Chap1see4113
Upcoming SlideShare
Loading in...5
×
 

Chap1see4113

on

  • 909 views

 

Statistics

Views

Total Views
909
Views on SlideShare
909
Embed Views
0

Actions

Likes
0
Downloads
9
Comments
0

0 Embeds 0

No embeds

Accessibility

Categories

Upload Details

Uploaded via as Adobe PDF

Usage Rights

© All Rights Reserved

Report content

Flagged as inappropriate Flag as inappropriate
Flag as inappropriate

Select your reason for flagging this presentation as inappropriate.

Cancel
  • Full Name Full Name Comment goes here.
    Are you sure you want to
    Your message goes here
    Processing…
Post Comment
Edit your comment

Chap1see4113 Chap1see4113 Presentation Transcript

  • Chapter 1 Mathematical Modeling of Dynamic Systems in State SpaceSaturday, September PMDRMFRCIED 129, 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 techniqueSaturday, September PMDRMFRCIED 229, 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 informationSaturday, September PMDRMFRCIED 329, 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 conditionSaturday, September PMDRMFRCIED 529, 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 629, 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 _ matrixSaturday, September PMDRMFRCIED 729, 2012
  • Two types of equation• State equationx(t )  Ax (t )  Bu (t )• Output equationy(t )  Cx(t )  Du(t )Saturday, September PMDRMFRCIED 829, 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 equationSaturday, September PMDRMFRCIED 929, 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 variablesSaturday, September PMDRMFRCIED 1029, 2012
  • Modeling of Electrical Networks Voltage-current, voltage-charge, and impedance relationships for capacitors, resistors, and inductorsSaturday, September PMDRMFRCIED 1129, 2012
  • 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 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
  • 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
  • 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
  • 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
  • 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
  • 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 (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
  • 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
  • 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 CSaturday, September PMDRMFRCIED 2229, 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 2329, 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) dtSaturday, September PMDRMFRCIED 2429, 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 RCSaturday, September PMDRMFRCIED 2529, 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 2629, 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  1Saturday, September PMDRMFRCIED 2729, 2012
  • Modeling of Electrical Networks • Consider RLC serial network • RLC serial network – second order systemSaturday, September PMDRMFRCIED 2829, 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 2929, 2012
  • 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
  • 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
  • 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
  • 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
  • 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 ) CSaturday, September PMDRMFRCIED 3529, 2012
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • Electrical network LRC L node 1 VLVi 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 4629, 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) dtSaturday, September PMDRMFRCIED 4729, 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 4829, 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 LSaturday, September PMDRMFRCIED 4929, 2012
  • Solution : Step 5• Find the output equation 1 iR (t )  VC (t )    (11) RSaturday, September PMDRMFRCIED 5029, 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 5129, 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 5229, 2012
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • Solution• Therefore the state vector is,  x1   i2  x    x2  vc Saturday, September PMDRMFRCIED 6929, 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 2Saturday, September PMDRMFRCIED 7029, 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 2Saturday, September PMDRMFRCIED 7129, 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 2Saturday, September PMDRMFRCIED 7229, 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 7329, 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 = vCSaturday, September PMDRMFRCIED 7429, 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 = 1FSaturday, September PMDRMFRCIED 7529, 2012
