# __Signal
Flow Graph__

__Signal Flow Graph__Signal flow
graph is the graphical representation of control system in which nodes
representing each of system variables are connected through the direct
branches. It may be regarded as the simplified version of Block Diagram. It
applies only to linear system.

**Basic Terminology of Signal Flow Graph **

**Node :-**It represent a system variable.

**Branch :- **Nodes are connected by line
segments called branches, these branches have associated branch gain and
direction, direction is represented by arrow. A signal can transmit through the
branch only in the direction of arrow.

**Input Node / Source node :- **It is a
node having outgoing branches only.

**Output Node / Sink Node :- **It is the
node having only incoming branches.

**Note :- **In a signal flow graph output can
be defined from any other node by extending the it with a gain of unity.

**Mixed / Chain Node :- **It is a
node having both incoming and outgoing branches.

**Path :- **It is the traversal of connected
branches in the direction of branch arrow such that no node is traversed more
than once. A signal flow graph can have one or more than one paths.

**Forward Path :- **It is a path starts from
input node and end at an output node.

**Loop :- **It is a path which originates and
terminates at the same node.

**Non Touching Loop :- **Two or more
loops are said to be non-touching if they do not have a common node.

**Mason Gain Formula**

Transfer
Function = ^{n}Æ©_{k=1} P_{k }.Î”_{k }/ Î”

Where,

n = number of forward paths

P_{k} = Path gain of the
forward K^{th} path

Î” = 1 – {Sum of loop gains all
individual loops} + {Sum of gain of product of two non-touching loops} – {Sum
of gain products of three non-touching loop}...........

Î”_{K}
= It is that value of “Î”” obtained by removing all the nodes touching K^{th}
forward path.

**Calculation of transfer function by using Mason
Gain formula**

** **

Let us take an example

In this Signal Flow Graph,

There are seven nodes these are Y_{1},
Y_{2}, Y_{3,} Y_{4}, Y_{5,} Y_{6}, Y_{7}.
In which Y_{1} is input node or source node and Y_{7} is output
node or sink node and rest of nodes are mixed or chain nodes.

These seven are connected through the
branches having forward transfer gain G_{1}, G_{2}, G_{3},
G_{4}, G_{5}, 1 respectively and Y_{2}, Y_{6}
is connected through another branch have forward transfer gain G_{6}.

Similarly Y_{2} and Y_{3 }is
connected through a branch having negative gain H_{1} and Y_{4}
and Y_{6} is also connected through a branch having negative gain H_{2}.

So according to formula first we have
to find the number of forward path and product of the gains of those forward
paths.

There are two forward paths

P_{1} = G_{1}G_{2}G_{3}G_{4}G_{5}1

P_{2} = G_{1}G_{6}1

Then, we have to find product of gain
of individual loops. So there are two individual loops these are

L_{1} = -G_{2}H_{1}

L_{2} = -G_{4}G_{5}H_{2}

_{}

Then we have to find product of gain
of two non-touching loop. There are two non-touching loop these are

L_{1} = -G_{2}H_{1}

L_{2} = -G_{4}G_{5}H_{2}

_{}

Product of gain of these two loops is
I_{1} = G_{2}G_{4}G_{5}H_{1}H_{2}

_{}

Then we have to find Î”_{k}.

Î”_{1} =1

Î”_{2} = 1

So put all those value in Mason Gain
formula after putting these values we get

So In this way we apply the Mason Gain formula for a particular problem.

**How to
convert Block Diagram into Signal Flow Graph**

There are some steps which we have to
follow for the conversion of Block Diagram into Signal Flow Graph

**Step :- 1** Convert all the variables, summing point,
take-off point, junctions into nodes

**Step :- 2 **Represent all the blocks of the block
diagram as branches with given directions.

**Step :- 3 **The transfer function of the particular
block represent the gain of the corresponding branch of that particular block.

**Step :- 4** Connect all the nodes with
branches according to the block diagram convention and direction.

Let us take an example

Block diagram of feedback control system. |

In the above diagram there are three
summing points S_{1} S_{2} and S_{3 }respectively, two
take-off points T_{1} and T_{2} and one junction point J_{1}.

So, according to above steps convert

S_{1} ――> N_{2}

S_{2} ――> N_{4}

S_{3} ――> N_{6}

T_{1} ――> N_{3}

T_{2} ――> N_{5}

J_{1} ――> N_{1}

_{}

Connects all the nodes with their
respective branches and assign the gain for all the branches accordingly.

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