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Chapter 1
         Mathematical Modeling
         of Dynamic Systems in
              State Space




Saturday, September     PMDRMFRCIED   1
29, 2012
Introduction to State Space
              analysis
• Two approaches are available for the
  analysis and design of feedback
  control systems
     – Classical or Frequency domain technique
     – Modern or Time domain technique




Saturday, September   PMDRMFRCIED                2
29, 2012
Introduction to State Space
              analysis
• Classical technique is based on converting a
  system’s differential equation to a transfer
  function
• Disadvantage
     – Can be applied only to Linear Time Invariant system
     – Restricted to Single Input and Single output system
• Advantage
     – Rapidly provide stability and transient response
       information

Saturday, September       PMDRMFRCIED                        3
29, 2012
Introduction to State Space analysis
 • Modern technique or state space approach is
   a unified method for modeling, analyzing and
   designing a wide range of systems
 • Advantages :
      –   Can be used to nonlinear system
      –   Applicable to time varying system
      –   Applicable to Multi Input and Multi Output system
      –   Easily tackled by the availability of advanced digital
          computer




 Saturday, September          PMDRMFRCIED                          4
 29, 2012
Time varying
• A time-varying control system is a
  system in which one or more of the
  parameters of the system may vary as a
  function of time
• Dynamic system: input, state, output
  and initial condition




Saturday, September      PMDRMFRCIED   5
29, 2012
The state variables of a dynamic
              system
• The state of a system is a set of variables
  whose values, together with the input
  signals and the equations describing
  the dynamics , will provide the future
  state and output of the system
• The state variables describe the present
  configuration of a system and can be
  used to determine the future response,
  given the excitation inputs and the
  equations describing the dynamics.
Saturday, September   PMDRMFRCIED           6
29, 2012
The State Space Equations
                  
                  x(t )  Ax (t )  Bu (t )
                  y (t )  Cx(t )  Du(t )
                      
                      x(t )  derivative _ of _ the _ state _ vector
                      x(t )  state _ vector
                      y (t )  output _ vector
                      u (t )  input _ of _ control _ vector
                      A  system _ matrix
                      B  input _ matrix
                      C  output _ matrix
                      D  feedfoward _ matrix

Saturday, September                    PMDRMFRCIED                     7
29, 2012
Two types of equation
• State equation

x(t )  Ax (t )  Bu (t )
• Output equation


y(t )  Cx(t )  Du(t )
Saturday, September   PMDRMFRCIED   8
29, 2012
Terms
• State equations: a set of n simultaneous,
  first order differential equations with n
  variables, where the n variables to be
  solved are the state variables
• State space: The n-dimensional space
  whose axes are the state variables
• State space representation: A
  mathematical model for a system that
  consists of simultaneous, first order
  differential equations and output equation
Saturday, September   PMDRMFRCIED              9
29, 2012
Terms
• State variables: the smallest set of
  linearly independent system variables
  such that the value of the members of the
  set
• State vector: a vector whose elements
  are the state variables



Saturday, September   PMDRMFRCIED             10
29, 2012
Modeling of Electrical Networks
  Voltage-current, voltage-charge, and
  impedance relationships for capacitors,
  resistors, and inductors




Saturday, September   PMDRMFRCIED       11
29, 2012
An RLC circuit




Saturday, September       PMDRMFRCIED   12
29, 2012
State variable characterization
• The state of the RLC system described
  a set of state variables x1 and x2
• X1 = capacitor voltage = vc(t)
• X2 = inductor current = iL(t)
• This choice of state variables is
  intuitively satisfactory because the
  stored energy of the network can be
  described in terms of these variables
                         1 2 1
                      E  LiL  Cvc
                                    2

                         2     2
Saturday, September        PMDRMFRCIED    13
29, 2012
Utilizing Kirchhoff’s current law
• At the junction
• First order differential equation
• Describing the rate of change of
  capacitor voltage

                 dvc
          ic  C      u (t )  iL
                 dt
Saturday, September   PMDRMFRCIED     14
29, 2012
Utilizing Kirchhoff’s voltage law
• Right hand loop
• Provide the equation describing the
  rate of change of inductor current
                        diL
                      L       Ri L  vc
                        dt
• Output of the system, linear algebraic
  equation
                       vo  RiL (t )
Saturday, September         PMDRMFRCIED     15
29, 2012
State space representation
• A set of two first order differential equation and
  output signal in terms of the state variables x1 and x2

             dx1        1        1
                     x2  u (t )
             dt        C         C
             dx2        1        R
                      x1  x2
              dt        L        L
             y (t )  vo (t )  Rx 2
                                1
                  0   x   1 
                      
                   x
             x   1         C . 1   .u
                                    C
                  x   1  R   x2   0 
                  2                     
                          L     L
                           x1 
             y  0 R . 
                           x2 
Saturday, September                PMDRMFRCIED         16
29, 2012
Example 1 : RL serial network

  • Figure below shows an RL serial
    network with an input voltage vi(t) and
    voltage drop at inductance, L as an
    output voltage vo(t). Form a state space
    model for this system using the current
    i(t) in the loop as the state variable.


Saturday, September   PMDRMFRCIED              17
29, 2012
Modeling of Electrical Networks
• RL serial network – first order system




Saturday, September   PMDRMFRCIED          18
29, 2012
RL serial network
• Write the loop equation for the system
  using Kirchhoff’s voltage law,
         Vi (t )  VR (t )  VL (t )  VR (t )  Vo (t )
                       di (t )
         VL (t )  L            Vo (t )
                         dt
         VR (t )  i (t ) R
                                di (t )
         Vi (t )  i (t ) R  L
                                 dt
Saturday, September         PMDRMFRCIED                    19
29, 2012
RL serial network
• State variable is given only one, therefore
  the system is a first order system
• A state equation involving i is required
                                              di (t )
                      Vi (t )  i (t ) R  L
                                               dt
                          di (t )
                      L            i (t ) R  Vi (t )
                            dt
                      di (t )        R         1
                                 i (t )  Vi (t )
                         dt          L         L
                      
                                 R           1
                      i (t )   i (t )   Vi (t )
                                 L           L
Saturday, September                 PMDRMFRCIED           20
29, 2012
RL serial network
 • The output equation,

Vi (t )  VR (t )  VL (t )  VR (t )  Vo (t )
Vo (t )  VR (t )  Vi (t )
Vo (t )  i (t ) R  Vi (t )
y (t )   R i (t )  1Vi (t )
 Saturday, September         PMDRMFRCIED    21
 29, 2012
Example 2 : RC serial network
• Figure below shows an RC circuit with
  input voltage vi(t) and output voltage at
  resistor ie vo(t). Form a state space model
  for this system using the voltage vc(t)
  across the capacitor as the state variable
                          R             V0
                          VR
       Vi             i     VC           C

Saturday, September       PMDRMFRCIED        22
29, 2012
RC serial network
• Write the equations for the system using
  Kirchhoff’s voltage law,
 vi (t )  vR (t )  vc (t )  vc (t )  vo (t )      (1)
  for _ the _ capacitor
             dvc (t )
 i (t )  C                (2)
                dt
  for _ the _ resistor
 vo (t )  i (t ) R      (3)
Saturday, September      PMDRMFRCIED                        23
29, 2012
RC serial network
• State variable is given only one
• Therefore the system is a first order
  system
• Therefore a state equation involving vc is
  required
• Combine equation (2) and (3) yields
               vo (t )               dvc (t )
                        i (t )  C
                 R                       dt
                                dvc (t )
               vo (t )  RC                   (4)
                                  dt
Saturday, September              PMDRMFRCIED             24
29, 2012
RC serial network
• Eliminate vo(t) from equation (4) and
  combine with equation (1) and rearrange
  gives vi (t )  vc (t )  vo (t )
                                        dvc (t )
                vi (t )  vc (t )  RC
                                           dt
                       dvc (t )
                RC               vc (t )  vi (t )
                         dt
                dvc (t )                  1             1
                           vc (t )         vc (t )     vi (t )    (5)
                   dt                    RC             RC
Saturday, September             PMDRMFRCIED                              25
29, 2012
RC serial network
• Output of the system
              vo (t )  vc (t )  vi (t )      (6)
• Rearrange equation (5) and (6) in matrix
  form yields
   
               1                  1 
   v c (t )          vc (t )       vi (t )
               RC                 RC 
   y (t )   1vc (t )  1vi (t )
Saturday, September          PMDRMFRCIED                   26
29, 2012
RC serial network
                      x(t )  state _ vector  vc (t )
• Where,                                                       
                      x(t )  derivative _ state _ vector  v c (t )
                      u (t )  input _ vector  vi (t )
                      y (t )  output _ vector  vo (t )  vr (t )
                                               1
                      A  state _ matrix  
                                             RC
                                             1
                      B  input _ matrix 
                                            RC
                      C  ouput _ matrix  1
                      D  direct _ transmission _ matrix  1

Saturday, September              PMDRMFRCIED                           27
29, 2012
Modeling of Electrical Networks
 • Consider RLC serial network
 • RLC serial network – second order
   system




Saturday, September   PMDRMFRCIED      28
29, 2012
State Variables and output
• Select two state variables,
        x1 (t )  q (t )
        x2 (t )  i (t )
       output  y (t )  VL (t )
       input  u (t )  Vi (t )
Saturday, September    PMDRMFRCIED   29
29, 2012
Loop equation
 • Using Kirchoff’s Voltage Law,
vi (t )  vR (t )  vL (t )  vc (t )


  di (t )            1
L          Ri (t )   i (t )dt  vi (t )
   dt                C
 Saturday, September       PMDRMFRCIED   30
 29, 2012
Converting to charge
 • Using equation,

          dq(t )
i (t ) 
            dt
       2
    d q(t )      dq(t ) 1
L         2
             R         q(t )  vi (t )
       dt         dt    C
 Saturday, September   PMDRMFRCIED   31
 29, 2012
Derivatives of state vector
       x1 (t )  q (t )
                dq (t )
       x1 (t )           i (t )  x2 (t )
                    dt
       x2 (t )  i (t )
                di (t )
       x2 (t ) 
                  dt
Saturday, September       PMDRMFRCIED         32
29, 2012
State equation
• First state equation
                   dq(t )
          x1 (t )          i(t )  x2 (t )
                     dt
• Second state equation, using
               q (t )   i (t )dt
                   di (t )              1
                L           Ri (t )   i (t )dt v(t )
                     dt                 C
                di (t )      q (t ) Ri (t ) v(t )
                                          
                   dt        LC           L     L
                             1             R         1
                x2 (t )          x1 (t )  x2 (t )  u (t )
                             LC             L         L
Saturday, September                  PMDRMFRCIED                33
29, 2012
State equation in matrix form

x(t )  Ax (t )  Bu (t )
          0                    1   x (t )   0 
          x1 (t ) 
x(t )             1             R    1    1 u (t )
         x (t )  LC               x2 (t )  
                                                   L
         2                        L
         dq (t ) 
        dt   0                   1  q (t )  0 
x(t )  
          di (t )    1            R          1  v(t )
                                       i (t )   
                   LC
                                     L             L
         dt 
    Saturday, September     PMDRMFRCIED                    34
    29, 2012
Output equation
 • Output system is VL
     VL (t )  VR (t )  VC (t )  vi (t )
     VL (t )  VC (t )  VR (t )  vi (t )
                1
     V L(t )    i (t )dt  i (t ) R  vi (t )
                C
                1
     VL (t )   q (t )  Ri (t )  vi (t )
                C
                1
     VL (t )   x1 (t )  Rx 2 (t )  u (t )
                C
Saturday, September        PMDRMFRCIED             35
29, 2012
Output equation in matrix form

y (t )  Cx(t )  Du(t )
          1                x1 (t ) 
y (t )               R           1u (t )
          C                x2 (t )
           1               q(t )
VL (t )               R          1v(t )
           C                i (t ) 
 Saturday, September   PMDRMFRCIED              36
 29, 2012
Change State Variables but
        output still same
                  x1 (t )  VR (t )
                  x2 (t )  VC (t )
                      y (t )  VL (t )
                  u (t )  Vi (t )
Saturday, September          PMDRMFRCIED   37
29, 2012
Voltage formula for R, L and C
     VR (t )  i (t ) R
               1
     VC (t )   i (t )dt
               C
                 di (t )
     VL (t )  L
                  dt
Saturday, September   PMDRMFRCIED   38
29, 2012
Derivative of first state equation


    x1 (t )  VR (t )
     
                                   v(t )  VR (t )  VC (t )
               dVR (t )    di (t ) R
    x1` (t )           R
                 dt          dt     L
               R          R          R
    x1 (t )   VR (t )  VC (t )  v(t )
                L          L          L
               R          R         R
    x1` (t )   x1 (t )  x2 (t )  u (t )
                L          L         L
Saturday, September        PMDRMFRCIED                        39
29, 2012
Derivative of second state
               equation


       x2 (t )  VC (t )
                dVC (t ) 1           1
       x2 (t )            i (t )     VR (t )
                   dt        C       RC
                 1
       x2 (t )      x1 (t )
                 RC
Saturday, September        PMDRMFRCIED            40
29, 2012
State equation in matrix form

       
       x(t )  Ax (t )  Bu (t )
                              R    R
                    
                                  x (t )   R 
                x1 (t )    L
       x(t )                      L  1
                                                 L u (t )
                x (t )  1       0   x2 (t )  0 
                2                                  
                              RC    
                dVR (t )   R       R
               dt   L  L  VR (t )   R 
       x(t )              1                   L  v(t )
                                        VC (t )  0 
                 dVC (t )
                                  0                  
                dt   RC             
Saturday, September                PMDRMFRCIED                       41
29, 2012
Output equation


  VL (t )  VR (t )  VC (t )  v(t )
  VL (t )  VR (t )  VC (t )  v(t )
    y (t )   x1 (t )  x2 (t )  u (t )

Saturday, September        PMDRMFRCIED      42
29, 2012
Output equation in matrix form

   y (t )  Cx(t )  Du(t )
                      x1 (t ) 
 y (t )   1  1            1 u (t )
                      x2 (t )
                       VR (t ) 
 VL (t )   1  1             1 v(t )
                       VC (t )
Saturday, September   PMDRMFRCIED             43
29, 2012
Example 3 : 2 loop
• Find a state space representation if the
  output is the current through the resistor.
• State variables VC(t) and iL(t)
• Output is iR(t)
• Input is Vi(t)




Saturday, September    PMDRMFRCIED              44
29, 2012
Electrical network LRC
                                L
                                                  node 1
                           VL


Vi                         iL
                                         VR
                                                  R             C
                                    iR
                                                      iC   VC




     Saturday, September            PMDRMFRCIED                     45
     29, 2012
Solution : Step 1
• Label all of the branch currents in the
  network.
• iL(t), iR(t) and iC(t)




Saturday, September        PMDRMFRCIED      46
29, 2012
Solution : Step 2
 • Select the state variables by writing
   the derivative equation for all
   energy-storage elements i.e.
   inductor and capacitor
                                   1
                        VC (t )   iC (t )dt
                                   C
                                       dVC (t )
                         iC (t )  C               (1)
                                          dt
                                     diL (t )
                        VL (t )  L              ( 2)
                                       dt
Saturday, September              PMDRMFRCIED                 47
29, 2012
Solution : Step 3
• Apply network theory, such as Kirchoff’s
  voltage and current laws to obtain iC(t)
  and VL(t) in terms of the state variable
  VC(t) and iL(t)
• At node 1, iL (t )  iR (t )  iC (t )
                           iC (t )  iL (t )  iR (t )
                                      1
                           iC (t )   VC (t )  iL (t )    (3)
                                      R
• Around the outer loop,
                      Vi (t )  VL (t )  VC (t )
                      VL (t )  VC (t )  Vi (t )    (4)
Saturday, September                 PMDRMFRCIED                      48
29, 2012
Solution : Step 4
  • Substitute the result of equation (3) and
    equation (4) into equation (1) and (2)
               dVC (t )    1
              C           VC (t )  iL (t )    (7)
                  dt       R
               di (t )
              L L  VC (t )  Vi (t )    (8)
                 dt
  • Rearrange
                  dVC (t )      1          1
                                VC (t )  iL (t )    (9)
                    dt         RC          C
                  diL (t )    1          1
                             VC (t )  Vi (t )    (10)
                    dt        L          L
Saturday, September                PMDRMFRCIED                  49
29, 2012
Solution : Step 5
• Find the output equation