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • RLC network with 2 input R L1 L2 i1 + i2Va iC C - VC Vb DC DCSaturday, September PMDRMFRCIED 8329, 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  vcSaturday, September PMDRMFRCIED 8429, 2012
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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 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
  • 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
  • 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
  • Force-velocity, force-displacement, and impedance translational relationships for springs, viscous dampers, and massSaturday, September PMDRMFRCIED 9729, 2012
  • Torque-angular velocity, torque-angular displacement, and impedance rotational relationships for springs, viscous dampers, and inertiaSaturday, September PMDRMFRCIED 9829, 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 9929, 2012
  • 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 )   dt dt input  u  f (t ) output  y  y (t )Saturday, September PMDRMFRCIED 10129, 2012
  • 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
  • Mass, spring and damper system• a. Free-body diagram of mass, spring, and damper system; b. transformed free-body diagramSaturday, September PMDRMFRCIED 10329, 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 2d y (t ) B dy (t ) K 1 2  .  . y (t )  . f (t ) dt M dt M MSaturday, September PMDRMFRCIED 10429, 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 10529, 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 10629, 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 10729, 2012
  • Mechanical system diagramSaturday, September PMDRMFRCIED 10829, 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 10929, 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  x1Saturday, September PMDRMFRCIED 11029, 2012
  • Mass, spring and damper system• a. Two-degrees-of-freedom translational mechanical system• b. block diagramSaturday, September PMDRMFRCIED 11129, 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 M1Saturday, September PMDRMFRCIED 11229, 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 M2Saturday, September PMDRMFRCIED 11329, 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 11429, 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 11529, 2012
  • Saturday, September PMDRMFRCIED 11629, 2012
  • Constant value• Bus body mass, M1 = 2500 kg• Suspension mass, M2 = 320 kg• Spring constant of suspension system, K1 = 80,000 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
  • 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
  • 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
  • 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 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
  • 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
  • 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
  • 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
  • Mechanical system consist of 2 mass, 2 spring and 1 damper f(t) X1 X2 K1 B K2 M1 M2Saturday, September PMDRMFRCIED 12529, 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 12629, 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 x2Saturday, September PMDRMFRCIED 12729, 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 12829, 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 12929, 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 13029, 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 13129, 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 13229, 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  x3Saturday, September PMDRMFRCIED 13329, 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 x3Saturday, September PMDRMFRCIED 13429, 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 13529, 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  x2Saturday, September PMDRMFRCIED 13629, 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  z5Saturday, September PMDRMFRCIED 13729, 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zSaturday, September PMDRMFRCIED 13829, 2012
  • Modeling of Electro-Mechanical System• NASA flight simulator robot arm with electromechanical control system componentsSaturday, September PMDRMFRCIED 13929, 2012
  • 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 )  dt d m (t ) VB (t )  K B .    (1) dt K B  Back _ emf _ cons tan tSaturday, September PMDRMFRCIED 14229, 2012
