L11-14 Chapter 4

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Chapter 4

Transfer Functions

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Chapter Objectives

End of this chapter, you should be able to:

• Define what is a transfer function

• Develop transfer functions from mathematical models

• Use properties of transfer functions in simplifying and

analyzing models

• Use linearization to derive transfer functions for

nonlinear processes

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Transfer Functions

• An algebraic expression

• Convenient representation of a linear,

dynamic model

• A transfer function (TF) relates one

input and one output:

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system

x t y t

X s Y s

Transfer Functions

• Independent of initial conditions

• Independent of particular choice of forcing

functions

The following terminology is used:

x

input

forcing function

“cause”

y

output

response

“effect”

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where:

Y sG s

X s

Y s y t

X s x t

= L

= L

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Definition of transfer function

• Let G(s) denote the transfer function

between an input, x, and an output, y. Then,

by definition

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

Development of Transfer Functions

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Example: Stirred Tank Heating System

Recall the previous dynamic model, assuming constant liquid

holdup and flow rates:

(2-36)i

dTV C wC T T Q

dt

Suppose the process is at steady state:

0 (2)iwC T T Q

Subtract (2) from (2-36):

(3)i i

dTV C wC T T T T Q Q

dt

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But,

(4)i

dTV C wC T T Q

dt

where the “deviation variables” are

, ,i i iT T T T T T Q Q Q

0 (5)iV C sT s T wC T s T s Q s

Take L of (4):

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At the initial steady state, T′(0) = 0.

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Rearrange (5) to solve for

1

(6)1 1

i

KT s Q s T s

s s

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1and

VK

wC w

)()()()()( 21 sTsGsQs=GsT i

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G1 and G2 are transfer functions and independent of the

inputs, Q′ and Ti′.

Note G1 (process) has gain K and time constant τ.

G2 (disturbance) has gain=1 and time constant τ.

Gain = G(s=0). Both are first order processes.

If there is no change in inlet temperature (Ti′= 0),

then Ti′(s) = 0.

System can be forced by a change in either Ti or Q

(see Example 4.1).

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Conclusions about TFs

1. Note that (6) 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 is

valid.

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.

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Example: Stirred Tank Heater

0.05K 2.0

0.05

2 1T Q

s

No change in Ti′

Step change in Q(t): 1500 cal/sec to 2000 cal/sec

500Q

s

0.05 500 25

2 1 (2 1)T

s s s s

What is T′(t)?

/ 25( ) 25[1 ] ( )

( 1)

tT t e T ss s

/ 2( ) 25[1 ]tT t e

From line 13, Table 3.1

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Properties of Transfer Function Models

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 the input. 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

Ku u

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

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

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Calculation of K from the TF Model:

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)

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2. Order of a TF Model

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

dtdt dt dt

d u dub b b u

dtdt

Take L, assuming the initial conditions are all zero. Rearranging

gives the TF:

0

0

(4-40)

mi

i

in

ii

i

b sY s

G sU s

a s

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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.

Definition:

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

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General 2nd order ODE:

Laplace Transform:

2 roots

: real roots

: imaginary roots

y=Kudt

dy+b

dt

yda

2

2

KU(s)Y(s)bs+as 12

1)(

)()(

2

bsas

K

sU

sYsG

a

abbs ,

2

42

21

2b 1

4a

2b 1

4a

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2nd order process

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Examples

1. 2

2

3 4 1s s

2 161.333 1

4 12

b

a

2 13 4 1 (3 1)( 1) 3( )( 1)3

s s s s s s

3 , ( )t

ttransforms to e e real roots

(no oscillation)

2. 2

2

1s s

2 11

4 4

b

a

2 3 31 ( 0.5 )( 0.5 )

2 2s s s j s j

0.5 0.53 3cos , sin

2 2

t ttransforms to e t e t

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From Table 3.1, line 17

2 2

2

( )

1 3

2

es b

s s

L- bt

2

2

sin t

2 2=

(s+ 0.5)

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Two IMPORTANT properties (L.T.)

