Web & Social Media Analytics Previous Year Question Paper.pdf
Chap1see4113
1. Chapter 1
Mathematical Modeling
of Dynamic Systems in
State Space
Saturday, September PMDRMFRCIED 1
29, 2012
2. Introduction to State Space
analysis
• Two approaches are available for the
analysis and design of feedback
control systems
– Classical or Frequency domain technique
– Modern or Time domain technique
Saturday, September PMDRMFRCIED 2
29, 2012
3. Introduction to State Space
analysis
• Classical technique is based on converting a
system’s differential equation to a transfer
function
• Disadvantage
– Can be applied only to Linear Time Invariant system
– Restricted to Single Input and Single output system
• Advantage
– Rapidly provide stability and transient response
information
Saturday, September PMDRMFRCIED 3
29, 2012
4. Introduction to State Space analysis
• Modern technique or state space approach is
a unified method for modeling, analyzing and
designing a wide range of systems
• Advantages :
– Can be used to nonlinear system
– Applicable to time varying system
– Applicable to Multi Input and Multi Output system
– Easily tackled by the availability of advanced digital
computer
Saturday, September PMDRMFRCIED 4
29, 2012
5. Time varying
• A time-varying control system is a
system in which one or more of the
parameters of the system may vary as a
function of time
• Dynamic system: input, state, output
and initial condition
Saturday, September PMDRMFRCIED 5
29, 2012
6. The state variables of a dynamic
system
• The state of a system is a set of variables
whose values, together with the input
signals and the equations describing
the dynamics , will provide the future
state and output of the system
• The state variables describe the present
configuration of a system and can be
used to determine the future response,
given the excitation inputs and the
equations describing the dynamics.
Saturday, September PMDRMFRCIED 6
29, 2012
7. The State Space Equations
x(t ) Ax (t ) Bu (t )
y (t ) Cx(t ) Du(t )
x(t ) derivative _ of _ the _ state _ vector
x(t ) state _ vector
y (t ) output _ vector
u (t ) input _ of _ control _ vector
A system _ matrix
B input _ matrix
C output _ matrix
D feedfoward _ matrix
Saturday, September PMDRMFRCIED 7
29, 2012
8. Two types of equation
• State equation
x(t ) Ax (t ) Bu (t )
• Output equation
y(t ) Cx(t ) Du(t )
Saturday, September PMDRMFRCIED 8
29, 2012
9. Terms
• State equations: a set of n simultaneous,
first order differential equations with n
variables, where the n variables to be
solved are the state variables
• State space: The n-dimensional space
whose axes are the state variables
• State space representation: A
mathematical model for a system that
consists of simultaneous, first order
differential equations and output equation
Saturday, September PMDRMFRCIED 9
29, 2012
10. Terms
• State variables: the smallest set of
linearly independent system variables
such that the value of the members of the
set
• State vector: a vector whose elements
are the state variables
Saturday, September PMDRMFRCIED 10
29, 2012
11. Modeling of Electrical Networks
Voltage-current, voltage-charge, and
impedance relationships for capacitors,
resistors, and inductors
Saturday, September PMDRMFRCIED 11
29, 2012
13. State variable characterization
• The state of the RLC system described
a set of state variables x1 and x2
• X1 = capacitor voltage = vc(t)
• X2 = inductor current = iL(t)
• This choice of state variables is
intuitively satisfactory because the
stored energy of the network can be
described in terms of these variables
1 2 1
E LiL Cvc
2
2 2
Saturday, September PMDRMFRCIED 13
29, 2012
14. Utilizing Kirchhoff’s current law
• At the junction
• First order differential equation
• Describing the rate of change of
capacitor voltage
dvc
ic C u (t ) iL
dt
Saturday, September PMDRMFRCIED 14
29, 2012
15. Utilizing Kirchhoff’s voltage law
• Right hand loop
• Provide the equation describing the
rate of change of inductor current
diL
L Ri L vc
dt
• Output of the system, linear algebraic
equation
vo RiL (t )
Saturday, September PMDRMFRCIED 15
29, 2012
16. State space representation
• A set of two first order differential equation and
output signal in terms of the state variables x1 and x2
dx1 1 1
x2 u (t )
dt C C
dx2 1 R
x1 x2
dt L L
y (t ) vo (t ) Rx 2
1
0 x 1
x
x 1 C . 1 .u
C
x 1 R x2 0
2
L L
x1
y 0 R .
x2
Saturday, September PMDRMFRCIED 16
29, 2012
17. Example 1 : RL serial network
• Figure below shows an RL serial
network with an input voltage vi(t) and
voltage drop at inductance, L as an
output voltage vo(t). Form a state space
model for this system using the current
i(t) in the loop as the state variable.
Saturday, September PMDRMFRCIED 17
29, 2012
18. Modeling of Electrical Networks
• RL serial network – first order system
Saturday, September PMDRMFRCIED 18
29, 2012
19. RL serial network
• Write the loop equation for the system
using Kirchhoff’s voltage law,
Vi (t ) VR (t ) VL (t ) VR (t ) Vo (t )
di (t )
VL (t ) L Vo (t )
dt
VR (t ) i (t ) R
di (t )
Vi (t ) i (t ) R L
dt
Saturday, September PMDRMFRCIED 19
29, 2012
