Signal flow graph is an important topic from gate point of view. One or two question is always expected from this topic. Experts at GATE - Prepladder has devised tips to draw Signal Flow Graph without mistakes.

**Signal Flow Graph **

Signal flow graph of the control system is a directed graph in which nodes represent system variables and edges represent functional associations between the pair of nodes. The Signal flow graph is derived from the simplification of the block diagram of the control system.

By means of aa signal flow graph, the equation y = Kx can be represented as

Here x denotes the input variable node, y denotes the output variable node and a denotes the transmittance of the edge connecting the two nodes.

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### **Simple Procedure For calculating Transfer function from Signal Flow Graph**

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- Firstly, you need to calculate the input signal at each node of the graph.
- After calculation of the input signal at all the nodes, we will get a large number of equations relating node variables and transmittance. We will get the equation for each of the input variable node.
- Solving these equations will give us the ultimate input and output of the entire signal flow graph of the control system
- Lastly, for calculating the expiration of transfer function of the signal flow graph, we divide the inspiration of ultimate output with the expression of initial input.

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**Mason's Gain Formula**

Mason’s Gain Formula gives the overall transmittance or gain of a signal flow graph of the control system.

As per the Mason’s gain formula, The overall transmittance is given by

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**Solved Examples from GATE**

Q1. For the signal flow graph shown below, calculate the value of

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**Correct Answer: (B)**

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**Solution: **We have to calculate C(s)/R(s) which is the transfer function.

This will be calculated using the Mason’s Gain formula

T = y_{out}/y_{in} where T denotes the Transfer function

∆ = 1 – Ʃ L_{i} + Ʃ L_{i }L_{j }– Ʃ L_{i }L_{j }L_{k }+..............+ (-1)^{m} Ʃ.........+............

Where

∆ = determinant of the graph

y_{in} = input node variable

y_{out}= output node variable

T = complete gain between y_{in} and y_{out}

K = Total number of forward paths between y_{in} and y_{out}

P_{k}= path gain of the kth forward path between y_{in} and y_{out}

L_{i} = loop gain of each closed loop in the system

L_{i} L_{j} = product of the loop gains of any two non-touching loops (non-common nodes)

L_{i} L_{j} L_{k} = product of the loop gains of any three pairwise non-touching loops

∆_{k} = the cofactor value of ∆ for the k^{th} forward path, with the loops touching the k^{th }forward path removed

Since the graph consist of only one forward path, so k = 1

Therefore, T = P_{1}.∆_{1}/∆ ................... (1)

Now,

∆ = 1 – (L_{1} + L_{2 }+L_{3}) + (L_{1} L_{2}) – 0

Since there are no 3 non-touching loops

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

L_{2} = - G_{3} G_{4} H_{2}

L_{3} = - G_{2} G_{3} H_{3}

L_{1 }L_{2} = G_{1 }G_{2} G_{3} G_{4} H_{1} H_{2}

_{ }

_{ }So, ∆ = 1+ G_{1} G_{2} H_{1} + G_{3} G_{4} H_{2 }+ G_{2} G_{3} H_{3} + G_{1 }G_{2} G_{3} G_{4} H_{1} H_{2}

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

_{ }∆_{1} = 1 – 0 = 1

Therefore, substituting the values in the equation (1)

We arrive at the correct answer as (B)

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