CHBE306
LECTURE V
LAPLACE TRANSFORM AND TRANSFER FUNCTION Professor Dae Ryook Yang Fall 2006 Dept. of Chemical and Biological Engineering Korea University CHBE306 Process Dynamics and Control
Korea University
5-1
Road Map of the Lecture V •
Laplace Transform and Transfer functions – Definition of Laplace transform – Properties of Laplace transform – Inverse Laplace transform – Definition of transfer function – How to get the transfer functions – Properties of transfer function
+-
Controller
Actuator
Sensor
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PROCESS
Lectures IV to VII
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5-2
SOLUTION OF LINEAR ODE •
1st-order linear ODE – Integrating factor:
For
a (t ) dt a (t )dt [ xe ] f (t )e •
dx dt
a(t ) x f ( x), I.F. exp( a(t )dt) a ( t ) dt a (t )dt x(t ) [ f (t )e dt C ]e
High-order linear ODE with constant coeffs. – Modes: roots of characteristic equation
For a2 x a1 x a0 x f (t ) ,
a2 p 2 a1 p a0 a2 ( p p1 )( p p2 ) 0 – Depending on the roots, modes are
p t p t Solution is a linear combination of • Distinct roots: (e 1 , e 2 ) modes and the coefficients are p t p t • Double roots: (e 1 , te 1 ) decided by the initial conditions . t t • Imaginary roots: (e cos t , e sin t )
•
Many other techniques for different cases
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5-3
LAPLACE TRANSFORM FOR LINEAR ODE AND PDE •
Laplace Transform – Not in time domain, rather in frequency domain – Derivatives and integral become some operators. – ODE is converted into algebraic equation – PDE is converted into ODE in spatial coordinate – Need inverse transform to recover time-domain solution (D.E. calculation) u(t )
y(t )
ODE or PDE
L
L- 1
L
L- 1
L
Transfer Function
U ( s)
L- 1 Y ( s)
(Algebraic calculation) CHBE306 Process Dynamics and Control
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5-4
DEFINITION OF LAPLACE TRANSFORM •
Definition
F ( s) L f (t )
0
st
f (t )e dt
– F(s) is called Laplace transform of f(t). – – – – •
f(t) must be piecewise continuous. F(s) contains no information on f(t) for t < 0. The past information on f(t) (for t < 0) is irrelevant. The s is a complex variable called “ Laplace transform variable”
Inverse Laplace transform
f (t ) L -1
-1
F (s)
– L and L are linear.
L af1 (t ) bf 2 (t ) aF1 (s ) bF2 (s)
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5-5
LAPLACE TRANSFORM OF FUNCTIONS •
Constant function, a L a
•
0
st
ae dt
a s
st
e
0
f(t)
a a 0 s s
t
Step function, S(t )
1 for t 0 f (t ) S(t ) 0 for t 0
L S(t )
•
a
0
f(t) 1
1 1 1 e st dt e st 0 s s s 0
Exponential function, e-bt L e
bt
0
1 (b s )t e e dt e s b bt st
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f(t)
0
t
1 s b
1
b<0 b>0 t
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5-6
•
Trigonometric functions j t
cos t j sin t
sin t
– Euler’s Identity: e cos t
1
e 2
jt
e
1
j t
•
1
e 2 j
e jt L
jt
e j t
1
1 2
e j t
t
2 2 2 j 2 j 2 j s j s j s 1 s 1 1 1 1 L cos t L e jt + L e j t 2 2 2 2 2 s j s j s L sin t L
1
sin(t)
1
1
Rectangular pulse, P(t ) 0 for t t w f (t ) P(t ) h for t w t 0 0 for t 0 L P(t )
t w
0
he st dt
h s
e st
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f(t) h t w
t w
0
h
t
1 e s tw s
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5-7
•
Impulse function, (t ) for t t w 0 f (t ) (t ) lim 1/ tw for t w t 0 t 0 0 for t 0
f(t) 1 /t w
w
L (t ) lim
t w
t w 0 0
•
1 tw
e dt lim
t w 0
tw
t
1 e 1 s tw s
f (t ) f (t ) lim L'Hospital's rule: lim t 0 g (t ) t 0 g (t ) Ramp function, t
L t
t
0
te st dt
s
f(t)
e st 0
0
e st
s
Integration by part:
dt
0
•
1
st
t w
1 s
0
e st dt
1
1
1
s2
f ' gdt f g 0
0
t
f g dt
Refer the Table 3.1 (Seborg et al.) for other functions
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5-10
PROPERTIES OF LAPLACE TRANSFORM •
Differentiation df st st st f e dt f ( t ) e f ( s ) e dt 0 0 0 dt
L
(by i .b. p.)
