Chapter 1         Mathematical Modeling         of Dynamic Systems in              State SpaceSaturday, September     PMDR...
Introduction to State Space              analysis• Two approaches are available for the  analysis and design of feedback  ...
Introduction to State Space              analysis• Classical technique is based on converting a  system’s differential equ...
Introduction to State Space analysis • Modern technique or state space approach is   a unified method for modeling, analyz...
Time varying• A time-varying control system is a  system in which one or more of the  parameters of the system may vary as...
The state variables of a dynamic              system• The state of a system is a set of variables  whose values, together ...
The State Space Equations                                    x(t )  Ax (t )  Bu (t )                  y (t )  Cx(t ) ...
Two types of equation• State equationx(t )  Ax (t )  Bu (t )• Output equationy(t )  Cx(t )  Du(t )Saturday, September...
Terms• State equations: a set of n simultaneous,  first order differential equations with n  variables, where the n variab...
Terms• State variables: the smallest set of  linearly independent system variables  such that the value of the members of ...
Modeling of Electrical Networks  Voltage-current, voltage-charge, and  impedance relationships for capacitors,  resistors,...
An RLC circuitSaturday, September       PMDRMFRCIED   1229, 2012
State variable characterization• The state of the RLC system described  a set of state variables x1 and x2• X1 = capacitor...
Utilizing Kirchhoff’s current law• At the junction• First order differential equation• Describing the rate of change of  c...
Utilizing Kirchhoff’s voltage law• Right hand loop• Provide the equation describing the  rate of change of inductor curren...
State space representation• A set of two first order differential equation and  output signal in terms of the state variab...
Example 1 : RL serial network  • Figure below shows an RL serial    network with an input voltage vi(t) and    voltage dro...
Modeling of Electrical Networks• RL serial network – first order systemSaturday, September   PMDRMFRCIED          1829, 2012
RL serial network• Write the loop equation for the system  using Kirchhoff’s voltage law,         Vi (t )  VR (t )  VL (...
RL serial network• State variable is given only one, therefore  the system is a first order system• A state equation invol...
RL serial network • The output equation,Vi (t )  VR (t )  VL (t )  VR (t )  Vo (t )Vo (t )  VR (t )  Vi (t )Vo (t )...
Example 2 : RC serial network• Figure below shows an RC circuit with  input voltage vi(t) and output voltage at  resistor ...
RC serial network• Write the equations for the system using  Kirchhoff’s voltage law, vi (t )  vR (t )  vc (t )  vc (t ...
RC serial network• State variable is given only one• Therefore the system is a first order  system• Therefore a state equa...
RC serial network• Eliminate vo(t) from equation (4) and  combine with equation (1) and rearrange  gives vi (t )  vc (t )...
RC serial network• Output of the system              vo (t )  vc (t )  vi (t )      (6)• Rearrange equation (5) an...
RC serial network                      x(t )  state _ vector  vc (t )• Where,                                          ...
Modeling of Electrical Networks • Consider RLC serial network • RLC serial network – second order   systemSaturday, Septem...
State Variables and output• Select two state variables,        x1 (t )  q (t )        x2 (t )  i (t )       output  y (...
Loop equation • Using Kirchoff’s Voltage Law,vi (t )  vR (t )  vL (t )  vc (t )  di (t )            1L          Ri (t ...
Converting to charge • Using equation,          dq(t )i (t )             dt       2    d q(t )      dq(t ) 1L         2  ...
Derivatives of state vector       x1 (t )  q (t )                dq (t )       x1 (t )           i (t )  x2 (t )     ...
State equation• First state equation                   dq(t )          x1 (t )          i(t )  x2 (t )                ...
State equation in matrix formx(t )  Ax (t )  Bu (t )          0                    1   x (t )   0           x...
Output equation • Output system is VL     VL (t )  VR (t )  VC (t )  vi (t )     VL (t )  VC (t )  VR (t )  vi (t )...
Output equation in matrix formy (t )  Cx(t )  Du(t )          1                x1 (t ) y (t )               R ...
Change State Variables but        output still same                  x1 (t )  VR (t )                  x2 (t )  VC (t ) ...
Voltage formula for R, L and C     VR (t )  i (t ) R               1     VC (t )   i (t )dt               C            ...
Derivative of first state equation    x1 (t )  VR (t )                                        v(t )  VR (t )  VC (t ...
Derivative of second state               equation       x2 (t )  VC (t )                dVC (t ) 1           1       x2 ...
State equation in matrix form              x(t )  Ax (t )  Bu (t )                              R    R              ...
Output equation  VL (t )  VR (t )  VC (t )  v(t )  VL (t )  VR (t )  VC (t )  v(t )    y (t )   x1 (t )  x2 (t )...
Output equation in matrix form   y (t )  Cx(t )  Du(t )                      x1 (t )  y (t )   1  1           ...
Example 3 : 2 loop• Find a state space representation if the  output is the current through the resistor.• State variables...
