Transfer Functions

• Convenient representation of a linear, dynamic model.

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

The following terminology is used:

u

input

forcing function

“cause”

y

output

response

“effect”

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)(

)(system

)(

)(

sY

ty

sU

tu

Definition of the transfer function:

Let G(s) denote the transfer function between an input, x, and an output, y. Then, by definition

where:

Development of Transfer Functions

Example: Stirred Tank Heating System

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)()(

)(sUsY

sG

)(L)(

)(L)(

tusU

tysY

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

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Recall the previous dynamic model, assuming constant liquid holdup and flow rates:

(2-36)idT

V C wC T T Qdt

Suppose the process is at steady state:

0 (2)iwC T T Q

Subtract (2) from (2-36):

(3)i idT

V C wC T T T T Q Qdt

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But,

(4)idT

V C wC T T Qdt

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.

Rearrange (5) to solve for

1(6)

1 1 iK

T s Q s T ss s

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1and

VK

wC w

G1 and G2 are transfer functions and independent of the inputs, Q′ and Ti′.Note G1 (process) has gain K and time constant G2 (disturbance) has gain=1 and time constant 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 4.3).

(s)T(s)G(s)Q(s)(s)=GT i 21

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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 s

s s

/ 2( ) 25[1 ]tT t e

From line 13, Table 3.1

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

iin

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

Ku=ydt

dyb+

dt

yda

2

2

)()(1+bsas2 sKUsY

1)(

)()(

2

bsas

K

sU

sYsG

a2

a4bbs

2

2,1

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)3s 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 2t ttransforms to e t e t

(oscillation)

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From Table 3.1, line 17

2 2

2

( )

1 32

es b

s s

L- bt

22

sin t

2 2=

(s+0.5)

Two IMPORTANT properties (L.T.)

A. Multiplicative Rule

B. Additive Rule

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Example 1:Example 1:

Place sensor for temperature downstream from heatedtank (transport lag)

Distance L for plug flow,

Dead time

V = fluid velocity

Tank:

Sensor:

Overall transfer function:

11

1

KT(s)G = =

U(s) 1+ s

V

L

222

s-2s

2 1,K s+1

eK=

T(s)

(s)T=G

is very small(neglect)

s1

eKKGG

U

T

T

T

U

T

1

s21

12ss

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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:Example 3:

q0: control, qi: disturbance

0 0 at s.s.i i

dhA q q q q

dt

Use L.T.

0( ) ( ) ( )iAsH s q s q s (deviation variables)

suppose q0 is constant

H (s) 1( ) ( ),

q ( )i

AsH s q si Ass

pure integrator (ramp) for step change in qi

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0i

dhA q q

dt

If q0 is manipulated by a flow control valve,

hCq v0

nonlinear element

Figure 2.5

R: line and valve resistance

linear ODE : eq. (4-74)

Linear model

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0if Vq C hC

hap

ter

4

i v

dhA q C h

dt

Perform Taylor series of right hand side

1i

dhA q h

dt R

0.52 / vR h C

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