20. RL serial network
• State variable is given only one, therefore
the system is a first order system
• A state equation involving i is required
di (t )
Vi (t ) i (t ) R L
dt
di (t )
L i (t ) R Vi (t )
dt
di (t ) R 1
i (t ) Vi (t )
dt L L
R 1
i (t ) i (t ) Vi (t )
L L
Saturday, September PMDRMFRCIED 20
29, 2012
21. RL serial network
• The output equation,
Vi (t ) VR (t ) VL (t ) VR (t ) Vo (t )
Vo (t ) VR (t ) Vi (t )
Vo (t ) i (t ) R Vi (t )
y (t ) R i (t ) 1Vi (t )
Saturday, September PMDRMFRCIED 21
29, 2012
22. Example 2 : RC serial network
• Figure below shows an RC circuit with
input voltage vi(t) and output voltage at
resistor ie vo(t). Form a state space model
for this system using the voltage vc(t)
across the capacitor as the state variable
R V0
VR
Vi i VC C
Saturday, September PMDRMFRCIED 22
29, 2012
23. RC serial network
• Write the equations for the system using
Kirchhoff’s voltage law,
vi (t ) vR (t ) vc (t ) vc (t ) vo (t ) (1)
for _ the _ capacitor
dvc (t )
i (t ) C (2)
dt
for _ the _ resistor
vo (t ) i (t ) R (3)
Saturday, September PMDRMFRCIED 23
29, 2012
24. RC serial network
• State variable is given only one
• Therefore the system is a first order
system
• Therefore a state equation involving vc is
required
• Combine equation (2) and (3) yields
vo (t ) dvc (t )
i (t ) C
R dt
dvc (t )
vo (t ) RC (4)
dt
Saturday, September PMDRMFRCIED 24
29, 2012
25. RC serial network
• Eliminate vo(t) from equation (4) and
combine with equation (1) and rearrange
gives vi (t ) vc (t ) vo (t )
dvc (t )
vi (t ) vc (t ) RC
dt
dvc (t )
RC vc (t ) vi (t )
dt
dvc (t ) 1 1
vc (t ) vc (t ) vi (t ) (5)
dt RC RC
Saturday, September PMDRMFRCIED 25
29, 2012
26. RC serial network
• Output of the system
vo (t ) vc (t ) vi (t ) (6)
• Rearrange equation (5) and (6) in matrix
form yields
1 1
v c (t ) vc (t ) vi (t )
RC RC
y (t ) 1vc (t ) 1vi (t )
Saturday, September PMDRMFRCIED 26
29, 2012
27. RC serial network
x(t ) state _ vector vc (t )
• Where,
x(t ) derivative _ state _ vector v c (t )
u (t ) input _ vector vi (t )
y (t ) output _ vector vo (t ) vr (t )
1
A state _ matrix
RC
1
B input _ matrix
RC
C ouput _ matrix 1
D direct _ transmission _ matrix 1
Saturday, September PMDRMFRCIED 27
29, 2012
28. Modeling of Electrical Networks
• Consider RLC serial network
• RLC serial network – second order
system
Saturday, September PMDRMFRCIED 28
29, 2012
29. State Variables and output
• Select two state variables,
x1 (t ) q (t )
x2 (t ) i (t )
output y (t ) VL (t )
input u (t ) Vi (t )
Saturday, September PMDRMFRCIED 29
29, 2012
30. Loop equation
• Using Kirchoff’s Voltage Law,
vi (t ) vR (t ) vL (t ) vc (t )
di (t ) 1
L Ri (t ) i (t )dt vi (t )
dt C
Saturday, September PMDRMFRCIED 30
29, 2012
31. Converting to charge
• Using equation,
dq(t )
i (t )
dt
2
d q(t ) dq(t ) 1
L 2
R q(t ) vi (t )
dt dt C
Saturday, September PMDRMFRCIED 31
29, 2012
32. Derivatives of state vector
x1 (t ) q (t )
dq (t )
x1 (t ) i (t ) x2 (t )
dt
x2 (t ) i (t )
di (t )
x2 (t )
dt
Saturday, September PMDRMFRCIED 32
29, 2012
33. State equation
• First state equation
dq(t )
x1 (t ) i(t ) x2 (t )
dt
• Second state equation, using
q (t ) i (t )dt
di (t ) 1
L Ri (t ) i (t )dt v(t )
dt C
di (t ) q (t ) Ri (t ) v(t )
dt LC L L
1 R 1
x2 (t ) x1 (t ) x2 (t ) u (t )
LC L L
Saturday, September PMDRMFRCIED 33
29, 2012
35. Output equation
• Output system is VL
VL (t ) VR (t ) VC (t ) vi (t )
VL (t ) VC (t ) VR (t ) vi (t )
1
V L(t ) i (t )dt i (t ) R vi (t )
C
1
VL (t ) q (t ) Ri (t ) vi (t )
C
1
VL (t ) x1 (t ) Rx 2 (t ) u (t )
C
Saturday, September PMDRMFRCIED 35
29, 2012
36. Output equation in matrix form
y (t ) Cx(t ) Du(t )
1 x1 (t )
y (t ) R 1u (t )
C x2 (t )
1 q(t )
VL (t ) R 1v(t )
C i (t )
Saturday, September PMDRMFRCIED 36
29, 2012
37. Change State Variables but
output still same
x1 (t ) VR (t )
x2 (t ) VC (t )
y (t ) VL (t )
u (t ) Vi (t )
Saturday, September PMDRMFRCIED 37
29, 2012
38. Voltage formula for R, L and C
VR (t ) i (t ) R
1
VC (t ) i (t )dt
C
di (t )
VL (t ) L
dt
Saturday, September PMDRMFRCIED 38
29, 2012
39. Derivative of first state equation
x1 (t ) VR (t )
v(t ) VR (t ) VC (t )
dVR (t ) di (t ) R
x1` (t ) R
dt dt L
R R R
x1 (t ) VR (t ) VC (t ) v(t )
L L L
R R R
x1` (t ) x1 (t ) x2 (t ) u (t )
L L L
Saturday, September PMDRMFRCIED 39
29, 2012
40. Derivative of second state
equation
x2 (t ) VC (t )
dVC (t ) 1 1
x2 (t ) i (t ) VR (t )
dt C RC
1
x2 (t ) x1 (t )
RC
Saturday, September PMDRMFRCIED 40
29, 2012
41. State equation in matrix form
x(t ) Ax (t ) Bu (t )
R R
x (t ) R
x1 (t ) L
x(t ) L 1
L u (t )
x (t ) 1 0 x2 (t ) 0
2
RC
dVR (t ) R R
dt L L VR (t ) R
x(t ) 1 L v(t )
VC (t ) 0
dVC (t )
0
dt RC
Saturday, September PMDRMFRCIED 41
29, 2012
44. Example 3 : 2 loop
• Find a state space representation if the
output is the current through the resistor.
• State variables VC(t) and iL(t)
• Output is iR(t)
• Input is Vi(t)
Saturday, September PMDRMFRCIED 44
29, 2012
45. Electrical network LRC
L
node 1
VL
Vi iL
VR
R C
iR
iC VC
Saturday, September PMDRMFRCIED 45
29, 2012
46. Solution : Step 1
• Label all of the branch currents in the
network.
