Post on 26-Mar-2018
CHE302 Process Dynamics and Control Korea University 5-1
CHE302 LECTURE VLAPLACE TRANSFORM AND
TRANSFER FUNCTION
Professor Dae Ryook Yang
Fall 2001Dept. of Chemical and Biological Engineering
Korea University
CHE302 Process Dynamics and Control Korea University 5-2
SOLUTION OF LINEAR ODE
• 1st-order linear ODE– Integrating factor:
• High-order linear ODE with constant coeffs.– Modes: roots of characteristic equation
– Depending on the roots, modes are• Distinct roots:• Double roots:• Imaginary roots:
• Many other techniques for different cases
For ( ) ( ), I.F. exp( ( ) )dx
a t x f x a t dtdt
+ = = ∫( ) ( )
[ ] ( )a t dt a t dt
xe f t e∫ ∫′ =( ) ( )
( ) [ ( ) ]a t dt a t dt
x t f t e dt C e−∫ ∫= +∫
2 1 0For ( ),a x a x a x f t′′ ′+ + =2
2 1 0 2 1 1( )( ) 0a p a p a a p p p p+ + = − − =
1 2( , )p t p te e− −
( cos , sin )t te t e tα αβ β− −
1 1( , )p t p te te− −
Solution is a linear combination of modes and the coefficients are
decided by the initial conditions.
CHE302 Process Dynamics and Control Korea University 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
ODE or PDEu(t) y(t)
Transfer FunctionU(s) Y(s)
(Algebraic calculation)
(D.E. calculation)
L -1L L -1LL -1L
CHE302 Process Dynamics and Control Korea University 5-4
DEFINITION OF LAPLACE TRANSFORM
• Definition
– 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
– and are linear.
{ }0
( ) ( ) ( ) stF s f t f t e dt∞ −= ∫@L
{ }-1( ) ( )f t F s= LL -1L { }1 2 1 2( ) ( ) ( ) ( )af t bf t aF s bF s+ = +L
CHE302 Process Dynamics and Control Korea University 5-5
LAPLACE TRANSFORM OF FUNCTIONS
• Constant function, a
• Step function, S(t)
• Exponential function, e-bt
{ }0
0
0st sta a aa ae dt e
s s s
∞∞ − − = = − = − − =
∫Lt
f(t)
a
t
f(t)
1
1 for 0( ) S( )
0 for 0
tf t t
t
≥= = <
{ }0
0
1 1 1S( ) 0st stt e dt e
s s s
∞∞ − − = = − = − − =
∫L
{ } ( )
00
1 1bt bt st b s te e e dt es b s b
∞∞− − − − +−
= = =+ +∫L
t
f(t)
1b>0
b<0
CHE302 Process Dynamics and Control Korea University 5-6
• Trigonometric functions– Euler’s Identity:
• Rectangular pulse, P(t)
cos sinj te t j tω ω ω+@( )1
cos2
j t j tt e eω ωω −= + ( )1sin
2j t j tt e e
jω ωω −= −
{ } 2 2
1 1 1 1 1sin
2 2 2j t j tt e e
j j j s j s j sω ω ω
ωω ω ω
− = − = − = − + +
L L L
{ } 2 2
1 1 1 1 1cos
2 2 2j t j t s
t e es j s j s
ω ωωω ω ω
− = = + = − + + L L +L
0 for
( ) P( ) for 0
0 for 0
w
w
t t
f t t h t t
t
>= = ≥ ≥ <
{ } ( )0
0
P( ) 1w
ww
tt t sst sth h
t he dt e es s
−− −= = − = −∫Lt
f(t)
h
tw
t
sin(t)
1
2πω
CHE302 Process Dynamics and Control Korea University 5-7
• Impulse function,
• Ramp function, t
• Refer the Table 3.1 (Seborg et al.) for other functions
0
0 for
( ) ( ) lim 1/ for 0
0 for 0 w
w
w wt
t t
f t t t t t
t
δ→
>= = ≥ ≥ <
{ } ( )00 0
1 1( ) lim lim 1 1
ww
w w
t t sst
t tw w
t e dt et t s
δ −−
→ →= = − =∫L
0 0
( ) ( )L'Hospital's rule: lim lim
( ) ( )t t
f t f tg t g t→ →
′ = ′
t
f(t)1/tw
tw
t
f(t)
1
1
( )tδ
{ }0
20 00
1 1
st
stst st
t te dt
t ee dt e dt
s s s s
∞ −
∞ −∞ ∞− −
=
= − = =− −
∫
∫ ∫
L
( )00 0Integration by part: 'f gdt f g f g dt
∞ ∞∞ ′⋅ = ⋅ − ⋅∫ ∫
CHE302 Process Dynamics and Control Korea University 5-8
CHE302 Process Dynamics and Control Korea University 5-9
CHE302 Process Dynamics and Control Korea University 5-10
PROPERTIES OF LAPLACE TRANSFORM
• Differentiation
00 0
0
( ) ( ) (by . . .)
