Chapter 4 Transfer Functions

7
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.
Y (s) G (s) U (s)
Chapter 4
where:
U ( s) L u (t )
Y ( s) L y (t )
Development of Transfer Functions
Example: Stirred Tank Heating System
2
Chapter 4
At the initial steady state, T′(0) = 0.
5
Rearrange (5) to solve for
K 1 T s Q s Ti s s 1 s 1 (6)
Chapter 4
T
0.05 500 25 2s 1 s s(2s 1)
What is T′(t)?
From line 13, Table 3.1
T (t ) 25[1 e
t /
25 ] T (s) s( s 1)
9
t / 2 T (t ) 25[1 e ]
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 .
10
Calculation of K from the TF Model: If a TF model has a steady-state gain, then:
Suppose the process is at steady state:
Chapter 4
0 wC Ti T Q
Subtract (2) from (2-36):
(2)
dT Q Q V C wC T T T T i i dt
Transfer Functions
• Convenient representation of a linear, dynamic model. • A transfer function (TF) relates one input and one output:
Chapter 4
u (t ) y(t ) system U ( s) Y ( s)
A. Multiplicative Rule
Chapter 4
B. Additive Rule
16
Example 1:
Place sensor for temperature downstream from heated tank (transport lag) Distance L for plug flow,
Figure 2.3 Stirred-tank heating process with constant holdup, V.
3
Recall the previous dynamic model, assuming constant liquid holdup and flow rates:
dT V C wC Ti T Q dt (2-36)

i 0 n
(4-40)
Definition:
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.
K lim G s
s0
(14)
Chapter 4
• This important result is a consequence of the Final Value Theorem • Note: Some TF models do not have a steady-state gain.
d m1u dt
du b1 b0u dt
Take L, assuming the initial conditions are all zero. Rearranging gives the TF:
Y s
i b s i i a s i i 0
12
m
G s
U s
Chapter 4
8
Example: Stirred Tank Heater
K 0.05 2.0
0.05 T Q 2s 1
No change in Ti′ 1500 cal/sec to 2000 cal/sec
Chapter 4
Step change in Q(t): 500 Q s
The following terminology is used: u input forcing function y output response
“cause”
“effect”
1
Definition of the transfer function:
Let G(s) denote the transfer function between an input, x, and an output, y. Then, by definition
as
2
bs +1 Y (s) KU (s)

Y (s) K 2 U ( s) as bs 1
b b 2 4a 2a 2 b 1 4a
b2 1 4a
2 roots
s1, 2
: real roots : imaginary roots
14
Examples
1.
(no oscillation)
2.
2 s2 s 1
s 2 s 1 ( s 0.5
transforms to e
0.5t
3 3 j )( s 0.5 j) 2 2
3 3 0.5 t cos t , e sin t 2 2
15
(oscillation)
Two IMPORTANT properties (L.T.)
Chapter 4
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:
dy f ( y, u ) dt
f ( y, u ) f ( y, u ) f y ( y y)
Chapter 4
G2 (disturbance) has gain=1 and time constant . 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 .
Properties of Transfer Function Models
1. Steady-State Gain
Chapter 4
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, f 2 16 1.333 1 4a 12
3s 2 4s 1 (3s 1)( s 1) 3( s 1 )( s 1) 3
Chapter 4
transforms to e
t
3
, e t (real roots )
b2 1 1 4a 4
(3)
4
But,
dT V C wC Ti T Q dt (4)
Chapter 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)
du a0 y b1 b0u and step change in u dt (4-41)