesis

reactor of Example 15.1

b) From the stoichiometric equation, it follows that the desired

i i /d 1/3 S b i i i i 1 3 i



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5

ratio is Rd = u/d = 1/3. Substitution into Equation 15-3 gives:

1 30 L / min 2

3 15 L / min 3 R K = =

Feedforward Controller Design Based onSteady-State Models

• A useful interpretation of feedforward control is that itcontinually attempts to balance the material or energy that must

be delivered to the process against the demands of the load.

• For example, the level control system in Fig. 15.3 adjusts thefeedwater flow so that it balances the steam demand.

• Thus, it is natural to base the feedforward control calculations

on material and energy balances.



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5

Figure 15.8 A simple schematic diagram of a distillation

column.

• To illustrate the design procedure, consider the distillation

column shown in Fig 15 8 which is used to separate a binary



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5

column shown in Fig. 15.8 which is used to separate a binary

mixture.

• In Fig. 15.8, the symbols B, D, and F denote molar flow rates,

whereas x, y, and z are the mole fractions of the more volatile

component.

• The objective is to control the distillation composition, y,

despite measurable disturbances in feed flow rate F and feed

composition z , by adjusting distillate flow rate, D.• It is assumed that measurements of x and y are not available.

The steady-state mass balances for the distillation column can bewritten as

(15-4)

(15-5) z

F D B

F Dy Bx

= +

= +

Solving (15-4) for D and substituting into (15-5) gives



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5

( )

(15-6)

F z x

D y x

−=

−

Because x and y are not measured, we replace these variables by

their set points to yield the feedforward control law:

( )(15-7)

sp

sp sp

F z x D

y x

−=

−

Blending System



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• Consider the blending system and feedforward controller shown

in Fig. 15.9.

• We wish to design a feedforward control scheme to maintain

exit composition x at a constant set point x sp

, despite

disturbances in inlet composition, x1.

• Suppose that inlet flow rate w1 and the composition of the other

inlet stream, x2, are constant.• It is assumed that x1 is measured but x is not.



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5

Figure 15.9 Feedforward control of exit composition in the

blending system.

The starting point for the feedforward controller design is the

steady-state mass and component balances,



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5

steady state mass and component balances,

1 2

(15-8)w w w= +

1 1 2 2 (15-9)w x w x w x= +

where the bar over the variable denotes a steady-state value.

Substituting Eq. 15-8 into 15-9 and solving for gives:2w

1 12

2

( )(15-10)

w x xw

x x

−=

−

In order to derive a feedforward control law, we replace by x sp,

and and , by w2(t ) and x1(t ), respectively:

x

2w 1 x

1 1

2

2

( )( ) (15-11)

sp

sp

w x x t w t

x x

− =−

Note that this feedforward control law is based on the physicalvariables rather than on the deviation variables.

• The feedforward control law in Eq. 15-11 is not in the final

form required for actual implementation because it ignores



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5

form required for actual implementation because it ignores

two important instrumentation considerations:

• First, the actual value of x1 is not available but its measured

value, x1m, is.

• Second, the controller output signal is p rather than inlet flowrate, w2.

• Thus, the feedforward control law should be expressed in

terms of x1m and p, rather than x1 and w2.

• Consequently, a more realistic feedforward control law should

incorporate the appropriate steady-state instrument relations

for the w2 flow transmitter and the control valve. (See text.)

Feedforward Controller Design Based on



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Dynamic Models

In this section, we consider the design of feedforward control

systems based on dynamic, rather than steady-state, process

models.

• As a starting point for our discussion, consider the block

diagram shown in Fig. 15.11.

• This diagram is similar to Fig. 11.8 for feedback control butan additional signal path through Gt and G f has been added.



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Figure 15.11 A block diagram of a feedforward-feedback controlsystem.

The closed-loop transfer function for disturbance changes is:

( ) G G G G GY +



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5

( )

( )(15-20)

1

d t f v p

c v p m

G G G G GY s

D s G G G G

+=

+Ideally, we would like the control system to produce perfect

control where the controlled variable remains exactly at the set

point despite arbitrary changes in the disturbance variable, D.Thus, if the set point is constant (Y sp( s) = 0), we want Y ( s) = 0,

even though D( s)

(15-21)d f

t v p

GGG G G= −

• Figure 15.11 and Eq. 15-21 provide a useful interpretation of the

ideal feedforward controller. Figure 15.11 indicates that adisturbance has two effects.

• It upsets the process via the disturbance transfer function, Gd ;

however, a corrective action is generated via the path throughGt G f GvG p.

• Ideally, the corrective action compensates exactly for the upset

so that signals Yd and Y cancel each other and Y(s) = 0.



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so that signals Y d and Y u cancel each other and Y ( s) 0.

Example 15.2

Suppose that

, (15-22)τ 1 τ 1

pd d p

d p

K K G G s s= =+ +

Then from (15-22), the ideal feedforward controller is

τ 1(15-23)

τ 1

pd f

t v p d

s K G

K K K s

+ = − +

This controller is a lead-lag unit with a gain given by

K f = - K d / K t K v K p. The dynamic response characteristics of lead-

lag units were considered in Example 6.1 of Chapter 6.

Example 15.3



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5

Now consider

θ

, (15-24)τ 1 τ 1

s pd

d p

d p

K e K G G

s s

−

= =+ +

From (15-21),

θτ 1

(15-25)τ 1

p sd f

t v p d

s K G e

K K K s

+ +

= +

Because the term is a negative time delay, implying a

predictive element, the ideal feedforward controller in (15-25)

is physically unrealizable. However, we can approximate it byomitting the term and increasing the value of the lead time

constant from to .