               1
      iR (t )  VC (t )    (11)
               R

Saturday, September        PMDRMFRCIED    50
29, 2012
Solution : Step 6
• State space representation in vector
  matrix form are
     dVC (t )   1            1
     dt   RC C  VC (t )  0 
     di (t )    1             .            1  v(t )    (12)
     L                      0   iL (t )   L 
                                                   
     dt   L                    
              1     VC (t )
    iR (t )      0.             (13)
              R      iL (t ) 

Saturday, September           PMDRMFRCIED                              51
29, 2012
Example 4 : 2 loop
• Find the state space representation of the
  electrical network shown in figure below
• Input vi(t)
• Output vo(t)
• State variables x1(t) = vC1(t), x2(t) = iL(t)
  and x3(t) = vC2(t)



Saturday, September    PMDRMFRCIED                52
29, 2012
RLC two loop network
 • Identifying appropriate variables on the
   circuit yields C1
                                       node           R


                                                 iR
                            iC1
                  Vi                                  iC2    Vo
                       DC




                                       L                C2

                                  iL


Saturday, September                PMDRMFRCIED                    53
29, 2012
RLC two loop network
• Represent the electrical network shown in
  figure in state space where
• Output is v0(t)
• Input is vi(t)
• State variables :-
     X1(t) = vC1(t)
     X2(t) = iL(t)
     X3(t) = vC2(t)

Saturday, September   PMDRMFRCIED         54
29, 2012
Solution
 • Writing the derivative relations for
   energy storage elements i.e. C1, C2 and
   L
                         dvC1 (t )
                      C1            iC1 (t )
                            dt
                        diL (t )
                      L           vL (t )
                          dt
                         dVC 2 (t )
                      C2             iC 2 (t )
                             dt
Saturday, September             PMDRMFRCIED       55
29, 2012
Solution
  • Using Kirchhoff’s current and voltage
    laws                             C
                                       node
                                                               1
                                                                             R

iC1 (t )  iL (t )  iR (t )                             iC1
                                                                        iR
                                                                             iC2         Vo
iC1 (t )  iL (t )  ic 2 (t )
                                               Vi




                                                    DC
                                                                    L          C2

                                                               iL
                    1
iC1 (t )  iL (t )  (vL (t )  vC 2 (t ))
                    R
vL (t )  vC1 (t )  vi (t )
                     1
iC 2 (t )  iR (t )  (vL (t )  vC 2 (t ))
  Saturday, September
                     R           PMDRMFRCIED                                        56
  29, 2012
Solution
 • Substituting these relations and
   simplifying yields the state equations as
    dvC1     1       1      1          1
              vC1  iL      vC 2      vi
     dt     RC1      C1    RC1        RC1
    diL    1       1
          vC1  vi
    dt     L       L
    dvC 2     1          1           1
                vC1       vC 2       vi
     dt      RC 2       RC 2        RC 2
   vo  vC 2
Saturday, September    PMDRMFRCIED             57
29, 2012
Solution
 • Putting the equations in vector matrix
   form
       1             1          1   1 
        RC          C1
                              
                                RC1    RC 
   
           1
                                       1
   x  1            0         0     x   1 v
       L                              L      i

       1                        1   1 
                    0                     
       RC 2                    RC 2   RC 2 
   y  0 0 1x
Saturday, September        PMDRMFRCIED               58
29, 2012
Tutorial 1 : Number 1

  • Represent the electrical network shown in
    figure in state space where
  • Output is v0(t) and Input is vi(t)
  • State variables :-
       x1 = v 1
       x2 = i4
       x3 = v 0

Saturday, September   PMDRMFRCIED          59
29, 2012
Electrical network 1
  • Add the branch current and node
    voltages to the network
            R1 = 1 Ohm          R2 = 1 Ohm          R3 = 1 Ohm
                           V1                  V2



                 i1                 i3                      i5

   Vi           C1 = 1 F
                                        L=1H
                                                       C2 = 1 F   Vo
                                   i2                  i4




Saturday, September                PMDRMFRCIED                         60
29, 2012
Solution
• Write the differential equation for each
  energy storage element
             dv1
                  i2 ; because _ C1  1F
             dt
             di4
                  v2 ; because _ L  1H
             dt
             dv0
                  i5 ; because _ C2  1F
             dt
Saturday, September       PMDRMFRCIED        61
29, 2012
Solution
• Therefore the state vector is ,
                x1   v1 
               x   i 
           x   2  4 
                x3  vo 
                  
• Derivative state vector is ,
                           
                      
                           x1   v1 
                                     
                      x   x2    i4 
                           
                           x3  vo 
                             
                             
Saturday, September          PMDRMFRCIED    62
29, 2012
Solution
• Now obtain i2, v2 and i5 in terms of the
  state variables,
i2  i1  i3  vi  v1  (v1  v2 )  vi  2v1  v2
v2  i5  vo  i3  i4  v0  v1  v2  i4  v0
Therefore,
    1      1  1
v2  v1  i4  vo
    2      2  2
Saturday, September    PMDRMFRCIED                63
29, 2012
Solution
• Substituting v2 in i2,
               3      1  1
      i2  vi  v1  i4  v0
               2      2  2
      also,
              i5  i3  i4  v1  v2  i4
              substituti ng _ v2 ,
                  1    1    1
              i5  v1  i4  vo
                  2    2    2
Saturday, September           PMDRMFRCIED   64
29, 2012
Solution
• Therefore rearrange i2, v2 and i5 in matrix
  form yields
                                      3      1    1 
                                  2   
                      

               x1   v1   i2            2    2   v1  1
                       
               x    i   v    1      1    1    
          x 2                                       . i4  0 vi
                     2  2
                         4
                                              2    2    
               x3  vo   i5   1
                                            1     1  vo  0
                                                           
                 
                                             
                                      2
                                             2     2
                            v1 
          y  0 0 1. i4 
                            
                           vo 
                            
Saturday, September             PMDRMFRCIED                          65
29, 2012
Tutorial 1 : Number 2
• Represent the electrical network shown in
  figure in state space where
• Output is iR(t)
• Input is vi(t)
• State variables :-
     x1 = i2
     x2 = vC


Saturday, September   PMDRMFRCIED         66
29, 2012
Electrical network 2
   • Add the branch currents and node
     voltages to the schematic and obtain
                                C = 1F
            R1 = 1 Ohm
                      node V1                    node V2



                  i1
                                      i3
   Vi                                                 R2=1 Ohm
                                           4V1                   iR
           DC




                       L = 1H    i2
                                                     i4




Saturday, September              PMDRMFRCIED                          67
29, 2012
Solution
• Write the differential equation for each
  energy storage element
                 di2
                      v1 ; because _ L  1H
                 dt
                 dvc
                      i3 : because _ C  1F
                 dt
Saturday, September      PMDRMFRCIED           68
29, 2012
Solution
• Therefore the state vector is,



      x1   i2 
   x  
      x2  vc 
Saturday, September    PMDRMFRCIED   69
29, 2012
Solution
• Now obtain v1 in terms of the state
  variables v1  vc  v2
             v1  vc  iR
             v1  vc  i3  4v1
             v1  vc  (i1  i2 )  4v1
             v1  vc  vi  v1  i2  4v1
                  1       1       1
             v1  i2  vc  vi
                  2       2       2
Saturday, September    PMDRMFRCIED          70
29, 2012
Solution
• Now obtain i3 in terms of the state
  variables i  i  i
                      3    1     2

                      i3  vi  v1  i2
                                1    1    1
                      i3  vi  i2  vc  vi  i2
                                2    2    2
                              3    1    3
                      i3   i2  vc  vi
                              2    2    2
Saturday, September            PMDRMFRCIED          71
29, 2012
Solution
• Now obtain the output iR in terms of the
  state variables

          iR  i3  4v1
              1    3    1
          iR  i2  vc  vi
              2    2    2
Saturday, September    PMDRMFRCIED           72
29, 2012
Solution
• Hence the state space representation
                        1    1       1
        v  
               
                               i   
        i2 
           1   2       2 . 2  2 v
    x
       v   i3       3 1  vc   3  i
        c                       
                        2 2          2 
       1     3   i2   1 
    y       .     vi
       2     2  vc   2 
Saturday, September    PMDRMFRCIED              73
29, 2012
Tutorial 1 : Number 3
  • Find the state space representation of the
    network shown in figure if
  • Output is v0(t)
  • Input is vi(t)
  • State variables :-
       x1 = iL1
       x2 = iL2
       x3 = vC

Saturday, September   PMDRMFRCIED           74
29, 2012
Electrical network 3
  • Add the branch currents and node
    voltages to the schematic and obtain
                                R3 = 1 Ohm




                      L1 = 1H         i3     L2 = 1H

       node                                                    node
        Vi                                                      Vo

          Vi                                   i2
                       i1                           R2=1 Ohm
                 DC




                                                               Vo
                            C = 1F




Saturday, September                PMDRMFRCIED                        75
29, 2012
Solution
• Write the differential equation for each
  energy storage element
                      diL1
                             vc  v1
                       dt
                      diL 2
                             vc  i2
                       dt
                      dvc
                             i1  i2
                      dt
Saturday, September          PMDRMFRCIED     76
29, 2012
Solution

  •   where,
  •   L1 is the inductor in the loop with i1
  •   L2 is the inductor in the loop with i2
  •   iL1 = i1 –i3
  •   iL2 = i2 – i3
  •   Now,
  •   i1 – i2 = ic = iL1 – iL2 -----------------(1)
Saturday, September       PMDRMFRCIED                 77
29, 2012
Solution
•   Also writing the node equation at vo,
•   i2 = i3 + iL2 ----------------------(2)
•   Writing KVL around the outer loop yields
•   i2 + i3 = vi -----------------------(3)
•   Solving (2) and (3) for i2 and i3 yields
               1      1
          i2  iL 2  vi            (4)
               2      2
                 1      1
          i3   iL 2  vi          (5)
                 2      2
Saturday, September           PMDRMFRCIED            78
29, 2012
Solution
• Substituting (1) and (4) into the state
  equations.
• To find the output equation,
• vo = -i3 + vi
• Using equation (5),

                     1      1
                 vo  iL 2  vi
                     2      2
Saturday, September     PMDRMFRCIED         79
29, 2012
Solution
• Summarizing the results in vector matrix
  form        diL1 
          
                      
                    x1      dt  0 0  1  iL1   1 
                                   
                     x    diL 2   0  1 1 .i    1  v
                
                x 2
                        dt                  2         L2   2  i
                     x3   dvC  1  1 0   vC   0 
                                                            
                      
                     
                              dt  
                                          iL1 
                              1          1
                y  vo  0           0.iL 2     vi
                              2                  2
                                          vC 
                                          
Saturday, September                PMDRMFRCIED                               80
29, 2012
Tutorial 1 : Number 4
• An RLC network is shown in figure. Define
  the state variable as :-
• X1 = i1
• X2 = i2
• X3 = Vc
• Let voltage across capacitor, Vc is the
  output from the network. Input of the
  system is Va and Vb
Saturday, September   PMDRMFRCIED         81
29, 2012
Tutorial 1 : Number 4
• Determine the state space representation
  of the RLC network in matrix form
• Determine the range of resistor R in order
  to maintain the system’s stability, if C = 0.1
  F and L1=L2=0.1 H. The characteristic
  equation of the system is,

 s  10Rs  200s  1000R  0
    3                 2

Saturday, September       PMDRMFRCIED         82
29, 2012
RLC network with 2 input
                      R        L1                 L2

                          i1                 +    i2

Va                             iC     C -
                                                 VC     Vb
            DC




                                                       DC
Saturday, September            PMDRMFRCIED                  83
29, 2012
Solution
• State variables and their derivatives
                          di1
            x1  i1  x1 
                           dt
                           di2
            x2  i2  x2 
                            dt
                           dvc
            x3  vc  x3 
                             dt
            u1  va
                      u 2  vb
                      y  vc
Saturday, September              PMDRMFRCIED   84
29, 2012
Solution
• The derivatives equations for energy
  storage elements
                  di1
               L1       vL1      (1)
                   dt
                   di2
               L2       vL 2      (2)
                   dt
                  dvC
               C        iC      (3)
                   dt
Saturday, September        PMDRMFRCIED        85
29, 2012
Solution                          L1          L2
                                               R
                                                   i1          +    i2
• For loop (1) ;                     Va                 iC         VC          Vb

         va  i1R  vL1  vC
                                                             C -




                                          DC




                                                                              DC
         vL1  va  i1 R  vC      (4)
• For loop (2) ;

          vb  vL 2  vC
          vL 2  vb  vC      (5)
Saturday, September    PMDRMFRCIED                                       86
29, 2012
Solution
• For current iC ;
               iC  i1  i2      (6)
• Substituting equation (4), (5) and (6) into
  equation (1), (2) and (3) yields
                                                           L1          L2
       di1                                        R

    L1      va  i1 R  vC                           i1          +    i2
        dt                              Va                 iC   C -
                                                                      VC     Vb




                                             DC




                                                                            DC
    di1      R       1      1
           i1  vC  va
    dt      L1       L1     L1
         di1    R    1    1
     x1        x1  x3  va      (7)
          dt     L1   L1   L1
Saturday, September       PMDRMFRCIED                                            87
29, 2012
Solution
• Substituting equation (4), (5) and (6) into
  equation (1), (2) and (3) yields
                di2
            L2       vb  vC
                dt
            di2       1       1
                   vC  vb
            dt       L2       L2
                di2    1    1
            x2        x3  vb      (8)
                 dt     L2   L2
Saturday, September      PMDRMFRCIED             88
29, 2012
Solution
• Substituting equation (4), (5) and (6) into
  equation (1), (2) and (3) yields
            dvC
         C        i1  i2
             dt
         dvC 1           1
               i1  i2
          dt     C       C
              dvC 1       1
         x3         x1  x2      (9)
                dt     C   C
Saturday, September    PMDRMFRCIED              89
29, 2012
Solution
• Rewrite equation (7), (8) and (9) in state
  space representation matrix form
                        R            1          1      
                 L
                              0    
                                     L1  x   L1
                                                       0
               x1   1                 1               v
                                    1            1   a
          x   x2    0     0        . x2   0       . 
                                 L2              L2  vb 
               x3   1       1          x3   0
                                                     0
                
                                   0                  
                        C     C                        
                       x1 
          y  0 0 1. x2 
                       
                       x3 
                       
Saturday, September            PMDRMFRCIED                          90
29, 2012
Solution
• Characteristic equation
       s  10Rs  200s  1000R  0
         3            2

• Routh Hurwitz table
             s3            1              200   0

             s2           10R        1000R      0

             s1       (2000 R  1000 R)
                                           0
                            10 R
             s0           200