• iL(t), iR(t) and iC(t)
Saturday, September PMDRMFRCIED 46
29, 2012
47. Solution : Step 2
• Select the state variables by writing
the derivative equation for all
energy-storage elements i.e.
inductor and capacitor
1
VC (t ) iC (t )dt
C
dVC (t )
iC (t ) C (1)
dt
diL (t )
VL (t ) L ( 2)
dt
Saturday, September PMDRMFRCIED 47
29, 2012
48. Solution : Step 3
• Apply network theory, such as Kirchoff’s
voltage and current laws to obtain iC(t)
and VL(t) in terms of the state variable
VC(t) and iL(t)
• At node 1, iL (t ) iR (t ) iC (t )
iC (t ) iL (t ) iR (t )
1
iC (t ) VC (t ) iL (t ) (3)
R
• Around the outer loop,
Vi (t ) VL (t ) VC (t )
VL (t ) VC (t ) Vi (t ) (4)
Saturday, September PMDRMFRCIED 48
29, 2012
49. Solution : Step 4
• Substitute the result of equation (3) and
equation (4) into equation (1) and (2)
dVC (t ) 1
C VC (t ) iL (t ) (7)
dt R
di (t )
L L VC (t ) Vi (t ) (8)
dt
• Rearrange
dVC (t ) 1 1
VC (t ) iL (t ) (9)
dt RC C
diL (t ) 1 1
VC (t ) Vi (t ) (10)
dt L L
Saturday, September PMDRMFRCIED 49
29, 2012
50. Solution : Step 5
• Find the output equation
1
iR (t ) VC (t ) (11)
R
Saturday, September PMDRMFRCIED 50
29, 2012
51. Solution : Step 6
• State space representation in vector
matrix form are
dVC (t ) 1 1
dt RC C VC (t ) 0
di (t ) 1 . 1 v(t ) (12)
L 0 iL (t ) L
dt L
1 VC (t )
iR (t ) 0. (13)
R iL (t )
Saturday, September PMDRMFRCIED 51
29, 2012
52. Example 4 : 2 loop
• Find the state space representation of the
electrical network shown in figure below
• Input vi(t)
• Output vo(t)
• State variables x1(t) = vC1(t), x2(t) = iL(t)
and x3(t) = vC2(t)
Saturday, September PMDRMFRCIED 52
29, 2012
53. RLC two loop network
• Identifying appropriate variables on the
circuit yields C1
node R
iR
iC1
Vi iC2 Vo
DC
L C2
iL
Saturday, September PMDRMFRCIED 53
29, 2012
54. RLC two loop network
• Represent the electrical network shown in
figure in state space where
• Output is v0(t)
• Input is vi(t)
• State variables :-
X1(t) = vC1(t)
X2(t) = iL(t)
X3(t) = vC2(t)
Saturday, September PMDRMFRCIED 54
29, 2012
55. Solution
• Writing the derivative relations for
energy storage elements i.e. C1, C2 and
L
dvC1 (t )
C1 iC1 (t )
dt
diL (t )
L vL (t )
dt
dVC 2 (t )
C2 iC 2 (t )
dt
Saturday, September PMDRMFRCIED 55
29, 2012
56. Solution
• Using Kirchhoff’s current and voltage
laws C
node
1
R
iC1 (t ) iL (t ) iR (t ) iC1
iR
iC2 Vo
iC1 (t ) iL (t ) ic 2 (t )
Vi
DC
L C2
iL
1
iC1 (t ) iL (t ) (vL (t ) vC 2 (t ))
R
vL (t ) vC1 (t ) vi (t )
1
iC 2 (t ) iR (t ) (vL (t ) vC 2 (t ))
Saturday, September
R PMDRMFRCIED 56
29, 2012
57. Solution
• Substituting these relations and
simplifying yields the state equations as
dvC1 1 1 1 1
vC1 iL vC 2 vi
dt RC1 C1 RC1 RC1
diL 1 1
vC1 vi
dt L L
dvC 2 1 1 1
vC1 vC 2 vi
dt RC 2 RC 2 RC 2
vo vC 2
Saturday, September PMDRMFRCIED 57
29, 2012
58. Solution
• Putting the equations in vector matrix
form
1 1 1 1
RC C1
RC1 RC
1
1
x 1 0 0 x 1 v
L L i
1 1 1
0
RC 2 RC 2 RC 2
y 0 0 1x
Saturday, September PMDRMFRCIED 58
29, 2012
59. Tutorial 1 : Number 1
• Represent the electrical network shown in
figure in state space where
• Output is v0(t) and Input is vi(t)
• State variables :-
x1 = v 1
x2 = i4
x3 = v 0
Saturday, September PMDRMFRCIED 59
29, 2012
60. Electrical network 1
• Add the branch current and node
voltages to the network
R1 = 1 Ohm R2 = 1 Ohm R3 = 1 Ohm
V1 V2
i1 i3 i5
Vi C1 = 1 F
L=1H
C2 = 1 F Vo
i2 i4
Saturday, September PMDRMFRCIED 60
29, 2012
61. Solution
• Write the differential equation for each
energy storage element
dv1
i2 ; because _ C1 1F
dt
di4
v2 ; because _ L 1H
dt
dv0
i5 ; because _ C2 1F
dt
Saturday, September PMDRMFRCIED 61
29, 2012
66. Tutorial 1 : Number 2
• Represent the electrical network shown in
figure in state space where
• Output is iR(t)
• Input is vi(t)
• State variables :-
x1 = i2
x2 = vC
Saturday, September PMDRMFRCIED 66
29, 2012
67. Electrical network 2
• Add the branch currents and node
voltages to the schematic and obtain
C = 1F
R1 = 1 Ohm
node V1 node V2
i1
i3
Vi R2=1 Ohm
4V1 iR
DC
L = 1H i2
i4
Saturday, September PMDRMFRCIED 67
29, 2012
68. Solution
• Write the differential equation for each
energy storage element
di2
v1 ; because _ L 1H
dt
dvc
i3 : because _ C 1F
dt
Saturday, September PMDRMFRCIED 68
29, 2012
69. Solution
• Therefore the state vector is,
x1 i2
x
x2 vc
Saturday, September PMDRMFRCIED 69
29, 2012
70. Solution
• Now obtain v1 in terms of the state
variables v1 vc v2
v1 vc iR
v1 vc i3 4v1
v1 vc (i1 i2 ) 4v1
v1 vc vi v1 i2 4v1
1 1 1
v1 i2 vc vi
2 2 2
Saturday, September PMDRMFRCIED 70
29, 2012
71. Solution
• Now obtain i3 in terms of the state
variables i i i
3 1 2
i3 vi v1 i2
1 1 1
i3 vi i2 vc vi i2
2 2 2
3 1 3
i3 i2 vc vi
2 2 2
Saturday, September PMDRMFRCIED 71
29, 2012
72. Solution
• Now obtain the output iR in terms of the
state variables
iR i3 4v1
1 3 1
iR i2 vc vi
2 2 2
Saturday, September PMDRMFRCIED 72
29, 2012
74. Tutorial 1 : Number 3
• Find the state space representation of the
network shown in figure if
• Output is v0(t)
• Input is vi(t)
• State variables :-
x1 = iL1
x2 = iL2
x3 = vC
Saturday, September PMDRMFRCIED 74
29, 2012
75. Electrical network 3
• Add the branch currents and node
voltages to the schematic and obtain
R3 = 1 Ohm
L1 = 1H i3 L2 = 1H
node node
Vi Vo
Vi i2
i1 R2=1 Ohm
DC
Vo
C = 1F
Saturday, September PMDRMFRCIED 75
29, 2012
76. Solution
• Write the differential equation for each
energy storage element
diL1
vc v1
dt
diL 2
vc i2
dt
dvc
i1 i2
dt
Saturday, September PMDRMFRCIED 76
29, 2012
77. Solution
• where,
• L1 is the inductor in the loop with i1
• L2 is the inductor in the loop with i2
• iL1 = i1 –i3
• iL2 = i2 – i3
• Now,
• i1 – i2 = ic = iL1 – iL2 -----------------(1)
Saturday, September PMDRMFRCIED 77
29, 2012
78. Solution
• Also writing the node equation at vo,
• i2 = i3 + iL2 ----------------------(2)
• Writing KVL around the outer loop yields
• i2 + i3 = vi -----------------------(3)
• Solving (2) and (3) for i2 and i3 yields
1 1
i2 iL 2 vi (4)
2 2
1 1
i3 iL 2 vi (5)
2 2
Saturday, September PMDRMFRCIED 78
29, 2012
79. Solution
• Substituting (1) and (4) into the state
equations.
