Modeling & Simulation of Dynamic
Systems
Lecture-2
Block Diagram & Signal Flow Graph
Representation of Control Systems
Dr. Imtiaz Hussain
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
Introduction
• A Block Diagram is a shorthand pictorial representation of
the cause-and-effect relationship of a system.
• The interior of the rectangle representing the block usually
contains a description of or the name of the element, or the
symbol for the mathematical operation to be performed on
the input to yield the output.
• The arrows represent the direction of information or signal
flow.
x
d
dt
y
Introduction
• The operations of addition and subtraction have a special
representation.
• The block becomes a small circle, called a summing point, with
the appropriate plus or minus sign associated with the arrows
entering the circle.
• The output is the algebraic sum of the inputs.
• Any number of inputs may enter a summing point.
• Some books put a cross in the circle.
Introduction
• In order to have the same signal or variable be an input
to more than one block or summing point, a takeoff
point is used.
• This permits the signal to proceed unaltered along
several different paths to several destinations.
Example-1
• Consider the following equations in which x1, x2,. . . , xn, are
variables, and a1, a2,. . . , an , are general coefficients or
mathematical operators.
x n  a1 x1  a 2 x 2  a n 1 x n 1
Exercise-1
• Draw the Block Diagrams of the following equations.
(1 )
(2)
x 2  a1
dx 1
x 3  a1
d

dt
2
dt
1
b
x2
2
 x1 dt
 3
dx 1
dt
 bx 1
Canonical Form of A Feedback Control System
Characteristic Equation
• The control ratio is the closed loop transfer function of the system.
C(s)
R( s )

G(s)
1  G ( s )H ( s )
• The denominator of closed loop transfer function determines the
characteristic equation of the system.
• Which is usually determined as:
1  G ( s )H ( s )  0
Example-2
B( s )
1. Open loop transfer function
E(s)
2. Feed Forward Transfer function
C(s)
3. control ratio
R( s )
4. feedback ratio
5. error ratio

B( s )
R( s )
E(s)
R( s )

 G ( s )H ( s )
C(s)
 G(s)
E(s)
G(s)
G(s)
1  G ( s )H ( s )

G ( s )H ( s )
1  G ( s )H ( s )
H (s)
1
1  G ( s )H ( s )
6. closed loop transfer function
C(s)
R( s )

G(s)
1  G ( s )H ( s )
7. characteristic equation 1  G ( s ) H ( s )  0
8. closed loop poles and zeros if K=10.
Reduction techniques
1. Combining blocks in cascade
G2
G1
G 1G 2
2. Combining blocks in parallel
G1
G2
G1  G 2
Reduction techniques
3. Moving a summing point behind a block
G
G
G
3. Moving a summing point ahead of a block
G
G
1
G
4. Moving a pickoff point behind a block
G
G
1
G
5. Moving a pickoff point ahead of a block
G
G
G
6. Eliminating a feedback loop
G
G
1  GH
H
G
G
1 G
H 1
7. Swap with two neighboring summing points
A
B
B
A
Example-3
• For the system represented by the following block diagram
determine:
1.
2.
3.
4.
5.
6.
7.
8.
Open loop transfer function
Feed Forward Transfer function
control ratio
feedback ratio
error ratio
closed loop transfer function
characteristic equation
closed loop poles and zeros if K=10.
Example-3
– First we will reduce the given block diagram to canonical form
K
s 1
Example-3
K
s 1
K
G
1  GH

1
s 1
K
s 1
s
Example-3
B( s )
1. Open loop transfer function
E(s)
2. Feed Forward Transfer function
C(s)
3. control ratio
R( s )
4. feedback ratio
5. error ratio

B( s )
R( s )
E(s)
R( s )

 G ( s )H ( s )
C(s)
 G(s)
E(s)
G(s)
G(s)
1  G ( s )H ( s )

G ( s )H ( s )
1  G ( s )H ( s )
H (s)
1
1  G ( s )H ( s )
6. closed loop transfer function
C(s)
R( s )

