Engineering Metabolism in Plant Cell Tissue Cultures for Optimizing

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```					             Transfer Functions

1. The Laplace transform
2. Solution of linear differential equations
3. Transient response example
The Laplace Transform
   Definition

F ( s )  L[ f (t )]   f (t )e  st dt
0

» Time (t) is replaced by a new independent variable (s)
» We call s the Laplace transform variable

   The Laplace domain
» Often more convenient to work in Laplace domain than time domain
» Time domain  ordinary differential equations in t
» Laplace domain  algebraic equations in s

   General solution approach
»   Formulate model in time domain
»   Convert model to Laplace domain
»   Solve problem in Laplace domain
»   Invert solution back to time domain
Laplace Transform of Selected Functions
   Constant function: f(t) = a

            a  st                           a a
F ( s )  L( a )                st
ae dt   e                          0   
0             s                      0         s s

   Exponential function: f(t) = e-bt

F ( s)  L(e )   e e dt  e
bt
0

bt  st


0
( b  s ) t
dt  
1
bs

e ( b  s ) t      

0

1
sb
   Derivatives and integrals
 df   df  st            
 st 
L           e dt   f (t )e sdt  f (t )e
 st
 sF ( s )  f (0)
 dt  0 dt                0                               0

 dn f  n
L n   s F ( s )  s n 1 f (0)  s n  2 f (1) (0)    sf ( n  2) (0)  f ( n 1) (0)
 dt 
      
 t f (t*)dt *    t f (t*)dt *e  st dt  1 F ( s )
L 
0
                0  0
                         
           s
Properties of Laplace Transforms
   Superposition
L[af (t )  bg (t )]  aF (s)  bG(s)

   Final value theorem
If lim y(t ) exists  lim y(t )  lim[sY (s)]
t                          t      s 0

   Initial value theorem
lim y(t )  lim[sY (s)]
t 0             s 

   Time delay

f d (t )  f (t  t0 )S (t  t0 )    L
             Fd (s)  e st0 F (s)
 st 0               L1
Fd ( s)  e            F ( s)         f d (t )  f (t  t0 )S (t  t0 )
Linear ODE Example
   ODE
dy
5  4y  2         y (0)  1
dt
   Laplace transform
2
5[ sY ( s)  y (0)]  4Y ( s) 
s
   Substitute y(0) & rearrange
5s  2     s  0.4
Y ( s)            
s (5s  4) s ( s  0.8)
   Inverse Laplace transform
1            s  0.4 
1
y(t )  L [Y (s)]  L              
 s( s  0.8) 
Linear ODE Example cont.
   Table 3.1


1       s  b3         b3  b1 b1t b3  b2 b2t
L                              e           e
 ( s  b1 )(s  b2 )  b2  b1      b1  b2

   Our problem
 s  0.4 
1
y(t )  L                b1  0 b2  0.8 b3  0.4
 s( s  0.8) 

   Substitute and simplify


1     s  0.4      0.4  0 0t 0.4  0.8 0.8t
L                            e                   0.5  0.5e0.8t
(s  0)(s  0.8)  0.8  0
e
                                0  0.8
General ODE Solution Procedure
   Procedure
» Transform to Laplace
domain
» Solve resulting algebraic
equations
» Transform solution back
to time domain
   Partial fraction
expansion
» Necessary when inverse
Laplace transform not
tabularized
» Break complex functions
into simpler tabularized
functions
Partial Fraction Example
   Partial fraction expansion

s5           s5         1    2
Y ( s)  2                             
s  5s  4 ( s  1)( s  4) s  1 s  4

   Determination of coefficients

s5           4      s5             1
1                 2               
s  4 s  1 3       s  1 s  4    3

   Inverse Laplace transform

1           1   4 / 3 1 / 3  4 t 1  4 t
y (t )  L [Y ( s )]  L               3e  3e
 s 1 s  4 
Repeated Factor Example

s 1        1       2       3
Y ( s)                               
s( s  2) 2
s  2 ( s  2) 2
s

s 1        1         s 1                1        1
2               3                          1  
s s  2 2       ( s  2) 2
s 0
4        4

s 1        1       2       3 1/ 4      1/ 2      1/ 4
Y ( s)                                                    
s( s  2) 2
s  2 ( s  2) 2
s   s  2 ( s  2) 2
s

1             1  2t 1  2t 1
y (t )  L [Y ( s)]   e  te  S (t )
4      2      4
s 1       1  2  3 s   4
Y (s)  2 2              2  2
s ( s  4 s  5) s s    s  4s  5

s  1  1s ( s 2  4 s  5)   2 ( s 2  4 s  5)  ( 3 s   4 ) s 2

(1   3 ) s 3  (41   2   4 ) s 2  (51  4 2  1) s  (5 2  1)  0

s 1       1  2  3 s   4   0.04 0.2  0.04s  0.36
Y ( s)  2 2              2  2                 2  2
s ( s  4 s  5) s s    s  4s  5      s  s     s  4s  5

 0.04s  0.36  0.04( s  2)     0.28
               
s  4s  5
2
( s  2)  1 ( s  2) 2  1
2

y (t )  L1[Y ( s )]  0.04S (t )  0.2t  0.04e  2t cost  0.28e  2t sin t
Transient Response Example

   Component balance
dc1                   dc1
V      q(ci  c1 )  4      2c1  2ci   c1 (0)  0
dt                    dt
   Step input
0 t  0            5
ci (t )           Ci (s) 
5 t  0            s
Transient Response Example cont.
   Laplace transform
2
4[ sC1 ( s)  0]  2C1 ( s)  2Ci ( s)  C1 ( s)         Ci ( s)
4s  2

   Substitute input
2 5        5
C1 ( s)          
4s  2 s s (2s  1)

   Inverse Laplace transform

 5 
1
c1 (t )  L                  
 5 1  e t / 2   
 s(2s  1) 

inlet

Inlet 1

2
outlet
4s+2
Step               Tank                To Workspace

6
Input
Output
>> plot(tout,inlet)
5
>> hold
Concentration (kmol/m3)

4                                           >> plot(tout,outlet,'r')
3
>> axis([0 15 0 6])
>> ylabel('Concentration (kmol/m^3)')
2
>> xlabel('Time (min)')
1
>> legend(‘Input','Output')
0
0   5                10            15
Time (min)

```
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