• To find the output equation,
• vo = -i3 + vi
• Using equation (5),
1 1
vo iL 2 vi
2 2
Saturday, September PMDRMFRCIED 79
29, 2012
81. Tutorial 1 : Number 4
• An RLC network is shown in figure. Define
the state variable as :-
• X1 = i1
• X2 = i2
• X3 = Vc
• Let voltage across capacitor, Vc is the
output from the network. Input of the
system is Va and Vb
Saturday, September PMDRMFRCIED 81
29, 2012
82. Tutorial 1 : Number 4
• Determine the state space representation
of the RLC network in matrix form
• Determine the range of resistor R in order
to maintain the system’s stability, if C = 0.1
F and L1=L2=0.1 H. The characteristic
equation of the system is,
s 10Rs 200s 1000R 0
3 2
Saturday, September PMDRMFRCIED 82
29, 2012
83. RLC network with 2 input
R L1 L2
i1 + i2
Va iC C -
VC Vb
DC
DC
Saturday, September PMDRMFRCIED 83
29, 2012
84. Solution
• State variables and their derivatives
di1
x1 i1 x1
dt
di2
x2 i2 x2
dt
dvc
x3 vc x3
dt
u1 va
u 2 vb
y vc
Saturday, September PMDRMFRCIED 84
29, 2012
85. Solution
• The derivatives equations for energy
storage elements
di1
L1 vL1 (1)
dt
di2
L2 vL 2 (2)
dt
dvC
C iC (3)
dt
Saturday, September PMDRMFRCIED 85
29, 2012
86. Solution L1 L2
R
i1 + i2
• For loop (1) ; Va iC VC Vb
va i1R vL1 vC
C -
DC
DC
vL1 va i1 R vC (4)
• For loop (2) ;
vb vL 2 vC
vL 2 vb vC (5)
Saturday, September PMDRMFRCIED 86
29, 2012
87. Solution
• For current iC ;
iC i1 i2 (6)
• Substituting equation (4), (5) and (6) into
equation (1), (2) and (3) yields
L1 L2
di1 R
L1 va i1 R vC i1 + i2
dt Va iC C -
VC Vb
DC
DC
di1 R 1 1
i1 vC va
dt L1 L1 L1
di1 R 1 1
x1 x1 x3 va (7)
dt L1 L1 L1
Saturday, September PMDRMFRCIED 87
29, 2012
92. Solution
• For stability, all coefficients in first column
of Routh Hurwitz table must be positive ;
(2000 R 1000 R)
0
10 R
1000 R 0
R 0
Saturday, September PMDRMFRCIED 92
29, 2012
93. Modeling of Mechanical
• Mass
Networks
f (t ) M .a (t )
f (t ) M .
d 2 y (t ) y(t)
dt 2
dv(t )
f (t ) M .
dt
a (t ) accelerati on
v(t ) velocity
M f(t)
y (t ) displaceme nt
f (t ) force
M mass
Saturday, September PMDRMFRCIED 93
29, 2012
94. Modeling of Mechanical
• Linear Spring
Networks
f (t ) K . y (t )
K y(t)
f (t ) force
y (t ) displaceme nt f(t)
K spring _ cons tan t
Saturday, September PMDRMFRCIED 94
29, 2012
95. Modeling of Mechanical
Networks
• Damper
dy (t ) B y(t)
f (t ) B.
dt
f (t ) force f(t)
y (t ) displaceme nt
B viscous _ frictional
Saturday, September PMDRMFRCIED 95
29, 2012
96. Modeling of Mechanical
Networks
• Inertia
d (t )
T (t ) J .
dt
T(t)
d 2 (t ) (t )
T (t ) J .
dt 2
T (t ) Torque J
(t ) angular _ velocity
(t ) angular _ displaceme nt
J Inertia
Saturday, September PMDRMFRCIED 96
29, 2012
97. Force-velocity, force-displacement, and
impedance translational relationships
for springs, viscous dampers, and mass
Saturday, September PMDRMFRCIED 97
29, 2012
98. Torque-angular velocity, torque-angular
displacement, and impedance rotational
relationships for springs, viscous dampers,
and inertia
Saturday, September PMDRMFRCIED 98
29, 2012
99. Example 5
• Determine the state space
representation of the mechanical
system below if the state variables are
y(t) and dy(t)/dt. Input system is force
f(t) and output system is y(t)
K
y(t)
B M
f(t)
Saturday, September PMDRMFRCIED 99
29, 2012
100. Example 5
• a. Mass, spring, and damper system;
b. block diagram
Saturday, September PMDRMFRCIED 100
29, 2012
101. State variables, input and output
x1 (t ) y (t )
dy (t ) dx1 (t )
x2 (t )
dt dt
input u f (t )
output y y (t )
Saturday, September PMDRMFRCIED 101
29, 2012
102. Mass, spring and damper
system
• Draw the free body diagram
y(t)
d 2 y (t )
M
dt 2
Ky (t ) M f(t)
dy (t )
B
dt
Saturday, September PMDRMFRCIED 102
29, 2012
103. Mass, spring and damper system
• a. Free-body diagram of mass, spring, and
damper system;
b. transformed free-body diagram
Saturday, September PMDRMFRCIED 103
29, 2012
104. Mass, spring and damper
system
• The force equation of the system is
2
d y(t ) dy(t )
f (t ) M . 2
B. K . y(t )
dt dt
• Rearranged the equation yields
2
d y (t ) B dy (t ) K 1
2
. . y (t ) . f (t )
dt M dt M M
Saturday, September PMDRMFRCIED 104
29, 2012
105. Mass, spring and damper
system
• State equations and output equation
x1 (t ) x2 (t )
K B 1
x 2 (t ) .x1 (t ) .x2 (t ) . f (t )
M M M
y (t ) x1 (t )
Saturday, September PMDRMFRCIED 105
29, 2012
106. Mass, spring and damper
system
• State space representation in vector
matrix form are
0 1 x (t ) 0
x1 (t )
K B . 1 1 . f (t )
x (t ) M x2 (t )
2 M M
K
x1 (t )
y (t ) 1 0.
y(t)
B M
x2 (t ) f(t)
Saturday, September PMDRMFRCIED 106
29, 2012
107. Example: The mechanical
system
• Consider the mechanical system shown in
Figure below by assuming that the system
is linear. The external force u(t) is the
input to the system and the displacement
y(t) of the mass is the output. The
displacement y(t) is measured from the
equilibrium position in the absence of the
external force. This system is a single
input and single output system.
