1) The document discusses signal flow graphs, which provide a graphical representation of linear algebraic equations in a system. Nodes represent variables and branches show relationships. 2) It compares block diagrams and signal flow graphs, noting that signal flow graphs only need to be drawn once, saving time and space over repeatedly redrawing block diagrams. 3) Mason's gain formula allows directly calculating the overall transfer function of a system from its signal flow graph representation, without lengthy reduction steps.

1. Gandhinagar institute
of technology(012)
ALA
Subject :- CE(2151908)
Topic name:- Signal flow graph
Prepared by :- Jani Parth U. (150120119051)
guided by :-Prof. Samir Raval

2. content
1. Introduction
2. Comparison of BD and SFG
3. SFG terms representation
4. Mason’s gain formula
5. Example 1
6. Example 2

3. 1. introduction
The graphical representation of the variables of set of linear
algebraic equation representing the system is called signal flow
graph
There are two important elements constituting signal flow graph
Nodes:- as variable of system are represented first in signal flow
graph by small circle called nodes
Branches:- the lines joining the nodes are called branches. The
relationship between various nodes are represented by joining the
nodes as per the equation
NODES
BRANCHES

4. 2. Comparison of BD and SFG
)(sR
)(sG
)(sC
)(sG
)(sR )(sC
Block diagram Signal flow graph
 In this case at each step block
diagram is to be redrawn.
 That’s why it is tedious method.
 So wastage of time and space.
 Only one time SFG is to be
drawn and then mason’s
gain formula is to be
evaluated.
 So time and space is saved.

5. 3. SFG terms representation
Source node
b1x
2x
c
1
3x
3x
Dummy node
Feedback loop OR
Individual loop
branch
node
Sink node
Loop gain= b x c
Chain node
1

6. 4. Mason’s gain formula
 We know that in block diagram representation ,reduction method is
time consuming.
 In signal flow graph approach, direct use of one formula leads to the
overall system transfer function
𝐶(𝑠)
𝑅(𝑠)
.
 This formula is given by mason and hence it is know as mason’s gain
formula.
T =
𝐶(𝑠)
𝑅(𝑠)
=
1
∆
𝑃 𝐾 ∆ 𝐾
 STEP of mason’s gain formula
1. Step 1 : calculating forward path gains
2. Step 1 : individual loop gain
3. Step 3 :gain product of non touching
loops
4. Step 4:Calculate ∆, ∆ 𝐾.
5. STEP 5 : Calculate T

7. 5. EXAMPLE 1
1.FIND
𝐶(𝑠)
𝑅(𝑠)
BY USING MASON’S GAIN FORMULA.
Solution:- STEP 1 : GAIN OF FORWARD PATH = 𝑃1= 𝐺1 𝐺2 𝐺3 𝐺4
STEP 2 : GAIN OF INDIVIDUAL LOOP
𝑃11= -𝐺2 𝐺3 𝐻2
-𝑯 𝟐
-1
-𝑯 𝟏
2 3 4 5 6
1 1
1 7
𝑮 𝟏
𝑮 𝟐
𝑮 𝟑
𝑮 𝟒

8. …
𝑃21= −𝐻1 𝐺4 𝐺3
𝑃31= -𝐺1 𝐺2 𝐺3 𝐺4
STEP 3: GAIN OF NON – TOUGHING LOOP
NO NON-TOUGHING LOOP
STEP 4: TO CALCULATE ∆, ∆ 𝐾.
∆ = 1- (SUM OF ALL INDIVIDUAL LOOPS GAIN ) + (MULTIPLICATION OF NON
TOUGHING LOOPS GAIN)
= 1- (-𝐺2 𝐺3 𝐻2 −𝐻1 𝐺4 𝐺3-𝐺1 𝐺2 𝐺3 𝐺4)+ 0
= 1+ 𝐺2 𝐺3 𝐻2 +𝐻1 𝐺4 𝐺3 + 𝐺1 𝐺2 𝐺3 𝐺4

9. In our case K =1
∆ 1=1-(If L 1 , P 1 is non touching than put L 1)
But in this case there are no non-toughing loop
=1
T =
𝐶(𝑠)
𝑅(𝑠)
=
1
∆
𝑃 𝐾 ∆ 𝐾
=
𝑃1∆1
∆
=
𝐺1 𝐺2 𝐺3 𝐺4 x 1
1+ 𝐺2 𝐺3 𝐻2+𝐻1 𝐺4 𝐺3+𝐺1 𝐺2 𝐺3 𝐺4

10. 6. Example 2
2. Find
𝐶(𝑠)
𝑅(𝑠)
by using mason’s gain formula.
 STEP 1 : Gain of forward path = 𝑃1= 𝐺1 𝐺2 𝐺3 𝐺4, 𝑃2= 𝐺5 𝐺4
 STEP 2 : Gain of individual loop
𝑃11= -𝐺2 𝐻1
𝑮 𝟓
-𝑯 𝟐
-𝑯 𝟏
2 3 4 5 6
1 1
1 7
𝑮 𝟏 𝑮 𝟐
𝑮 𝟑
𝑮 𝟒

11. ...
𝑃21= -𝐺1 𝐺2 𝐺3 𝐺4 𝐻2
𝑃31= −𝐺5 𝐺4 𝐻2
 STEP 3: Gain of non – toughing loop
𝑃11= -𝐺2 𝐻1
𝑃31= −𝐺5 𝐺4 𝐻2
 STEP 4: To calculate ∆, ∆ 𝐾.
∆ = 1- (sum of all individual loops gain ) + (multiplication of non toughing
loops gain)
= 1-(-𝐺2 𝐻1-𝐺1 𝐺2 𝐺3 𝐺4 𝐻2-𝐺5 𝐺4 𝐻2) + (𝐺2 𝐻1 𝐺5 𝐺4 𝐻2)
=1+𝐺2 𝐻1 + 𝐺1 𝐺2 𝐺3 𝐺4 𝐻2 + 𝐺5 𝐺4 𝐻2 + 𝐺2 𝐻1 𝐺5 𝐺4 𝐻2

12. ...
In our case K =2
For 𝑇1 all loops are touching
∆ 1= 1-(If 𝑃11, 𝑃21, are non touching to P 1)
= 1-0
= 0
∆ 2= 1-(If 𝑃11, 𝑃21, are non touching to P 2)
= 1-(L 1)
= 1+𝐺2 𝐻1
T =
𝐶(𝑠)
𝑅(𝑠)
=
1
∆
𝑃 𝐾 ∆ 𝐾
=
𝑃1∆1+𝑃2∆2
∆
=
𝐺1 𝐺2 𝐺3 𝐺4 x 1+𝐺5 𝐺4(1+ 𝐺2 𝐻1)
1+ 𝐺2 𝐻1+𝐺1 𝐺2 𝐺3 𝐺4 𝐻2+𝐺5 𝐺4 𝐻2+ 𝐺2 𝐻1 𝐺5 𝐺4 𝐻2