Feedback Control Systems (FCS)
Lecture-16-17-18
Signal Flow Graphs
Dr. Imtiaz Hussain
email: imtiaz.hussain@faculty.muet.edu.pk
URL :http://imtiazhussainkalwar.weebly.com/
Outline
• Introduction to Signal Flow Graphs
– Definitions
– Terminologies
– Examples
• Mason’s Gain Formula
– Examples
• Signal Flow Graph from Block Diagrams
• Design Examples
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.
3
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
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
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
xo is input and x4 is output
x1  ax0  bx1  cx2
x 2  dx1  ex3
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
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
• There are two forward path gains;
Consider the signal flow graph below and identify the following
• There are four loops
Consider the signal flow graph below and identify the following
• Nontouching loop gains;
Consider the signal flow graph below and identify the
following
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
a) Input node
b) Output node
(c) Forward Paths
(d) Feedback Paths or Loops
(d) Feedback Paths or Loops
(d) Feedback Paths or Loops
(d) Feedback Paths or Loops
(e) Self Loop(s)
(f) Loop Gains of the Feedback Loops
(g) Path Gains of the Forward Paths
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
28
Example#1: 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#1: 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#1: 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#1: Continue
Example#2: Apply Mason’s Rule to calculate the transfer function
of the system represented by following Signal Flow Graph
P1
P2
1. Calculate forward path gains for each forward path.
P1  G1G 2 G 3 G 4 (path
1)
and
P2  G 5 G 6 G 7 G 8 (path
2)
2. Calculate all loop gains.
L1  G 2 H 2 ,
L2  H 3G 3 ,
L3  G 6 H 6 ,
3. Consider two non-touching loops.
L1L3 L1L4
L2L4 L2L3
L4  G 7 H 7
33
Example#2: continue
4. Consider three non-touching loops.
None.
5. Calculate Δ from steps 2,3,4.
  1   L1  L 2  L 3  L 4    L1 L 3  L1 L 4  L 2 L 3  L 2 L 4 
  1  G 2 H 2  H 3 G 3  G 6 H 6  G 7 H 7
G 2 H 2 G 6 H 6

 G 2 H 2 G 7 H 7  H 3G 3G 6 H 6  H 3G 3G 7 H 7 
34
Example#2: continue
Eliminate forward path-1
 1  1  L3  L 4 
 1  1  G 6 H 6  G 7 H 7 
Eliminate forward path-2
 2  1   L1  L 2 
 2  1  G 2 H 2  G 3 H 3 
35
Example#2: continue
Y (s)
R( s )
Y (s)
R( s )


P1  1  P2  2

G 1G 2 G 3 G 4 1  G 6 H 6  G 7 H 7   G 5 G 6 G 7 G 8 1  G 2 H 2  G 3 H 3 
1  G 2 H 2  H 3 G 3  G 6 H 6  G 7 H 7
  G 2 H 2 G 6 H 6
 G 2 H 2 G 7 H 7  H 3G 3G 6 H 6  H 3G 3G 7 H 7 
36
Example#3
• Find the transfer function, C(s)/R(s), for the signal-flow
graph in figure below.
Example#3
• There is only one forward Path.
P1  G 1 ( s )G 2 ( s )G 3 ( s )G 4 ( s )G 5 ( s )
Example#3
• There are four feedback loops.
Example#3
• Non-touching loops taken two at a time.
Example#3
• Non-touching loops taken three at a time.
Example#3
Eliminate forward path-1
Example#4: 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#4: Forward Paths
P3  A42 A54 A65 A76
P1  A32 A43 A54 A65 A76
P2  A72
Example#4: 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#4: 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#4: 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#5
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#5
－H1
R(s)
1
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
Example#6
－
R(s)
E(s)
－
－
R(s) 1
Y1
G1
+
+
－X
2
+
－
-1
X1
X1
G2
Y2
-1
G1
Y1
-1
E(s)
1
X2
-1
C(s)
G2
1
-1
Y2
1
1
C(s)
Example#6
-1
R(s) 1
E(s)
-1
1
X1
X2
-1
7 loops:
3 ‘2 non-touching loops’ :
G1
G2
Y1
-1
1
-1
Y2
1
1
C(s)
Example#6
-1
E(s)
R(s) 1
-1
1
X1
X2
G1
Y1
-1
G2
-1
Then:
4 forward paths:
1
Y2
1
C(s)
1
-1
Δ  1  2G 2  4G1G 2
p1  (  1)  G1  1
p2  (  1)  G1  (  1)  G2  1
p3  1  G 2  1
p4  1  G 2  1  G1  1
Δ1  1  G2
Δ2  1
Δ3  1  G1
Δ4  1
Example#6
We have
C( s )
R( s )

 pk k


G2  G1  2G1G2
1  2G2  4G1G2
Example-7: Determine the transfer function C/R for the block diagram below
by signal flow graph techniques.
•
The signal flow graph of the above block diagram is shown below.
•
There are two forward paths. The path gains are
•
The three feedback loop gains are
•
No loops are non-touching, hence
•
Because the loops touch the nodes of P1,
hence
•
Since no loops touch the nodes of P2,
therefore
•
Hence the control ratio T = C/R is
Example-6: Find the control ratio C/R for the system given below.
•
The signal flow graph is shown in the figure.
•
The two forward path gains are
•
The five feedback loop gains are
•
•
All feedback loops touches the two forward
paths, hence
•
Hence the control ratio T =
There are no non-touching loops, hence
Design Example#1
V1 ( s ) 
1
Cs
I1 ( s )  I1 ( s )R
CsV 1 ( s )  CsV 2 ( s )  I 1 ( s )
V 2 ( s )  I1 ( s )R
 Cs
Cs
V1 ( s )
R
I1 ( s )
V2 (s)
Design Example#2
2
F  M 1 s X 1  k1 ( X 1  X 2 )
2
0  M 2s X
2
 k1 ( X
2
 X1)  k2 X
2
Design Example#2
Design Example#2
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END OF LECTURES-16-17-18