Electrical network LRC                                L                                                  node 1           ...
Solution : Step 1• Label all of the branch currents in the  network.• iL(t), iR(t) and iC(t)Saturday, September        PMD...
Solution : Step 2 • Select the state variables by writing   the derivative equation for all   energy-storage elements i.e....
Solution : Step 3• Apply network theory, such as Kirchoff’s  voltage and current laws to obtain iC(t)  and VL(t) in terms ...
Solution : Step 4  • Substitute the result of equation (3) and    equation (4) into equation (1) and (2)               dVC...
Solution : Step 5• Find the output equation               1      iR (t )  VC (t )    (11)               RSaturday, Sep...
Solution : Step 6• State space representation in vector  matrix form are     dVC (t )   1            1     dt   RC...
Example 4 : 2 loop• Find the state space representation of the  electrical network shown in figure below• Input vi(t)• Out...
RLC two loop network • Identifying appropriate variables on the   circuit yields C1                                       ...
RLC two loop network• Represent the electrical network shown in  figure in state space where• Output is v0(t)• Input is vi...
Solution • Writing the derivative relations for   energy storage elements i.e. C1, C2 and   L                         dvC1...
Solution  • Using Kirchhoff’s current and voltage    laws                             C                                   ...
Solution • Substituting these relations and   simplifying yields the state equations as    dvC1     1       1      1      ...
Solution • Putting the equations in vector matrix   form       1             1          1   1         RC          C1...
Tutorial 1 : Number 1  • Represent the electrical network shown in    figure in state space where  • Output is v0(t) and I...
Electrical network 1  • Add the branch current and node    voltages to the network            R1 = 1 Ohm          R2 = 1 O...
Solution• Write the differential equation for each  energy storage element             dv1                  i2 ; because ...
Solution• Therefore the state vector is ,                x1   v1                x   i            x   2  4   ...
Solution• Now obtain i2, v2 and i5 in terms of the  state variables,i2  i1  i3  vi  v1  (v1  v2 )  vi  2v1  v2v2 ...
Solution• Substituting v2 in i2,               3      1  1      i2  vi  v1  i4  v0               2      2  2      also...
Solution• Therefore rearrange i2, v2 and i5 in matrix  form yields                                      3      1    1   ...
Tutorial 1 : Number 2• Represent the electrical network shown in  figure in state space where• Output is iR(t)• Input is v...
Electrical network 2   • Add the branch currents and node     voltages to the schematic and obtain                        ...
Solution• Write the differential equation for each  energy storage element                 di2                      v1 ; ...
Solution• Therefore the state vector is,      x1   i2    x        x2  vc Saturday, September    PMDRMFRCIED ...
Solution• Now obtain v1 in terms of the state  variables v1  vc  v2             v1  vc  iR             v1  vc  i3  ...
Solution• Now obtain i3 in terms of the state  variables i  i  i                      3    1     2                      ...
Solution• Now obtain the output iR in terms of the  state variables          iR  i3  4v1              1    3    1       ...
Solution• Hence the state space representation                        1    1       1        v                  ...
Tutorial 1 : Number 3  • Find the state space representation of the    network shown in figure if  • Output is v0(t)  • In...
Electrical network 3  • Add the branch currents and node    voltages to the schematic and obtain                          ...
Solution• Write the differential equation for each  energy storage element                      diL1                      ...
Solution  •   where,  •   L1 is the inductor in the loop with i1  •   L2 is the inductor in the loop with i2  •   iL1 = i1...
Solution•   Also writing the node equation at vo,•   i2 = i3 + iL2 ----------------------(2)•   Writing KVL around the out...
Solution• Substituting (1) and (4) into the state  equations.• To find the output equation,• vo = -i3 + vi• Using equation...
Solution• Summarizing the results in vector matrix  form        diL1                                                ...
Tutorial 1 : Number 4• An RLC network is shown in figure. Define  the state variable as :-• X1 = i1• X2 = i2• X3 = Vc• Let...
Tutorial 1 : Number 4• Determine the state space representation  of the RLC network in matrix form• Determine the range of...
RLC network with 2 input                      R        L1                 L2                          i1                 +...
Solution• State variables and their derivatives                          di1            x1  i1  x1                    ...
Solution• The derivatives equations for energy  storage elements                  di1               L1       vL1     ...
Solution                          L1          L2                                               R                          ...
Solution• For current iC ;               iC  i1  i2      (6)• Substituting equation (4), (5) and (6) into  equation...
Solution• Substituting equation (4), (5) and (6) into  equation (1), (2) and (3) yields                di2            L2  ...
Solution• Substituting equation (4), (5) and (6) into  equation (1), (2) and (3) yields            dvC         C        i...
Solution• Rewrite equation (7), (8) and (9) in state  space representation matrix form                        R          ...
Solution• Characteristic equation       s  10Rs  200s  1000R  0         3            2• Routh Hurwitz table           ...