Saturday, September             PMDRMFRCIED         91
29, 2012
Solution
• For stability, all coefficients in first column
  of Routh Hurwitz table must be positive ;
         (2000 R  1000 R)
                           0
               10 R
         1000 R 0
         R 0
Saturday, September    PMDRMFRCIED                  92
29, 2012
Modeling of Mechanical
• Mass
                    Networks
   f (t )  M .a (t )

   f (t )  M .
                d 2 y (t )                 y(t)
                  dt 2
                dv(t )
   f (t )  M .
                 dt
   a (t )  accelerati on
   v(t )  velocity
                                    M         f(t)
   y (t )  displaceme nt
   f (t )  force
   M  mass
 Saturday, September         PMDRMFRCIED          93
 29, 2012
Modeling of Mechanical
  •   Linear Spring
                    Networks

f (t )  K . y (t )
                                        K   y(t)
f (t )  force
y (t )  displaceme nt                             f(t)
K  spring _ cons tan t




  Saturday, September     PMDRMFRCIED               94
  29, 2012
Modeling of Mechanical
                  Networks
• Damper


            dy (t )                 B   y(t)
f (t )  B.
             dt
f (t )  force                                 f(t)
y (t )  displaceme nt
B  viscous _ frictional

Saturday, September   PMDRMFRCIED              95
29, 2012
Modeling of Mechanical
                    Networks
 • Inertia
             d (t )
T (t )  J .
               dt
                                   T(t)
             d 2 (t )                            (t )
T (t )  J .
               dt 2
T (t )  Torque                              J
 (t )  angular _ velocity
 (t )  angular _ displaceme nt
J  Inertia

 Saturday, September           PMDRMFRCIED         96
 29, 2012
Force-velocity, force-displacement, and
   impedance translational relationships
  for springs, viscous dampers, and mass




Saturday, September   PMDRMFRCIED       97
29, 2012
Torque-angular velocity, torque-angular
     displacement, and impedance rotational
   relationships for springs, viscous dampers,
                    and inertia




Saturday, September   PMDRMFRCIED                98
29, 2012
Example 5
   • Determine the state space
     representation of the mechanical
     system below if the state variables are
     y(t) and dy(t)/dt. Input system is force
     f(t) and output system is y(t)
                      K
                                          y(t)

                      B          M
                                          f(t)


Saturday, September         PMDRMFRCIED          99
29, 2012
Example 5
• a. Mass, spring, and damper system;
  b. block diagram




Saturday, September     PMDRMFRCIED     100
29, 2012
State variables, input and output


              x1 (t )  y (t )
                       dy (t ) dx1 (t )
             x2 (t )         
                        dt       dt
             input  u  f (t )
             output  y  y (t )
Saturday, September         PMDRMFRCIED   101
29, 2012
Mass, spring and damper
                   system
    • Draw the free body diagram


                                        y(t)
   d 2 y (t )
M
      dt 2
Ky (t )                       M         f(t)
  dy (t )
B
    dt
    Saturday, September   PMDRMFRCIED     102
    29, 2012
Mass, spring and damper system
• a. Free-body diagram of mass, spring, and
  damper system;
  b. transformed free-body diagram




Saturday, September   PMDRMFRCIED        103
29, 2012
Mass, spring and damper
               system
• The force equation of the system is
                      2
             d y(t )      dy(t )
f (t )  M .     2
                      B.         K . y(t )
              dt           dt
• Rearranged the equation yields
  2
d y (t )   B dy (t ) K          1
    2
          .        . y (t )  . f (t )
 dt        M dt      M          M
Saturday, September       PMDRMFRCIED      104
29, 2012
Mass, spring and damper
                system
• State equations and output equation

    
    x1 (t )  x2 (t )
                 K        B          1
    x 2 (t )   .x1 (t )  .x2 (t )  . f (t )
                 M         M          M
    y (t )  x1 (t )
Saturday, September     PMDRMFRCIED           105
29, 2012
Mass, spring and damper
                system
• State space representation in vector
  matrix form are
   
             0         1   x (t )   0 
 x1 (t ) 
              K         B . 1    1 . f (t )
  x (t )  M            x2 (t )  
  2                   M             M 
                                        K

                 x1 (t ) 
 y (t )  1 0.
                                                y(t)

                                       B   M
                 x2 (t )                      f(t)


Saturday, September       PMDRMFRCIED                  106
29, 2012
Example: The mechanical
              system
• Consider the mechanical system shown in
  Figure below by assuming that the system
  is linear. The external force u(t) is the
  input to the system and the displacement
  y(t) of the mass is the output. The
  displacement y(t) is measured from the
  equilibrium position in the absence of the
  external force. This system is a single
  input and single output system.
Saturday, September   PMDRMFRCIED          107
29, 2012
Mechanical system diagram




Saturday, September   PMDRMFRCIED   108
29, 2012
Mechanical system diagram
• From the diagram, the system equation is
                         
                m y  b y  ky  u
• The system is of second order. This
  means that the system involves two
  integrators. Define the state variables x1(t)
  and x2(t) as     x1 (t )  y (t )
                                       
                           x2 ( t )  y ( t )
Saturday, September         PMDRMFRCIED         109
29, 2012
• Then we obtain,
                      
                      x 1  x2
                           1         
                                          1 u
                      x 2    ky  b y 
                            m            m
                            k     b       1
                      x 2   x1  x2  u
                             m     m       m
                      y  x1
Saturday, September              PMDRMFRCIED      110
29, 2012
Mass, spring and damper system
• a. Two-degrees-of-freedom translational
  mechanical system
• b. block diagram




Saturday, September   PMDRMFRCIED           111
29, 2012
Mass, spring and damper system
• a. Forces on M1 due only to motion of M1
  b. forces on M1 due only to motion of M2
  c. all forces on M1




Saturday, September   PMDRMFRCIED            112
29, 2012
Mass, spring and damper system
• a. Forces on M2 due only to motion of M2;
  b. forces on M2 due only to motion of M1;
  c. all forces on M2




Saturday, September   PMDRMFRCIED             113
29, 2012
Exercise 1
• Figure below shows a diagram for a quarter car
  model (one of the four wheels) of an automatic
  suspension system for a long distance express
  bus. A good bus suspension system should
  have satisfactory road handling capability, while
  still providing comfort when riding over bumps
  and holes in the road. When the coach is
  experiencing any road disturbance, such as
  potholes, cracks, and uneven pavement, the bus
  body should not have large oscillations, and the
  oscillations should be dissipate quickly.
Saturday, September     PMDRMFRCIED              114
29, 2012
Exercise 1
     (i). Draw the free-body diagrams of the system
     (ii). Determine the state space representation
        of the quarter car system by considering the
        state vector
                                                       T
                                           
                                                     
              z(t)  x1 (t ) x2 (t ) x1 (t ) x 2 (t )
                  
                                                    
     And the displacement of bus body mass M1 as
        the output of the system.


Saturday, September     PMDRMFRCIED                    115
29, 2012
Saturday, September   PMDRMFRCIED   116
29, 2012
Constant value
• Bus body mass, M1 = 2500 kg
• Suspension mass, M2 = 320 kg
• Spring constant of suspension system, K1 =
  80,000 N/m
• Spring constant of wheel and tire, K2 =
  500,000 N/m
• Damping constant of suspension system, B1
  = 350 Ns/m
• Damping constant of wheel and tire, B2 =
  15,020 Ns/m
Saturday, September       PMDRMFRCIED      117
29, 2012
Solution
• Free body diagram for M1
     – Forces on M1 due to motion of M1
             K1X1
             M1s2X1                     u
                            M1
             B1sX1

     – Forces on M1 due to motion of M2
              K1X2
                             M1
              B1sX2

     – All forces on M1
            K1X1                            K1X2
            M1s2X1          M1              u
            B1sX1                           B1sX2

Saturday, September       PMDRMFRCIED               118
29, 2012
Solution
• Free body diagram for M2
     – Forces on M2 due to motion of M2
             K2X2
             M2s2X2                     K1X2
                            M2
             B2sX2                      B1sX2

     – Forces on M2 due to motion of M1
                                         K1X1
                             M2
                                         B1sX1
     – All forces on M2
      (K1+K2)X2
      M2s2X2                M2            K1X1
      (B1+B2)sX2
                                          B1sX1
Saturday, September       PMDRMFRCIED             119
29, 2012
Solution
    • State variables
                                                             
z1  x1; z2  x2 ; z3  x1; z4  x 2
    • Derivative state variables
                                                          
z1  x1  z3 ; z 2  x 2  z4 ; z 3  x1; z 4  x 2
    Saturday, September        PMDRMFRCIED                        120
    29, 2012
Solution
• Total force for M1
               dx2               d 2 x1    dx1
u  K1 x2  B1      K1 x1  M 1 2  B1
                dt                dt        dt
  2
d x1                          dx1        dx2
    2
       32 x1  32 x2  0.14      0.14       0.0004u
 dt                           dt          dt

z 3  32 z1  32 z2  0.14 z3  0.14 z4  0.0004u


Saturday, September    PMDRMFRCIED                   121
29, 2012
Solution
• Total force for M2
                                        2
           dx1                          d x2               dx2
K1 x1  B1      ( K1  K 2 ) x2  M 2        ( B1  B2 )
           dt                            dt                 dt
d 2 x2                                 dx1           dx2
    2
        250 x1  1812.5 x2  1.094         48.031
 dt                                    dt             dt

z 4  250 z1  1812.5 z 2  1.094 z3  48.031z 4


Saturday, September      PMDRMFRCIED                      122
29, 2012
Solution
• State space representation
       0      0         1    0   z1   0 
      0      0         0        z   0 
                              1  2  
  z                                            u
       32   32       0.14 0.14   z3  0.0004
                                             
       250 1812.5 1.094 48.031  z 4   0 
                  z1 
                 z 
  y  1 0 0 0   2
                  z3 
                  
                  z4 
Saturday, September    PMDRMFRCIED                 123
29, 2012
Tutorial 1 : Number 5
• Figure shows a mechanical system
  consisting of mass M1 and M2, damper
  constant B, spring stiffness K1 and K2.
  When force f(t) acts on mass M1, it
  moves to position x1(t) while mass M2
  moves to position x2(t). Find the state
  space representation of the system
  using x1(t), x2(t) and their first
  derivatives as state variables. Let x2(t)
  be the output.
Saturday, September   PMDRMFRCIED         124
29, 2012
Mechanical system consist of 2
   mass, 2 spring and 1 damper

        f(t)               X1                  X2

           K1                   B               K2
                      M1                  M2




Saturday, September         PMDRMFRCIED              125
29, 2012
Mechanical system consist of 2
   mass, 2 spring and 1 damper
• State variables and their derivatives :-
                                    
     z1 (t )  x1 (t )  z1 (t )  x1 (t )  z 2 (t )
                                       
     z 2 (t )  x1 (t )  z 2 (t )  x1 (t )
                                     
     z3 (t )  x2 (t )  z3 (t )  x 2 (t )  z 4 (t )
                                       
     z 4 (t )  x 2 (t )  z 4 (t )  x2 (t )
     input  u (t )  f (t )
     output  y (t )  x2 (t )
Saturday, September                PMDRMFRCIED           126
29, 2012
Mechanical system consist of 2
   mass, 2 spring and 1 damper
• Draw the free body diagram

              K1 x1
                         
                                            f (t )
              M 1 x1                           
                                M1
                 
              B x1
                                            B x2

               K 2 x2
                                                
                         
               M 2 x2
                               M2           B x1
                     
               B x2

Saturday, September           PMDRMFRCIED            127
29, 2012
Mechanical system consist of 2
   mass, 2 spring and 1 damper
• Differential equation in mass M1
                                                       
      f (t )  M 1 x1  B x1  K1 x1  B x2
                                           
      f (t )  M 1 x1  B( x1  x2 )  K1 x1      (1)
• Differential equation in mass M2
                                                   
       0  M 2 x2  B x2  K 2 x2  B x1
                                       
       0  M 2 x2  B( x2  x1 )  K 2 x2      (2)
Saturday, September                     PMDRMFRCIED           128
29, 2012
Mechanical system consist of 2
   mass, 2 spring and 1 damper
• Substitute all state variables and their first
  derivatives in equation (1) and (2) yields
              B  B  K1              f (t )
         x1      x1     x2     x1 
                M1      M1      M1      M1
                 B       B       K       1
         z2        z2     z 4  1 z1     u      (3)
                  M1      M1      M1      M1
              B      B  K2
         x2      x2     x1     x2
                M2      M2      M2
                 B       B       K
         z4        z4     z 2  2 z3      (4)
                  M2      M2      M2
          
         z1  z 2      (5)
          
         z 3  z 4      ( 6)
Saturday, September                 PMDRMFRCIED                129
29, 2012
Mechanical system consist of 2
   mass, 2 spring and 1 damper
• Rearrange equation 3, 4, 5 and 6 in matrix
  form     0    1     0    0 
                                        0 
          z         1   K1      z  B              B   1
                    z   M                  0          z   1 
               
                                      M1             M1   2   M 
               z   2    1
                                                         .        1 u
                    z3   0         0      0        1   z3     0
                     0            B      K2        B    
                                                          z4   0 
                    z4  
                                  M2      M2       M2 
                                                                   
                              z1 
                             z 
               y  0 0 1 0. 2 
                              z3 
                              
                              z4 
Saturday, September               PMDRMFRCIED                          130
29, 2012
Tutorial 1 : Number 6
• Represent the translational mechanical
  system shown in figure in state space
  where x3(t) is the output and f(t) is the
  input.
                      X1                X2          X3
      K1                   B1             K2             B2
                M1                 M2          M3
       f(t)


Saturday, September             PMDRMFRCIED                   131
29, 2012
Example : 3M, 2K and 2B
• Represent the translational mechanical
  system shown in figure in state space
  where x3(t) is the output and f(t) is the
  input.