Saturday, September PMDRMFRCIED 107
29, 2012
109. Mechanical system diagram
• From the diagram, the system equation is
m y b y ky u
• The system is of second order. This
means that the system involves two
integrators. Define the state variables x1(t)
and x2(t) as x1 (t ) y (t )
x2 ( t ) y ( t )
Saturday, September PMDRMFRCIED 109
29, 2012
110. • Then we obtain,
x 1 x2
1
1 u
x 2 ky b y
m m
k b 1
x 2 x1 x2 u
m m m
y x1
Saturday, September PMDRMFRCIED 110
29, 2012
111. Mass, spring and damper system
• a. Two-degrees-of-freedom translational
mechanical system
• b. block diagram
Saturday, September PMDRMFRCIED 111
29, 2012
112. Mass, spring and damper system
• a. Forces on M1 due only to motion of M1
b. forces on M1 due only to motion of M2
c. all forces on M1
Saturday, September PMDRMFRCIED 112
29, 2012
113. Mass, spring and damper system
• a. Forces on M2 due only to motion of M2;
b. forces on M2 due only to motion of M1;
c. all forces on M2
Saturday, September PMDRMFRCIED 113
29, 2012
114. Exercise 1
• Figure below shows a diagram for a quarter car
model (one of the four wheels) of an automatic
suspension system for a long distance express
bus. A good bus suspension system should
have satisfactory road handling capability, while
still providing comfort when riding over bumps
and holes in the road. When the coach is
experiencing any road disturbance, such as
potholes, cracks, and uneven pavement, the bus
body should not have large oscillations, and the
oscillations should be dissipate quickly.
Saturday, September PMDRMFRCIED 114
29, 2012
115. Exercise 1
(i). Draw the free-body diagrams of the system
(ii). Determine the state space representation
of the quarter car system by considering the
state vector
T
z(t) x1 (t ) x2 (t ) x1 (t ) x 2 (t )
And the displacement of bus body mass M1 as
the output of the system.
Saturday, September PMDRMFRCIED 115
29, 2012
117. Constant value
• Bus body mass, M1 = 2500 kg
• Suspension mass, M2 = 320 kg
• Spring constant of suspension system, K1 =
80,000 N/m
• Spring constant of wheel and tire, K2 =
500,000 N/m
• Damping constant of suspension system, B1
= 350 Ns/m
• Damping constant of wheel and tire, B2 =
15,020 Ns/m
Saturday, September PMDRMFRCIED 117
29, 2012
118. Solution
• Free body diagram for M1
– Forces on M1 due to motion of M1
K1X1
M1s2X1 u
M1
B1sX1
– Forces on M1 due to motion of M2
K1X2
M1
B1sX2
– All forces on M1
K1X1 K1X2
M1s2X1 M1 u
B1sX1 B1sX2
Saturday, September PMDRMFRCIED 118
29, 2012
119. Solution
• Free body diagram for M2
– Forces on M2 due to motion of M2
K2X2
M2s2X2 K1X2
M2
B2sX2 B1sX2
– Forces on M2 due to motion of M1
K1X1
M2
B1sX1
– All forces on M2
(K1+K2)X2
M2s2X2 M2 K1X1
(B1+B2)sX2
B1sX1
Saturday, September PMDRMFRCIED 119
29, 2012
120. Solution
• State variables
z1 x1; z2 x2 ; z3 x1; z4 x 2
• Derivative state variables
z1 x1 z3 ; z 2 x 2 z4 ; z 3 x1; z 4 x 2
Saturday, September PMDRMFRCIED 120
29, 2012
121. Solution
• Total force for M1
dx2 d 2 x1 dx1
u K1 x2 B1 K1 x1 M 1 2 B1
dt dt dt
2
d x1 dx1 dx2
2
32 x1 32 x2 0.14 0.14 0.0004u
dt dt dt
z 3 32 z1 32 z2 0.14 z3 0.14 z4 0.0004u
Saturday, September PMDRMFRCIED 121
29, 2012
122. Solution
• Total force for M2
2
dx1 d x2 dx2
K1 x1 B1 ( K1 K 2 ) x2 M 2 ( B1 B2 )
dt dt dt
d 2 x2 dx1 dx2
2
250 x1 1812.5 x2 1.094 48.031
dt dt dt
z 4 250 z1 1812.5 z 2 1.094 z3 48.031z 4
Saturday, September PMDRMFRCIED 122
29, 2012
124. Tutorial 1 : Number 5
• Figure shows a mechanical system
consisting of mass M1 and M2, damper
constant B, spring stiffness K1 and K2.
When force f(t) acts on mass M1, it
moves to position x1(t) while mass M2
moves to position x2(t). Find the state
space representation of the system
using x1(t), x2(t) and their first
derivatives as state variables. Let x2(t)
be the output.