A. Multiplicative Rule

B. Additive Rule

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Example 1:

Place sensor for temperature downstream from heated

tank (transport lag)

Distance L for plug flow,

Delay time:

V = fluid velocity

Tank:

Sensor:

V

Lθ

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1)(

)(

1

11

s

K

sU

sTG

s+τ

eK=

T(s)

(s)T=G

s-

s

2

22

1

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Tank:

Sensor:

Overall transfer function:

2 is very small – can be neglected

sτ

eKKGG

U

T

T

T

U

T θs

ss

1

2112

1

Overall Transfer Function C

ha

pte

r 4 1)(

)(

1

11

s

K

sU

sTG

1 1

2

2

22 K

s+τ

eK=

T(s)

(s)T=G

s-

s

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Example 2: Consider the system shown below.

The system consists of two liquid surge tanks in

series so that the outflow from the first tank is the

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For tank 1,

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48)-(4 11

1 qqdt

dhA i

1

1

1

1h

Rq (4 - 49)

1

1

1

1h

Rq

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

(4 -52)

51)-(4 1

1

1

11 h

Rq

dt

hdA i

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

50)-(4 1

1

1

11 h

Rq

dt

dhA i

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The transfer function relating )(1 sH to )(sQi is found

by transforming (4-51) and rearranging to obtain

(4-53)

11)(

)(

1

1

11

11

s

K

sRA

R

sQ

sH

i

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Similarly, the transfer function relating )(1 sQ to

)(1 sH is obtained by transforming (4-52).

(4-54) 111

1 11

)(

)(

KRsH

sQ

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• 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).

The same procedure leads to the corresponding

transfer functions for Tank 2,

11)(

)(

2

2

22

2

1

2

s

K

sRA

R

sQ

sH

(4-55)

222

2 11

)(

)(

KRsH

sQ

(4.56)

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or

which can be simplified to give

11

1

)(

)(

21

2

sssQ

sQ

i (4-59)

)(

)(

)(

)(

)(

)(

)(

)(

)(

)( 1

1

1

1

2

2

22

sQ

sH

sH

sQ

sQ

sH

sH

sQ

sQ

sQ

ii

(4-57)

1

1

1

1

)(

)(

1

1

12

2

2

2

s

K

Ks

K

KsQ

sQ

i (4-58)

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Linearization of Nonlinear Models

• Required to derive transfer function.

• Good approximation near a given operating point.

• Gain, time constants may change with operating point.

• Use 1st order Taylor series.

),( uyfdt

dy

)()(),(),(,,

uuu

fyy

y

fuyfuyf

uyuy

uu

fy

y

f

dt

yd

ss

Subtract steady-state equation from dynamic equation

(4-60)

(4-61)

(4-62)

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Example 3:

q0: control variable,

qi: disturbance variable

...(1) 0qqdt

dhA i

Use L.T.

0( ) ( ) ( )iAsH s q s q s

(in deviation variables)

Suppose q0 is constant: 00 Δq

...(2) 1

As(s)q

H(s)

(s)qAsH(s)

i

i

• Pure integrator (ramp) for step

change in qi

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• Example of non-self regulating

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Nonlinear element

More realistically, if q0 is manipulated by a flow

control valve, ...(3) 0 hCq v

nonlinear element

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Figure 2.5

RV: line resistance

linear ODE : eq. (4-74)

hR

qV

1

...(4) 1

hR

qdt

dhA

V

i

...(5) hCq Vif

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Substitute linearized expression of

(5) into (1):

The steady-state version of (3) is:

0 (7)i vq C h

1(8)i

dhA q h

dt R

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Linearised version of (5) is

(6) 1

hR

hCqdt

dhA vi

hR

hC

hhh

ChC

hhdh

dqq

hCq

v

vv

h

v

1

1

2

1

)(

Subtract (7) from (6) and let , noting that

gives the linearized model:

dh dh

dt dt

iii qqq

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Conclusions

• Definition of transfer functions

• Development of transfer functions

• Properties of transfer functions

• 1st order process

• 2nd order process

• Integrating process (Non-self regulating)

• Examples

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