: (Transfer Function)

: : (Transfer Function)(Transfer Function)

(Transfer Function)

•

• (TF)

:

( )( )

( )( )system

x t y t

X s Y s→ →

x

input ( )

forcing function( )

“cause” ( )

y

output ( )

response ( )

“effect” ( )

• G(s)

( ) ( )( )

( )( )

Y s sG s

X s s=≜輸出

輸入

( ) ( )( ) ( )

Y s y t

X s x t

≜

≜

L

L

•

Stirred-tank heating process with constant holdup, V.

T Ti

T Q

• (wi=w)

• (steady-state)

•

• (1)-(3)

( ) (1)idT

V C wC T T Qdt

ρ = − +

( ) ( ) ( ) ( )0 , 0 , 0 2s i is sT T T T Q Q= = =

( )0 (3)is s swC T T Q= − +

( ) ( ) ( ) (4)i is s sdT

V C wC T T T T Q Qdt

ρ = − − − + −

( s )

( ) ( )in out

dUwH wH Q W

dt= − + +

⌢ ⌢

• (deviation variable)

• Ts

• (4)

• (6) Laplace transform

( )(5)sd T TdT

dt dt

−=

, ,s i i is sT T T T T T Q Q Q− − −≜ ≜ ≜

( = - )

( ) (6)idT

V C wC T T Qdt

ρ = − +

( ) ( ) ( ) ( ) ( )0 (7)iV C sT s T t wC T s T s Q sρ − = = − +

L

•

• (7)

( ) ( ) ( )0 0 0 0 (8)sT t T T T= = = − =

( )0 sT T=

( ) ( ) ( )1(10)

1 1iK

T s T s Q ss sτ τ

= + + +

( ) ( ) ( ) ( ) (9)iV Cs wC T s wCT s Q sρ + = +

( )1and 11

VK

w wC

ρτ ≜ ≜

( )T s

•– Q ( )

•– Q ( )

T iT

( ) ( ) ( )0 0sQ t Q Q t Q s= ⇒ = ⇒ =

( )( )

1(12)

1i

T s

T s sτ=

+

T

( ) ( ) ( )0 0i is i iT t T T t T s= ⇒ = ⇒ =

( )( )

(13)1

T s K

Q s sτ=

+

iT

iT

Q

• Q (10)

•

•

iT

( ) ( ) ( )1

1 1iK

T s T s Q ss sτ τ

= + + + ( )

Example: stirred-tank heating process

• =62.4 lb/ft3, C=0.32 Btu/lboF, V=1.6ft3

• Initial steady-state: Ti=70 oF, Q=1920 Btu/min, wi=w=200 lb/min

• Ti changes to 90 oF, and Q changes to 1600 Btu/min. Calculate the output

temperature response.

ρ

• Steady-state energy balance

� Ts = 100oF

• The input changes are

• Parameters and K

( )0 is s swC T T Q= − +

( )

( )

90 70 20

1600 1920 320

iT ss s

Q ss s

−= =

−= = −

(1.6)(62.4)0.5 min

2001 F

0.0156(200)(0.32) Btu/min

K

τ = =

= =

τ

Example: stirred-tank heating process

•

• inverse Laplace transform

( ) ( ) ( )1

1 1iK

T s T s Q ss sτ τ

= + + +

( ) 1 20 0.0156 320

0.5 1 0.5 1T s

s s s s = + − + +

( ) ( ) ( ) ( )20 5 15

0.5 1 0.5 1 0.5 1T s

s s s s s s

−= + =+ + +

( ) ( )215 1 tT t e−= −

( ) ( ) ( )2100 15 1 tsT t T t T e−= + = + −

�

： (Cv = constant)

Example:

A = area

qi

h

q

idh

A q qdt

= −

i vdh

A q C hdt

⇒ = −

( )i

d Vq q

dt

ρρ ρ= −

vq C h=

ρ

Ex. A = 20cm2, Cv = 2Initial steady-state: qi = 30cm3/secIf qi changes to 40cm3/sec, h =?

1. Steady-State Gain ( )

The steady-state gain 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

s s

s s

y yK

u u

−=

−

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

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

( ) � K

Calculation of K from the TF Model:

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

( )0

lims

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 )

( )

( ) 1G s

s=

2. Order of a TF Model ( )

Consider a general n-th order, linear ODE:

1 1

1 0 1 01 1

n n m m

n n m mn n m m

d y dy d u d ua a a y b b b u

dt dt dt dt

− −

− −− −+ + + = + + +… …

Take , assuming the initial conditions are all zero.