Solution• For stability, all coefficients in first column  of Routh Hurwitz table must be positive ;         (2000 R  100...
Modeling of Mechanical• Mass                    Networks   f (t )  M .a (t )   f (t )  M .                d 2 y (t )    ...
Modeling of Mechanical  •   Linear Spring                    Networksf (t )  K . y (t )                                  ...
Modeling of Mechanical                  Networks• Damper            dy (t )                 B   y(t)f (t )  B.           ...
Modeling of Mechanical                    Networks • Inertia             d (t )T (t )  J .               dt             ...
Force-velocity, force-displacement, and   impedance translational relationships  for springs, viscous dampers, and massSat...
Torque-angular velocity, torque-angular     displacement, and impedance rotational   relationships for springs, viscous da...
Example 5   • Determine the state space     representation of the mechanical     system below if the state variables are  ...
Example 5• a. Mass, spring, and damper system;  b. block diagramSaturday, September     PMDRMFRCIED     10029, 2012
State variables, input and output              x1 (t )  y (t )                       dy (t ) dx1 (t )             x2 (t )...
Mass, spring and damper                   system    • Draw the free body diagram                                        y(...
Mass, spring and damper system• a. Free-body diagram of mass, spring, and  damper system;  b. transformed free-body diagra...
Mass, spring and damper               system• The force equation of the system is                      2             d y(t...
Mass, spring and damper                system• State equations and output equation        x1 (t )  x2 (t )             ...
Mass, spring and damper                system• State space representation in vector  matrix form are                0 ...
Example: The mechanical              system• Consider the mechanical system shown in  Figure below by assuming that the sy...
Mechanical system diagramSaturday, September   PMDRMFRCIED   10829, 2012
Mechanical system diagram• From the diagram, the system equation is                                         m y  b y ...
• Then we obtain,                                            x 1  x2                           1                     ...
Mass, spring and damper system• a. Two-degrees-of-freedom translational  mechanical system• b. block diagramSaturday, Sept...
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...
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. a...
Exercise 1• Figure below shows a diagram for a quarter car  model (one of the four wheels) of an automatic  suspension sys...
Exercise 1     (i). Draw the free-body diagrams of the system     (ii). Determine the state space representation        of...
Saturday, September   PMDRMFRCIED   11629, 2012
Constant value• Bus body mass, M1 = 2500 kg• Suspension mass, M2 = 320 kg• Spring constant of suspension system, K1 =  80,...
Solution• Free body diagram for M1     – Forces on M1 due to motion of M1             K1X1             M1s2X1             ...
Solution• Free body diagram for M2     – Forces on M2 due to motion of M2             K2X2             M2s2X2             ...
Solution    • State variables                                                             z1  x1; z2  x2 ; z3  x1; z4...
Solution• Total force for M1               dx2               d 2 x1    dx1u  K1 x2  B1      K1 x1  M 1 2  B1         ...
Solution• Total force for M2                                        2           dx1                          d x2         ...
Solution• State space representation       0      0         1    0   z1   0       0      0         0        z  ...
Tutorial 1 : Number 5• Figure shows a mechanical system  consisting of mass M1 and M2, damper  constant B, spring stiffnes...
Mechanical system consist of 2   mass, 2 spring and 1 damper        f(t)               X1                  X2           K1...
Mechanical system consist of 2   mass, 2 spring and 1 damper• State variables and their derivatives :-                    ...
Mechanical system consist of 2   mass, 2 spring and 1 damper• Draw the free body diagram              K1 x1               ...
Mechanical system consist of 2   mass, 2 spring and 1 damper• Differential equation in mass M1                           ...
Mechanical system consist of 2   mass, 2 spring and 1 damper• Substitute all state variables and their first  derivatives ...
Mechanical system consist of 2   mass, 2 spring and 1 damper• Rearrange equation 3, 4, 5 and 6 in matrix  form     0  ...
Tutorial 1 : Number 6• Represent the translational mechanical  system shown in figure in state space  where x3(t) is the o...
Example : 3M, 2K and 2B• Represent the translational mechanical  system shown in figure in state space  where x3(t) is the...
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 representat...
Example : 3M, 2K and 2B• Draw the free body diagram                                                               M 1 x...
Example : 3M, 2K and 2B• Writing the equations of motion                                     M 1 x1  B1 x1  K1 x1  ...
Example : 3M, 2K and 2B• Substitute the value of K, M and B.• Rearrange equation (1), (2) and (3)                       ...
Example : 3M, 2K and 2B• From the state variables                                     z1  x1  z1  x1  z 2           ...
Example : 3M, 2K and 2B• In vector matrix form        0 1 0 0 0 0  0         1  1 0 1 0 0  1                   ...
Modeling of Electro-Mechanical System• NASA flight simulator robot arm with  electromechanical control system componentsSa...