Saturday, September   PMDRMFRCIED             132
29, 2012
Example : 3M, 2K and 2B
•   K1 = K2 = 1 N/m
•   M1 = M2 = M3 = 1 kg
•   B1 = B2 = 1 N-s/m
•   Find the state space representation of the
    system using x1, x2, x3 and their first
    derivatives as state variables.
                      
                 z1  x1 ; z2  x1 ; z3  x2 ;
                                           
                 z4  x2 ; z5  x3 ; z6  x3
Saturday, September             PMDRMFRCIED      133
29, 2012
Example : 3M, 2K and 2B
• Draw the free body diagram
                                               
                M 1 x1                       B1 x2
                B1 x1
                      
                                  M1         f (t )
                K1 x1
                      
                                                 
              M 2 x2
                  
                                             B1 x1
              B1 x2               M2         K 2 x3
              K 2 x2
                      
               M 3 x3

               B2 x3
                      
                                  M3         K 2 x2
               K 2 x3
Saturday, September            PMDRMFRCIED            134
29, 2012
Example : 3M, 2K and 2B
• Writing the equations of motion

                                     
M 1 x1  B1 x1  K1 x1  B1 x2  f (t )    (1)
                                         
M 2 x2  B1 x2  K 2 x2  B1 x1  K 2 x3    (2)
                        
M 3 x3  B2 x3  K 2 x3  K 2 x2    (3)
Saturday, September           PMDRMFRCIED       135
29, 2012
Example : 3M, 2K and 2B
• Substitute the value of K, M and B.
• Rearrange equation (1), (2) and (3)
                                    
           x1   x1  x1  x2  f
                                
           x2  x1  x2  x2  x3
                        
           x3   x3  x3  x2
Saturday, September           PMDRMFRCIED   136
29, 2012
Example : 3M, 2K and 2B
• From the state variables
                             
        z1  x1  z1  x1  z 2
                            
        z 2  x1  z 2  x1   z 2  z1  z 4  f
                                 
        z 3  x 2  z 3  x2  z 4
                            
        z 4  x2  z 4  x2  z 2  z 4  z 3  z 5
                                 
        z5  x3  z5  x3  z6
                            
        z6  x3  z6  x3   z6  z5  z3
        y  x3  z5
Saturday, September                   PMDRMFRCIED     137
29, 2012
Example : 3M, 2K and 2B
• In vector matrix form
        0 1 0 0 0 0  0 
        1  1 0 1 0 0  1
                          
       0 0 0 1 0 0  0 
   z                    z    f (t )
        0 1  1  1 1 0  0 
        0 0 0 0 0 1  0 
                          
        0 0 1 0  1  1 0
                          
   y  0 0 0 0 1 0z
Saturday, September   PMDRMFRCIED           138
29, 2012
Modeling of Electro-Mechanical System
• NASA flight simulator robot arm with
  electromechanical control system components




Saturday, September   PMDRMFRCIED               139
29, 2012
Modeling of Electro-Mechanical System

• Armature Controlled DC Motor




Saturday, September   PMDRMFRCIED     140
29, 2012
Armature Controlled DC Motor




Saturday, September   PMDRMFRCIED   141
29, 2012
DC motor armature control
• The back electromotive force(back emf),
  VB
                        d m (t )
              VB (t ) 
                           dt
                              d m (t )
              VB (t )  K B .              (1)
                                dt
              K B  Back _ emf _ cons tan t
Saturday, September      PMDRMFRCIED                142
29, 2012
DC motor armature control
  • Kirchoff’s voltage equation around the
    armature circuit
ea (t )  ia (t ) Ra  Vb (t )
                           d m (t )
ea (t )  ia (t ) Ra  K b              (2)
                             dt
ia  armature _ current
 m  angular _ displaceme nt _ of _ the _ armature
Ra  armature _ resis tan ce
ignore _ La
 Saturday, September         PMDRMFRCIED         143
 29, 2012
DC motor armature control
• The torque, Tm(t) produced by the motor
 Tm (t )  ia (t )
 Tm (t )  K t ia (t )
               d 2 m      d m
 Tm (t )  J m     2
                       Dm         (3)
                dt          dt
 K t  Torque _ cons tan t
 J m  equivalent _ inertia _ by _ the _ motor
 Dm  equivalent _ viscous _ density _ by _ the _ motor
Saturday, September      PMDRMFRCIED                 144
29, 2012
DC motor armature control
• Solving equation (3) for ia(t)


          J m d  m Dm d m
                      2
ia (t )         2
                              (4)
          K t dt     K t dt


Saturday, September       PMDRMFRCIED   145
29, 2012
DC motor armature control
 • Substituting equation (4) into equation (2)
   yields

              J m d 2 m Dm d m           d m
ea (t )  Ra          2
                                     Kb
              K t dt       K t dt           dt
           Ra J m  d 2 m  Ra Dm         d m
ea (t )  
           K  dt . 2     K      K b .
                                            dt    (5)
           t                  t         
 Saturday, September    PMDRMFRCIED                    146
 29, 2012
DC motor armature control
• Define the state variables, input and ouput
                       x1   m    (6a )
                           d m
                      x2            (6b)
                             dt
                      u  ea (t )
                       y  0.1 m
• Substituting equation (6) into equation (5)
  yields       Ra J m  dx2  Ra Dm 
                  ea (t )  
                             K . dt   K  K b .x2    (7)
                                                
                             t          t      
Saturday, September             PMDRMFRCIED                        147
29, 2012
DC motor armature control
• Solving for x2 dot yields,


                      Ra Dm     
           ea (t )  
                      K  K b .x2
                                 
     dx2
                         t     
      dt               Ra J m 
                      
                       K     
                       t 
     dx2  K t              Dm K b K t 
         
           R J .ea (t )   J  R J .x2
      dt  a m  
                            
                             m       a m 
                                           

     dx2     1        Kb Kt          Kt 
              Dm 
                              .x2  
                                      R J .ea (t )    (8)
                                             
      dt    Jm           Ra          a m
Saturday, September            PMDRMFRCIED                        148
29, 2012
DC motor armature control
• Using equation (6) and (8), the state
  equations are written as

dx1 d m
          x2
dt    dt
dx2     1       Kt Kb        Kt 
         Dm 
                      .x2  
                                  .ea (t )
                              R J 
 dt    Jm        Ra          a m
Saturday, September   PMDRMFRCIED         149
29, 2012
DC motor armature control
• Assuming that the output o(t) is 0.1 the
  displacement of the armature m(t) as x1.
  Hence the output equation is
                       y  0.1x1
• State space representation in vector
  matrix form are
             0         1           
                                          x1   K 
                                                   0
            x1  
                   1     K t K b .    t .ea (t )
           x  0  J  Dm  R   x2   R J 
                                    
           2        m        a             a m
                        x1 
           y  0.1 0. 
                        x2 
Saturday, September             PMDRMFRCIED                   150
29, 2012
Tutorial 1 : Number 7
• The representation of the positioning system
  using an armature-controlled dc motor is shown
  in figure.
• The input is the applied reference voltage, r(t)
  and the output is the shaft’s angular position,
  o(t).
• The dynamic of the system can be described
  through the Kirchoff equation for the armature
  circuit, the Newtonian equation for the
  mechanical load and the torque field current
  relationship.
Saturday, September   PMDRMFRCIED                151
29, 2012
Figure : DC motor armature control




Saturday, September   PMDRMFRCIED   152
29, 2012
Example : ex-exam question
• The Newtonian equation for the
  mechanical load is
                                  
              J  o (t )    o (t )   (t )
• The back e.m.f voltage induced in the
  armature circuit, eb(t) is proportional to
  the motor shaft speed,
                                        
                           eb  K b  o
Saturday, September             PMDRMFRCIED      153
29, 2012
Example : ex-exam question
• A potentiometer was installed to measure
  the motor output position. Its output
  voltage, v(t) is then compared with the
  system reference input voltage, r(t)
  through an op-amp.
• Determine the complete state-space
  representation of the system by
  considering the following state variables.

Saturday, September   PMDRMFRCIED          154
29, 2012
Example : ex-exam question
                                                        x1(t)  ia (t)
• State variables :-                                    x 2 (t)   o (t)
                                                                  
                                                        x 3 (t)   o (t)
• State variables derivative
                               dia (t) 
                      x 1 (t)             ia
                                  dt
                               d 
                                    o (t)
                      x 2 (t)               o (t)
                                   dt
                               d 2 o (t) 
                      x 3 (t)         2
                                              o (t)
                                   dt
Saturday, September              PMDRMFRCIED                                155
29, 2012
Example : ex-exam question
• Mechanical load
                    
J o (t)    o (t)   (t)  K t ia
    
J x 3  x 3  K t x1
      Kt      
x3       x1  x 3          (1)
        J       J
Saturday, September       PMDRMFRCIED   156
29, 2012
Example : ex-exam question
• Electrical (armature) circuit
• Using Kirchoff Voltage Law
                dia
            uL      Ria  eb
                dt
            but
                          
            eb  K b  o ( given)
                  dia               
           u  L       Ria  K b  o
                  dt
                      
            u  L x1  Rx1  K b x3
                 R     Kb     1
            x1   x1     x3  u        (2)
                  L     L      L
Saturday, September                 PMDRMFRCIED       157
29, 2012
Example : ex-exam question
• From the state variable defination
     x2   o
                     
            x 2   o  x3      (3)
            For _ the _ input _ part
           u  r v
           u  r  K s o
           u  r  K s x2        (4)
Saturday, September         PMDRMFRCIED       158
29, 2012
Example : ex-exam question
• Substituting (4) into (2)

     R     Kb      1
x1   x1     x3  (r  K s x2 )
      L     L       L
     R     Ks      Kb     1
x1   x1     x2     x3  r      (5)
      L     L        L     L

Saturday, September   PMDRMFRCIED       159
29, 2012
Example : ex-exam question
• Writing equations (1), (3) and (5) in the
  vector matrix form gives :-

   R
  
                      
                        Ks
                                
                                  Kb          1
x1  
           L             L             x1   L 
                                   L  
x2    0             0         1 . x2    0  r
    KT                           
                                       
 x3                 0           x3   0 
   J                           J 
                                               
                                                
Saturday, September     PMDRMFRCIED                160
29, 2012
Example : ex-exam question
• The output


                      x1 
                     x 
    y   o  0 1 0 2 
                      x3 
                      
Saturday, September   PMDRMFRCIED   161
29, 2012
Modelling of Electro-Mechanical
             System
• Field Controlled DC
                                                                       +
                                                          Rf
                                                                             ef (t)
  Motor                                     if (t)
                                                                                      -
                                                               Lf                     Gelung Medan
                      Ra               La
                                                 +
                ea
                                                     Ba
                                             Ja
                             ia
                                                 -                                        TL(t)
                                                               Tm(t)
                       Gelung Angker                                        m (t )
                           Tetap



                       RAJAH 7.11 : MOTOR SERVO A.T. TERUJA BERASINGAN
                                     DALAM KAWALAN MEDAN
Saturday, September                PMDRMFRCIED                                                       162
29, 2012
DC motor field control
• For field circuit
                                  di f
          e(t )  i f R f  L f             (1)
                                  dt
• For mechanical load, torque

              d o    d o
                      2
   T (t )  J      B         (2)
               dt      dt
Saturday, September        PMDRMFRCIED               163
29, 2012
DC motor field control
• For torque and field current relationship
                      T (t )  i f (t )
              T (t )  K t i f (t )    (3)
• Define the state variables, input and
  output        x   (t )    (4)
                         1     o

                              d o (t )
                        x2                (5)
                                  dt
                        x3  i f (t )    (6)
                        u  e(t )
                        y   o (t )
Saturday, September            PMDRMFRCIED          164
29, 2012
DC motor field control
• From equation (4) and (5), we can
  determine the first state equation as :
                           d o
        x1 (t )  x2 (t )          ( 7)
                             dt
• Another two state equations are :
                          d 2 o
                      x2  2    (8)
                            dt
                          di f
                      x3          (9)
                           dt
Saturday, September          PMDRMFRCIED      165
29, 2012
DC motor field control
• Substituting x3 and x3 dot into equation (1)
  yields                              
               e(t )  x3 R f  L f x3
• Substituting equation (3) into equation (2)
                       d o  d o
  yields       2
                      J 2 B       Kt i f
                        dt    dt
Saturday, September         PMDRMFRCIED      166
29, 2012
DC motor field control
• Substituting x2 dot, x2 and x3, hence
                          
                      J x2  Bx2  Kt x3
• Rewrite equations
                              Rf          1
                      x3   x3     e(t )
                                Lf         Lf
                           B     Kt
                      x2   x2     x3
                            J     J
Saturday, September            PMDRMFRCIED      167
29, 2012
DC motor field control
• Matrix form
                             
                                      
       
           0 1     0          0       
       x  0  B   Kt        x   0   u
               J   J          1       
                   Rf                 
           0 0                Lf
                                        
                                         
           
                   Lf        
                              
       y  1 0 0x
Saturday, September   PMDRMFRCIED             168
29, 2012
Block diagrams
• The block diagram is a useful tool for
  simplifying the representation of a
  system.
• Simple block diagrams only have one
  feedback loop.
• Complex block diagram consist of more
  than one feedback loop, more than 1 input
  and more than 1 output i.e. inter-coupling
  exists between feedback loops
Saturday, September       PMDRMFRCIED     169
29, 2012
Block diagrams
• Integrator
                                                 x2   x1dt
                      x1




• Amplifier or gain                 x1                          x2 = Kx1

                                              K
                               x1
                                              +                   x4 = x1-x2+x3
• Summer                       x2        -
                               x3
                                          +
Saturday, September        PMDRMFRCIED                                     170
29, 2012
Signal flow graphs
• Having the block diagram simplifies the
  analysis of a complex system.
• Such an analysis can be further simplified
  by using a signal flow graphs (SFG) which
  looks like a simplified block diagram
• An SFG is a diagram which represents a
  set of simultaneous equation.
• It consist of a graph in which nodes are
  connected by directed branches.
Saturday, September    PMDRMFRCIED         171
29, 2012
Signal flow graphs
• The nodes represent each of the system
  variables.
• A branch connected between two nodes
  acts as a one way signal multiplier: the
  direction of signal flow is indicated by an
  arrow placed on the branch, and the
  multiplication factor(transmittance or
  transfer function) is indicated by a letter
  placed near the arrow.
Saturday, September    PMDRMFRCIED              172
29, 2012
Signal flow graphs
• A node performs two functions:
     1. Addition of the signals on all incoming
        branches
     2. Transmission of the total node signal(the
        sum of all incoming signals) to all outgoing
        branches




Saturday, September    PMDRMFRCIED                     173
29, 2012
Signal flow graphs
• There are three types of nodes:
     1. Source nodes (independent nodes) – these
        represent independent variables and have
        only outgoing branches. u and v are source
        nodes
     2. Sink nodes (dependent nodes) - these
        represent dependent variables and have
        only incoming branches. x and y are source
        nodes
     3. Mixed nodes (general nodes) – these have
        both incoming and outgoing branch. W is a
        mixed node.
Saturday, September     PMDRMFRCIED             174
29, 2012
Signal flow graphs
• x2 = ax1

                  x1      a          x2 = ax1




Saturday, September    PMDRMFRCIED              175
29, 2012
Signal flow graphs
• w = au + bv
• x = cw
• y = dw
                          a        w        c   x
        u

       v              b                     d   y

Saturday, September           PMDRMFRCIED           176
29, 2012
Signal flow graphs
• x = au + bv +cw

           u
                          a          x
                      c                     1     x
          w

            v             b         Mixed       Sink
                                    node        node
Saturday, September           PMDRMFRCIED              177
29, 2012
Signal flow graphs
• A path is any connected sequence of
  branches whose arrows are in the same
  direction
• A forward path between two nodes is one
  which follows the arrows of successive
  branches and in which a node appears
  only once.
• The path uwx is a forward path between
  the nodes u and x

Saturday, September    PMDRMFRCIED      178
29, 2012
Signal flow graphs
• Series path (cascade nodes) – series path
  can be combined into a single path by
  multiplying the transmittances
• Path gain – the product of the
  transmittance in a series path
• Parallel paths – parallel paths can be
  combined by adding the transmittances
• Node absorption – a node representing a
  variable other than a source or sink can be
  eliminated
Saturday, September    PMDRMFRCIED         179
29, 2012
Signal flow graphs
• Feedback loop – a closed path which
  starts at a node and ends at the same
  node.
• Loop gain – the product of the
  transmittances of a feedback loop




Saturday, September    PMDRMFRCIED        180
29, 2012
Signal flow graphs
                   simplification
Original graph                      Equivalent graph




        a                 b                     ab
x                     y       z             x          z

Saturday, September           PMDRMFRCIED                  181
29, 2012
Signal flow graphs
                       simplification
    Original graph                  Equivalent graph

                      a

                                             (a+b)
x                              y
                                       x               y
                          b
    Saturday, September       PMDRMFRCIED                  182
    29, 2012
Signal flow graphs
                   simplification
Original graph                      Equivalent graph



   w                                          ac
            a                 z       w
                          c                            z
 x                    y               x         bc
             b
Saturday, September           PMDRMFRCIED              183
29, 2012
Block diagram of feedback
              system


    R                 E                 C
                              G

                      B
                              H

Saturday, September       PMDRMFRCIED       184
29, 2012
Block diagram of feedback system
•   R=reference input
•   E=actuating signal
•   G=control elements and controlled system
•   C=controlled variable
•   B=primary feedback
•   H=feedback elements
•   C = GE
•   B = HC
•   E = R-B
Saturday, September   PMDRMFRCIED          185
29, 2012
Successive reduction of SFG
first                                 second
• 4 nodes                             • Node B eliminated



R 1                   E G       C       R 1       E G C

              -1
                            H
                      B                                -H
Saturday, September             PMDRMFRCIED                 186
29, 2012
Successive reduction of SFG
third                             fourth
• Node E eliminated, self         • Self loop eliminated
   loop of value -GH

R            G        C             R                 C

                                          G/(1+GH)

                      -GH
Saturday, September         PMDRMFRCIED                    187
29, 2012
SIGNAL FLOW GRAPHS OF
           STATE EQUATIONS
    • demonstrate how to draw signal flow
      graphs from state equations.
    • Consider the following state and output
      equations:

x1  2 x1  5x2  3x3  2r          (1a)

x2  6 x1  2 x2  2 x3  5r          (1b)

x3  x1  3x2  4 x3  7r          (1c)
y  4x1  6x2  9x3          (1d)
 Saturday, September   PMDRMFRCIED                188
 29, 2012
SIGNAL FLOW GRAPHS OF
            STATE EQUATIONS
       • Step 1 : Identify three nodes to be the
         three state variables, , and three nodes,
         placed to the left of each respective
         state variables. Also identify a node as
         the input, r, and another node as the
         output, y.
R(s)                                                                 Y(s)
              sX3 (s)    X3 (s)   sX (s)   X2 (s)   sX (s)   X (s)
                                    2                 1       1




   Saturday, September                PMDRMFRCIED                      189
   29, 2012
SIGNAL FLOW GRAPHS OF
             STATE EQUATIONS
   • Step 2 : Interconnect the state variables
     and their derivatives with the defining
     integration, 1/s.