G(s)
1  G ( s )H ( s )
7. characteristic equation 1  G ( s ) H ( s )  0
8. closed loop poles and zeros if K=10.
Exercise-2
• For the system represented by the following block diagram
determine:
1.
2.
3.
4.
5.
6.
7.
8.
Open loop transfer function
Feed Forward Transfer function
control ratio
feedback ratio
error ratio
closed loop transfer function
characteristic equation
closed loop poles and zeros if K=100.
Example-4
H
2
_
R
+_
+
+
G1
+
H1
C
G2
G3
Example-4
H2
G1
_
R
+_
+
+
+
C
G1
H1
G2
G3
Example-4
H2
G1
_
R
+_
+
+
C
+
G 1G 2
H1
G3
Example-4
H2
G1
_
R
+_
+
C
+
G 1G 2
+
H1
G3
Example-4
H2
G1
_
R
+_
+
G 1G 2
1  G 1G 2 H 1
C
G3
Example-4
H2
G1
_
R
+_
+
G 1G 2 G 3
1  G 1G 2 H 1
C
Example-4
R
+_
G 1G 2 G 3
1  G 1G 2 H 1  G 2 G 3 H 2
C
Example-5
Find the transfer function of the following block diagrams
H
4
R (s)
Y (s)
G1
G2
G3
H3
H
2
H1
G4
Example-5
Solution:
1. Moving pickoff point A behind block
G4
I
H
R (s)
4
Y (s)
G1
G2
G3
H
H3
H
2
3
1
G4 G4
1
H
2
G4 G 4
H1
A
G4
B
Example-5
2. Eliminate loop I and Simplify
R (s)
II
G 2 G 3G 4
G1
Y (s)
B
1  G 3G 4 H 4
H
3
G4
H2
III
G4
H1
II
feedback
III
Not feedback
G 2 G 3G 4
H 2  G4H1
1  G 3G 4 H 4  G 2 G 3 H 3
G4
Example-5
3. Eliminate loop II & IIII
R (s)
G 1G 2 G 3 G 4
Y (s)
1  G 3G 4 H 4  G 2 G 3 H 3
H 2  G4H1
G4
Y (s)
R( s )

