Transcript Transfer Functions • Convenient representation of a linear, dynamic model. Chapter 4 • A transfer function (TF) relates one input and one output: u.

Transfer Functions
• Convenient representation of a linear, dynamic model.
Chapter 4
• A transfer function (TF) relates one input and one output:
u (t )
U ( s)
 system
y(t )
Y ( s)
The following terminology is used:
u
y
input
output
forcing function
response
“cause”
“effect”
Definition of the transfer function:
Let G(s) denote the transfer function between an input, x, and an
output, y. Then, by definition
Chapter 4
Y (s)
G (s) 
U (s)
where:
Y ( s)  L y (t ) 
U ( s)  L u (t ) 
Development of Transfer Functions
Example: Stirred Tank Heating System
Chapter 4
Figure 2.3 Stirred-tank heating process with constant holdup, V.
Recall the previous dynamic model, assuming constant liquid
holdup and flow rates:
dT
V C
 wC Ti  T   Q
dt
(2-36)
Chapter 4
Suppose the process is at steady state:
0  wC Ti  T   Q
(2)
Subtract (2) from (2-36):
dT
V C
 wC Ti  Ti   T  T     Q  Q 
dt
(3)
But,
Chapter 4
dT 
V C
 wC Ti  T    Q
dt
(4)
where the “deviation variables” are
T   T  T , Ti  Ti  Ti , Q  Q  Q
Take L of (4):
V C  sT   s   T   0   wC Ti s   T   s   Q  s  (5)
At the initial steady state, T′(0) = 0.
Rearrange (5) to solve for
Chapter 4
K 
1 


T  s   
 Q  s   
 Ti s 
  s 1 
  s 1
where
1
V
K
and  
wC
w
(6)

T (s)=G1(s)Q(s)  G2(s)Ti(s)
Chapter 4
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).
Chapter 4
Conclusions about TFs
1. Note that (6) shows that the effects of changes in both Q
and Ti are additive. This always occurs for linear, dynamic
models (like TFs) because the Principle of Superposition is
valid.
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.
Example: Stirred Tank Heater
K  0.05   2.0
Chapter 4
0.05
T 
Q
2s  1
No change in Ti′
Step change in Q(t):
500

Q 
s
T 
1500 cal/sec to 2000 cal/sec
0.05 500
25

2s  1 s
s(2s  1)
What is T′(t)?
T (t )  25[1  e
t /
From line 13, Table 3.1
25
] 
 T (s) 
s( s  1)
t / 2

T (t )  25[1  e ]
Properties of Transfer Function Models
Chapter 4
1. Steady-State Gain
The steady-state of a TF can be used to calculate the steadystate 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 steadystate gain, K, from:
y2  y1
K
u2  u1
(4-38)
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:
If a TF model has a steady-state gain, then:
K  lim G  s 
Chapter 4
s0
(14)
• 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:
Chapter 4
an
dny
dt
n
 an1
bm1
dy n1
dt
n 1
d m1u
dt
m 1

dy
d mu
a1  a0 y  bm m 
dt
dt
du
 b1
 b0u
dt

(4-39)
Take L, assuming the initial conditions are all zero. Rearranging
gives the TF:
m
G s 
Y s
U s

i
b
s
i
i 0
n
i
a
s
i
i 0
(4-40)
Definition:
The order of the TF is defined to be the order of the denominator
polynomial.
Chapter 4
Note: The order of the TF is equal to the order of the ODE.
Physical Realizability:
For any physical system, n  m in (4-38). Otherwise, the system
response to a step input will be an impulse. This can’t happen.
Example:
du
a0 y  b1
 b0u and step change in u
dt
(4-41)
2nd order process
General 2nd order ODE:
Chapter 4
d2y
dy
a 2 + b  y = Ku
dt
dt
Laplace Transform:
G( s) 
2 roots
as
2

 bs +1  Y (s)  KU (s)
Y ( s)
K
 2
U ( s) as  bs  1
s1, 2
 b  b 2  4a

2a
2
b
1
4a
b2
1
4a
: real roots
: imaginary roots
Examples
1.
2
3s 2  4 s  1
b 2 16

 1.333  1
4a 12
Chapter 4
3s 2  4s  1  (3s  1)( s  1)  3( s  1 )( s  1)
3
transforms to e
t
3
, e  t (real roots )
(no oscillation)
2.
2
s2  s  1
b2 1
 1
4a 4
s 2  s  1  ( s  0.5 
transforms to e
(oscillation)
0.5t
3
3
j )( s  0.5 
j)
2
2
3
3
0.5t
cos
t , e sin
t
2
2
From Table 3.1, line 17
e
- bt
sin t 
Chapter 4
2
s  s 1
2

L
=
( s  b) 2   2
2
 3
(s+ 0.5)  

2


2
2
Two IMPORTANT properties (L.T.)
Chapter 4
A. Multiplicative Rule
B. Additive Rule
Example 1:
Place sensor for temperature downstream from heated
tank (transport lag)
Chapter 4
Distance L for plug flow,
Dead time

L
V
V = fluid velocity
Tank:
K1
T(s)
G1 =
=
U(s) 1+ 1s
Ts (s) K 2e-s
=
Sensor: G 2 =
T (s) 1 + 2s
K 2  1,
Overall transfer function:
Ts Ts T
K1K 2e s
   G 2  G1 
U T U
1  1s
2 is very small
(neglect)
Chapter 4
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.
dy
 f ( y, u )
dt
f ( y, u )  f ( y, u ) 
f
y
( y  y) 
y ,u
(4-60)
f
u
(u  u )
(4-61)
y ,u
Subtract steady-state equation from dynamic equation
dy f
f


y
u
dt y s
u s
(4-62)
Example 3:
q0: control,
qi: disturbance
Chapter 4
A
Use L.T.
dh
 qi  q0
dt
qi  q0 at s.s.
dh
A
 qi  q0
dt
AsH ( s )  qi ( s )  q0 ( s ) (deviation variables)
suppose q0 is constant
H(s)
1


AsH (s)  qi (s),

q  ( s) As
i
pure integrator (ramp) for
step change in qi
If q0 is manipulated by a flow control valve,
q0  Cv h
Chapter 4
nonlinear element
Figure 2.5
Linear model
R: line and valve resistance
linear ODE : eq. (4-74)
if q0  CV h
Chapter 4
dh
A  qi  Cv h
dt
Perform Taylor series of right hand side
dh
1
A
 qi  h
dt
R
R  2h
0.5
/ Cv
Chapter 4
Chapter 4
Chapter 4
Chapter 4
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