                     1                    1                    1
                     s                    s                    s
R(s)                                                                       Y(s)
              sX (s)     X (s)   sX (s)       X (s)   sX (s)       X (s)
                3         3        2           2        1           1


   Saturday, September             PMDRMFRCIED                             190
   29, 2012
SIGNAL FLOW GRAPHS OF
            STATE EQUATIONS
  • Step 3 : Using Eqn (1a), feed to each node
    the indicated signals.
                            2




                    1                   1                    1
                    s                   s            -5      s
R(s)                                                                       Y(s)
                        X (s)               X2 (s)        sX (s)   X (s)
              sX3 (s)    3    sX2 (s)                       1       1

                                                             2


                                        3
  Saturday, September            PMDRMFRCIED                               191
  29, 2012
SIGNAL FLOW GRAPHS OF
                 STATE EQUATIONS
       • Step 4 : Using Eqn (1b), feed to each node
         the indicated signals.
                                  2

                          5

                      1                   1                   1
                      s          2        s       -5          s
R(s)                                                                        Y(s)
                              X (s)   sX (s)   X (s)       sX (s)   X (s)
                sX3 (s)        3        2       2            1       1
                                         -2                   2


                                          3           -6
       Saturday, September              PMDRMFRCIED                          192
       29, 2012
SIGNAL FLOW GRAPHS OF
            STATE EQUATIONS
• Step 5 : Using Eqn (1c), feed to each node
  the indicated signals.
                                  2

                          5

                      1                  1                 1
            7         s          2       s        -5       s
  R(s)                                                                    Y(s)
          sX (s)              X (s)   sX (s)   X2 (s)   sX (s)   X1 (s)
            3                  3        2                 1
                      -4                 -2                2


                                 -3       3        -6


                                          1
Saturday, September                    PMDRMFRCIED                               193
29, 2012
SIGNAL FLOW GRAPHS OF
            STATE EQUATIONS
• Step 6 : Finally, use Eqn (1d) to complete
  the signal flow2 graph. 9
                          5                                  6

                      1                   1                 1
            7         s          2        s         -5      s          -4
  R(s)                                                                      Y(s)
           sX (s)             X (s)    sX (s)   X2 (s)    sX (s)   X1 (s)
             3                 3         2                  1
                      -4                  -2                 2


                                  -3       3         -6


                                           1
Saturday, September                    PMDRMFRCIED                                 194
29, 2012
Example 7
• Draw a signal-flow graph for each of the
  following state equations :
                         0    1      0   x1  0
                
                x(t )  0    0        . x   0 r (t )
                                      1   2  
                          2  4  6  x3  1
                                          
                                  x1 
                y (t )  1 1 0. x2 
                                  
                                  x3 
                                  
Saturday, September            PMDRMFRCIED                     195
29, 2012
Solution
• State and output equations
    
   x1 (t )  x2 (t )
     
   x2 (t )  x3 (t )
     
   x3 (t )  2 x1 (t )  4 x2 (t )  6 x3 (t )  r (t )
   y (t )  x1 (t )  x2 (t )
Saturday, September     PMDRMFRCIED                  196
29, 2012
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Chap1see4113