(0) ( ) (0)
st st st
st
dff e dt f t e f s e dt i b p
dt
s f e dt f sF s f
∞ ∞∞− − −
∞ −
′= ⋅ = − ⋅ −
= ⋅ − = −
∫ ∫
∫
L
( )
2
2 00 0 0
2
( ) ( ) (0)
( ) (0) (0) ( ) (0) (0)
st st st std ff e dt f t e f s e dt s f e dt f
dt
s sF s f f s F s sf f
∞ ∞ ∞∞− − − − ′′ ′ ′ ′ ′= ⋅ = − ⋅ − = ⋅ −
′ ′= − − = − −
∫ ∫ ∫L
( ) ( 1) ( 1)
00 0
1( 1) ( 1) ( 1)
10
1 ( 2) ( 1)
( ) ( )
(0) (0)
( ) (0) (0) (0)
nn st n st n st
n
nn st n n
n
n n n n
d ff e dt f t e f s e dt
dt
d fs f e dt f s f
dt
s F s s f sf f
∞ ∞∞− − − − −
−∞ − − − −−
− − −
= ⋅ = − ⋅ −
= ⋅ − = − = − − − −
∫ ∫
∫L
L
L
M
CHE302 Process Dynamics and Control Korea University 5-11
• 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
initial condition vanish.
• Transforms of linear differential equations.
22
2
( )
( )
( )n
n
n
dfsF s
dt
d fs F s
dt
d fs F s
dt
=
=
=
M
L
L
L
( ) ( ), ( ) ( )
( )( ) (if (0) 0)
y t Y s u t U s
dy tsY s y
dt
→ →
→ =
L L
L
( )( ) ( ) ( (0) 0) ( 1) ( ) ( )
dy ty t Ku t y s Y s KU s
dtτ τ= − + = → + =L
1 ( )( ) ( 1) ( ) ( )L L L
w L HL HL L wHL
T T T sv T T v s T s T s
t z zτ τ
τ∂ ∂ ∂
= − + − → + + =∂ ∂ ∂
% % %L
CHE302 Process Dynamics and Control Korea University 5-12
• Integration
• Time delay (Translation in time)
• Derivative of Laplace transform
{ } ( )0 0 0
0 00
( ) ( )
1 ( )( ) (by . . .)
t t st
stt
st
f d f d e dt
e F sf d f e dt i b p
s s s
ξ ξ ξ ξ
ξ ξ
∞ −
∞− ∞ −
=
= + ⋅ =−
∫ ∫ ∫
∫ ∫
L0
( )
( )
( ) ( )Leibniz rule: ( ) ( ( )) ( ( ))
b t
a t
d db t da tf d f b t f a t
dt dt dtτ τ = −
∫
in ( ) ( )S( )tf t f t tθ θ θ+→ − −
{ } ( )
0
0
( )S( ) ( ) ( ) (let )
( ) ( )
st s
s s s
f t t f t e dt f e d t
e f e d e F s
τ θ
θ
θ τ θ
θ θ θ τ τ τ θ
τ τ
∞ ∞− − +
∞− − −
− − = − = = −
= =
∫ ∫∫
Lt
f(t)
θ
[ ]0 0 0
( )( ) ( )st st stdF s d d
f e dt f e dt t f e dt t f tds ds ds
∞ ∞ ∞− − −= ⋅ = ⋅ = − ⋅ = − ⋅∫ ∫ ∫ L
CHE302 Process Dynamics and Control Korea University 5-13
• Final value theorem– From the LT of differentiation, as s approaches to zero
– Limitation: 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
[ ]00 0
lim lim ( ) (0)st
s s
dfe dt sF s f
dt
∞ −
→ →⋅ = −∫
0 0 0( ) (0) lim ( ) (0) ( ) lim ( )
s s
dfdt f f sF s f f sF s
dt
∞
→ →= ∞ − = − ⇒ ∞ =∫
[ ]0
lim lim ( ) (0)st
s s
dfe dt sF s f
dt
∞ −
→∞ →∞⋅ = −∫
0lim 0 lim ( ) (0) (0) lim ( )st
s s s
dfe dt sF s f f sF s
dt
∞ −
→∞ →∞ →∞= = − ⇒ =∫
( )f ∞
CHE302 Process Dynamics and Control Korea University 5-14
EXAMPLE ON LAPLACE TRANSFORM (1)
•
•
– Using the initial and final value theorems
– But the final value theorem is not valid because
t
f(t)
3
2 6
1.5 for 0 2
3 for 2 6( )
0 for 6
0 for 0
t t
tf t
t
t
≤ < ≤ <= ≤ <
( ) 1.5 S( ) 1.5( 2)S( 2) 3S( 6)f t t t t t t= − − − − −
{ } 2 62
1.5 3( ) ( ) (1 )s sF s f t e e
s s− −∴ = = − −L
2For ( ) , find (0) and ( ).