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • Figure : DC motor armature controlSaturday, September PMDRMFRCIED 15229, 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  oSaturday, September PMDRMFRCIED 15329, 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 15429, 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) dtSaturday, September PMDRMFRCIED 15529, 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 JSaturday, September PMDRMFRCIED 15629, 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 LSaturday, September PMDRMFRCIED 15729, 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 15829, 2012
  • 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
  • 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
  • Example : ex-exam question• The output  x1  x  y   o  0 1 0 2   x3   Saturday, September PMDRMFRCIED 16129, 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 MEDANSaturday, September PMDRMFRCIED 16229, 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 dtSaturday, September PMDRMFRCIED 16329, 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 16429, 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) dtSaturday, September PMDRMFRCIED 16529, 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 dtSaturday, September PMDRMFRCIED 16629, 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 JSaturday, September PMDRMFRCIED 16729, 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xSaturday, September PMDRMFRCIED 16829, 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 loopsSaturday, September PMDRMFRCIED 16929, 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 17029, 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 17129, 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 17229, 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 branchesSaturday, September PMDRMFRCIED 17329, 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 17429, 2012
  • Signal flow graphs• x2 = ax1 x1 a x2 = ax1Saturday, September PMDRMFRCIED 17529, 2012
  • Signal flow graphs• w = au + bv• x = cw• y = dw a w c x u v b d ySaturday, September PMDRMFRCIED 17629, 2012
  • Signal flow graphs• x = au + bv +cw u a x c 1 x w v b Mixed Sink node nodeSaturday, September PMDRMFRCIED 17729, 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 xSaturday, September PMDRMFRCIED 17829, 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 eliminatedSaturday, September PMDRMFRCIED 17929, 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 loopSaturday, September PMDRMFRCIED 18029, 2012
  • Signal flow graphs simplificationOriginal graph Equivalent graph a b abx y z x zSaturday, September PMDRMFRCIED 18129, 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 simplificationOriginal graph Equivalent graph w ac a z w c z x y x bc bSaturday, September PMDRMFRCIED 18329, 2012
  • Block diagram of feedback system R E C G B HSaturday, September PMDRMFRCIED 18429, 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-BSaturday, September PMDRMFRCIED 18529, 2012
  • 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
  • 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
  • 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 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
  • 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
  • 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
  • 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 1Saturday, September PMDRMFRCIED 19329, 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 1Saturday, September PMDRMFRCIED 19429, 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 19529, 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 19629, 2012
  • Solution• Signal flow graph 1 1 1/s 1 1/s 1 1/s 1 r y x3 x2 x1 -6 -4 -2Saturday, September PMDRMFRCIED 19729, 2012
  • Example 8• Draw a signal-flow graph for each of the following state equations : 0 1 0 0 x(t )  0 3 1   x(t )  1 r (t )    3   4  5  1  y (t )  1 2 0x(t )Saturday, September PMDRMFRCIED 19829, 2012
  • Solution• State and output equations  x1 (t )  x2 (t )  x2 (t )  3x2 (t )  x3 (t )  r (t )  x3 (t )  3 x1 (t )  4 x2 (t )  5 x3 (t )  r (t ) y (t )  x1 (t )  2 x2 (t )Saturday, September PMDRMFRCIED 19929, 2012
  • Solution• Signal flow graph 1 2 1 1/s 1 1/s 1 1/s 1 r y x3 x2 x1 -5 -3 -4 -3Saturday, September PMDRMFRCIED 20029, 2012
  • Example 9• Draw a signal-flow graph for each of the following state equations : 7 1 0 1   x(t )   3 2  1  x(t )  2 r (t )    1  0 2  1    y (t )  1 3 2x(t )Saturday, September PMDRMFRCIED 20129, 2012