Modeling of Electro-Mechanical System• Armature Controlled DC MotorSaturday, September   PMDRMFRCIED     14029, 2012
Armature Controlled DC MotorSaturday, September   PMDRMFRCIED   14129, 2012
DC motor armature control• The back electromotive force(back emf),  VB                        d m (t )              VB (t...
DC motor armature control  • Kirchoff’s voltage equation around the    armature circuitea (t )  ia (t ) Ra  Vb (t )     ...
DC motor armature control• The torque, Tm(t) produced by the motor Tm (t )  ia (t ) Tm (t )  K t ia (t )               d...
DC motor armature control• Solving equation (3) for ia(t)          J m d  m Dm d m                      2ia (t )       ...
DC motor armature control • Substituting equation (4) into equation (2)   yields              J m d 2 m Dm d m        ...
DC motor armature control• Define the state variables, input and ouput                       x1   m    (6a )         ...
DC motor armature control• Solving for x2 dot yields,                      Ra Dm                ea (t )               ...
DC motor armature control• Using equation (6) and (8), the state  equations are written asdx1 d m          x2dt    dtdx...
DC motor armature control• Assuming that the output o(t) is 0.1 the  displacement of the armature m(t) as x1.  Hence the...
Tutorial 1 : Number 7• The representation of the positioning system  using an armature-controlled dc motor is shown  in fi...
Figure : DC motor armature controlSaturday, September   PMDRMFRCIED   15229, 2012
Example : ex-exam question• The Newtonian equation for the  mechanical load is                                         ...
Example : ex-exam question• A potentiometer was installed to measure  the motor output position. Its output  voltage, v(t)...
Example : ex-exam question                                                        x1(t)  ia (t)• State variables :-      ...
Example : ex-exam question• Mechanical load                    J o (t)    o (t)   (t)  K t ia    J x 3  x 3 ...
Example : ex-exam question• Electrical (armature) circuit• Using Kirchoff Voltage Law                dia            uL   ...
Example : ex-exam question• From the state variable defination     x2   o                                 x 2   o  ...
Example : ex-exam question• Substituting (4) into (2)     R     Kb      1x1   x1     x3  (r  K s x2 )      L     L  ...
Example : ex-exam question• Writing equations (1), (3) and (5) in the  vector matrix form gives :-   R               ...
Example : ex-exam question• The output                      x1                      x     y   o  0 1 0 2       ...
Modelling of Electro-Mechanical             System• Field Controlled DC                                                   ...
DC motor field control• For field circuit                                  di f          e(t )  i f R f  L f           ...
DC motor field control• For torque and field current relationship                      T (t )  i f (t )              T (t...
DC motor field control• From equation (4) and (5), we can  determine the first state equation as :                       ...
DC motor field control• Substituting x3 and x3 dot into equation (1)  yields                                             ...
DC motor field control• Substituting x2 dot, x2 and x3, hence                                                J x2  Bx2 ...
DC motor field control• Matrix form                                                                                ...
Block diagrams• The block diagram is a useful tool for  simplifying the representation of a  system.• Simple block diagram...
Block diagrams• Integrator                                                 x2   x1dt                      x1• Amplifier...
Signal flow graphs• Having the block diagram simplifies the  analysis of a complex system.• Such an analysis can be furthe...
Signal flow graphs• The nodes represent each of the system  variables.• A branch connected between two nodes  acts as a on...
Signal flow graphs• A node performs two functions:     1. Addition of the signals on all incoming        branches     2. T...
Signal flow graphs• There are three types of nodes:     1. Source nodes (independent nodes) – these        represent indep...
Signal flow graphs• x2 = ax1                  x1      a          x2 = ax1Saturday, September    PMDRMFRCIED              1...
Signal flow graphs• w = au + bv• x = cw• y = dw                          a        w        c   x        u       v         ...
Signal flow graphs• x = au + bv +cw           u                          a          x                      c              ...
Signal flow graphs• A path is any connected sequence of  branches whose arrows are in the same  direction• A forward path ...
Signal flow graphs• Series path (cascade nodes) – series path  can be combined into a single path by  multiplying the tran...
Signal flow graphs• Feedback loop – a closed path which  starts at a node and ends at the same  node.• Loop gain – the pro...
Signal flow graphs                   simplificationOriginal graph                      Equivalent graph        a          ...
Signal flow graphs                       simplification    Original graph                  Equivalent graph               ...
Signal flow graphs                   simplificationOriginal graph                      Equivalent graph   w               ...
Block diagram of feedback              system    R                 E                 C                              G     ...
Block diagram of feedback system•   R=reference input•   E=actuating signal•   G=control elements and controlled system•  ...
Successive reduction of SFGfirst                                 second• 4 nodes                             • Node B elim...
Successive reduction of SFGthird                             fourth• Node E eliminated, self         • Self loop eliminate...
SIGNAL FLOW GRAPHS OF           STATE EQUATIONS    • demonstrate how to draw signal flow      graphs from state equations....
