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Unformatted text preview: Transfer Functions Definition of the transfer function:
of the transfer function: • Convenient representation of a linear, dynamic model. Let G(s) denote the transfer function between an input, x, and an
output, y. Then, by definition: x ( t ) input
y ( t ) output in time domain
→ system →
X (s)
Y ( s ) Laplace domain
The following terminology (from mathematics) is used:
x y input response “cause” Y (s)
a "ratio" of functions of s
X (s) G (s) = Y ( s ) = L ⎡ y ( t )⎤
⎣
⎦ where: X ( s ) = L ⎡ x ( t )⎤
⎣
⎦ output forcing function Chapter 4 Chapter 4 • A transfer function (TF) relates one input and one output:
transfer function (TF) relates one
and one “effect” Development of Transfer Functions Recall the previous dynamic model, assuming constant liquid
holdup and flow rates simplified dynamic energy balance
holdup and flow rates, a simplified dynamic energy balance is: Example: Stirred Tank Heating System
Feed temperature Ti and flow rate wi are disturbances to output T. Chapter 4 Chapter 4 V ρC dT (t )
= wC [Ti (t ) − T (t )] + Q (t )
dt (236) Suppose the process is at steady state: 0 = wC (Ti − T ) + Q (2) Subtract (2) from (236):
Rate of heat “in”, Q, is the manipulated variable.
Figure 2.3 Stirredtank heating process with constant holdup, V. V ρC dT
= wC ⎡(Ti − Ti ) − (T − T ) ⎤ + ( Q − Q )
⎣
⎦
dt (3) where the “deviation variables” are denoted by
primes ( primes are not time derivatives !) T ′ = T − T , Ti′ = Ti − Ti , Q′ = Q − Q
Take Laplace transform of (4):
transform of (4):
V ρ C ⎡ sT ′ ( s ) − T ′ ( 0 ) ⎤ = wC ⎡Ti′( s ) − T ′ ( s ) ⎤ − Q ′ ( s ) (5)
⎣
⎦
⎣
⎦ Rearrange (5) to solve for two transfer functions of the process
since there is only one output but two inputs: Chapter 4 Chapter 4 Then because the derivative of a variable (T) and the
derivative of its deviation variable ( T ′ = T − T ) are the
same:
dT ′
(4)
V ρC
= wC ( Ti′ − T ′) + Q ′
dt ⎛K⎞′
⎛ 1 ⎞′
T ′( s ) = ⎜
⎟Q (s) + ⎜
⎟ Ti ( s )
⎝τ s +1⎠
⎝τ s +1⎠ (6) T ′(s) = G1(s) Q ′(s) + G2 (s) Ti′(s)
where K= 1
Vρ
and τ =
wC
w Since at t < 0 we assume we are at the nominal
steady state, the initial value of the output deviation
variable is zero, e.g. T′(0) = 0. T ′(s)= G1(s)Q′(s) + G2 (s)Ti′(s)
Conclusions about TFs : Note G1 (process TF) has gain K and time constant τ.
G2 (disturbance TF) has gain=1 and time constant τ.
gain = G(with s=0) for stable TFs. Both these are first order
“processes” = ouput/input relationship.
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 Chapter 4 Chapter 4 G1 and G2 are transfer functions (TFs) and independent of the
specific input functions of time, Q′(t) and Ti′(t). 1. Note that (6) shows that the effects of changes in both Q
(6)
bo
and Ti are additive. This always occurs for linear, dynamic
models (like TFs) because the Principle of Superposition is
valid for linear ODEs
valid for linear ODEs.
2. The TF model enables us to determine the output response to
any change in an input
any change in an input.
3. Use deviation variables to eliminate initial conditions for TF
models.