Rearranging gives the TF:

( ) ( )( )

0

0

mi

iin

ii

i

b sY s

G s n mU s

a s

=

=

= = ≥∑

∑

( ) n

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

L

(Physical realizability)

3. Additive Property ( )

Suppose that an output is influenced by two inputs

and that the transfer functions are known:

( )( ) ( ) ( )

( ) ( )1 21 2

andY s Y s

G s G sU s U s

= =

Then the response to changes in both U1 and U2 can be

written as:

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

U1(s)

U2(s)

G1(s)

G2(s)

Y(s)

The graphical representation (or block diagram) is:

4. Multiplicative Property ( )

Suppose that,

( )( ) ( ) ( )

( ) ( )11 2

1 2and

Y s U sG s G s

U s U s= =

Then,

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

Substitute,

( ) ( ) ( ) ( )1 22G s GY sss U=Or,

( )( ) ( ) ( ) ( ) ( ) ( ) ( )1 2 2 2 1

2

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

U s=

( )1U s

Ex: 1

1 1 1vq C h=

2 2 2vq C h=

Non-interacting system

( )

Ex: 2

( )1 1 1 2vq C h h= − 2 2 2vq C h=

Interacting system

( )

( )1H s ( )2H s( )iQ s1

1 1

K

sτ +2

2 1

K

sτ ++

11 K

Linearization of Nonlinear Models

( )• Linear models can be transformed into TF models.

• But most physical processes and physical models are

nonlinear.

- But over a small range of operating conditions, the

behavior may 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.

Linearization (continued)

• Consider a nonlinear, dynamic model relating two

process variables, u and y:

( ), (A)dy

f y udt

=

• Perform a Taylor Series Expansion about and

and truncate after the first order terms,

( ) ( ) ( ) ( ), , (B)s s s ss s

f ff y u f y u y y u u

y u

∂ ∂= + − + −∂ ∂

where the subscript s denotes the steady state, Note

that the partial derivative terms are actually constants because

they have been evaluated at the nominal operating point, s.

su u=sy y=

( ), .s sy u

Substitute (B) into (A) gives:

( ), 0 (D)s ss

dyf y u

dt= =

(F)s s

dy f fy u

dt y u

∂ ∂= +∂ ∂

( ), (C)s ss s

dy f ff y u y u

dt y u

∂ ∂= + +∂ ∂

Also, because it follows that,

(E)

Substitute (D) and (E) into (C) gives thelinearized model :

sy y - y ,

dy dy=

dt dt

≜

Linearization (continued)

Because is a steady state, it follows from (A) that( ),s su y

Example: Liquid Storage System

Mass balance:

Valve relation:

A = area, Cv = constant

qi

h

q

(1)idh

A q qdt

= −

(2)vq C h=

Combine (1) and (2) and rearrange:

1 ( , ) (3)

Eq. (3) is in the form of (A) with and . Thus,we can linearize Eq. (3) around and = ,

vi i

i

s i is

iis s

Cdhq h f h q

dt A A

y h u qh h q q

dh f fh q

dt h q

= − =

= ==

∂ ∂ = + ∂ ∂ (4)

where:

(5)2

1 (6)

Substitute into (4) gives the linearized

v

s s

i s

Cf

h A h

f

q A

∂ = − ∂

∂ = ∂

model:

1 (7)

2

This model can be expressed in terms of a valve resistance,

21 1 with (8)

vi

s

si

v

Cdhh q

dt AA h

R

hdhh q R

dt AR A C

= − +

= − + ≜

•

•

( ) ( ) ( )1 1 2 21 2

s s s n nsns s s

f f fy y x x x x x x

x x x

∂ ∂ ∂⇒ − = − + − + + − ∂ ∂ ∂

⋯線性化

1 2, , , ,s s s nsy x x x⋯ 1 2, , , , ny x x x⋯

( )1 2, , ,s s s nsy f x x x= ⋯

1 21 2

nns s s

f f fy x x x

x x x

∂ ∂ ∂= ⋅ + ⋅ + + ⋅ ∂ ∂ ∂ ⋯

( )1 2, , , ny f x x x= ⋯

1 2, , , , ny x x x⋯1 2, , , , ny x x x⋯

1 1 1 2 2 2, , , ,s s s n n nsy y y x x x x x x x x x= − = − = − = −⋯

Exercise - 1

•1

22 32y x x x= +

1 2 31, 2, 4s s sx x x= = =

1 2 3

16 8( 1) 2( 2) ( 4)

4y x x x− = − + − + −

(1.1,2.1,4.1) 7.1068

7.025 (linearized)

y

y

==

•

•

•

( )1 2, , , , n

dyf y x x x

dt= ⋯

( )1 2, , , , 0ss s s ns

dyf y x x x

dt= =⋯

( )1

1n

ns s s

s dy f f fy x x

dt

d y ydy

dt d y xt x

∂ ∂ ∂= ⋅ + ⋅ + + ⋅ ∂ ∂ ∂ = =

−⋯

( )1 2, , , , nf y x x x⋯

( ) ( )1 2 1 2

11

, , , , , , , ,n s s s ns

nns s s

f y x x x f y x x x

f f fy x x

y x x

− =

∂ ∂ ∂⋅ + ⋅ + + ⋅ ∂ ∂ ∂

⋯ ⋯

⋯

Exercise - 2

• 22dy

x ydt

= − 2sx = y

4 2dy

y xdt

= − +