Saturday, September PMDRMFRCIED 124
29, 2012
125. Mechanical system consist of 2
mass, 2 spring and 1 damper
f(t) X1 X2
K1 B K2
M1 M2
Saturday, September PMDRMFRCIED 125
29, 2012
126. Mechanical system consist of 2
mass, 2 spring and 1 damper
• State variables and their derivatives :-
z1 (t ) x1 (t ) z1 (t ) x1 (t ) z 2 (t )
z 2 (t ) x1 (t ) z 2 (t ) x1 (t )
z3 (t ) x2 (t ) z3 (t ) x 2 (t ) z 4 (t )
z 4 (t ) x 2 (t ) z 4 (t ) x2 (t )
input u (t ) f (t )
output y (t ) x2 (t )
Saturday, September PMDRMFRCIED 126
29, 2012
127. Mechanical system consist of 2
mass, 2 spring and 1 damper
• Draw the free body diagram
K1 x1
f (t )
M 1 x1
M1
B x1
B x2
K 2 x2
M 2 x2
M2 B x1
B x2
Saturday, September PMDRMFRCIED 127
29, 2012
128. Mechanical system consist of 2
mass, 2 spring and 1 damper
• Differential equation in mass M1
f (t ) M 1 x1 B x1 K1 x1 B x2
f (t ) M 1 x1 B( x1 x2 ) K1 x1 (1)
• Differential equation in mass M2
0 M 2 x2 B x2 K 2 x2 B x1
0 M 2 x2 B( x2 x1 ) K 2 x2 (2)
Saturday, September PMDRMFRCIED 128
29, 2012
129. Mechanical system consist of 2
mass, 2 spring and 1 damper
• Substitute all state variables and their first
derivatives in equation (1) and (2) yields
B B K1 f (t )
x1 x1 x2 x1
M1 M1 M1 M1
B B K 1
z2 z2 z 4 1 z1 u (3)
M1 M1 M1 M1
B B K2
x2 x2 x1 x2
M2 M2 M2
B B K
z4 z4 z 2 2 z3 (4)
M2 M2 M2
z1 z 2 (5)
z 3 z 4 ( 6)
Saturday, September PMDRMFRCIED 129
29, 2012
130. Mechanical system consist of 2
mass, 2 spring and 1 damper
• Rearrange equation 3, 4, 5 and 6 in matrix
form 0 1 0 0
0
z 1 K1 z B B 1
z M 0 z 1
M1 M1 2 M
z 2 1
. 1 u
z3 0 0 0 1 z3 0
0 B K2 B
z4 0
z4
M2 M2 M2
z1
z
y 0 0 1 0. 2
z3
z4
Saturday, September PMDRMFRCIED 130
29, 2012
131. Tutorial 1 : Number 6
• Represent the translational mechanical
system shown in figure in state space
where x3(t) is the output and f(t) is the
input.
X1 X2 X3
K1 B1 K2 B2
M1 M2 M3
f(t)
Saturday, September PMDRMFRCIED 131
29, 2012
132. Example : 3M, 2K and 2B
• Represent the translational mechanical
system shown in figure in state space
where x3(t) is the output and f(t) is the
input.
Saturday, September PMDRMFRCIED 132
29, 2012
133. Example : 3M, 2K and 2B
• K1 = K2 = 1 N/m
• M1 = M2 = M3 = 1 kg
• B1 = B2 = 1 N-s/m
• Find the state space representation of the
system using x1, x2, x3 and their first
derivatives as state variables.
z1 x1 ; z2 x1 ; z3 x2 ;
z4 x2 ; z5 x3 ; z6 x3
Saturday, September PMDRMFRCIED 133
29, 2012
134. Example : 3M, 2K and 2B
• Draw the free body diagram
M 1 x1 B1 x2
B1 x1
M1 f (t )
K1 x1
M 2 x2
B1 x1
B1 x2 M2 K 2 x3
K 2 x2
M 3 x3
B2 x3
M3 K 2 x2
K 2 x3
Saturday, September PMDRMFRCIED 134
29, 2012
135. Example : 3M, 2K and 2B
• Writing the equations of motion
M 1 x1 B1 x1 K1 x1 B1 x2 f (t ) (1)
M 2 x2 B1 x2 K 2 x2 B1 x1 K 2 x3 (2)
M 3 x3 B2 x3 K 2 x3 K 2 x2 (3)
Saturday, September PMDRMFRCIED 135
29, 2012
136. Example : 3M, 2K and 2B
• Substitute the value of K, M and B.
• Rearrange equation (1), (2) and (3)
x1 x1 x1 x2 f
x2 x1 x2 x2 x3
x3 x3 x3 x2
Saturday, September PMDRMFRCIED 136
29, 2012
137. Example : 3M, 2K and 2B
• From the state variables
z1 x1 z1 x1 z 2
z 2 x1 z 2 x1 z 2 z1 z 4 f
z 3 x 2 z 3 x2 z 4
z 4 x2 z 4 x2 z 2 z 4 z 3 z 5
z5 x3 z5 x3 z6
z6 x3 z6 x3 z6 z5 z3
y x3 z5
Saturday, September PMDRMFRCIED 137
29, 2012
139. Modeling of Electro-Mechanical System
• NASA flight simulator robot arm with
electromechanical control system components
Saturday, September PMDRMFRCIED 139
29, 2012
142. DC motor armature control
• The back electromotive force(back emf),
VB
d m (t )
VB (t )
dt
d m (t )
VB (t ) K B . (1)
dt
K B Back _ emf _ cons tan t
Saturday, September PMDRMFRCIED 142
29, 2012
143. DC motor armature control
• Kirchoff’s voltage equation around the
armature circuit
ea (t ) ia (t ) Ra Vb (t )
d m (t )
ea (t ) ia (t ) Ra K b (2)
dt
ia armature _ current
m angular _ displaceme nt _ of _ the _ armature
Ra armature _ resis tan ce
ignore _ La
Saturday, September PMDRMFRCIED 143
29, 2012
144. DC motor armature control
• The torque, Tm(t) produced by the motor
Tm (t ) ia (t )
Tm (t ) K t ia (t )
d 2 m d m
Tm (t ) J m 2
Dm (3)
dt dt
K t Torque _ cons tan t
J m equivalent _ inertia _ by _ the _ motor
Dm equivalent _ viscous _ density _ by _ the _ motor
Saturday, September PMDRMFRCIED 144
29, 2012
145. DC motor armature control
• Solving equation (3) for ia(t)
J m d m Dm d m
2
ia (t ) 2
(4)
K t dt K t dt
Saturday, September PMDRMFRCIED 145
29, 2012
146. DC motor armature control
• Substituting equation (4) into equation (2)
yields
J m d 2 m Dm d m d m
ea (t ) Ra 2
Kb
K t dt K t dt dt
Ra J m d 2 m Ra Dm d m
ea (t )
K dt . 2 K K b .
dt (5)
t t
Saturday, September PMDRMFRCIED 146
29, 2012
147. DC motor armature control
• Define the state variables, input and ouput
x1 m (6a )
d m
x2 (6b)
dt
u ea (t )
y 0.1 m
• Substituting equation (6) into equation (5)
yields Ra J m dx2 Ra Dm
ea (t )
K . dt K K b .x2 (7)
t t
Saturday, September PMDRMFRCIED 147
29, 2012
148. DC motor armature control
• Solving for x2 dot yields,
Ra Dm
ea (t )
K K b .x2
dx2
t
dt Ra J m
K
t
dx2 K t Dm K b K t
R J .ea (t ) J R J .x2
dt a m
m a m
dx2 1 Kb Kt Kt
Dm
.x2
R J .ea (t ) (8)
dt Jm Ra a m
Saturday, September PMDRMFRCIED 148
29, 2012
149. DC motor armature control
• Using equation (6) and (8), the state
equations are written as
dx1 d m
x2
dt dt
dx2 1 Kt Kb Kt
Dm
.x2
.ea (t )
R J
dt Jm Ra a m
Saturday, September PMDRMFRCIED 149
29, 2012
150. DC motor armature control
• Assuming that the output o(t) is 0.1 the
displacement of the armature m(t) as x1.