SIGNAL FLOW GRAPHS OF            STATE EQUATIONS       • Step 1 : Identify three nodes to be the         three state varia...
SIGNAL FLOW GRAPHS OF             STATE EQUATIONS   • Step 2 : Interconnect the state variables     and their derivatives ...
SIGNAL FLOW GRAPHS OF            STATE EQUATIONS  • Step 3 : Using Eqn (1a), feed to each node    the indicated signals.  ...
SIGNAL FLOW GRAPHS OF                 STATE EQUATIONS       • Step 4 : Using Eqn (1b), feed to each node         the indic...
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Chap1see4113
Upcoming SlideShare
Loading in …5
×

Chap1see4113

1,753
-1

Published on

Published in: Education
0 Comments
3 Likes
Statistics
Notes
  • Be the first to comment

No Downloads
Views
Total Views
1,753
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
Downloads
79
Comments
0
Likes
3
Embeds 0
No embeds

No notes for slide

Chap1see4113

  1. 1. Chapter 1 Mathematical Modeling of Dynamic Systems in State SpaceSaturday, September PMDRMFRCIED 129, 2012
  2. 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 techniqueSaturday, September PMDRMFRCIED 229, 2012
  3. 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 informationSaturday, September PMDRMFRCIED 329, 2012
  4. 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. 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 conditionSaturday, September PMDRMFRCIED 529, 2012
  6. 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 629, 2012
  7. 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 _ matrixSaturday, September PMDRMFRCIED 729, 2012
  8. 8. Two types of equation• State equationx(t )  Ax (t )  Bu (t )• Output equationy(t )  Cx(t )  Du(t )Saturday, September PMDRMFRCIED 829, 2012
  9. 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 equationSaturday, September PMDRMFRCIED 929, 2012
  10. 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 variablesSaturday, September PMDRMFRCIED 1029, 2012
  11. 11. Modeling of Electrical Networks Voltage-current, voltage-charge, and impedance relationships for capacitors, resistors, and inductorsSaturday, September PMDRMFRCIED 1129, 2012
  12. 12. An RLC circuitSaturday, September PMDRMFRCIED 1229, 2012
  13. 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 2Saturday, September PMDRMFRCIED 1329, 2012
  14. 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 dtSaturday, September PMDRMFRCIED 1429, 2012
  15. 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 1529, 2012
  16. 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 1629, 2012
  17. 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 1729, 2012
  18. 18. Modeling of Electrical Networks• RL serial network – first order systemSaturday, September PMDRMFRCIED 1829, 2012
  19. 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 dtSaturday, September PMDRMFRCIED 1929, 2012
  20. 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 2029, 2012
  21. 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. 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 CSaturday, September PMDRMFRCIED 2229, 2012
  23. 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 2329, 2012
  24. 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) dtSaturday, September PMDRMFRCIED 2429, 2012
  25. 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 RCSaturday, September PMDRMFRCIED 2529, 2012
  26. 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 2629, 2012
  27. 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  1Saturday, September PMDRMFRCIED 2729, 2012
  28. 28. Modeling of Electrical Networks • Consider RLC serial network • RLC serial network – second order systemSaturday, September PMDRMFRCIED 2829, 2012
  29. 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 2929, 2012
  30. 30. Loop equation • Using Kirchoff’s Voltage Law,vi (t )  vR (t )  vL (t )  vc (t ) di (t ) 1L  Ri (t )   i (t )dt  vi (t ) dt C Saturday, September PMDRMFRCIED 30 29, 2012
  31. 31. Converting to charge • Using equation, dq(t )i (t )  dt 2 d q(t ) dq(t ) 1L 2 R  q(t )  vi (t ) dt dt C Saturday, September PMDRMFRCIED 31 29, 2012
  32. 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 )  dtSaturday, September PMDRMFRCIED 3229, 2012
  33. 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 LSaturday, September PMDRMFRCIED 3329, 2012
  34. 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. 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 ) CSaturday, September PMDRMFRCIED 3529, 2012
  36. 36. Output equation in matrix formy (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. 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 3729, 2012
  38. 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 dtSaturday, September PMDRMFRCIED 3829, 2012
  39. 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 LSaturday, September PMDRMFRCIED 3929, 2012
  40. 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 ) RCSaturday, September PMDRMFRCIED 4029, 2012
  41. 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 4129, 2012
  42. 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 4229, 2012
  43. 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 4329, 2012
  44. 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 4429, 2012
  45. 45. Electrical network LRC L node 1 VLVi iL VR R C iR iC VC Saturday, September PMDRMFRCIED 45 29, 2012
  46. 46. Solution : Step 1• Label all of the branch currents in the network.• iL(t), iR(t) and iC(t)Saturday, September PMDRMFRCIED 4629, 2012