s f e st dt f (0) sF (s ) f (0) 0
d 2 f st st st f ( s )e dt s f e st dt f (0) L 2 f e dt f (t )e 0 0 0 dt 0 s sF ( s) f (0) f (0) s 2 F (s ) sf (0) f (0)
d n f ( n) st ( n 1) st e f ( n1) (s )e stdt L n f e dt f (t ) 0 0 dt 0 d n1 f st ( n 1) ( n1) (0) s L n1 f ( n1) (0) s f e dt f 0 dt s n F (s ) s n1 f (0) sf ( n 2) (0) f ( n1) (0)
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•
If f (0) = f ’ (0) = f ”(0) =…= f (n-1)(0) = 0, – Initial condition effects are vanished. – It is very convenient to use deviation variables so that all the effects of
df sF ( s ) dt d 2 f 2 L 2 s F (s ) dt L
initial condition vanish.
d n f n L n s F (s ) dt •
Transforms of linear differential equations. L Y (s ), y (t )
dy(t )
L U (s ) u (t )
L sY (s ) (if y (0) 0)
dt dy(t ) L y(t ) Ku (t ) ( y (0) 0) ( s 1)Y (s ) KU (s ) dt 1 T L TL TL (s ) L (Tw TL ) v HL v ( HL s 1)TL ( s) Tw (s) t z HL z CHBE306 Process Dynamics and Control
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5-12
•
Integration L
t
0
f ( )d
t
0
0
st f ( )d e dt
e st
s
t
0
f ( )d 0
1
s
0
f e st dt
F (s)
(by i.b. p.)
s
d b (t ) db(t ) da (t ) Leibniz rule: f ( ) d f ( b ( t )) f ( a ( t )) dt a (t ) dt dt •
Time delay (Translation in time)
f(t)
in t f (t ) f (t )S(t )
L f (t )S(t )
e •
st
f (t )e dt
s
0
0
f ( )e s ( )d
t
(let t )
f ( )e s d e s F (s )
Derivative of Laplace transform dF ( s) ds
d ds
0
st
f e dt
0
f
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d ds
st
e dt
0
(t f )e st dt L t f (t ) Korea University
5-13
•
Final value theorem – From the LT of differentiation, as s approaches to zero
lim s 0
df
0
dt
df
0
dt
e st dt lim sF (s) f (0) s 0
dt f () f (0) lim sF ( s) f (0) f () lim sF ( s) s 0
s 0
– Limitation: f () has to exist. If it diverges or oscillates, this theorem is not valid. •
Initial value theorem – From the LT of differentiation, as s approaches to infinity
lim
lim
df
s 0
dt df
s 0
dt
e st dt lim sF (s) f (0) s
e st dt 0 lim sF (s) f (0) f (0) lim sF ( s) s
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s
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5-14
EXAMPLE ON LAPLACE TRANSFORM (1) •
f(t) 3
1.5t for 0 t 2 3 for 2 t 6 f (t ) 0 for 6 t 2 6 t 0 for t 0 f (t ) 1.5t S(t ) 1.5(t 2)S(t 2) 3S(t 6)
F ( s) L f (t ) •
For F ( s)
2 s 5
1.5
3 6 s 2 s (1 e ) e 2 s s
, find f (0) and f ().