models. Example: Stirred Tank Heater Properties of Transfer Function Models K = 0.05 τ = 2.0 0.05
Q′
2s + 1 No change in Ti′ 1. SteadyState Gain Step change in Q(t) : from 1500 cal/sec to 2000 cal/sec Q′( s ) = 500
→ deviation step from 0 to 500
s T ′( s ) = 0.05 ⎛ 500 ⎞
25
⎜
⎟=
2 s + 1 ⎝ s ⎠ s(2 s + 1) Use case 13, Table 3.1
13 What is T′(t) ? ⎯
T ′(t ) = 25[1 − e − t /τ ] ←⎯ T ′( s ) = Chapter 4 Chapter 4 T′ = 25
s(τ s + 1) The steadystate of a TF can be used to calculate the steadystate change in an output due to a steadystate change in the
input For example suppose we know two steady states for an
input. For example, suppose we know two steady states for an
input/output = u and y = {u1,y1} , {u2,y2}. Then we can
calculate the average steadystate gain, K, from:
y2 − y1
u2 − u1 K= (438) For a linear system, K is a constant. But for a nonlinear
system, K will depend on the operating conditions( u , y ) . T ′(t ) = 25[1 − e − t / 2 ] = T (t ) − T 2. Order of a TF Model Calculation of process gain ( K ) from the TF Model: Consider a general nth order, linear ODE:
general
order linear ODE: If a TF model has a steadystate gain (it’s stable), then: an K = lim G ( s )
• This important result is a consequence of the Final Value
th Fi
Theorem (consider the case when input = unit step)
• Note: Some TF models do not have a steadystate gain (e.g.,
integrating process in Ch. 5, because they are unstable TFs)
integrating process in Ch. 5, because they are “unstable” TFs) Chapter 4 Chapter 4 s →0 dny
dt n + an−1 bm−1 dy n−1
dt n−1 d m−1u
dt m −1 + … a1 dy
d mu
+ a0 y = bm m +
dt
dt + … + b1 du
+ b0u
dt (439) Take Laplace Transform, assuming the initial conditions are all
zero. Rearranging gives the TF:
m ∑ bi si
Y ( s ) i =0
=
G (s) =
U (s) n ∑ ai s
i =0 i (440) Definition: 2nd order process The order of the TF is defined to be the order of the denominator
polynomial. General 2nd order ODE:
a Physical Realizability:
For any physical system, n ≥ m in (438). The denominator is
higher than or same order as the numerator. Otherwise, the system
response to a step input will be an impulse. This can’t happen. Chapter 4 Chapter 4 Note: The order of the TF is equal to the order of the ODE. du
+ b0u and step change in u
dt ⎡ as 2 + bs + 1⎤ ⋅ Y ( s) = KU ( s )
⎣
⎦ Laplace Transform:
G(s) = Y (s)
K
both a and b > 0 for stability
=2
U ( s ) as + bs + 1 2 roots Example of nonphysical system: a0 y = b1 d2y
dy
+ b + y = Ku
2
dt
dt s1,2 = −b ± b 2 − 4a
2a b2 − 4ac > 0 (441) : real roots b2 − 4ac < 0 : complex roots Example:
IMPORTANT properties of Transfer Functions
properties of Transfer Functions Place sensor for temperature downstream from heated
sensor for temperature downstream from heated
tank (transport lag) B. Additive Rule Y = (G1 x U1 ) + (G2 x U2 ) Chapter 4 Chapter 4 A. Multiplicative Rule Y = (G2 x G1 ) U Distance L for plug flow,
L
Dead time θ =
v
v = fluid velocity
U ′ = Q ′ = the input Tank: G1 = T ′( s )
K1
=
U ′( s ) 1 + τ 1s Sensor: G2 = Ts′ ( s ) K 2eθ s
=
, τ 2 << τ 1 , neglect → τ 2 ≈ 0
neglect
T ′( s ) 1 + τ 2 s Overall transfer function:
T′ T′ T′
KKe
= ⋅ = G ⋅G =
U′ T′ U′
1+τ s −θs s s 1 2 2 1 1 Linearization of Nonlinear Models
Required to derive transfer function, output = y(t), input = u(t).
Good approximation “near” a steadystate operating point. Fig. 4.3 Schematic diagram of two
liquid surge tanks in series.
surge tanks in series Use 1st order Taylor series in y and u around steady state
dy
⎛ dy ⎞
= f ( y, u ) and f ( y, u ) = ⎜ ⎟
and
=0
dt
⎝ dt ⎠at steadystate conditions
f ( y, u ) ≅ f ( y, u ) + ∂f
∂y ( y − y) +
y ,u ∂f
∂u (u − u ) (461)
(462) y ,u Chapter 4 Chapter 4 y,u The derivative wrt t of y and y’ are the same so that:
dy ′ ∂f
∂f
y′ +
u′ = C y y ′ + Cu u′ where C y and Cu are constants.
=
dt ∂y s
∂u s (463) Chapter 4 Chapter 4 Fig. 4.4 Inputoutput model for
two liquid surge tanks in series. Fig. 4.5 General procedure for
developing transfer function models. Page 1 of 3 Linear vs. Nonlinear Tank Simulation: MATLAB ode45 (CHE 361)
Here we simulate the dynamics of a linear and a nonlinear tank by using ode45 in
MATLAB to solve the ODE for a step change in feed flow rate to each tank. Since
the two tanks are independent, we can simply write one program file and one RHS
file to evaluate the derivatives with respect to time for each tank : TwoTanks.m and
RHS2Tanks.m respectively. See pgs 2526 in SEMD3 for details of the models.
dh1 (t )
dt q (t )
q (t ) K
1
= i1 −
h1 (t ) = i1 − 1 h1 (t )
A
ARv
A
A (257) for Tank 1, the Linear Tank dh2 (t )
dt q (t ) C
=i 2 − v
A
A (261) for Tank 2, the Nonlinear Tank For both tanks, q (t ) K
h2 (t ) =i 2 − 2 h2 (t )
A
A A = 0.4 m 2 . m2
= 2 m3 /min.