5F s f f
s= ∞
−
2(0) lim ( ) lim 2
5s s
sf sF s
s→∞ →∞= = =
− 0 0
2( ) lim ( ) lim 0
5s s
sf sF s
s→ →∞ = = =
−
5lim ( ) lim2 t
t tf t e
→∞ →∞= = ∞
CHE302 Process Dynamics and Control Korea University 5-15
EXAMPLE ON LAPLACE TRANSFORM (2)
• What is the final value of the following system?
– Actually, cannot be defined due to sin t term.
• Find the Laplace transform for ?
sin ; (0) (0) 0x x x t x x′′ ′ ′+ + = = =2
2 2 2
1 1 ( ) ( ) x(s)=
1 ( 1)( 1)s X s sX s X
s s s s⇒ + + = ⇒
+ + + +
2 20( )=lim 0
( 1)( 1)s
sx
s s s→∞ =
+ + +( )x ∞
[ ]( )From ( )
dF st f t
ds= − ⋅L
( sin )t tω
[ ] 2 2 2 2 2
2sin
( )
d st t
ds s s
ω ωω
ω ω ⋅ = − = + +
L
CHE302 Process Dynamics and Control Korea University 5-16
INVERSE LAPLACE TRANSFORM
• Used to recover the solution in time domain
– From the table– By partial fraction expansion
– By inversion using contour integral
• Partial fraction expansion– After the partial fraction expansion, it requires to know some
simple formula of inverse Laplace transform such as
{ } -1 ( ) ( )F s f t=L
{ } -1 1( ) ( ) ( )
2st
Cf t F s e F s ds
jπ= = ∫ÑL
2 2 2 2
1 ( 1)!, , , , etc.
( 1) ( ) 2 1
s
n
s n e
s s b s s s
θ
τ ω τ ζτ
−−+ + + + +
CHE302 Process Dynamics and Control Korea University 5-17
PARTIAL FRACTION EXPANSION
• 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.
– Inverse LT:
1
1 1
( ) ( )( )
( ) ( ) ( ) ( ) ( )n
n n
N s N sF s
D s s p s p s p s p
αα= = = + +
+ + + +LL
( )( )
( )i
i i
s p
N ss p
D sα
=−
= +
1 21 2( ) np tp t p t
nf t e e eα α α −− −= + + +L
CHE302 Process Dynamics and Control Korea University 5-18
• Case II: Some roots are repeated
– Each repeated factors have to be separated first.
– Same methods as Case I can be applied.– Heaviside expansion for repeated factors
– Inverse LT
11 0 1( ) ( )
( )( ) ( ) ( ) ( ) ( )
rr r
r r r
b s bN s N sF s
D s s p s p s p s p
α α−− + +
= = = = + ++ + + +
L L
( )
( )
1 ( )( ) ( 0, , 1)
! ( )
ir
r i i
s p
d N ss p i r
i ds D sα −
=−
= + = −
L
11 2( )
( 1)!pt pt r ptrf t e te t e
rα
α α− − − −= + + +−
L
CHE302 Process Dynamics and Control Korea University 5-19
• Case III: Some roots are complex
– Each repeated factors have to be separated first.
