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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
4. Simulink 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
                     Quadratic Factor Example
                             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) 
                                      Simulink Solution: mixing.mdl

                                                                                 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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posted:5/8/2013
language:English
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