The signal flow graph for a system is given below.

The transfer function \(\frac{{Y\left( s \right)}}{{U\left( s \right)}}\) for this system is

This question was previously asked in

UJVNL AE EE 2016 Official Paper

Option 1 : \(\frac{{s + 1}}{{5{s^2} + 6s + 2}}\)

__Concept:__

Signal flow graph

- It is a graphical representation of a set of linear algebraic equations between input and output.
- The set of linear
**algebraic**equations represents the**systems**. - The signal flow graphs are developed to avoid mathematical calculation.

Maon gain formula is used to find the ratio of any two nodes or transfer function.

T F = \(\mathop \sum \limits_{k = 1}^i \frac{{{P_k}{{\rm{\Delta }}_k}}}{{\rm{\Delta }}}\)

Where P_{k} = k^{th }forward path gain

Δ = 1- ∑ individual loop gain + ∑ two non-touching loops gain - ∑ the gain product of three non-touching loops + ∑ gain of four non-touching loops

Shotcut: while writing Δ take the opposite sign for the odd number of non-touching loops snd the same sign for the even the number of non-touching loops.

Δ_{K} is obtained from Δ by removing the loops touching the K^{th }forward path.

__Calculation:__

For the given SFG two forward paths

\({P_{K1}} = 1\left( {{s^{ - 1}}} \right)\left( {{s^{ - 1}}} \right)\left( 1 \right) = {s^{ - 2}}\)

\({P_{k2}} = 1\left( {{s^{ - 1}}} \right)\left( 1 \right)\left( 1 \right) = {s^{ - 1}}\)

Since all loops are touching the paths P_{K1} and P_{K2} so Δ_{K1} = Δ_{K2} = 1

We have Δ = 1- ∑ individual loops + ∑ non-touching loops gain

Loops are

\({L_1} = \left( { - 4} \right)\left( 1 \right) = - 4\)

\({L_2} = \left( { - 4} \right)\left( {{s^{ - 1}}} \right) = - 4{s^{ - 1}}\)

\({L_3} = \; - 2\left( {{s^{ - 1}}} \right)\left( {{s^{ - 1}}} \right) = - 2{s^{ - 2}}\)

\({L_4} = - 2\left( {{s^{ - 1}}} \right)\left( 1 \right) = - 2{s^{ - 1}}\)

As all the loops are touching each other we have

Δ = 1 – ( L1 + L2 + L3 + L4)

Δ = 1 – ( - 4 – 4s^{-1} – 2s^{-2} -2s^{-1} )

Δ = 5 + 6s^{-1 }+ 2s^{-2}

\(T.F = \frac{{{s^{ - 2}} + {s^{ - 1}}}}{{5 + 6{s^{ - 1}} + 2{s^{ - 2}}}}\)

\( = \frac{{s + 1}}{{5{s^2} + 6s + 2}}\)