Hence the output equation is
y 0.1x1
• State space representation in vector
matrix form are
0 1
x1 K
0
x1
1 K t K b . t .ea (t )
x 0 J Dm R x2 R J
2 m a a m
x1
y 0.1 0.
x2
Saturday, September PMDRMFRCIED 150
29, 2012
151. Tutorial 1 : Number 7
• The representation of the positioning system
using an armature-controlled dc motor is shown
in figure.
• The input is the applied reference voltage, r(t)
and the output is the shaft’s angular position,
o(t).
• The dynamic of the system can be described
through the Kirchoff equation for the armature
circuit, the Newtonian equation for the
mechanical load and the torque field current
relationship.
Saturday, September PMDRMFRCIED 151
29, 2012
152. Figure : DC motor armature control
Saturday, September PMDRMFRCIED 152
29, 2012
153. Example : ex-exam question
• The Newtonian equation for the
mechanical load is
J o (t ) o (t ) (t )
• The back e.m.f voltage induced in the
armature circuit, eb(t) is proportional to
the motor shaft speed,
eb K b o
Saturday, September PMDRMFRCIED 153
29, 2012
154. Example : ex-exam question
• A potentiometer was installed to measure
the motor output position. Its output
voltage, v(t) is then compared with the
system reference input voltage, r(t)
through an op-amp.
• Determine the complete state-space
representation of the system by
considering the following state variables.
Saturday, September PMDRMFRCIED 154
29, 2012
155. Example : ex-exam question
x1(t) ia (t)
• State variables :- x 2 (t) o (t)
x 3 (t) o (t)
• State variables derivative
dia (t)
x 1 (t) ia
dt
d
o (t)
x 2 (t) o (t)
dt
d 2 o (t)
x 3 (t) 2
o (t)
dt
Saturday, September PMDRMFRCIED 155
29, 2012
156. Example : ex-exam question
• Mechanical load
J o (t) o (t) (t) K t ia
J x 3 x 3 K t x1
Kt
x3 x1 x 3 (1)
J J
Saturday, September PMDRMFRCIED 156
29, 2012
157. Example : ex-exam question
• Electrical (armature) circuit
• Using Kirchoff Voltage Law
dia
uL Ria eb
dt
but
eb K b o ( given)
dia
u L Ria K b o
dt
u L x1 Rx1 K b x3
R Kb 1
x1 x1 x3 u (2)
L L L
Saturday, September PMDRMFRCIED 157
29, 2012
158. Example : ex-exam question
• From the state variable defination
x2 o
x 2 o x3 (3)
For _ the _ input _ part
u r v
u r K s o
u r K s x2 (4)
Saturday, September PMDRMFRCIED 158
29, 2012
159. Example : ex-exam question
• Substituting (4) into (2)
R Kb 1
x1 x1 x3 (r K s x2 )
L L L
R Ks Kb 1
x1 x1 x2 x3 r (5)
L L L L
Saturday, September PMDRMFRCIED 159
29, 2012
160. Example : ex-exam question
• Writing equations (1), (3) and (5) in the
vector matrix form gives :-
R
Ks
Kb 1
x1
L L x1 L
L
x2 0 0 1 . x2 0 r
KT
x3 0 x3 0
J J
Saturday, September PMDRMFRCIED 160
29, 2012
161. Example : ex-exam question
• The output
x1
x
y o 0 1 0 2
x3
Saturday, September PMDRMFRCIED 161
29, 2012
162. Modelling of Electro-Mechanical
System
• Field Controlled DC
+
Rf
ef (t)
Motor if (t)
-
Lf Gelung Medan
Ra La
+
ea
Ba
Ja
ia
- TL(t)
Tm(t)
Gelung Angker m (t )
Tetap
RAJAH 7.11 : MOTOR SERVO A.T. TERUJA BERASINGAN
DALAM KAWALAN MEDAN
Saturday, September PMDRMFRCIED 162
29, 2012
163. DC motor field control
• For field circuit
di f
e(t ) i f R f L f (1)
dt
• For mechanical load, torque
d o d o
2
T (t ) J B (2)
dt dt
Saturday, September PMDRMFRCIED 163
29, 2012
164. DC motor field control
• For torque and field current relationship
T (t ) i f (t )
T (t ) K t i f (t ) (3)
• Define the state variables, input and
output x (t ) (4)
1 o
d o (t )
x2 (5)
dt
x3 i f (t ) (6)
u e(t )
y o (t )
Saturday, September PMDRMFRCIED 164
29, 2012
165. DC motor field control
• From equation (4) and (5), we can
determine the first state equation as :
d o
x1 (t ) x2 (t ) ( 7)
dt
• Another two state equations are :
d 2 o
x2 2 (8)
dt
di f
x3 (9)
dt
Saturday, September PMDRMFRCIED 165
29, 2012
166. DC motor field control
• Substituting x3 and x3 dot into equation (1)
yields
e(t ) x3 R f L f x3
• Substituting equation (3) into equation (2)
d o d o
yields 2
J 2 B Kt i f
dt dt
Saturday, September PMDRMFRCIED 166
29, 2012
167. DC motor field control
• Substituting x2 dot, x2 and x3, hence
J x2 Bx2 Kt x3
• Rewrite equations
Rf 1
x3 x3 e(t )
Lf Lf
B Kt
x2 x2 x3
J J
Saturday, September PMDRMFRCIED 167
29, 2012
168. DC motor field control
• Matrix form
0 1 0 0
x 0 B Kt x 0 u
J J 1
Rf
0 0 Lf
Lf
y 1 0 0x
Saturday, September PMDRMFRCIED 168
29, 2012
169. Block diagrams
• The block diagram is a useful tool for
simplifying the representation of a
system.
• Simple block diagrams only have one
feedback loop.
• Complex block diagram consist of more
than one feedback loop, more than 1 input
and more than 1 output i.e. inter-coupling
exists between feedback loops
Saturday, September PMDRMFRCIED 169
29, 2012
170. Block diagrams
• Integrator
x2 x1dt
x1
• Amplifier or gain x1 x2 = Kx1
K
x1
+ x4 = x1-x2+x3
• Summer x2 -
x3
+
Saturday, September PMDRMFRCIED 170
29, 2012
171. Signal flow graphs
• Having the block diagram simplifies the
analysis of a complex system.
• Such an analysis can be further simplified
by using a signal flow graphs (SFG) which
looks like a simplified block diagram
• An SFG is a diagram which represents a
set of simultaneous equation.
• It consist of a graph in which nodes are
connected by directed branches.