  47. 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) dtSaturday, September PMDRMFRCIED 4729, 2012
  48. 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 4829, 2012
  49. 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 LSaturday, September PMDRMFRCIED 4929, 2012
  50. 50. Solution : Step 5• Find the output equation 1 iR (t )  VC (t )    (11) RSaturday, September PMDRMFRCIED 5029, 2012
  51. 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 5129, 2012
  52. 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 5229, 2012
  53. 53. RLC two loop network • Identifying appropriate variables on the circuit yields C1 node R iR iC1 Vi iC2 Vo DC L C2 iLSaturday, September PMDRMFRCIED 5329, 2012
  54. 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 5429, 2012
  55. 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 ) dtSaturday, September PMDRMFRCIED 5529, 2012
  56. 56. Solution • Using Kirchhoff’s current and voltage laws C node 1 RiC1 (t )  iL (t )  iR (t ) iC1 iR iC2 VoiC1 (t )  iL (t )  ic 2 (t ) Vi DC L C2 iL 1iC1 (t )  iL (t )  (vL (t )  vC 2 (t )) RvL (t )  vC1 (t )  vi (t ) 1iC 2 (t )  iR (t )  (vL (t )  vC 2 (t )) Saturday, September R PMDRMFRCIED 56 29, 2012
  57. 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 2Saturday, September PMDRMFRCIED 5729, 2012
  58. 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xSaturday, September PMDRMFRCIED 5829, 2012
  59. 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 0Saturday, September PMDRMFRCIED 5929, 2012
  60. 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 i4Saturday, September PMDRMFRCIED 6029, 2012
  61. 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 dtSaturday, September PMDRMFRCIED 6129, 2012
  62. 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 6229, 2012
  63. 63. Solution• Now obtain i2, v2 and i5 in terms of the state variables,i2  i1  i3  vi  v1  (v1  v2 )  vi  2v1  v2v2  i5  vo  i3  i4  v0  v1  v2  i4  v0Therefore, 1 1 1v2  v1  i4  vo 2 2 2Saturday, September PMDRMFRCIED 6329, 2012
  64. 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 2Saturday, September PMDRMFRCIED 6429, 2012
  65. 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 6529, 2012
  66. 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 = vCSaturday, September PMDRMFRCIED 6629, 2012
  67. 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 i4Saturday, September PMDRMFRCIED 6729, 2012
  68. 68. Solution• Write the differential equation for each energy storage element di2  v1 ; because _ L  1H dt dvc  i3 : because _ C  1F dtSaturday, September PMDRMFRCIED 6829, 2012
  69. 69. Solution• Therefore the state vector is,  x1   i2  x    x2  vc Saturday, September PMDRMFRCIED 6929, 2012
  70. 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 2Saturday, September PMDRMFRCIED 7029, 2012
  71. 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 2Saturday, September PMDRMFRCIED 7129, 2012
  72. 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 2Saturday, September PMDRMFRCIED 7229, 2012
  73. 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 7329, 2012
  74. 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 = vCSaturday, September PMDRMFRCIED 7429, 2012
  75. 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 = 1FSaturday, September PMDRMFRCIED 7529, 2012
  76. 76. Solution• Write the differential equation for each energy storage element diL1  vc  v1 dt diL 2  vc  i2 dt dvc  i1  i2 dtSaturday, September PMDRMFRCIED 7629, 2012
  77. 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 7729, 2012
  78. 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 2Saturday, September PMDRMFRCIED 7829, 2012
  79. 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 2Saturday, September PMDRMFRCIED 7929, 2012
  80. 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 8029, 2012
  81. 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 VbSaturday, September PMDRMFRCIED 8129, 2012
  82. 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 2Saturday, September PMDRMFRCIED 8229, 2012
  83. 83. RLC network with 2 input R L1 L2 i1 + i2Va iC C - VC Vb DC DCSaturday, September PMDRMFRCIED 8329, 2012
  84. 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  vcSaturday, September PMDRMFRCIED 8429, 2012
  85. 85. Solution• The derivatives equations for energy storage elements di1 L1  vL1      (1) dt di2 L2  vL 2      (2) dt dvC C  iC      (3) dtSaturday, September PMDRMFRCIED 8529, 2012
  86. 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 8629, 2012
  87. 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 L1Saturday, September PMDRMFRCIED 8729, 2012
  88. 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 L2Saturday, September PMDRMFRCIED 8829, 2012
  89. 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 CSaturday, September PMDRMFRCIED 8929, 2012
  90. 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 9029, 2012
  91. 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 200Saturday, September PMDRMFRCIED 9129, 2012
  92. 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 9229, 2012
  93. 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. 94. Modeling of Mechanical • Linear Spring Networksf (t )  K . y (t ) K y(t)f (t )  forcey (t )  displaceme nt f(t)K  spring _ cons tan t Saturday, September PMDRMFRCIED 94 29, 2012
  95. 95. Modeling of Mechanical Networks• Damper dy (t ) B y(t)f (t )  B. dtf (t )  force f(t)y (t )  displaceme ntB  viscous _ frictionalSaturday, September PMDRMFRCIED 9529, 2012