– Then,
– Inverse LT
1 0 1 12 2 2
1 0
( ) ( )( )
( ) ( )c s cN s s b
F sD s s d s d s b
α β ωω
+ + += = =
+ + + +
1 11 12 2 2 2 2 2
( ) ( )( ) ( ) ( )
s b s b
s b s b s b
α β ω ωα β
ω ω ω+ + +
= ++ + + + + +
1 1( ) cos sinbt btf t e t e tα ω β ω− −= +
21 0 1where / 2, / 4b d d dω= = −
( )1 1 1 0 1, /c c bα β α ω= = −
CHE302 Process Dynamics and Control Korea University 5-20
EXAMPLES ON INVERSE LAPLACE TRANSFORM
•
– Multiply each factor and insert the zero value
( 5)( ) (distinct)
( 1)( 2)( 3) 1 2 3s A B C D
F ss s s s s s s s
+= = + + +
+ + + + + +
00
( 5)5 / 6
( 1)( 2)( 3) 1 2 3ss
s B C DA s s s A
s s s s s s==
+ = + + + ⇒ = + + + + + +
11
( 5) ( 1) ( 1) ( 1)2
( 2)( 3) 2 3ss
s A s C s D sB B
s s s s s s=−=−
+ + + + = + + + ⇒ = − + + + +
22
( 5) ( 2) ( 2) ( 2)3 / 2
( 1)( 3) 1 3ss
s A s B s D sC C
s s s s s s=−=−
+ + + + = + + + ⇒ = + + + +
33
( 5) ( 3) ( 3) ( 3)1/3
( 1)( 2) 1 2ss
s A s B s C sD D
s s s s s s=−=−
+ + + + = + + + ⇒ = − + + + +
{ } -1 2 35 3 1( ) ( ) 2
6 2 3t t tf t F s e e e− − −∴ = = − + −L
CHE302 Process Dynamics and Control Korea University 5-21
•
– Use of Heaviside expansion
2
3 3
1( ) (repeated)
( 1) ( 2) ( 1) ( 2)As Bs C D
F ss s s s
+ += = +
+ + + +
( )
( )
1 ( )( ) ( 0, , 1)
! ( )
ir
r i i
s p
d N ss p i r
i ds D sα −
=−
= + = −
L
2 3
3 2
1 ( )( 2) ( 1)
( ) (2 3 ) (2 3 ) (2 )
As Bs C s D s
A D s A B D s B C D s C D
= + + + + +
= + + + + + + + + +, 2 3 0, 2 3 0, 2 1A D A B D B C D C D∴ = − + + = + + = + =
1, 1, 1, 1A B C D⇒ = = = = −
21 2 3
3 2 3
1( 1) ( 1) ( 1) ( 1)
s ss s s s
α α α+ += + +
+ + + +
( )22
1
1( 1) : 1 1
1! s
di s s
dsα
=−
= = + + = −
( )2
21 2
1
1( 2) : 1 1
2!s
di s s
dsα
=−
= = + + =
( )23
1( 0) : 1 1
si s sα
=−= = + + =
{ } -1 2 21( ) ( )
2t t t tf t F s e te t e e− − − −∴ = = − + −L
CHE302 Process Dynamics and Control Korea University 5-22
•2 2 2 2
( 1) ( 2)( ) (complex)
( 4 5) ( 2) 1s A s B Cs D
F ss s s s s
+ + + += = +
+ + + +2 2 2
3 2
1 ( 2) ( )( 4 5)
( ) (2 4 ) (5 4 ) 5
s A s s Bs Cs D s s
A C s A B C D s C D s D
+ = + + + + + +
= + + + + + + + +
, 2 4 0, 5 4 1, 5 1A C A B C D C D D∴ = − + + + = + = =
1/25, 7 / 25, 1/25, 1 /5A B C D⇒ = − = − = =
2 2 2
( 2) 1 ( 2) 7( 2) 1 25 ( 2) 1 25 ( 2) 1A s B s B
s s s+ + +
= − −+ + + + + +
2 2
1 1 1 125 5
Cs D
s s s
+= +
{ } -1 2 21 7 1 1( ) ( ) cos sin
25 25 25 5t tf t F s e t e t t− −∴ = = − − + +L
CHE302 Process Dynamics and Control Korea University 5-23
•2
21( ) (1 ) (Time delay)
(4 1)(3 1) 4 1 3 1
sse A B
F s es s s s
−−+ = = + + + + + +
1/4 1/31/(3 1) 4, 1/(4 1) 3
s sA s B s
=− =−= + = = + = −
{ }
( )
2 21 1 1
/4 /3 ( 2 ) /4 ( 2)/3
4 3 4 3( ) ( )
4 1 3 1 4 1 3 1
S( 2)
s s
t t t t
e ef t F s
s s s s
e e e e t
− −− − −
− − − − − −
∴ = = − + − + + + +
= − + − −
L L L
It is a brain teaser!!!But you have to live with it.