– Using the initial and final value theorems 2 s 2 s f (0) lim sF ( s) lim 2 f () lim sF ( s) lim 0 s s s 5 s 0 s 0 s 5 – But the final value theorem is not valid because lim f (t ) lim 2e5t t
t
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EXAMPLE ON LAPLACE TRANSFORM (2) •
What is the final value of the following system?
x x x sin t; x(0) x(0) 0
s X ( s) sX (s) X
1
2
x()= lim
s
s 1 2
x(s)=
1 ( s 2 1)( s 2 s 1)
0
( s 1)( s s 1) – Actually, x() cannot be defined due to sin t term. s 0
•
2
2
Find the Laplace transform for From
dF ( s) ds
(t sin t )
?
L t f (t )
L t sin t
d 2 s ds s 2 2 ( s 2 2 ) 2
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INVERSE LAPLACE TRANSFORM •
Used to recover the solution in time domain
L -1 F ( s) f (t ) – From the table – By partial fraction expansion – By inversion using contour integral 1 -1 st f (t ) L F ( s) e F ( s) ds C 2 j
•
Partial fraction expansion – After the partial fraction expansion, it requires to know some simple formula of inverse Laplace transform such as 1 ( s 1)
,
s ( s b) 2
2
,
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(n 1)! s
n
,
e s s 2 s 1 2 2
, etc.
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5-17
PARTIAL FRACTION EXPANSION F ( s)
•
N ( s) D( s)
N ( s) ( s p1 )( s pn )
1 ( s p1 )
n ( s pn )
Case I: All pi ’s are distinct and real – By a root-finding technique, find all roots (time-consuming) – Find the coefficients for each fraction • Comparison of the coefficients after multiplying the denominator • Replace some values for s and solve linear algebraic equation • Use of Heaviside expansion – Multiply both side by a factor, ( s+ pi), and replace s with – pi.
i ( s pi )
N ( s) D( s) s p
i
– Inverse LT:
f (t ) 1e
p1t
2e p t ne p t
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2
n
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5-18
•
Case II: Some roots are repeated F (s )
N (s ) D(s )
N (s ) (s p )
r
br 1s r 1 b0 (s p )
r
1
(s p )
r
(s p )r
– Each repeated factors have to be separated first. – Same methods as Case I can be applied. – Heaviside expansion for repeated factors
N ( s) r ( s p) (i 0,, r 1) r i (i ) i ! ds D( s) s p 1 d
(i )
– Inverse LT
f (t ) 1e
pt
2te
pt
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r
(r 1)!
t r 1e pt
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5-19
•
Case III: Some roots are complex F ( s)
N ( s) D( s)
c1s c0 s 2 d1s d0
1 ( s b) 1 ( s b)2 2
– Each repeated factors have to be separated first. – Then,
1 ( s b) 1
( s b)2 2
1
(s b) ( s b )2 2
1
(s b )2 2
where b d1 / 2, d0 d1 / 4 2
1 c1 , 1 c0 1b /
– Inverse LT f (t ) 1ebt cos t 1e bt sin t
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EXAMPLES ON INVERSE LAPLACE TRANSFORM •
F (s)
( s 5)
s(s 1)( s 2)( s 3)
A s
B s 1
C s 2
D s 3
(distinct)
– Multiply each factor and insert the zero value ( s 5) ( s 1)( s 2)(s 3) s 0