  • 1. Chapter 1 Mathematical Modeling of Dynamic Systems in State Space Saturday, September PMDRMFRCIED 1 29, 2012
  • 2. Introduction to State Space analysis • Two approaches are available for the analysis and design of feedback control systems – Classical or Frequency domain technique – Modern or Time domain technique Saturday, September PMDRMFRCIED 2 29, 2012
  • 3. Introduction to State Space analysis • Classical technique is based on converting a system’s differential equation to a transfer function • Disadvantage – Can be applied only to Linear Time Invariant system – Restricted to Single Input and Single output system • Advantage – Rapidly provide stability and transient response information Saturday, September PMDRMFRCIED 3 29, 2012
  • 4. Introduction to State Space analysis • Modern technique or state space approach is a unified method for modeling, analyzing and designing a wide range of systems • Advantages : – Can be used to nonlinear system – Applicable to time varying system – Applicable to Multi Input and Multi Output system – Easily tackled by the availability of advanced digital computer Saturday, September PMDRMFRCIED 4 29, 2012
  • 5. Time varying • A time-varying control system is a system in which one or more of the parameters of the system may vary as a function of time • Dynamic system: input, state, output and initial condition Saturday, September PMDRMFRCIED 5 29, 2012
  • 6. The state variables of a dynamic system • The state of a system is a set of variables whose values, together with the input signals and the equations describing the dynamics , will provide the future state and output of the system • The state variables describe the present configuration of a system and can be used to determine the future response, given the excitation inputs and the equations describing the dynamics. Saturday, September PMDRMFRCIED 6 29, 2012
  • 7. The State Space Equations  x(t )  Ax (t )  Bu (t ) y (t )  Cx(t )  Du(t )  x(t )  derivative _ of _ the _ state _ vector x(t )  state _ vector y (t )  output _ vector u (t )  input _ of _ control _ vector A  system _ matrix B  input _ matrix C  output _ matrix D  feedfoward _ matrix Saturday, September PMDRMFRCIED 7 29, 2012
  • 8. Two types of equation • State equation  x(t )  Ax (t )  Bu (t ) • Output equation y(t )  Cx(t )  Du(t ) Saturday, September PMDRMFRCIED 8 29, 2012
  • 9. Terms • State equations: a set of n simultaneous, first order differential equations with n variables, where the n variables to be solved are the state variables • State space: The n-dimensional space whose axes are the state variables • State space representation: A mathematical model for a system that consists of simultaneous, first order differential equations and output equation Saturday, September PMDRMFRCIED 9 29, 2012
  • 10. Terms • State variables: the smallest set of linearly independent system variables such that the value of the members of the set • State vector: a vector whose elements are the state variables Saturday, September PMDRMFRCIED 10 29, 2012
  • 11. Modeling of Electrical Networks Voltage-current, voltage-charge, and impedance relationships for capacitors, resistors, and inductors Saturday, September PMDRMFRCIED 11 29, 2012
  • 12. An RLC circuit Saturday, September PMDRMFRCIED 12 29, 2012
  • 13. State variable characterization • The state of the RLC system described a set of state variables x1 and x2 • X1 = capacitor voltage = vc(t) • X2 = inductor current = iL(t) • This choice of state variables is intuitively satisfactory because the stored energy of the network can be described in terms of these variables 1 2 1 E  LiL  Cvc 2 2 2 Saturday, September PMDRMFRCIED 13 29, 2012
  • 14. Utilizing Kirchhoff’s current law • At the junction • First order differential equation • Describing the rate of change of capacitor voltage dvc ic  C  u (t )  iL dt Saturday, September PMDRMFRCIED 14 29, 2012
  • 15. Utilizing Kirchhoff’s voltage law • Right hand loop • Provide the equation describing the rate of change of inductor current diL L   Ri L  vc dt • Output of the system, linear algebraic equation vo  RiL (t ) Saturday, September PMDRMFRCIED 15 29, 2012
  • 16. State space representation • A set of two first order differential equation and output signal in terms of the state variables x1 and x2 dx1 1 1   x2  u (t ) dt C C dx2 1 R   x1  x2 dt L L y (t )  vo (t )  Rx 2  1    0   x   1   x x   1    C . 1   .u   C  x   1  R   x2   0   2   L L  x1  y  0 R .   x2  Saturday, September PMDRMFRCIED 16 29, 2012
  • 17. Example 1 : RL serial network • Figure below shows an RL serial network with an input voltage vi(t) and voltage drop at inductance, L as an output voltage vo(t). Form a state space model for this system using the current i(t) in the loop as the state variable. Saturday, September PMDRMFRCIED 17 29, 2012
  • 18. Modeling of Electrical Networks • RL serial network – first order system Saturday, September PMDRMFRCIED 18 29, 2012
  • 19. RL serial network • Write the loop equation for the system using Kirchhoff’s voltage law, Vi (t )  VR (t )  VL (t )  VR (t )  Vo (t ) di (t ) VL (t )  L  Vo (t ) dt VR (t )  i (t ) R di (t ) Vi (t )  i (t ) R  L dt Saturday, September PMDRMFRCIED 19 29, 2012
  • 20. RL serial network • State variable is given only one, therefore the system is a first order system • A state equation involving i is required di (t ) Vi (t )  i (t ) R  L dt di (t ) L  i (t ) R  Vi (t ) dt di (t ) R 1   i (t )  Vi (t ) dt L L   R 1 i (t )   i (t )   Vi (t )  L L Saturday, September PMDRMFRCIED 20 29, 2012
  • 21. RL serial network • The output equation, Vi (t )  VR (t )  VL (t )  VR (t )  Vo (t ) Vo (t )  VR (t )  Vi (t ) Vo (t )  i (t ) R  Vi (t ) y (t )   R i (t )  1Vi (t ) Saturday, September PMDRMFRCIED 21 29, 2012
  • 22. Example 2 : RC serial network • Figure below shows an RC circuit with input voltage vi(t) and output voltage at resistor ie vo(t). Form a state space model for this system using the voltage vc(t) across the capacitor as the state variable R V0 VR Vi i VC C Saturday, September PMDRMFRCIED 22 29, 2012
  • 23. RC serial network • Write the equations for the system using Kirchhoff’s voltage law, vi (t )  vR (t )  vc (t )  vc (t )  vo (t )      (1) for _ the _ capacitor dvc (t ) i (t )  C      (2) dt for _ the _ resistor vo (t )  i (t ) R      (3) Saturday, September PMDRMFRCIED 23 29, 2012
  • 24. RC serial network • State variable is given only one • Therefore the system is a first order system • Therefore a state equation involving vc is required • Combine equation (2) and (3) yields vo (t ) dvc (t )  i (t )  C R dt dvc (t ) vo (t )  RC      (4) dt Saturday, September PMDRMFRCIED 24 29, 2012
  • 25. RC serial network • Eliminate vo(t) from equation (4) and combine with equation (1) and rearrange gives vi (t )  vc (t )  vo (t ) dvc (t ) vi (t )  vc (t )  RC dt dvc (t ) RC  vc (t )  vi (t ) dt dvc (t )  1 1  vc (t )   vc (t )  vi (t )    (5) dt RC RC Saturday, September PMDRMFRCIED 25 29, 2012
  • 26. RC serial network • Output of the system vo (t )  vc (t )  vi (t )      (6) • Rearrange equation (5) and (6) in matrix form yields   1   1  v c (t )    vc (t )    vi (t )  RC   RC  y (t )   1vc (t )  1vi (t ) Saturday, September PMDRMFRCIED 26 29, 2012
  • 27. RC serial network x(t )  state _ vector  vc (t ) • Where,   x(t )  derivative _ state _ vector  v c (t ) u (t )  input _ vector  vi (t ) y (t )  output _ vector  vo (t )  vr (t ) 1 A  state _ matrix   RC 1 B  input _ matrix  RC C  ouput _ matrix  1 D  direct _ transmission _ matrix  1 Saturday, September PMDRMFRCIED 27 29, 2012
  • 28. Modeling of Electrical Networks • Consider RLC serial network • RLC serial network – second order system Saturday, September PMDRMFRCIED 28 29, 2012
  • 29. State Variables and output • Select two state variables, x1 (t )  q (t ) x2 (t )  i (t ) output  y (t )  VL (t ) input  u (t )  Vi (t ) Saturday, September PMDRMFRCIED 29 29, 2012
  • 30. Loop equation • Using Kirchoff’s Voltage Law, vi (t )  vR (t )  vL (t )  vc (t ) di (t ) 1 L  Ri (t )   i (t )dt  vi (t ) dt C Saturday, September PMDRMFRCIED 30 29, 2012
  • 31. Converting to charge • Using equation, dq(t ) i (t )  dt 2 d q(t ) dq(t ) 1 L 2 R  q(t )  vi (t ) dt dt C Saturday, September PMDRMFRCIED 31 29, 2012
  • 32. Derivatives of state vector x1 (t )  q (t )  dq (t ) x1 (t )   i (t )  x2 (t ) dt x2 (t )  i (t )  di (t ) x2 (t )  dt Saturday, September PMDRMFRCIED 32 29, 2012
  • 33. State equation • First state equation  dq(t ) x1 (t )   i(t )  x2 (t ) dt • Second state equation, using q (t )   i (t )dt di (t ) 1 L  Ri (t )   i (t )dt v(t ) dt C di (t ) q (t ) Ri (t ) v(t )    dt LC L L  1 R 1 x2 (t )   x1 (t )  x2 (t )  u (t ) LC L L Saturday, September PMDRMFRCIED 33 29, 2012
  • 34. State equation in matrix form  x(t )  Ax (t )  Bu (t )     0 1   x (t )   0  x1 (t )  x(t )     1 R    1    1 u (t )  x (t )  LC    x2 (t )   L  2   L  dq (t )    dt   0 1  q (t )  0  x(t )   di (t )    1 R    1  v(t )    i (t )       LC  L L  dt  Saturday, September PMDRMFRCIED 34 29, 2012
  • 35. Output equation • Output system is VL VL (t )  VR (t )  VC (t )  vi (t ) VL (t )  VC (t )  VR (t )  vi (t ) 1 V L(t )    i (t )dt  i (t ) R  vi (t ) C 1 VL (t )   q (t )  Ri (t )  vi (t ) C 1 VL (t )   x1 (t )  Rx 2 (t )  u (t ) C Saturday, September PMDRMFRCIED 35 29, 2012
  • 36. Output equation in matrix form y (t )  Cx(t )  Du(t )  1   x1 (t )  y (t )    R     1u (t )  C   x2 (t )  1  q(t ) VL (t )    R     1v(t )  C   i (t )  Saturday, September PMDRMFRCIED 36 29, 2012
  • 37. Change State Variables but output still same x1 (t )  VR (t ) x2 (t )  VC (t ) y (t )  VL (t ) u (t )  Vi (t ) Saturday, September PMDRMFRCIED 37 29, 2012
  • 38. Voltage formula for R, L and C VR (t )  i (t ) R 1 VC (t )   i (t )dt C di (t ) VL (t )  L dt Saturday, September PMDRMFRCIED 38 29, 2012
  • 39. Derivative of first state equation x1 (t )  VR (t )   v(t )  VR (t )  VC (t ) dVR (t ) di (t ) R x1` (t )  R dt dt L  R R R x1 (t )   VR (t )  VC (t )  v(t ) L L L  R R R x1` (t )   x1 (t )  x2 (t )  u (t ) L L L Saturday, September PMDRMFRCIED 39 29, 2012
  • 40. Derivative of second state equation x2 (t )  VC (t )  dVC (t ) 1 1 x2 (t )   i (t )  VR (t ) dt C RC  1 x2 (t )  x1 (t ) RC Saturday, September PMDRMFRCIED 40 29, 2012
  • 41. State equation in matrix form  x(t )  Ax (t )  Bu (t )  R R         x (t )   R   x1 (t )    L x(t )   L  1      L u (t )  x (t )  1 0   x2 (t )  0   2     RC   dVR (t )   R R   dt   L  L  VR (t )   R  x(t )    1     L  v(t )  VC (t )  0  dVC (t )    0    dt   RC  Saturday, September PMDRMFRCIED 41 29, 2012
  • 42. Output equation VL (t )  VR (t )  VC (t )  v(t ) VL (t )  VR (t )  VC (t )  v(t ) y (t )   x1 (t )  x2 (t )  u (t ) Saturday, September PMDRMFRCIED 42 29, 2012
  • 43. Output equation in matrix form y (t )  Cx(t )  Du(t )  x1 (t )  y (t )   1  1    1 u (t )  x2 (t ) VR (t )  VL (t )   1  1    1 v(t ) VC (t ) Saturday, September PMDRMFRCIED 43 29, 2012
  • 44. Example 3 : 2 loop • Find a state space representation if the output is the current through the resistor. • State variables VC(t) and iL(t) • Output is iR(t) • Input is Vi(t) Saturday, September PMDRMFRCIED 44 29, 2012
  • 45. Electrical network LRC L node 1 VL Vi iL VR R C iR iC VC Saturday, September PMDRMFRCIED 45 29, 2012
  • 46. Solution : Step 1 • Label all of the branch currents in the network. • iL(t), iR(t) and iC(t) Saturday, September PMDRMFRCIED 46 29, 2012
  • 47. Solution : Step 2 • Select the state variables by writing the derivative equation for all energy-storage elements i.e. inductor and capacitor 1 VC (t )   iC (t )dt C dVC (t )  iC (t )  C    (1) dt diL (t ) VL (t )  L    ( 2) dt Saturday, September PMDRMFRCIED 47 29, 2012
  • 48. Solution : Step 3 • Apply network theory, such as Kirchoff’s voltage and current laws to obtain iC(t) and VL(t) in terms of the state variable VC(t) and iL(t) • At node 1, iL (t )  iR (t )  iC (t )  iC (t )  iL (t )  iR (t ) 1 iC (t )   VC (t )  iL (t )    (3) R • Around the outer loop, Vi (t )  VL (t )  VC (t ) VL (t )  VC (t )  Vi (t )    (4) Saturday, September PMDRMFRCIED 48 29, 2012
  • 49. Solution : Step 4 • Substitute the result of equation (3) and equation (4) into equation (1) and (2) dVC (t ) 1 C   VC (t )  iL (t )    (7) dt R di (t ) L L  VC (t )  Vi (t )    (8) dt • Rearrange dVC (t ) 1 1  VC (t )  iL (t )    (9) dt RC C diL (t ) 1 1   VC (t )  Vi (t )    (10) dt L L Saturday, September PMDRMFRCIED 49 29, 2012
  • 50. Solution : Step 5 • Find the output equation 1 iR (t )  VC (t )    (11) R Saturday, September PMDRMFRCIED 50 29, 2012
  • 51. Solution : Step 6 • State space representation in vector matrix form are  dVC (t )   1 1  dt   RC C  VC (t )  0   di (t )    1 .    1  v(t )    (12)  L    0   iL (t )   L     dt   L  1  VC (t ) iR (t )   0.     (13) R   iL (t )  Saturday, September PMDRMFRCIED 51 29, 2012
  • 52. Example 4 : 2 loop • Find the state space representation of the electrical network shown in figure below • Input vi(t) • Output vo(t) • State variables x1(t) = vC1(t), x2(t) = iL(t) and x3(t) = vC2(t) Saturday, September PMDRMFRCIED 52 29, 2012
  • 53. RLC two loop network • Identifying appropriate variables on the circuit yields C1 node R iR iC1 Vi iC2 Vo DC L C2 iL Saturday, September PMDRMFRCIED 53 29, 2012
  • 54. RLC two loop network • Represent the electrical network shown in figure in state space where • Output is v0(t) • Input is vi(t) • State variables :- X1(t) = vC1(t) X2(t) = iL(t) X3(t) = vC2(t) Saturday, September PMDRMFRCIED 54 29, 2012
  • 55. Solution • Writing the derivative relations for energy storage elements i.e. C1, C2 and L dvC1 (t ) C1  iC1 (t ) dt diL (t ) L  vL (t ) dt dVC 2 (t ) C2  iC 2 (t ) dt Saturday, September PMDRMFRCIED 55 29, 2012
  • 56. Solution • Using Kirchhoff’s current and voltage laws C node 1 R iC1 (t )  iL (t )  iR (t ) iC1 iR iC2 Vo iC1 (t )  iL (t )  ic 2 (t ) Vi DC L C2 iL 1 iC1 (t )  iL (t )  (vL (t )  vC 2 (t )) R vL (t )  vC1 (t )  vi (t ) 1 iC 2 (t )  iR (t )  (vL (t )  vC 2 (t )) Saturday, September R PMDRMFRCIED 56 29, 2012
  • 57. Solution • Substituting these relations and simplifying yields the state equations as dvC1 1 1 1 1  vC1  iL  vC 2  vi dt RC1 C1 RC1 RC1 diL 1 1   vC1  vi dt L L dvC 2 1 1 1  vC1  vC 2  vi dt RC 2 RC 2 RC 2 vo  vC 2 Saturday, September PMDRMFRCIED 57 29, 2012
  • 58. Solution • Putting the equations in vector matrix form  1 1 1   1    RC C1  RC1   RC    1   1 x  1 0 0  x   1 v  L   L  i  1 1   1   0      RC 2 RC 2   RC 2  y  0 0 1x Saturday, September PMDRMFRCIED 58 29, 2012
  • 59. Tutorial 1 : Number 1 • Represent the electrical network shown in figure in state space where • Output is v0(t) and Input is vi(t) • State variables :- x1 = v 1 x2 = i4 x3 = v 0 Saturday, September PMDRMFRCIED 59 29, 2012
  • 60. Electrical network 1 • Add the branch current and node voltages to the network R1 = 1 Ohm R2 = 1 Ohm R3 = 1 Ohm V1 V2 i1 i3 i5 Vi C1 = 1 F L=1H C2 = 1 F Vo i2 i4 Saturday, September PMDRMFRCIED 60 29, 2012
  • 61. Solution • Write the differential equation for each energy storage element dv1  i2 ; because _ C1  1F dt di4  v2 ; because _ L  1H dt dv0  i5 ; because _ C2  1F dt Saturday, September PMDRMFRCIED 61 29, 2012
  • 62. Solution • Therefore the state vector is ,  x1   v1  x   i  x   2  4   x3  vo      • Derivative state vector is ,     x1   v1    x   x2    i4     x3  vo          Saturday, September PMDRMFRCIED 62 29, 2012
  • 63. Solution • Now obtain i2, v2 and i5 in terms of the state variables, i2  i1  i3  vi  v1  (v1  v2 )  vi  2v1  v2 v2  i5  vo  i3  i4  v0  v1  v2  i4  v0 Therefore, 1 1 1 v2  v1  i4  vo 2 2 2 Saturday, September PMDRMFRCIED 63 29, 2012
  • 64. Solution • Substituting v2 in i2, 3 1 1 i2  vi  v1  i4  v0 2 2 2 also, i5  i3  i4  v1  v2  i4 substituti ng _ v2 , 1 1 1 i5  v1  i4  vo 2 2 2 Saturday, September PMDRMFRCIED 64 29, 2012