Hang on!!!You are half way there.
CHE302 Process Dynamics and Control Korea University 5-24
SOLVING ODE BY LAPLACE TRANSFORM
• Procedure1. Given linear ODE with initial condition,2. Take Laplace transform and solve for output3. Inverse Laplace transform
• Example: Solve for 5 4 2; (0) 1dy
y ydt
+ = =
{ } { } 25 4 2 5( ( ) (0)) 4 ( )
dyy sY s y Y s
dt s = ⇒ − + =
L +L L
2 5 2(5 4) ( ) 5 ( )
(5 4)s
s Y s Y ss s s
++ = + ⇒ =
+
{ }1 1 0.80.5 2.5( ) ( ) 0.5 0.5
5 4ty t Y s e
s s− − − ∴ = = + = + +
L L
CHE302 Process Dynamics and Control Korea University 5-25
TRANSFER FUNCTION (1)
• Definition– An algebraic expression for the dynamic relation between the
input and output of the process model
• How to find transfer function1. Find the equilibrium point2. If the system is nonlinear, then linearize around equil. point3. Introduce deviation variables4. Take Laplace transform and solve for output5. Do the Inverse Laplace transform and recover the original
variables from deviation variables
TransferFunction, G(s)
5 4 ; (0) 1
Let 1 and 4
( ) 1 0.25(5 4) ( ) ( ) ( )
( ) 5 4 1.25 1
dyy u y
dty y u u
Y ss Y s U s G s
U s s s
+ = =
= − = −
+ = ⇒ = = =+ +
% %%% % %
( )U s% ( )Y s%
CHE302 Process Dynamics and Control Korea University 5-26
TRANSFER FUNCTION (2)
• Benefits– Once TF is known, the output response to various given inputs
can be obtained easily.
– Interconnected system can be analyzed easily. • By block diagram algebra
– 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
{ } { } { } { }1 1 1 1( ) ( ) ( ) ( ) ( ) ( )y t Y s G s U s G s U s− − − −= = ≠L L L L
G1 G2
G3
+-
X Y ( ) 1( ) 2( )( ) 1 1( ) 2( ) 3( )
Y s G s G s
X s G s G s G s=
+
CHE302 Process Dynamics and Control Korea University 5-27
TRANSFER FUNCTION (3)
• Steady-state Gain: The ratio between ultimate changes in input and output
– 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
– The transfer function itself is an impulse response of the system
ouput ( ( ) (0))Gain= =
input ( ( ) (0))y y
Ku u
∆ ∞ −=
∆ ∞ −
0 0 0
( ) 1lim ( ) lim ( ) lim ( )
1 s s s
yK sY s sG s G s
s→ → →
∞= = = =
{ }( ) ( ) ( ) ( ) ( )Y s G s U s G s G sδ= = =L (t)
CHE302 Process Dynamics and Control Korea University 5-28
EXAMPLE
• Horizontal cylindrical storage tank (Ex4.7)
– Equilibrium point:(if , can be any value in .)