B C D A s s s A 5/ 6 s 1 s 2 s 3 s 0
( s 5)
C ( s 1) D(s 1) A( s 1) B B 2 s( s 2)( s 3) s 1 s s2 s 3 s 1 ( s 5) s( s 1)( s 3) s 2
D(s 2) A( s 2) B(s 2) C C 3 / 2 s s 1 s 3 s 2
( s 5)
A( s 3) B( s 3) C (s 3) D D 1/ 3 s( s 1)( s 2) s 3 s s 1 s2 s 3
5
3
1
6
2
3
f (t ) L -1 F ( s) 2et e2t e3t CHBE306 Process Dynamics and Control
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•
F (s)
1 ( s 1) ( s 2) 3
As 2 Bs C ( s 1)
3
D ( s 2)
(repeated)
1 ( As2 Bs C )( s 2) D( s 1)3
( A D)s3 (2 A B 3D) s 2 (2 B C 3D) s (2C D)
A D, 2 A B 3D 0, 2 B C 3 D 0, 2 C D 1 A 1, B 1, C 1, D 1 r i
– Use of Heaviside expansion
s s 1 2
( s 1)
3
1
( s 1)
2
( s 1)
(i 0) : 3 s 2 s 1 (i 1) : 2 (i 2) : 1
1 d
s 1! ds 1 d 2
2! ds
2
2
2
s 1
2
i ! ds D(s ) (i )
3
( s 1)
( s p) r
s p
(i 0,, r 1)
3
1
s 1
s
1 d (i ) N (s )
1 s 1
s 1
1 s 1
1
f (t ) L -1 F ( s) et tet t 2 et e2t
2
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•
F (s)
( s 1) s (s 4s 5) 2
2
A(s 2) B
(s 2) 1 2
Cs D s
(complex)
2
s 1 A( s 2) s 2 Bs2 (Cs D)( s2 4 s 5)
( A C )s3 (2 A B 4C D) s 2 (5C 4 D) s 5 D
A C, 2 A B 4C D 0, 5C 4 D 1, 5 D 1 A 1/ 25, B 7 / 25, C 1/ 25, D 1/ 5
A(s 2) B ( s 2) 2 1 Cs D s
2
1 1 25 s
( s 2)
1
25 ( s 2) 2 1
7
B
25 ( s 2) 2 1
1 1 5s
f (t ) L -1 F ( s)
2
1 25
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e2t cos t
7 25
e2t sin t
1 25
1
t 5
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5-23
•
F (s)
1 e2 s
B A 2 s (1 e ) (Time delay) (4 s 1)(3s 1) 4 s 1 3s 1
A 1/(3s 1) s 1/ 4 4, B 1/(4 s 1) s 1/ 3 3 2 s 2 s 4 3 4 e 3 e 1 1 1 f (t ) L F (s) L L 4 s 1 3 s 1 4 s 1 3s 1
et / 4 et / 3 e (t 2) / 4 e (t 2) / 3 S(t 2)
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SOLVING ODE BY LAPLACE TRANSFORM •
Procedure 1. Given linear ODE with initial condition, 2. Take Laplace transform and solve for output 3. Inverse Laplace transform
•
Example: Solve for 5
dy dt
4 y 2; y(0) 1
2 dy + L 4 y L 2 5( sY ( s ) y (0)) 4 Y ( s ) dt s 2 5 s 2 (5 s 4)Y (s) 5 Y ( s) s s(5s 4) 0.5 2.5 0.8t y(t ) L 1 Y ( s) L1 0.5 0.5e s 5s 4
L 5
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TRANSFER FUNCTION (1) •
Definition – An algebraic expression for the dynamic relation between the input and output of the process model 5
dy
U ( s)
4 y u; y (0) 1
dt Let y y 1 and u u 4
(5 s 4)Y (s ) U (s ) •
Y ( s) U (s )
1 5s 4
0.25 1.25s 1
Transfer Function, G(s)
Y ( s)
G (s )
How to find transfer function 1. 2. 3. 4. 5.
Find the equilibrium point If the system is nonlinear, then linearize around equil. point Introduce deviation variables Take Laplace transform and solve for output Do the Inverse Laplace transform and recover the original variables from deviation variables
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TRANSFER FUNCTION (2) •
Benefits – Once TF is known, the output response to various given inputs can be obtained easily.