G1G 2 G 3 G 4
1  G 2 G 3 H 3  G 3 G 4 H 4  G 1G 2 G 3 H 2  G 1G 2 G 3 G 4 H 1
Example-5
2. Eliminate loop I & Simplify
H
2
G3
G2
H1
R (s)
1
H1
G3
G3
G1
B
G 2G3
B
 H2
II
G 2G 3
1  G 2 H 1  G 2G 3 H 2
H1
G3
G4
Y (s)
Example-5
3. Eliminate loop II
R (s)
Y (s)
G 1G 2 G 3
1  G 2 H 1  G 2 G 3 H 2  G 1G 2 H 1
G4
T (s) 
Y (s)
R (s)
 G4 
G 1G 2 G 3
1  G 2 H 1  G 2 G 3 H 2  G 1G 2 H 1
Superposition of Multiple Inputs
Multiple Input System. Determine the output C due to
inputs R and U using the Superposition Method.
Example-6
Example-6
Example-6
Exercise-3: Multi-Input Multi-Output System. Determine C1
and C2 due to R1 and R2.
Introduction
• Alternative method to block diagram representation,
developed by Samuel Jefferson Mason.
• Advantage: the availability of a flow graph gain formula,
also called Mason’s gain formula.
• A signal-flow graph consists of a network in which nodes
are connected by directed branches.
• It depicts the flow of signals from one point of a system
to another and gives the relationships among the signals.
37
Fundamentals of Signal Flow Graphs
• Consider a simple equation below and draw its signal flow graph:
y  ax
• The signal flow graph of the equation is shown below;
x
a
y
• Every variable in a signal flow graph is designed by a Node.
• Every transmission function in a signal flow graph is designed by a
Branch.
• Branches are always unidirectional.
• The arrow in the branch denotes the direction of the signal flow.
Signal-Flow Graph Models
Example-7: R1 and R2 are inputs and Y1 and Y2 are outputs
Y1 ( s )
G11 ( s )  R 1 ( s )  G12 ( s )  R 2 ( s )
Y2 ( s )
G21 ( s )  R 1 ( s )  G22 ( s )  R 2 ( s )
Signal-Flow Graph Models
Exercise-4: r1 and r2 are inputs and x1 and x2 are outputs
a 11  x1  a 12  x2  r1
x1
a 21  x1  a 22  x2  r2
x2
Signal-Flow Graph Models
Example-8:
xo is input and x4 is output
x1  ax0  bx1  cx 2
x 2  dx1  ex 3
f
c
x0
a
x1
d
x2
x3
g
x 3  fx0  gx 2
x4  hx 3
b
e
h
x4
Construct the signal flow graph for the following set of
simultaneous equations.
• There are four variables in the equations (i.e., x1,x2,x3,and x4) therefore four nodes are
required to construct the signal flow graph.
• Arrange these four nodes from left to right and connect them with the associated
branches.
• Another way to arrange this
graph is shown in the figure.
Terminologies
• An input node or source contain only the outgoing branches. i.e., X1
• An output node or sink contain only the incoming branches. i.e., X4
• A path is a continuous, unidirectional succession of branches along which no
node is passed more than ones. i.e.,
X1 to X2 to X3 to X4
X1 to X2 to X4
X2 to X3 to X4
• A forward path is a path from the input node to the output node. i.e.,
X1 to X2 to X3 to X4 , and X1 to X2 to X4 , are forward paths.
• A feedback path or feedback loop is a path which originates and terminates on
the same node. i.e.; X2 to X3 and back to X2 is a feedback path.
Terminologies
• A self-loop is a feedback loop consisting of a single branch. i.e.; A33 is a self
loop.
• The gain of a branch is the transmission function of that branch.
• The path gain is the product of branch gains encountered in traversing a path.
i.e. the gain of forwards path X1 to X2 to X3 to X4 is A21A32A43
• The loop gain is the product of the branch gains of the loop. i.e., the loop gain
of the feedback loop from X2 to X3 and back to X2 is A32A23.
• Two loops, paths, or loop and a path are said to be non-touching if they have
no nodes in common.
Consider the signal flow graph below and identify the following
Example-9:
a)
b)
c)
d)
e)
f)
g)
Input node.
Output node.
Forward paths.
Feedback paths (loops).
Determine the loop gains of the feedback loops.
Determine the path gains of the forward paths.
Non-touching loops
Consider the signal flow graph below and identify the following
Example-9:
• There are two forward path gains;
Consider the signal flow graph below and identify the following
Example-9:
• There are four loops
Consider the signal flow graph below and identify the following
Example-9:
• Nontouching loop gains;
Consider the signal flow graph below and identify the
following
Example-10:
a)
b)
c)
d)
e)
f)
g)
Input node.
Output node.
Forward paths.
Feedback paths.
Self loop.
Determine the loop gains of the feedback loops.
Determine the path gains of the forward paths.
Input and output Nodes
Example-10:
a) Input node
b) Output node
(c) Forward Paths
Example-10:
(d) Feedback Paths or Loops
Example-10:
(d) Feedback Paths or Loops
Example-10:
(d) Feedback Paths or Loops
Example-10:
(d) Feedback Paths or Loops
Example-10:
(e) Self Loop(s)
Example-10:
(f) Loop Gains of the Feedback Loops
Example-10:
(g) Path Gains of the Forward Paths
Example-10:
Mason’s Rule (Mason, 1953)
• The block diagram reduction technique requires successive
application of fundamental relationships in order to arrive at the
system transfer function.
• On the other hand, Mason’s rule for reducing a signal-flow graph
to a single transfer function requires the application of one
formula.
• The formula was derived by S. J. Mason when he related the
signal-flow graph to the simultaneous equations that can be
written from the graph.
Mason’s Rule:
• The transfer function, C(s)/R(s), of a system represented by a signal-flow graph
is;
n
C(s)
R( s )
 Pi  i

i 1

Where
n = number of forward paths.
Pi = the i th forward-path gain.
∆ = Determinant of the system
∆i = Determinant of the ith forward path
• ∆ is called the signal flow graph determinant or characteristic function. Since
∆=0 is the system characteristic equation.
Mason’s Rule:
n
C(s)
R( s )
 Pi  i