  96. 96. Modeling of Mechanical Networks • Inertia d (t )T (t )  J . dt T(t) d 2 (t )  (t )T (t )  J . dt 2T (t )  Torque J (t )  angular _ velocity (t )  angular _ displaceme ntJ  Inertia Saturday, September PMDRMFRCIED 96 29, 2012
  97. 97. Force-velocity, force-displacement, and impedance translational relationships for springs, viscous dampers, and massSaturday, September PMDRMFRCIED 9729, 2012
  98. 98. Torque-angular velocity, torque-angular displacement, and impedance rotational relationships for springs, viscous dampers, and inertiaSaturday, September PMDRMFRCIED 9829, 2012
  99. 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 9929, 2012
  100. 100. Example 5• a. Mass, spring, and damper system; b. block diagramSaturday, September PMDRMFRCIED 10029, 2012
  101. 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 10129, 2012
  102. 102. Mass, spring and damper system • Draw the free body diagram y(t) d 2 y (t )M dt 2Ky (t ) M f(t) dy (t )B dt Saturday, September PMDRMFRCIED 102 29, 2012
  103. 103. Mass, spring and damper system• a. Free-body diagram of mass, spring, and damper system; b. transformed free-body diagramSaturday, September PMDRMFRCIED 10329, 2012
  104. 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 2d y (t ) B dy (t ) K 1 2  .  . y (t )  . f (t ) dt M dt M MSaturday, September PMDRMFRCIED 10429, 2012
  105. 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 10529, 2012
  106. 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 10629, 2012
  107. 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 10729, 2012
  108. 108. Mechanical system diagramSaturday, September PMDRMFRCIED 10829, 2012
  109. 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 10929, 2012
  110. 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  x1Saturday, September PMDRMFRCIED 11029, 2012
  111. 111. Mass, spring and damper system• a. Two-degrees-of-freedom translational mechanical system• b. block diagramSaturday, September PMDRMFRCIED 11129, 2012
  112. 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 M1Saturday, September PMDRMFRCIED 11229, 2012
  113. 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 M2Saturday, September PMDRMFRCIED 11329, 2012
  114. 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 11429, 2012
  115. 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 11529, 2012
  116. 116. Saturday, September PMDRMFRCIED 11629, 2012
  117. 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/mSaturday, September PMDRMFRCIED 11729, 2012
  118. 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 B1sX2Saturday, September PMDRMFRCIED 11829, 2012
  119. 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 B1sX1Saturday, September PMDRMFRCIED 11929, 2012
  120. 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. 121. Solution• Total force for M1 dx2 d 2 x1 dx1u  K1 x2  B1  K1 x1  M 1 2  B1 dt dt dt 2d 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  0.0004uSaturday, September PMDRMFRCIED 12129, 2012
  122. 122. Solution• Total force for M2 2 dx1 d x2 dx2K1 x1  B1  ( K1  K 2 ) x2  M 2  ( B1  B2 ) dt dt dtd 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 4Saturday, September PMDRMFRCIED 12229, 2012
  123. 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 12329, 2012
  124. 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 12429, 2012
  125. 125. Mechanical system consist of 2 mass, 2 spring and 1 damper f(t) X1 X2 K1 B K2 M1 M2Saturday, September PMDRMFRCIED 12529, 2012
  126. 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 12629, 2012
  127. 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 x2Saturday, September PMDRMFRCIED 12729, 2012
  128. 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 12829, 2012
  129. 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 12929, 2012
  130. 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 13029, 2012
  131. 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 13129, 2012
  132. 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 13229, 2012
  133. 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  x3Saturday, September PMDRMFRCIED 13329, 2012
  134. 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 x3Saturday, September PMDRMFRCIED 13429, 2012
  135. 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 13529, 2012
  136. 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  x2Saturday, September PMDRMFRCIED 13629, 2012
  137. 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  z5Saturday, September PMDRMFRCIED 13729, 2012
  138. 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zSaturday, September PMDRMFRCIED 13829, 2012
  139. 139. Modeling of Electro-Mechanical System• NASA flight simulator robot arm with electromechanical control system componentsSaturday, September PMDRMFRCIED 13929, 2012
  140. 140. Modeling of Electro-Mechanical System• Armature Controlled DC MotorSaturday, September PMDRMFRCIED 14029, 2012
  141. 141. Armature Controlled DC MotorSaturday, September PMDRMFRCIED 14129, 2012
  142. 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 tSaturday, September PMDRMFRCIED 14229, 2012
  143. 143. DC motor armature control • Kirchoff’s voltage equation around the armature circuitea (t )  ia (t ) Ra  Vb (t ) d m (t )ea (t )  ia (t ) Ra  K b    (2) dtia  armature _ current m  angular _ displaceme nt _ of _ the _ armatureRa  armature _ resis tan ceignore _ La Saturday, September PMDRMFRCIED 143 29, 2012
  144. 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 _ motorSaturday, September PMDRMFRCIED 14429, 2012