y (t ) L1 Y (s ) L1 G (s )U (s ) L1 G (s ) L1 U (s ) – Interconnected system can be analyzed easily. • By block diagram algebra X + -
G1
G2
Y
G3
Y ( s) X ( s)
G1( s)G2( s) 1 G1( s)G2( s)G3( s)
– Easy to analyze the qualitative behavior of a process, such as stability, speed of response, oscillation, etc. • By inspecting “Poles” and “Zeros” • Poles: all s’s satisfying D( s)=0 • Zeros: all s’s satisfying N ( s)=0 CHBE306 Process Dynamics and Control
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5-27
TRANSFER FUNCTION (3) •
Steady-state Gain: The ratio between ultimate changes in input and output ouput ( y() y(0)) Gain= K = input (u() u (0)) – For a unit step change in input, the gain is the change in output – Gain may not be definable: for example, integrating processes and processes with sustaining oscillation in output – From the final value theorem, unit step change in input with zero initial condition gives
K
y() 1
1
lim sY ( s) lim sG( s) lim G( s) s 0
s 0
s
s 0
– The transfer function itself is an impulse response of the system Y ( s) G (s )U (s ) G (s )L (t) G (s ) CHBE306 Process Dynamics and Control
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EXAMPLE •
Horizontal cylindrical storage tank (Ex4.7) dm dt
V (h)
dt
h
0
dh dt
qi q
R
Lwi (h)dh
wi (h) / 2 wi L
qi
dV
dV dt
Lwi (h)
dh
h L
dt
wi
q
R 2 ( R h)2 (2 R h) h
qi q
dh dt
1 2 L ( D h) h
(qi q) (Nonlinear ODE)
– Equilibrium point: (qi , q , h ) 0 ( qi q ) /(2 L ( D h )h ) (if qi q, h can be any value in 0 h D .) – Linearization: dh dt
f (h, qi , q)
f f (h h ) h ( h ,q , q ) qi i
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(qi qi ) ( h , qi , q )
f (q q ) q ( h ,q ,q ) i
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f 1 (qi q ) 0 ( qi q ) h ( h ,q , q ) h 2 L ( D h)h i
f 1 , q ( h ,q ,q ) 2 L ( D h )h i
1 f qi ( h ,q ,q ) 2 L ( D h ) h
Let this term be k
i
i ( s) kQ ( s) sH (s) kQ • Transfer function between H (s )
(s) : and Q
• Transfer function between H (s ) and Qi (s) :
-
k
s k
(integrating) (integrating)
s
• If h is near 0 or D, k becomes very large and h is around h / 2 , k becomes minimum. The model could be quite different depending on the operating condition used for the linearization. The best suitable range for the linearization in this case is around h / 2 . (less change in gain) Linearized model would be valid in very narrow range near 0 CHBE306 Process Dynamics and Control
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5-30
PROPERTIES OF TRANSFER FUNCTION •
Additive property
G1(s)
Y (s) Y1 (s ) Y2 (s )
G1 ( s) X 1 (s ) G2 (s ) X 2 (s ) •
Y1(s)
X1(s)
Y(s) + +
X2(s)
G2(s)
Y2(s)
Multiplicative property X 3 (s ) G2 (s ) X 2 (s )
X1(s)
G1(s)
X2(s)
X3(s) G2(s)
G2 (s ) G1 (s ) X 1 (s ) G2 (s )G1 (s ) X 1 (s ) •
Physical realizability – In a transfer function, the order of numerator(m) is greater than that of denominator(n): called “physically unrealizable” – The order of derivative for the input is higher than that of output. (requires future input values for current output)
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EXAMPLES ON TWO TANK SYSTEM •
Two tanks in series (Ex3.7) – No reaction
V1
V2
dc1 dt dc2
dt
qc1 qci qc2 qc1
– Initial condition: c1(0)= c2(0)=1 kg mol/m3 (Use deviation var.) – Parameters: V 1/q=2 min., V 2/q=1.5 min. – Transfer functions
C1 ( s ) Ci ( s )
C 2 ( s) Ci ( s)
C 2 ( s )
1 (V1 / q)s 1
C 2 ( s) C1 ( s) C1 (s) C i (s)
C1 ( s )
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1 (V2 / q) s 1
1
(V2 / q)s 1 (V1 / q)s 1 Korea University
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•
Pulse input
P
ci
5 P 0.25 s Ci ( s) (1 e ) s
•
5 0.25
t
Equivalent impulse input Ci (s) L (5 0.25) (t ) 1.25
•
Pulse response vs. Impulse response
C1 P ( s )
1 2 s 1
CiP ( s)
5 s(2 s 1)
(1 e0.25 s )
5 10 (1 e0.25 s ) s 2s 1 c1 P (t ) 5(1 et / 2 ) 5(1 e(t 0.25) /2 )S(t 0.25) C1 ( s)
1 2 s 1
Ci ( s)
1.25 (2s 1)
c1 0.625et / 2 CHBE306 Process Dynamics and Control
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5-33