0.08
and q1
min
m 2.5
For Tank 2 : K 2 = 0.8
and q2 = 4 m3 /min.
min
Then= = 25 meters , both tanks have the same nominal steady state height (output).
h1
h2 =
For Tank 1 : K1 Contents of file TwoTanks.m
% TwoTanks.m
clear all, format compact
global A K1 K2 q1ss q2ss deltaq
A = 0.4; K1 = 0.08; K2 = 0.8; q1ss = 2; q2ss = 4;
h2initial = [25 25];
deltaq = input('Enter step for inlet flows, cubic meters/min (1) : ');
tstop = input('Enter simulation stop time in minutes (50) : ');
tspan = [0:0.5:tstop]; % calculate values at increments of 0.5
[t,heights] = ode45(@RHS2Tanks,tspan,h2initial);
h1 = heights(:,1); % h1 is first column in heights matrix
h2 = heights(:,2); % h2 is second column in heights matrix
figure
plot(t,h1,'b',t,h2,'r','Linewidth',2)
grid % puts a grid on plot to help read values
xlabel('Time (min)'), ylabel('Height of Liquid (m)')
title(['Step Responses of Two Tanks: delta q = ' num2str(deltaq) ' (m^3/min)'])
legend('Linear Tank','Nonlinear Tank','Location','Best')
%results = [t';heights']' % Prints out t,h1,h2 in columns Contents of file RHS2Tanks.m
function dhdt = RHS2Tanks(t,heights)
global A K1 K2 q1ss q2ss deltaq
h1 = heights(1,:);
h2 = heights(2,:);
dhdt(1,1) = (q1ss+deltaq)/A  K1*h1/A;
dhdt(2,1) = (q2ss+deltaq)/A  K2*sqrt(h2)/A; q1ss = qi1 and q2ss = qi 2
deltaq q= qi′2 for t ≥ 0
= i′1
=
qi1 (t ) qi1 + qi′1 (t )
qi 2 (= qi 2 + qi′2 (t )
t) Page 2 of 3 >> TwoTanks
Enter step for inlet flows, cubic meters/min (1) : 1
Enter simulation stop time in minutes (50) : 50
3 Step Responses of Two Tanks: delta q = 1 (m /min) Height of Liquid (m) 40 35
Linear Tank
Nonlinear Tank 30 25 0 5 10 15 30
25
20
Time (min) 35 40 45 50 >> TwoTanks
Enter step for inlet flows, cubic meters/min (1) : 0.1
Enter simulation stop time in minutes (50) : 50
>>
Step Responses of Two Tanks: delta q = 0.1 (m3/min)
26.4 Height of Liquid (m) 26.2
Linear Tank
Nonlinear Tank 26
25.8
25.6
25.4
25.2
25 0 5 10 15 20
25
30
Time (min) 35 40 45 50 For smaller changes in the input flow rate, the tank levels change less than for larger changes. The
dynamic plots of the heights are much more similar.
Thus the linearized model (the transfer function) of the nonlinear tank is a “good” model when the
“process” remains near the nominal operating point. Page 3 of 3 The addon toolbox “Simulink” can be used to simplify comparisons for positive or negative steps
to the two tanks as shown below in the Simulink model block diagram and resultant plot. snltnk qi_Step
Sum1 Level_upNL snltnk
qss_NL
Constant_levelNL snltnk
Sum2
Level_downNL Mux
PC_graph sltnk
Sum3
Level_upL sltnk
qss_Lin
Constant_levelL sltnk
Sum4
Level_downL
Mux L_vs_NL_Tank.mdl = Simulink model to compare
Linear versus Nonlinear Tank Responses.
For the Linear tank, flow out is proportional to h.
For the Nonlinear tank flow out is proportional to the square root of h. Responses of Linear and Nonlinear Tank Height to + or  Steps in qi h(m): Solid = Linear, Dashed = Nonlinear 40 35 30 25 20 15 10 0 10 20
30
Time (min) 40 50 ...
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This note was uploaded on 03/01/2012 for the course CHE 361 taught by Professor Staff during the Winter '08 term at Oregon State.
 Winter '08
 Staff

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