Saturday, September PMDRMFRCIED 171
29, 2012
172. Signal flow graphs
• The nodes represent each of the system
variables.
• A branch connected between two nodes
acts as a one way signal multiplier: the
direction of signal flow is indicated by an
arrow placed on the branch, and the
multiplication factor(transmittance or
transfer function) is indicated by a letter
placed near the arrow.
Saturday, September PMDRMFRCIED 172
29, 2012
173. Signal flow graphs
• A node performs two functions:
1. Addition of the signals on all incoming
branches
2. Transmission of the total node signal(the
sum of all incoming signals) to all outgoing
branches
Saturday, September PMDRMFRCIED 173
29, 2012
174. Signal flow graphs
• There are three types of nodes:
1. Source nodes (independent nodes) – these
represent independent variables and have
only outgoing branches. u and v are source
nodes
2. Sink nodes (dependent nodes) - these
represent dependent variables and have
only incoming branches. x and y are source
nodes
3. Mixed nodes (general nodes) – these have
both incoming and outgoing branch. W is a
mixed node.
Saturday, September PMDRMFRCIED 174
29, 2012
175. Signal flow graphs
• x2 = ax1
x1 a x2 = ax1
Saturday, September PMDRMFRCIED 175
29, 2012
176. Signal flow graphs
• w = au + bv
• x = cw
• y = dw
a w c x
u
v b d y
Saturday, September PMDRMFRCIED 176
29, 2012
177. Signal flow graphs
• x = au + bv +cw
u
a x
c 1 x
w
v b Mixed Sink
node node
Saturday, September PMDRMFRCIED 177
29, 2012
178. Signal flow graphs
• A path is any connected sequence of
branches whose arrows are in the same
direction
• A forward path between two nodes is one
which follows the arrows of successive
branches and in which a node appears
only once.
• The path uwx is a forward path between
the nodes u and x
Saturday, September PMDRMFRCIED 178
29, 2012
179. Signal flow graphs
• Series path (cascade nodes) – series path
can be combined into a single path by
multiplying the transmittances
• Path gain – the product of the
transmittance in a series path
• Parallel paths – parallel paths can be
combined by adding the transmittances
• Node absorption – a node representing a
variable other than a source or sink can be
eliminated
Saturday, September PMDRMFRCIED 179
29, 2012
180. Signal flow graphs
• Feedback loop – a closed path which
starts at a node and ends at the same
node.
• Loop gain – the product of the
transmittances of a feedback loop
Saturday, September PMDRMFRCIED 180
29, 2012
181. Signal flow graphs
simplification
Original graph Equivalent graph
a b ab
x y z x z
Saturday, September PMDRMFRCIED 181
29, 2012
182. Signal flow graphs
simplification
Original graph Equivalent graph
a
(a+b)
x y
x y
b
Saturday, September PMDRMFRCIED 182
29, 2012
183. Signal flow graphs
simplification
Original graph Equivalent graph
w ac
a z w
c z
x y x bc
b
Saturday, September PMDRMFRCIED 183
29, 2012
184. Block diagram of feedback
system
R E C
G
B
H
Saturday, September PMDRMFRCIED 184
29, 2012
185. Block diagram of feedback system
• R=reference input
• E=actuating signal
• G=control elements and controlled system
• C=controlled variable
• B=primary feedback
• H=feedback elements
• C = GE
• B = HC
• E = R-B
Saturday, September PMDRMFRCIED 185
29, 2012
186. Successive reduction of SFG
first second
• 4 nodes • Node B eliminated
R 1 E G C R 1 E G C
-1
H
B -H
Saturday, September PMDRMFRCIED 186
29, 2012
187. Successive reduction of SFG
third fourth
• Node E eliminated, self • Self loop eliminated
loop of value -GH
R G C R C
G/(1+GH)
-GH
Saturday, September PMDRMFRCIED 187
29, 2012
188. SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
• demonstrate how to draw signal flow
graphs from state equations.
• Consider the following state and output
equations:
x1 2 x1 5x2 3x3 2r (1a)
x2 6 x1 2 x2 2 x3 5r (1b)
x3 x1 3x2 4 x3 7r (1c)
y 4x1 6x2 9x3 (1d)
Saturday, September PMDRMFRCIED 188
29, 2012
189. SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
• Step 1 : Identify three nodes to be the
three state variables, , and three nodes,
placed to the left of each respective
state variables. Also identify a node as
the input, r, and another node as the
output, y.
R(s) Y(s)
sX3 (s) X3 (s) sX (s) X2 (s) sX (s) X (s)
2 1 1
Saturday, September PMDRMFRCIED 189
29, 2012
190. SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
• Step 2 : Interconnect the state variables
and their derivatives with the defining
integration, 1/s.
1 1 1
s s s
R(s) Y(s)
sX (s) X (s) sX (s) X (s) sX (s) X (s)
3 3 2 2 1 1
Saturday, September PMDRMFRCIED 190
29, 2012
191. SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
• Step 3 : Using Eqn (1a), feed to each node
the indicated signals.
2
1 1 1
s s -5 s
R(s) Y(s)
X (s) X2 (s) sX (s) X (s)
sX3 (s) 3 sX2 (s) 1 1
2
3
Saturday, September PMDRMFRCIED 191
29, 2012
192. SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
• Step 4 : Using Eqn (1b), feed to each node
the indicated signals.
2
5
1 1 1
s 2 s -5 s
R(s) Y(s)
X (s) sX (s) X (s) sX (s) X (s)
sX3 (s) 3 2 2 1 1
-2 2
3 -6
Saturday, September PMDRMFRCIED 192
29, 2012
193. SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
• Step 5 : Using Eqn (1c), feed to each node
the indicated signals.
2
5
1 1 1
7 s 2 s -5 s
R(s) Y(s)
sX (s) X (s) sX (s) X2 (s) sX (s) X1 (s)
3 3 2 1
-4 -2 2
-3 3 -6
1
Saturday, September PMDRMFRCIED 193
29, 2012
194. SIGNAL FLOW GRAPHS OF
STATE EQUATIONS
• Step 6 : Finally, use Eqn (1d) to complete
the signal flow2 graph. 9
5 6
1 1 1
7 s 2 s -5 s -4
R(s) Y(s)
sX (s) X (s) sX (s) X2 (s) sX (s) X1 (s)
3 3 2 1
-4 -2 2
-3 3 -6
1
Saturday, September PMDRMFRCIED 194
29, 2012
195. Example 7
• Draw a signal-flow graph for each of the
following state equations :
0 1 0 x1 0
x(t ) 0 0 . x 0 r (t )
1 2
2 4 6 x3 1
x1
y (t ) 1 1 0. x2
x3
Saturday, September PMDRMFRCIED 195
29, 2012