θ se+

θ se+

τ p τ θ p +

Example 15.4



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5

Finally, if

( )( )1 2

, (15-26)1 1 1

pd d p

d p p

K K G G

s s sτ τ τ

= =+ + +

then the ideal feedforward controller,

( )( )

( )

1 2τ 1 τ 1(15-27)

τ1

p pd f

t v p d

s s K G

K K K s

+ + = − +

is physically unrealizable because the numerator is a higher

order polynomial in s than the denominator. Again, we could

approximate this controller by a physically realizable one suchas a lead-lag unit, where the lead time constant is the sum of

the two time constants, 1 2τ τ . p p+

Stability Considerations

l h bili f h l d l i i



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• To analyze the stability of the closed-loop system in Fig. 15.11,

we consider the closed-loop transfer function in Eq. 15-20.

• Setting the denominator equal to zero gives the characteristic

equation,

1 0 (15-28)c v p mG G G G+ =

• In Chapter 11 it was shown that the roots of the characteristic

equation completely determine the stability of the closed-loopsystem.

• Because G f does not appear in the characteristic equation, the

feedforward controller has no effect on the stability of the

feedback control system.

• This is a desirable situation that allows the feedback and

feedforward controllers to be tuned individually.

Lead-Lag Units

Th h l i h i i h d d



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• The three examples in the previous section have demonstrated

that lead-lag units can provide reasonable approximations toideal feedforward controllers.

• Thus, if the feedforward controller consists of a lead-lag unit

with gain K f , we can write

( ) ( )

( )

( )1

2

τ 1(15-29)

τ 1

f

f

K sU sG s

D s s

+= =

+

Example 15.5

Consider the blending system of Section 15.3 and Fig. 15.9. A

feedforward-feedback control system is to be designed to reduce

the effect of disturbances in feed composition, x1, on the

controlled variable, produce composition, x. Inlet flow rate, w2,can be manipulated. (See text.)

Configurations for Feedforward-Feedback

Control



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Control

• In a typical control configuration, the outputs of the feedforward

and feedback controllers are added together, and the sum is sent

as the signal to the final control element.

• Another useful configuration for feedforward-feedback control

is to have the feedback controller output serve as the set point

for the feedforward controller.



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Figure 15.14 Feedforward-feedback control of exit composition inthe blending system.



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5Figure 15.15 The open-loop responses to step

changes in u and d .

Tuning Feedforward Controllers

Feedforward controllers like feedback controllers usually



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Feedforward controllers, like feedback controllers, usually

require tuning after installation in a plant.

Step 1. Adjust K f .

• The effort required to tune a controller is greatly reduced if goodinitial estimates of the controller parameters are available.

• An initial estimate of K f can be obtained from a steady-state

model of the process or steady-state data.

• For example, suppose that the open-loop responses to step

changes in d and u are available, as shown in Fig. 15.15.

• After K p and K d have been determined, the feedforward

controller gain can be calculated from the steady-state version

of Eq. 15-22:

(15-40)d f

t v p

K K K K K

= −

• To tune the controller gain, K f is set equal to an initial value, and

a small step change (3 to 5%) in the disturbance variable d is

i t d d if thi i f ibl



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introduced, if this is feasible.

• If an offset results, then K f is adjusted until the offset is

eliminated.

• While K f is being tuned, and should be set equal to theirminimum values, ideally zero.

1τ 2τ

Step 2. Determine initial values for and .1τ 2τ

• Theoretical values for and can be calculated if a dynamic

model of the process is available, as shown in Example 15.2.

• Alternatively, initial estimates can be determined from open-

loop response data.

• For example, if the step responses have the shapes shown inFigure 15.16, a reasonable process model is

1τ 2τ

( ) ( ), (15-41)τ 1 τ 1

p d p d

d

K K G s G s

s s= =

+ +



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τ 1 τ 1 p d s s+ +

where and can be calculated as shown in Fig. 15.16.

• A comparison of Eqs. 15-24 and 5-30 leads to the following

expression for and :

τ p τd

1τ 2τ

1τ τ (15-42) p=

2

τ τ (15-43)d

=

• These values can then be used as initial estimates for the fine

tuning of and in Step 3.

• If neither a process model nor experimental data are

available, the relations or may be used,

depending on whether the controlled variable responds faster

to the disturbance variable or to the manipulated variable.

1τ 2τ

1 2τ / τ 2= 1 2τ / τ 0.5=

• In view of Eq. 15-58, should be set equal to the estimated

dominant process time constant.1τ



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Step 3. Fine tune and .1τ 2τ

• The final step is to use a trial-and-error procedure to fine tune

and by making small step changes in d .

• The desired step response consists of small deviations in the

controlled variable with equal areas above and below the set

point, as shown in Fig. 15.17.

• For simple process models, it can be proved theoretically that

equal areas above and below the set point imply that the

difference, , is correct (Exercise 15.8).• In subsequent tuning to reduce the size of the areas, and

should be adjusted so that remains constant.

1τ

2τ

1 2τ τ−

1τ 2τ

1 2τ τ−



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Figure 15.16 The desired response for a well-tuned feedforward

controller. (Note approximately equal areas above and below theset point.)

• As a hypothetical illustration of this trial-and-error tuning

procedure, consider the set of responses shown in Fig. 15.17 for

iti t h i di t b i bl d



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positive step changes in disturbance variable d .

• It is assumed that K p > 0, K d < 0, and controller gain K f has

already been adjusted so that offset is eliminated.



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Figure 15.17 An example of feedforward controller tuning.

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