  • 65. Solution • Therefore rearrange i2, v2 and i5 in matrix form yields  3 1 1       2     x1   v1   i2   2 2   v1  1     x    i   v    1 1 1     x 2  . i4  0 vi        2  2 4 2 2      x3  vo   i5   1   1 1  vo  0                2  2 2  v1  y  0 0 1. i4    vo    Saturday, September PMDRMFRCIED 65 29, 2012
  • 66. Tutorial 1 : Number 2 • Represent the electrical network shown in figure in state space where • Output is iR(t) • Input is vi(t) • State variables :- x1 = i2 x2 = vC Saturday, September PMDRMFRCIED 66 29, 2012
  • 67. Electrical network 2 • Add the branch currents and node voltages to the schematic and obtain C = 1F R1 = 1 Ohm node V1 node V2 i1 i3 Vi R2=1 Ohm 4V1 iR DC L = 1H i2 i4 Saturday, September PMDRMFRCIED 67 29, 2012
  • 68. Solution • Write the differential equation for each energy storage element di2  v1 ; because _ L  1H dt dvc  i3 : because _ C  1F dt Saturday, September PMDRMFRCIED 68 29, 2012
  • 69. Solution • Therefore the state vector is,  x1   i2  x    x2  vc  Saturday, September PMDRMFRCIED 69 29, 2012
  • 70. Solution • Now obtain v1 in terms of the state variables v1  vc  v2 v1  vc  iR v1  vc  i3  4v1 v1  vc  (i1  i2 )  4v1 v1  vc  vi  v1  i2  4v1 1 1 1 v1  i2  vc  vi 2 2 2 Saturday, September PMDRMFRCIED 70 29, 2012
  • 71. Solution • Now obtain i3 in terms of the state variables i  i  i 3 1 2 i3  vi  v1  i2 1 1 1 i3  vi  i2  vc  vi  i2 2 2 2 3 1 3 i3   i2  vc  vi 2 2 2 Saturday, September PMDRMFRCIED 71 29, 2012
  • 72. Solution • Now obtain the output iR in terms of the state variables iR  i3  4v1 1 3 1 iR  i2  vc  vi 2 2 2 Saturday, September PMDRMFRCIED 72 29, 2012
  • 73. Solution • Hence the state space representation  1 1  1    v       i    i2      1   2 2 . 2  2 v x v   i3   3 1  vc   3  i  c      2 2   2  1 3   i2   1  y  .     vi 2 2  vc   2  Saturday, September PMDRMFRCIED 73 29, 2012
  • 74. Tutorial 1 : Number 3 • Find the state space representation of the network shown in figure if • Output is v0(t) • Input is vi(t) • State variables :- x1 = iL1 x2 = iL2 x3 = vC Saturday, September PMDRMFRCIED 74 29, 2012
  • 75. Electrical network 3 • Add the branch currents and node voltages to the schematic and obtain R3 = 1 Ohm L1 = 1H i3 L2 = 1H node node Vi Vo Vi i2 i1 R2=1 Ohm DC Vo C = 1F Saturday, September PMDRMFRCIED 75 29, 2012
  • 76. Solution • Write the differential equation for each energy storage element diL1  vc  v1 dt diL 2  vc  i2 dt dvc  i1  i2 dt Saturday, September PMDRMFRCIED 76 29, 2012
  • 77. Solution • where, • L1 is the inductor in the loop with i1 • L2 is the inductor in the loop with i2 • iL1 = i1 –i3 • iL2 = i2 – i3 • Now, • i1 – i2 = ic = iL1 – iL2 -----------------(1) Saturday, September PMDRMFRCIED 77 29, 2012
  • 78. Solution • Also writing the node equation at vo, • i2 = i3 + iL2 ----------------------(2) • Writing KVL around the outer loop yields • i2 + i3 = vi -----------------------(3) • Solving (2) and (3) for i2 and i3 yields 1 1 i2  iL 2  vi            (4) 2 2 1 1 i3   iL 2  vi          (5) 2 2 Saturday, September PMDRMFRCIED 78 29, 2012
  • 79. Solution • Substituting (1) and (4) into the state equations. • To find the output equation, • vo = -i3 + vi • Using equation (5), 1 1 vo  iL 2  vi 2 2 Saturday, September PMDRMFRCIED 79 29, 2012
  • 80. Solution • Summarizing the results in vector matrix form  diL1      x1  dt  0 0  1  iL1   1     x    diL 2   0  1 1 .i    1  v  x 2     dt   2   L2   2  i  x3   dvC  1  1 0   vC   0             dt    iL1   1    1 y  vo  0 0.iL 2     vi  2  2  vC    Saturday, September PMDRMFRCIED 80 29, 2012
  • 81. Tutorial 1 : Number 4 • An RLC network is shown in figure. Define the state variable as :- • X1 = i1 • X2 = i2 • X3 = Vc • Let voltage across capacitor, Vc is the output from the network. Input of the system is Va and Vb Saturday, September PMDRMFRCIED 81 29, 2012
  • 82. Tutorial 1 : Number 4 • Determine the state space representation of the RLC network in matrix form • Determine the range of resistor R in order to maintain the system’s stability, if C = 0.1 F and L1=L2=0.1 H. The characteristic equation of the system is, s  10Rs  200s  1000R  0 3 2 Saturday, September PMDRMFRCIED 82 29, 2012
  • 83. RLC network with 2 input R L1 L2 i1 + i2 Va iC C - VC Vb DC DC Saturday, September PMDRMFRCIED 83 29, 2012
  • 84. Solution • State variables and their derivatives  di1 x1  i1  x1  dt  di2 x2  i2  x2  dt  dvc x3  vc  x3  dt u1  va u 2  vb y  vc Saturday, September PMDRMFRCIED 84 29, 2012
  • 85. Solution • The derivatives equations for energy storage elements di1 L1  vL1      (1) dt di2 L2  vL 2      (2) dt dvC C  iC      (3) dt Saturday, September PMDRMFRCIED 85 29, 2012
  • 86. Solution L1 L2 R i1 + i2 • For loop (1) ; Va iC VC Vb va  i1R  vL1  vC C - DC DC vL1  va  i1 R  vC      (4) • For loop (2) ; vb  vL 2  vC vL 2  vb  vC      (5) Saturday, September PMDRMFRCIED 86 29, 2012
  • 87. Solution • For current iC ; iC  i1  i2      (6) • Substituting equation (4), (5) and (6) into equation (1), (2) and (3) yields L1 L2 di1 R L1  va  i1 R  vC i1 + i2 dt Va iC C - VC Vb DC DC di1 R 1 1   i1  vC  va dt L1 L1 L1  di1 R 1 1 x1    x1  x3  va      (7) dt L1 L1 L1 Saturday, September PMDRMFRCIED 87 29, 2012
  • 88. Solution • Substituting equation (4), (5) and (6) into equation (1), (2) and (3) yields di2 L2  vb  vC dt di2 1 1   vC  vb dt L2 L2  di2 1 1 x2    x3  vb      (8) dt L2 L2 Saturday, September PMDRMFRCIED 88 29, 2012
  • 89. Solution • Substituting equation (4), (5) and (6) into equation (1), (2) and (3) yields dvC C  i1  i2 dt dvC 1 1  i1  i2 dt C C  dvC 1 1 x3   x1  x2      (9) dt C C Saturday, September PMDRMFRCIED 89 29, 2012
  • 90. Solution • Rewrite equation (7), (8) and (9) in state space representation matrix form  R 1 1     L  0   L1  x   L1 0  x1   1  1   v   1    1   a x   x2    0 0  . x2   0 .    L2   L2  vb   x3   1 1   x3   0   0      0     C C     x1  y  0 0 1. x2     x3    Saturday, September PMDRMFRCIED 90 29, 2012
  • 91. Solution • Characteristic equation s  10Rs  200s  1000R  0 3 2 • Routh Hurwitz table s3 1 200 0 s2 10R 1000R 0 s1 (2000 R  1000 R) 0 10 R s0 200 Saturday, September PMDRMFRCIED 91 29, 2012
  • 92. Solution • For stability, all coefficients in first column of Routh Hurwitz table must be positive ; (2000 R  1000 R) 0 10 R 1000 R 0 R 0 Saturday, September PMDRMFRCIED 92 29, 2012
  • 93. Modeling of Mechanical • Mass Networks f (t )  M .a (t ) f (t )  M . d 2 y (t ) y(t) dt 2 dv(t ) f (t )  M . dt a (t )  accelerati on v(t )  velocity M f(t) y (t )  displaceme nt f (t )  force M  mass Saturday, September PMDRMFRCIED 93 29, 2012
  • 94. Modeling of Mechanical • Linear Spring Networks f (t )  K . y (t ) K y(t) f (t )  force y (t )  displaceme nt f(t) K  spring _ cons tan t Saturday, September PMDRMFRCIED 94 29, 2012
  • 95. Modeling of Mechanical Networks • Damper dy (t ) B y(t) f (t )  B. dt f (t )  force f(t) y (t )  displaceme nt B  viscous _ frictional Saturday, September PMDRMFRCIED 95 29, 2012
  • 96. Modeling of Mechanical Networks • Inertia d (t ) T (t )  J . dt T(t) d 2 (t )  (t ) T (t )  J . dt 2 T (t )  Torque J  (t )  angular _ velocity  (t )  angular _ displaceme nt J  Inertia Saturday, September PMDRMFRCIED 96 29, 2012
  • 97. Force-velocity, force-displacement, and impedance translational relationships for springs, viscous dampers, and mass Saturday, September PMDRMFRCIED 97 29, 2012
  • 98. Torque-angular velocity, torque-angular displacement, and impedance rotational relationships for springs, viscous dampers, and inertia Saturday, September PMDRMFRCIED 98 29, 2012
  • 99. Example 5 • Determine the state space representation of the mechanical system below if the state variables are y(t) and dy(t)/dt. Input system is force f(t) and output system is y(t) K y(t) B M f(t) Saturday, September PMDRMFRCIED 99 29, 2012
  • 100. Example 5 • a. Mass, spring, and damper system; b. block diagram Saturday, September PMDRMFRCIED 100 29, 2012
  • 101. State variables, input and output x1 (t )  y (t ) dy (t ) dx1 (t ) x2 (t )   dt dt input  u  f (t ) output  y  y (t ) Saturday, September PMDRMFRCIED 101 29, 2012
  • 102. Mass, spring and damper system • Draw the free body diagram y(t) d 2 y (t ) M dt 2 Ky (t ) M f(t) dy (t ) B dt Saturday, September PMDRMFRCIED 102 29, 2012
  • 103. Mass, spring and damper system • a. Free-body diagram of mass, spring, and damper system; b. transformed free-body diagram Saturday, September PMDRMFRCIED 103 29, 2012
  • 104. Mass, spring and damper system • The force equation of the system is 2 d y(t ) dy(t ) f (t )  M . 2  B.  K . y(t ) dt dt • Rearranged the equation yields 2 d y (t ) B dy (t ) K 1 2  .  . y (t )  . f (t ) dt M dt M M Saturday, September PMDRMFRCIED 104 29, 2012
  • 105. Mass, spring and damper system • State equations and output equation  x1 (t )  x2 (t )  K B 1 x 2 (t )   .x1 (t )  .x2 (t )  . f (t ) M M M y (t )  x1 (t ) Saturday, September PMDRMFRCIED 105 29, 2012
  • 106. Mass, spring and damper system • State space representation in vector matrix form are     0 1   x (t )   0  x1 (t )   K B . 1    1 . f (t )  x (t )  M    x2 (t )    2   M M  K  x1 (t )  y (t )  1 0. y(t)  B M  x2 (t ) f(t) Saturday, September PMDRMFRCIED 106 29, 2012
  • 107. Example: The mechanical system • Consider the mechanical system shown in Figure below by assuming that the system is linear. The external force u(t) is the input to the system and the displacement y(t) of the mass is the output. The displacement y(t) is measured from the equilibrium position in the absence of the external force. This system is a single input and single output system. Saturday, September PMDRMFRCIED 107 29, 2012
  • 108. Mechanical system diagram Saturday, September PMDRMFRCIED 108 29, 2012
  • 109. Mechanical system diagram • From the diagram, the system equation is   m y  b y  ky  u • The system is of second order. This means that the system involves two integrators. Define the state variables x1(t) and x2(t) as x1 (t )  y (t )  x2 ( t )  y ( t ) Saturday, September PMDRMFRCIED 109 29, 2012
  • 110. • Then we obtain,  x 1  x2  1   1 u x 2    ky  b y  m  m  k b 1 x 2   x1  x2  u m m m y  x1 Saturday, September PMDRMFRCIED 110 29, 2012
  • 111. Mass, spring and damper system • a. Two-degrees-of-freedom translational mechanical system • b. block diagram Saturday, September PMDRMFRCIED 111 29, 2012
  • 112. Mass, spring and damper system • a. Forces on M1 due only to motion of M1 b. forces on M1 due only to motion of M2 c. all forces on M1 Saturday, September PMDRMFRCIED 112 29, 2012
  • 113. Mass, spring and damper system • a. Forces on M2 due only to motion of M2; b. forces on M2 due only to motion of M1; c. all forces on M2 Saturday, September PMDRMFRCIED 113 29, 2012
  • 114. Exercise 1 • Figure below shows a diagram for a quarter car model (one of the four wheels) of an automatic suspension system for a long distance express bus. A good bus suspension system should have satisfactory road handling capability, while still providing comfort when riding over bumps and holes in the road. When the coach is experiencing any road disturbance, such as potholes, cracks, and uneven pavement, the bus body should not have large oscillations, and the oscillations should be dissipate quickly. Saturday, September PMDRMFRCIED 114 29, 2012
  • 115. Exercise 1 (i). Draw the free-body diagrams of the system (ii). Determine the state space representation of the quarter car system by considering the state vector T     z(t)  x1 (t ) x2 (t ) x1 (t ) x 2 (t )    And the displacement of bus body mass M1 as the output of the system. Saturday, September PMDRMFRCIED 115 29, 2012
  • 116. Saturday, September PMDRMFRCIED 116 29, 2012
  • 117. Constant value • Bus body mass, M1 = 2500 kg • Suspension mass, M2 = 320 kg • Spring constant of suspension system, K1 = 80,000 N/m • Spring constant of wheel and tire, K2 = 500,000 N/m • Damping constant of suspension system, B1 = 350 Ns/m • Damping constant of wheel and tire, B2 = 15,020 Ns/m Saturday, September PMDRMFRCIED 117 29, 2012
  • 118. Solution • Free body diagram for M1 – Forces on M1 due to motion of M1 K1X1 M1s2X1 u M1 B1sX1 – Forces on M1 due to motion of M2 K1X2 M1 B1sX2 – All forces on M1 K1X1 K1X2 M1s2X1 M1 u B1sX1 B1sX2 Saturday, September PMDRMFRCIED 118 29, 2012
  • 119. Solution • Free body diagram for M2 – Forces on M2 due to motion of M2 K2X2 M2s2X2 K1X2 M2 B2sX2 B1sX2 – Forces on M2 due to motion of M1 K1X1 M2 B1sX1 – All forces on M2 (K1+K2)X2 M2s2X2 M2 K1X1 (B1+B2)sX2 B1sX1 Saturday, September PMDRMFRCIED 119 29, 2012
  • 120. Solution • State variables   z1  x1; z2  x2 ; z3  x1; z4  x 2 • Derivative state variables         z1  x1  z3 ; z 2  x 2  z4 ; z 3  x1; z 4  x 2 Saturday, September PMDRMFRCIED 120 29, 2012
  • 121. Solution • Total force for M1 dx2 d 2 x1 dx1 u  K1 x2  B1  K1 x1  M 1 2  B1 dt dt dt 2 d x1 dx1 dx2 2  32 x1  32 x2  0.14  0.14  0.0004u dt dt dt  z 3  32 z1  32 z2  0.14 z3  0.14 z4  0.0004u Saturday, September PMDRMFRCIED 121 29, 2012
  • 122. Solution • Total force for M2 2 dx1 d x2 dx2 K1 x1  B1  ( K1  K 2 ) x2  M 2  ( B1  B2 ) dt dt dt d 2 x2 dx1 dx2 2  250 x1  1812.5 x2  1.094  48.031 dt dt dt  z 4  250 z1  1812.5 z 2  1.094 z3  48.031z 4 Saturday, September PMDRMFRCIED 122 29, 2012
  • 123. Solution • State space representation  0 0 1 0   z1   0    0 0 0 z   0  1  2   z   u  32 32  0.14 0.14   z3  0.0004       250 1812.5 1.094 48.031  z 4   0   z1  z  y  1 0 0 0  2  z3     z4  Saturday, September PMDRMFRCIED 123 29, 2012
  • 124. Tutorial 1 : Number 5 • Figure shows a mechanical system consisting of mass M1 and M2, damper constant B, spring stiffness K1 and K2. When force f(t) acts on mass M1, it moves to position x1(t) while mass M2 moves to position x2(t). Find the state space representation of the system using x1(t), x2(t) and their first derivatives as state variables. Let x2(t) be the output. Saturday, September PMDRMFRCIED 124 29, 2012
  • 125. Mechanical system consist of 2 mass, 2 spring and 1 damper f(t) X1 X2 K1 B K2 M1 M2 Saturday, September PMDRMFRCIED 125 29, 2012
  • 126. Mechanical system consist of 2 mass, 2 spring and 1 damper • State variables and their derivatives :-   z1 (t )  x1 (t )  z1 (t )  x1 (t )  z 2 (t )    z 2 (t )  x1 (t )  z 2 (t )  x1 (t )   z3 (t )  x2 (t )  z3 (t )  x 2 (t )  z 4 (t )    z 4 (t )  x 2 (t )  z 4 (t )  x2 (t ) input  u (t )  f (t ) output  y (t )  x2 (t ) Saturday, September PMDRMFRCIED 126 29, 2012
  • 127. Mechanical system consist of 2 mass, 2 spring and 1 damper • Draw the free body diagram K1 x1  f (t ) M 1 x1  M1  B x1 B x2 K 2 x2   M 2 x2 M2 B x1  B x2 Saturday, September PMDRMFRCIED 127 29, 2012
  • 128. Mechanical system consist of 2 mass, 2 spring and 1 damper • Differential equation in mass M1    f (t )  M 1 x1  B x1  K1 x1  B x2    f (t )  M 1 x1  B( x1  x2 )  K1 x1      (1) • Differential equation in mass M2    0  M 2 x2  B x2  K 2 x2  B x1    0  M 2 x2  B( x2  x1 )  K 2 x2      (2) Saturday, September PMDRMFRCIED 128 29, 2012
  • 129. Mechanical system consist of 2 mass, 2 spring and 1 damper • Substitute all state variables and their first derivatives in equation (1) and (2) yields  B  B  K1 f (t ) x1   x1  x2  x1  M1 M1 M1 M1  B B K 1 z2   z2  z 4  1 z1  u      (3) M1 M1 M1 M1  B  B  K2 x2   x2  x1  x2 M2 M2 M2  B B K z4   z4  z 2  2 z3      (4) M2 M2 M2  z1  z 2      (5)  z 3  z 4      ( 6) Saturday, September PMDRMFRCIED 129 29, 2012
  • 130. Mechanical system consist of 2 mass, 2 spring and 1 damper • Rearrange equation 3, 4, 5 and 6 in matrix form   0 1 0 0   0  z  1   K1 z B B   1  z   M  0 z   1   M1 M1   2   M  z   2    1 .   1 u  z3   0 0 0 1   z3  0   0 B K2 B        z4   0   z4      M2 M2 M2      z1  z  y  0 0 1 0. 2   z3     z4  Saturday, September PMDRMFRCIED 130 29, 2012
  • 131. Tutorial 1 : Number 6 • Represent the translational mechanical system shown in figure in state space where x3(t) is the output and f(t) is the input. X1 X2 X3 K1 B1 K2 B2 M1 M2 M3 f(t) Saturday, September PMDRMFRCIED 131 29, 2012