  • Solution• State and output equations  x1 (t )  7 x1 (t )  x2 (t )  r (t )  x2 (t )  3x1 (t )  2 x2 (t )  x1 (t )  2r (t )  x3 (t )   x1 (t )  2 x3 (t )  r (t ) y (t )  x1 (t )  3x2 (t )  2 x3 (t )Saturday, September PMDRMFRCIED 20229, 2012
  • Solution• Signal flow graph 1 2 2 3 1 1/s -1 1/s 1 1/s 1 r y x3 x2 x1 2 2 7 -3 -1Saturday, September PMDRMFRCIED 20329, 2012
  • Q1• For the circuit shown in figure, identify a set of state variables• Answer : one possible set of state variables is the current iL2 via L2, the voltage VC2 across C2 and the current iL1 via L1• VC1 the voltage across C1 can replace iL1 via L1 as the third state variableSaturday, September PMDRMFRCIED 20429, 2012
  • Saturday, September PMDRMFRCIED 20529, 2012
  • Q1• For the circuit shown in figure, determine the state space representation if :• (a). Input are V1 and V2, output is VC2 and state variables are define as x1=iL2, x2=VC2 and x3=iL1• (b). Input are V1 and V2, output is VC2 and state variables are define as x1=iL2, x2=VC2 and x3=VC1Saturday, September PMDRMFRCIED 20629, 2012
  • Q2• Use state variable model to describe the circuit of the figure.• Choose x1=VC and x2=i as state variables.• Determine the state equation only. iSaturday, September PMDRMFRCIED 20729, 2012
  • Tips di L  Ri  VC  Vin dt 1 VC   idt C  1 x1  x 2 C  R 1 1 x2   x2  x1  Vin L L L  1    0  0 C x   1 V x 1 R    in    L  L L   0 1000  0  x  x  10Vin  10  40   Saturday, September PMDRMFRCIED 20829, 2012
  • Q3• Determine a state variable differential matrix equation for the circuit shown in the figure. Choose x1=v1 and x2=v2 as state variables. Two inputs are u1=va and u2=vb. The output is y=v0=v2 V1 V2Saturday, September PMDRMFRCIED 20929, 2012
  • Tipsnode _ 1  Va  V1 V2  V1C1 V 1   R1 R2node _ 2  Vb  V2 V1  V2C2 V 2   R3 R2   1 1  1   1         0   R1C1 R2C1  R2C1 V1   R1C1 Va x . V    .  1  1 1   2   0 1  Vb        R C  R C    R2C2  3 2 2 2   R3C2    Saturday, September PMDRMFRCIED 210 29, 2012
  • Q4An RLC circuit is shown in figure, (a). identify a suitable set of state variables (b). obtain the set of first order differential equations in terms of the state variables x1=i and x2=VC (c). write the state differential equation.Saturday, September PMDRMFRCIED 21129, 2012
  • Tips di 1 R 1  V  i  VC dt L L L 1 VC   idt C  R 1    L    i  1 x L .     L V  1 0  VC   0     C Saturday, September PMDRMFRCIED 21229, 2012
  • Q5• Determine the state equation of the figure. State variables are define as x1=iL and x2=Vc. Input V1 and V2. Draw the corresponding block diagram and signal flow graph of the systemSaturday, September PMDRMFRCIED 21329, 2012
  • Q6• Determine the state space differential equation of the figure. Define the state variables as x1=iL and x2=Vc. System input v1 and v2. The output system is iR. Use KVL around the outer loop and KCL at the node. iRSaturday, September PMDRMFRCIED 21429, 2012
  • Tips diL L  VC  V2  V1  0 dt dVC C  iL  iR dt VC V2 iR    R R  1  1 1    0    V  x1   L . iL    L L . 1  1 1  VC   1  V2   x 2         0    C RC   RC Saturday, September PMDRMFRCIED 21529, 2012
  • Q7• Determine the state variable matrix equation for the circuit shown in the figure. Defined state variables as x1=v1, x2=v2 and x3=iL=i• System input are Vi and iSSaturday, September PMDRMFRCIED 21629, 2012
  • Tips Node _ equation dv1 vi  v1 0.00025  iL  0 dt 4000 dv2 v2 00005  iL   i3  0 dt 1000 diL 0.002  v2  v1  0 dt  1 0  4000  v1  1 0   x  0 2 .v   0 2000 vi  2000   2    i  500  500    i L  0 0    0  s Saturday, September PMDRMFRCIED 21729, 2012
  • Q8• Determine the state variable matrix differential equation for the circuit shown in the figure. The state variables are x1=i, x2=v1 and x3=v2. The output variable is vo(t) and input is V.Saturday, September PMDRMFRCIED 21829, 2012
  • Tips di L  V2 dt  (V1  V )  V1  V2   0 dV1 1 1 C1 dt R1 R2  V2  V1   i  dV2 1 V2 C2 0 dt R2 R3  1   0 0   L  i   0    1 1  1    1  x 0      .V1    V   R1 R2  C1 R2  V   R1C1   1  1 1     0    2 1       C2  R2C2  R2C2 R3C2  Saturday, September PMDRMFRCIED 21929, 2012
  • Q9• Determine the state equation for the two input and one output circuit shown in the figure where state variables are define as x1=iL and x2=Vc the output is y=i2 i1 i3 iCSaturday, September PMDRMFRCIED 22029, 2012
  • Tips dVC C  i2  i3 dt diL L  V1  i1 R1 dt R1i1  R2i2  VC  V1 i2  i1  iL R1  R2 i1  V1  VC  R2iL V2  VC i3  R3Saturday, September PMDRMFRCIED 22129, 2012