i 1

∆ = 1- (sum of all individual loop gains) + (sum of the products of the gains
of all possible two loops that do not touch each other) – (sum of the
products of the gains of all possible three loops that do not touch each
other) + … and so forth with sums of higher number of non-touching loop
gains
∆i = value of Δ for the part of the block diagram that does not touch the ith forward path (Δi = 1 if there are no non-touching loops to the i-th path.)
Systematic approach
1. Calculate forward path gain Pi for each forward
path i.
2. Calculate all loop transfer functions
3. Consider non-touching loops 2 at a time
4. Consider non-touching loops 3 at a time
5. etc
6. Calculate Δ from steps 2,3,4 and 5
7. Calculate Δi as portion of Δ not touching forward
path i
62
Example-11: Apply Mason’s Rule to calculate the transfer function of
the system represented by following Signal Flow Graph
Therefore,
C
R

P1  1  P2  2

There are three feedback loops
L1  G1G 4 H 1 ,
L 2   G1 G 2 G 4 H 2 ,
L 3   G1 G 3 G 4 H 2
Example-11: Apply Mason’s Rule to calculate the transfer function of
the system represented by following Signal Flow Graph
There are no non-touching loops, therefore
∆ = 1- (sum of all individual loop gains)
  1   L1  L 2  L 3 
  1  G1G 4 H 1  G1G 2 G 4 H 2  G1G 3 G 4 H 2 
Example-11: Apply Mason’s Rule to calculate the transfer function of
the system represented by following Signal Flow Graph
Eliminate forward path-1
∆1 = 1- (sum of all individual loop gains)+...
∆1 = 1
Eliminate forward path-2
∆2 = 1- (sum of all individual loop gains)+...
∆2 = 1
Example-11: Continue
Exercise-5: Apply Mason’s Rule to calculate the transfer function
of the system represented by following Signal Flow Graph
67
Exercise-6
• Find the transfer function, C(s)/R(s), for the signal-flow
graph in figure below.
Example-12: Apply Mason’s Rule to calculate the transfer function of
the system represented by following Signal Flow Graph
There are three forward paths, therefore n=3.
3
C(s)
R( s )
 Pi  i

i 1


P1  1  P2  2  P3  3

Example-12: Forward Paths
P3  A42 A54 A65 A76
P1  A32 A43 A54 A65 A76
P2  A72
Example-12: Loop Gains of the Feedback Loops
L1  A32 A 23
L 2  A43 A34
L 3  A54 A45
L 4  A65 A56
L 5  A76 A67
L 6  A77
L 7  A42 A34 A23
L 8  A65 A76 A67
L 9  A72 A57 A45 A34 A23
L10  A72 A67 A56 A45 A34 A23
Example-12: two non-touching loops
L1 L 3
L2 L4
L3 L5
L4 L6
L1 L 4
L 2 L5
L3 L6
L4 L7
L1 L 5
L2 L6
L1 L 6
L 2 L8
L1 L8
L5 L7
L 7 L8
Example-12: Three non-touching loops
L1 L 3
L2 L4
L3 L5
L4 L6
L1 L 4
L 2 L5
L3 L6
L4 L7
L1 L 5
L2 L6
L1 L 6
L 2 L8
L1 L8
L5 L7
L 7 L8
From Block Diagram to Signal-Flow Graph Models
Example-13:
H1
R(s)
E(s)
X1
G1
－
－
－
G2
X2
G3
X3
G4
H2
H3
R(s)
1
E(s)
G1
X1
G2
－H2
－H3
X2
G3
－H1
X3 G4 C(s)
C(s)
From Block Diagram to Signal-Flow Graph Models
Example-13:
R(s)
1
－H1
E(s)
G1
X1
G2
X2
G3
G4
X3
1
C(s)
－H2
－H3
  1  (G1G2G3G4 H 3  G2G3 H 2  G3G4 H 1 )
P1  G1G2G3G4 ;
G
C ( s)
R( s )

1  1
G1G2G3G4
1  G1G2G3G4 H 3  G2G3 H 2  G3G4 H 1
Exercise-7
－
R(s)
E(s)
－
－
X1
G1
Y1
+
+
－X
2
+
－
G2
Y2
C(s)
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END OF LECTURE