– Linearization:
Lq
qi
R
h
wi
idm dV
q qdt dt
ρ ρ ρ= = −
0( ) ( ) ( )
hi i
dV dhV h Lw h dh Lw h
dt dt= ⇒ =∫ % %
1 ( ) (Nonlinear ODE)
2 ( )i i i
dh dhw L q q q q
dt dt L D h h= − ⇒ = −
−
2 2( ) / 2 ( ) (2 )iw h R R h R h h= + − = −
0 ( ) /(2 ( ) )iq q L D h h= − −( , , )iq q hiq q= h 0 h D≤ ≤
( , , ) ( , , ) ( , , )
( , , ) ( ) ( ) ( )i i i
i i iih q q h q q h q q
dh f f ff h q q h h q q q q
dt h q q
∂ ∂ ∂= = − + − + −
∂ ∂ ∂
CHE302 Process Dynamics and Control Korea University 5-29
( ) ( ) ( )isH s kQ s kQ s= −% %%
( , , )
( , , ) ( , , )
1( ) 0 ( )
2 ( )
1 1,
2 ( ) 2 ( )
i
i i
i i
h q q
ih q q h q q
fq q q q
h h L D h h
f fq qL D h h L D h h
∂ ∂ −= − = =
∂ ∂ −
∂ − ∂= =
∂ ∂− −
∵Let this term be k
• Transfer function between (integrating)
• Transfer function between (integrating)
• If is near 0 or D, k becomes very large and is around , 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 . (less change in gain)
⇒ Linearized model would be valid in very narrow range near 0
( ) ( ) : -k
H s Q ss
and %%
( ) ( ) : ik
H s Q ss
and %%
h h / 2h
/ 2h
CHE302 Process Dynamics and Control Korea University 5-30
PROPERTIES OF TRANSFER FUNCTION
• Additive property
• Multiplicative property
• 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)
G1(s)
G2(s)
X1(s)
X2(s)
Y(s)
++
Y1(s)
Y2(s)
1 2
1 1 2 2
( ) ( ) ( )
( ) ( ) ( ) ( )
Y s Y s Y s
G s X s G s X s
= += +
G1(s) G2(s)X1(s) X2(s) X3(s)
[ ]3 2 2
2 1 1 2 1 1
( ) ( ) ( )
( ) ( ) ( ) ( ) ( ) ( )
X s G s X s
G s G s X s G s G s X s
=
= =
CHE302 Process Dynamics and Control Korea University 5-31
EXAMPLES ON TWO TANK SYSTEM
• Two tanks in series (Ex3.7)– No reaction
– Initial condition: c1(0)= c2(0)=1 kg mol/m3 (Use deviation var.)
– Parameters: V1/q=2 min., V2/q=1.5 min.– Transfer functions
11 1 i
dcV qc qc
dt+ =
22 2 1
dcV qc qc
dt+ =
1
1
( ) 1( / ) 1( )i
C sV q sC s
=+
%% 2
21
( ) 1( / ) 1( )
C sV q sC s
=+
%%
( ) ( )2 2 1
2 11
( ) ( ) ( ) 1( / ) 1 ( / ) 1( ) ( ) ( )i i
C s C s C sV q s V q sC s C s C s
= =+ +
% % %% % %
Ci
C1
C2
V1
V2
q
CHE302 Process Dynamics and Control Korea University 5-32
• Pulse input
• Equivalent impulse input
• Pulse response vs. Impulse response
t
5
0.25
Pic%
0.255( ) (1 )P s
iC s es
−= −%
{ }( ) (5 0.15) ( ) 1.25iC s tδ δ= × =% L
0.251
0.25
/ 21
( 0.25)/2
1 5( ) ( ) (1 )
2 1 (2 1)
5 10(1 )
2 1
( ) 5(1 )
5(1 )S( 0.25)
P P si
s
P t
t
C s C s es s s
es s
c t e
e t
−
−
−
− −
= = −+ +
= − − + ⇒ = −
− − −
% %
%
1
/ 21
1 1.25( ) ( )
2 1 (2 1)
0.625
i
t
C s C ss s
c e
δ δ
δ −
= =+ +
⇒ =
% %
%
CHE302 Process Dynamics and Control Korea University 5-33
2
/ 2 /1.52
1( ) ( )
(2 1)(1.5 1)
1.25(2 1)(1.5 1)
5 3.752 1 1.5 1
2.5 2.5
i
t t
C s C ss s
s s
s sc e e
δ δ
δ − −
=+ +
=+ +
= −+ +
⇒ = −
% %
%
0.252
0.25
1 5( ) ( ) (1 )
(2 1)(1.5 1) (2 1)(1.5 1)
5 40 22.5(1 )
2 1 1.5 1
P P si
s
C s C s es s s s s
es s s
−
−
= = −+ + + +
= − + − + +
% %
/ 2 /1.52
( 0 .25) /2 ( 0.25)/1.5
( ) (5 20 15 )
(5 20 15 )S( 0.25)
P t t
t t
c t e e
e e t
− −
− − − −
⇒ = − +
− − + −
%