  • 132. Example : 3M, 2K and 2B • Represent the translational mechanical system shown in figure in state space where x3(t) is the output and f(t) is the input. Saturday, September PMDRMFRCIED 132 29, 2012
  • 133. Example : 3M, 2K and 2B • K1 = K2 = 1 N/m • M1 = M2 = M3 = 1 kg • B1 = B2 = 1 N-s/m • Find the state space representation of the system using x1, x2, x3 and their first derivatives as state variables.  z1  x1 ; z2  x1 ; z3  x2 ;   z4  x2 ; z5  x3 ; z6  x3 Saturday, September PMDRMFRCIED 133 29, 2012
  • 134. Example : 3M, 2K and 2B • Draw the free body diagram   M 1 x1 B1 x2 B1 x1  M1 f (t ) K1 x1   M 2 x2  B1 x1 B1 x2 M2 K 2 x3 K 2 x2  M 3 x3 B2 x3  M3 K 2 x2 K 2 x3 Saturday, September PMDRMFRCIED 134 29, 2012
  • 135. Example : 3M, 2K and 2B • Writing the equations of motion    M 1 x1  B1 x1  K1 x1  B1 x2  f (t )    (1)    M 2 x2  B1 x2  K 2 x2  B1 x1  K 2 x3    (2)   M 3 x3  B2 x3  K 2 x3  K 2 x2    (3) Saturday, September PMDRMFRCIED 135 29, 2012
  • 136. Example : 3M, 2K and 2B • Substitute the value of K, M and B. • Rearrange equation (1), (2) and (3)    x1   x1  x1  x2  f    x2  x1  x2  x2  x3   x3   x3  x3  x2 Saturday, September PMDRMFRCIED 136 29, 2012
  • 137. Example : 3M, 2K and 2B • From the state variables   z1  x1  z1  x1  z 2    z 2  x1  z 2  x1   z 2  z1  z 4  f   z 3  x 2  z 3  x2  z 4    z 4  x2  z 4  x2  z 2  z 4  z 3  z 5   z5  x3  z5  x3  z6    z6  x3  z6  x3   z6  z5  z3 y  x3  z5 Saturday, September PMDRMFRCIED 137 29, 2012
  • 138. Example : 3M, 2K and 2B • In vector matrix form  0 1 0 0 0 0  0   1  1 0 1 0 0  1       0 0 0 1 0 0  0  z  z    f (t )  0 1  1  1 1 0  0   0 0 0 0 0 1  0       0 0 1 0  1  1 0     y  0 0 0 0 1 0z Saturday, September PMDRMFRCIED 138 29, 2012
  • 139. Modeling of Electro-Mechanical System • NASA flight simulator robot arm with electromechanical control system components Saturday, September PMDRMFRCIED 139 29, 2012
  • 140. Modeling of Electro-Mechanical System • Armature Controlled DC Motor Saturday, September PMDRMFRCIED 140 29, 2012
  • 141. Armature Controlled DC Motor Saturday, September PMDRMFRCIED 141 29, 2012
  • 142. DC motor armature control • The back electromotive force(back emf), VB d m (t ) VB (t )  dt d m (t ) VB (t )  K B .    (1) dt K B  Back _ emf _ cons tan t Saturday, September PMDRMFRCIED 142 29, 2012
  • 143. DC motor armature control • Kirchoff’s voltage equation around the armature circuit ea (t )  ia (t ) Ra  Vb (t ) d m (t ) ea (t )  ia (t ) Ra  K b    (2) dt ia  armature _ current  m  angular _ displaceme nt _ of _ the _ armature Ra  armature _ resis tan ce ignore _ La Saturday, September PMDRMFRCIED 143 29, 2012
  • 144. DC motor armature control • The torque, Tm(t) produced by the motor Tm (t )  ia (t ) Tm (t )  K t ia (t ) d 2 m d m Tm (t )  J m 2  Dm    (3) dt dt K t  Torque _ cons tan t J m  equivalent _ inertia _ by _ the _ motor Dm  equivalent _ viscous _ density _ by _ the _ motor Saturday, September PMDRMFRCIED 144 29, 2012
  • 145. DC motor armature control • Solving equation (3) for ia(t) J m d  m Dm d m 2 ia (t )  2     (4) K t dt K t dt Saturday, September PMDRMFRCIED 145 29, 2012
  • 146. DC motor armature control • Substituting equation (4) into equation (2) yields  J m d 2 m Dm d m  d m ea (t )  Ra  2    Kb  K t dt K t dt  dt  Ra J m  d 2 m  Ra Dm  d m ea (t )    K  dt . 2    K  K b .  dt    (5)  t   t  Saturday, September PMDRMFRCIED 146 29, 2012
  • 147. DC motor armature control • Define the state variables, input and ouput x1   m    (6a ) d m x2     (6b) dt u  ea (t ) y  0.1 m • Substituting equation (6) into equation (5) yields  Ra J m  dx2  Ra Dm  ea (t )    K . dt   K  K b .x2    (7)     t   t  Saturday, September PMDRMFRCIED 147 29, 2012
  • 148. DC motor armature control • Solving for x2 dot yields,  Ra Dm  ea (t )    K  K b .x2  dx2   t  dt  Ra J m    K    t  dx2  K t   Dm K b K t    R J .ea (t )   J  R J .x2 dt  a m     m a m   dx2 1  Kb Kt   Kt    Dm   .x2     R J .ea (t )    (8)  dt Jm  Ra   a m Saturday, September PMDRMFRCIED 148 29, 2012
  • 149. DC motor armature control • Using equation (6) and (8), the state equations are written as dx1 d m   x2 dt dt dx2 1  Kt Kb   Kt    Dm   .x2    .ea (t ) R J  dt Jm  Ra   a m Saturday, September PMDRMFRCIED 149 29, 2012
  • 150. DC motor armature control • Assuming that the output o(t) is 0.1 the displacement of the armature m(t) as x1. Hence the output equation is y  0.1x1 • State space representation in vector matrix form are    0 1   x1   K  0 x1     1  K t K b .    t .ea (t )  x  0  J  Dm  R   x2   R J     2  m  a   a m  x1  y  0.1 0.   x2  Saturday, September PMDRMFRCIED 150 29, 2012
  • 151. Tutorial 1 : Number 7 • The representation of the positioning system using an armature-controlled dc motor is shown in figure. • The input is the applied reference voltage, r(t) and the output is the shaft’s angular position, o(t). • The dynamic of the system can be described through the Kirchoff equation for the armature circuit, the Newtonian equation for the mechanical load and the torque field current relationship. Saturday, September PMDRMFRCIED 151 29, 2012
  • 152. Figure : DC motor armature control Saturday, September PMDRMFRCIED 152 29, 2012
  • 153. Example : ex-exam question • The Newtonian equation for the mechanical load is   J  o (t )    o (t )   (t ) • The back e.m.f voltage induced in the armature circuit, eb(t) is proportional to the motor shaft speed,  eb  K b  o Saturday, September PMDRMFRCIED 153 29, 2012
  • 154. Example : ex-exam question • A potentiometer was installed to measure the motor output position. Its output voltage, v(t) is then compared with the system reference input voltage, r(t) through an op-amp. • Determine the complete state-space representation of the system by considering the following state variables. Saturday, September PMDRMFRCIED 154 29, 2012
  • 155. Example : ex-exam question x1(t)  ia (t) • State variables :- x 2 (t)   o (t)  x 3 (t)   o (t) • State variables derivative  dia (t)  x 1 (t)   ia dt  d  o (t) x 2 (t)    o (t) dt  d 2 o (t)  x 3 (t)  2   o (t) dt Saturday, September PMDRMFRCIED 155 29, 2012
  • 156. Example : ex-exam question • Mechanical load   J o (t)    o (t)   (t)  K t ia  J x 3  x 3  K t x1  Kt  x3  x1  x 3          (1) J J Saturday, September PMDRMFRCIED 156 29, 2012
  • 157. Example : ex-exam question • Electrical (armature) circuit • Using Kirchoff Voltage Law dia uL  Ria  eb dt but  eb  K b  o ( given) dia  u  L   Ria  K b  o dt  u  L x1  Rx1  K b x3  R Kb 1 x1   x1  x3  u        (2) L L L Saturday, September PMDRMFRCIED 157 29, 2012
  • 158. Example : ex-exam question • From the state variable defination x2   o   x 2   o  x3      (3) For _ the _ input _ part u  r v u  r  K s o u  r  K s x2        (4) Saturday, September PMDRMFRCIED 158 29, 2012
  • 159. Example : ex-exam question • Substituting (4) into (2)  R Kb 1 x1   x1  x3  (r  K s x2 ) L L L  R Ks Kb 1 x1   x1  x2  x3  r      (5) L L L L Saturday, September PMDRMFRCIED 159 29, 2012
  • 160. Example : ex-exam question • Writing equations (1), (3) and (5) in the vector matrix form gives :-    R   Ks  Kb  1 x1   L L   x1   L  L   x2    0 0 1 . x2    0  r     KT        x3   0    x3   0     J  J       Saturday, September PMDRMFRCIED 160 29, 2012
  • 161. Example : ex-exam question • The output  x1  x  y   o  0 1 0 2   x3    Saturday, September PMDRMFRCIED 161 29, 2012
  • 162. Modelling of Electro-Mechanical System • Field Controlled DC + Rf ef (t) Motor if (t) - Lf Gelung Medan Ra La + ea Ba Ja ia - TL(t) Tm(t) Gelung Angker  m (t ) Tetap RAJAH 7.11 : MOTOR SERVO A.T. TERUJA BERASINGAN DALAM KAWALAN MEDAN Saturday, September PMDRMFRCIED 162 29, 2012
  • 163. DC motor field control • For field circuit di f e(t )  i f R f  L f    (1) dt • For mechanical load, torque d o d o 2 T (t )  J B    (2) dt dt Saturday, September PMDRMFRCIED 163 29, 2012
  • 164. DC motor field control • For torque and field current relationship T (t )  i f (t ) T (t )  K t i f (t )    (3) • Define the state variables, input and output x   (t )    (4) 1 o d o (t ) x2     (5) dt x3  i f (t )    (6) u  e(t ) y   o (t ) Saturday, September PMDRMFRCIED 164 29, 2012
  • 165. DC motor field control • From equation (4) and (5), we can determine the first state equation as :  d o x1 (t )  x2 (t )     ( 7) dt • Another two state equations are :  d 2 o x2  2    (8) dt  di f x3     (9) dt Saturday, September PMDRMFRCIED 165 29, 2012
  • 166. DC motor field control • Substituting x3 and x3 dot into equation (1) yields  e(t )  x3 R f  L f x3 • Substituting equation (3) into equation (2) d o d o yields 2 J 2 B  Kt i f dt dt Saturday, September PMDRMFRCIED 166 29, 2012
  • 167. DC motor field control • Substituting x2 dot, x2 and x3, hence  J x2  Bx2  Kt x3 • Rewrite equations  Rf 1 x3   x3  e(t ) Lf Lf  B Kt x2   x2  x3 J J Saturday, September PMDRMFRCIED 167 29, 2012
  • 168. DC motor field control • Matrix form        0 1 0  0  x 0  B Kt x   0 u  J J  1   Rf    0 0    Lf      Lf   y  1 0 0x Saturday, September PMDRMFRCIED 168 29, 2012
  • 169. Block diagrams • The block diagram is a useful tool for simplifying the representation of a system. • Simple block diagrams only have one feedback loop. • Complex block diagram consist of more than one feedback loop, more than 1 input and more than 1 output i.e. inter-coupling exists between feedback loops Saturday, September PMDRMFRCIED 169 29, 2012
  • 170. Block diagrams • Integrator  x2   x1dt x1 • Amplifier or gain x1 x2 = Kx1 K x1 + x4 = x1-x2+x3 • Summer x2 - x3 + Saturday, September PMDRMFRCIED 170 29, 2012
  • 171. Signal flow graphs • Having the block diagram simplifies the analysis of a complex system. • Such an analysis can be further simplified by using a signal flow graphs (SFG) which looks like a simplified block diagram • An SFG is a diagram which represents a set of simultaneous equation. • It consist of a graph in which nodes are connected by directed branches. Saturday, September PMDRMFRCIED 171 29, 2012
  • 172. Signal flow graphs • The nodes represent each of the system variables. • A branch connected between two nodes acts as a one way signal multiplier: the direction of signal flow is indicated by an arrow placed on the branch, and the multiplication factor(transmittance or transfer function) is indicated by a letter placed near the arrow. Saturday, September PMDRMFRCIED 172 29, 2012
  • 173. Signal flow graphs • A node performs two functions: 1. Addition of the signals on all incoming branches 2. Transmission of the total node signal(the sum of all incoming signals) to all outgoing branches Saturday, September PMDRMFRCIED 173 29, 2012
  • 174. Signal flow graphs • There are three types of nodes: 1. Source nodes (independent nodes) – these represent independent variables and have only outgoing branches. u and v are source nodes 2. Sink nodes (dependent nodes) - these represent dependent variables and have only incoming branches. x and y are source nodes 3. Mixed nodes (general nodes) – these have both incoming and outgoing branch. W is a mixed node. Saturday, September PMDRMFRCIED 174 29, 2012
  • 175. Signal flow graphs • x2 = ax1 x1 a x2 = ax1 Saturday, September PMDRMFRCIED 175 29, 2012
  • 176. Signal flow graphs • w = au + bv • x = cw • y = dw a w c x u v b d y Saturday, September PMDRMFRCIED 176 29, 2012
  • 177. Signal flow graphs • x = au + bv +cw u a x c 1 x w v b Mixed Sink node node Saturday, September PMDRMFRCIED 177 29, 2012
  • 178. Signal flow graphs • A path is any connected sequence of branches whose arrows are in the same direction • A forward path between two nodes is one which follows the arrows of successive branches and in which a node appears only once. • The path uwx is a forward path between the nodes u and x Saturday, September PMDRMFRCIED 178 29, 2012
  • 179. Signal flow graphs • Series path (cascade nodes) – series path can be combined into a single path by multiplying the transmittances • Path gain – the product of the transmittance in a series path • Parallel paths – parallel paths can be combined by adding the transmittances • Node absorption – a node representing a variable other than a source or sink can be eliminated Saturday, September PMDRMFRCIED 179 29, 2012
  • 180. Signal flow graphs • Feedback loop – a closed path which starts at a node and ends at the same node. • Loop gain – the product of the transmittances of a feedback loop Saturday, September PMDRMFRCIED 180 29, 2012
  • 181. Signal flow graphs simplification Original graph Equivalent graph a b ab x y z x z Saturday, September PMDRMFRCIED 181 29, 2012
  • 182. Signal flow graphs simplification Original graph Equivalent graph a (a+b) x y x y b Saturday, September PMDRMFRCIED 182 29, 2012
  • 183. Signal flow graphs simplification Original graph Equivalent graph w ac a z w c z x y x bc b Saturday, September PMDRMFRCIED 183 29, 2012
  • 184. Block diagram of feedback system R E C G B H Saturday, September PMDRMFRCIED 184 29, 2012
  • 185. Block diagram of feedback system • R=reference input • E=actuating signal • G=control elements and controlled system • C=controlled variable • B=primary feedback • H=feedback elements • C = GE • B = HC • E = R-B Saturday, September PMDRMFRCIED 185 29, 2012
  • 186. Successive reduction of SFG first second • 4 nodes • Node B eliminated R 1 E G C R 1 E G C -1 H B -H Saturday, September PMDRMFRCIED 186 29, 2012
  • 187. Successive reduction of SFG third fourth • Node E eliminated, self • Self loop eliminated loop of value -GH R G C R C G/(1+GH) -GH Saturday, September PMDRMFRCIED 187 29, 2012
  • 188. SIGNAL FLOW GRAPHS OF STATE EQUATIONS • demonstrate how to draw signal flow graphs from state equations. • Consider the following state and output equations:  x1  2 x1  5x2  3x3  2r          (1a)  x2  6 x1  2 x2  2 x3  5r          (1b)  x3  x1  3x2  4 x3  7r          (1c) y  4x1  6x2  9x3          (1d) Saturday, September PMDRMFRCIED 188 29, 2012
  • 189. SIGNAL FLOW GRAPHS OF STATE EQUATIONS • Step 1 : Identify three nodes to be the three state variables, , and three nodes, placed to the left of each respective state variables. Also identify a node as the input, r, and another node as the output, y. R(s) Y(s) sX3 (s) X3 (s) sX (s) X2 (s) sX (s) X (s) 2 1 1 Saturday, September PMDRMFRCIED 189 29, 2012
  • 190. SIGNAL FLOW GRAPHS OF STATE EQUATIONS • Step 2 : Interconnect the state variables and their derivatives with the defining integration, 1/s. 1 1 1 s s s R(s) Y(s) sX (s) X (s) sX (s) X (s) sX (s) X (s) 3 3 2 2 1 1 Saturday, September PMDRMFRCIED 190 29, 2012
  • 191. SIGNAL FLOW GRAPHS OF STATE EQUATIONS • Step 3 : Using Eqn (1a), feed to each node the indicated signals. 2 1 1 1 s s -5 s R(s) Y(s) X (s) X2 (s) sX (s) X (s) sX3 (s) 3 sX2 (s) 1 1 2 3 Saturday, September PMDRMFRCIED 191 29, 2012
  • 192. SIGNAL FLOW GRAPHS OF STATE EQUATIONS • Step 4 : Using Eqn (1b), feed to each node the indicated signals. 2 5 1 1 1 s 2 s -5 s R(s) Y(s) X (s) sX (s) X (s) sX (s) X (s) sX3 (s) 3 2 2 1 1 -2 2 3 -6 Saturday, September PMDRMFRCIED 192 29, 2012
  • 193. SIGNAL FLOW GRAPHS OF STATE EQUATIONS • Step 5 : Using Eqn (1c), feed to each node the indicated signals. 2 5 1 1 1 7 s 2 s -5 s R(s) Y(s) sX (s) X (s) sX (s) X2 (s) sX (s) X1 (s) 3 3 2 1 -4 -2 2 -3 3 -6 1 Saturday, September PMDRMFRCIED 193 29, 2012
  • 194. SIGNAL FLOW GRAPHS OF STATE EQUATIONS • Step 6 : Finally, use Eqn (1d) to complete the signal flow2 graph. 9 5 6 1 1 1 7 s 2 s -5 s -4 R(s) Y(s) sX (s) X (s) sX (s) X2 (s) sX (s) X1 (s) 3 3 2 1 -4 -2 2 -3 3 -6 1 Saturday, September PMDRMFRCIED 194 29, 2012
  • 195. Example 7 • Draw a signal-flow graph for each of the following state equations : 0 1 0   x1  0  x(t )  0 0 . x   0 r (t ) 1   2    2  4  6  x3  1       x1  y (t )  1 1 0. x2     x3    Saturday, September PMDRMFRCIED 195 29, 2012
  • 196. Solution • State and output equations  x1 (t )  x2 (t )  x2 (t )  x3 (t )  x3 (t )  2 x1 (t )  4 x2 (t )  6 x3 (t )  r (t ) y (t )  x1 (t )  x2 (t ) Saturday, September PMDRMFRCIED 196 29, 2012