  145. 145. DC motor armature control• Solving equation (3) for ia(t) J m d  m Dm d m 2ia (t )  2     (4) K t dt K t dtSaturday, September PMDRMFRCIED 14529, 2012
  146. 146. DC motor armature control • Substituting equation (4) into equation (2) yields  J m d 2 m Dm d m  d mea (t )  Ra  2    Kb  K t dt K t dt  dt  Ra J m  d 2 m  Ra Dm  d mea (t )    K  dt . 2    K  K b .  dt    (5)  t   t  Saturday, September PMDRMFRCIED 146 29, 2012
  147. 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 14729, 2012
  148. 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 14829, 2012
  149. 149. DC motor armature control• Using equation (6) and (8), the state equations are written asdx1 d m   x2dt dtdx2 1  Kt Kb   Kt    Dm   .x2    .ea (t ) R J  dt Jm  Ra   a mSaturday, September PMDRMFRCIED 14929, 2012
  150. 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 15029, 2012
  151. 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 15129, 2012
  152. 152. Figure : DC motor armature controlSaturday, September PMDRMFRCIED 15229, 2012
  153. 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  oSaturday, September PMDRMFRCIED 15329, 2012
  154. 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 15429, 2012
  155. 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) dtSaturday, September PMDRMFRCIED 15529, 2012
  156. 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 JSaturday, September PMDRMFRCIED 15629, 2012
  157. 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 LSaturday, September PMDRMFRCIED 15729, 2012
  158. 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 15829, 2012
  159. 159. Example : ex-exam question• Substituting (4) into (2) R Kb 1x1   x1  x3  (r  K s x2 ) L L L R Ks Kb 1x1   x1  x2  x3  r      (5) L L L LSaturday, September PMDRMFRCIED 15929, 2012
  160. 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 16029, 2012
  161. 161. Example : ex-exam question• The output  x1  x  y   o  0 1 0 2   x3   Saturday, September PMDRMFRCIED 16129, 2012
  162. 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 MEDANSaturday, September PMDRMFRCIED 16229, 2012
  163. 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 dtSaturday, September PMDRMFRCIED 16329, 2012
  164. 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 16429, 2012
  165. 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) dtSaturday, September PMDRMFRCIED 16529, 2012
  166. 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 dtSaturday, September PMDRMFRCIED 16629, 2012
  167. 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 JSaturday, September PMDRMFRCIED 16729, 2012
  168. 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xSaturday, September PMDRMFRCIED 16829, 2012
  169. 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 loopsSaturday, September PMDRMFRCIED 16929, 2012
  170. 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 17029, 2012
  171. 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 17129, 2012
  172. 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 17229, 2012
  173. 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 branchesSaturday, September PMDRMFRCIED 17329, 2012
  174. 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 17429, 2012
  175. 175. Signal flow graphs• x2 = ax1 x1 a x2 = ax1Saturday, September PMDRMFRCIED 17529, 2012
  176. 176. Signal flow graphs• w = au + bv• x = cw• y = dw a w c x u v b d ySaturday, September PMDRMFRCIED 17629, 2012
  177. 177. Signal flow graphs• x = au + bv +cw u a x c 1 x w v b Mixed Sink node nodeSaturday, September PMDRMFRCIED 17729, 2012
  178. 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 xSaturday, September PMDRMFRCIED 17829, 2012
  179. 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 eliminatedSaturday, September PMDRMFRCIED 17929, 2012
  180. 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 loopSaturday, September PMDRMFRCIED 18029, 2012
  181. 181. Signal flow graphs simplificationOriginal graph Equivalent graph a b abx y z x zSaturday, September PMDRMFRCIED 18129, 2012
  182. 182. Signal flow graphs simplification Original graph Equivalent graph a (a+b)x y x y b Saturday, September PMDRMFRCIED 182 29, 2012
  183. 183. Signal flow graphs simplificationOriginal graph Equivalent graph w ac a z w c z x y x bc bSaturday, September PMDRMFRCIED 18329, 2012
  184. 184. Block diagram of feedback system R E C G B HSaturday, September PMDRMFRCIED 18429, 2012
  185. 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-BSaturday, September PMDRMFRCIED 18529, 2012
  186. 186. Successive reduction of SFGfirst second• 4 nodes • Node B eliminatedR 1 E G C R 1 E G C -1 H B -HSaturday, September PMDRMFRCIED 18629, 2012
  187. 187. Successive reduction of SFGthird fourth• Node E eliminated, self • Self loop eliminated loop of value -GHR G C R C G/(1+GH) -GHSaturday, September PMDRMFRCIED 18729, 2012
  188. 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. 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. 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 sR(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. 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 sR(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. 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 sR(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
  1. Gostou de algum slide específico?

    Recortar slides é uma maneira fácil de colecionar informações para acessar mais tarde.

×