: (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 − = = − +
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•
• (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.
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(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
= − +