GATE 2022 IN Q.37: solution and answer

GATE 2022 · INQ.37 · 2 marks · MCQ

The signal flow graph of a system is shown. The expression for $Y (s) / X (s)$ is ________

GATE 2022 IN question 37: Signal flow graph with four nodes in a row: input node X(s), node A, node B and output node Y(s). — Figure from the GATE 2022 IN paper. It is part of the question.

(A)
$\frac{2 G _{1} ( s ) G _{2} ( s ) + 2 G _{1} ( s ) G _{3} ( s )}{1 + G _{2} ( s ) + G _{3} ( s )}$
(B)
$2 + G_{1} (s) + G_{3} (s) + \frac{G _{2} ( s )}{1 + G _{2} ( s )}$
(C)
$G_{1} (s) + G_{3} (s) - \frac{G _{2} ( s )}{2 + G _{2} ( s )}$
(D)
$\frac{2 G _{1} ( s ) G _{2} ( s ) + 2 G _{1} ( s ) G _{3} ( s ) - G _{1} ( s )}{1 + G _{2} ( s ) + G _{3} ( s )}$

Answer and solution

ANSWER

$\frac{Y ( s )}{X ( s )} = \frac{2 G _{1} ( s ) G _{2} ( s ) + 2 G _{1} ( s ) G _{3} ( s )}{1 + G _{2} ( s ) + G _{3} ( s )}$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2022 IN (IIT Kharagpur)

Read the branch directions first. Call the two unlabelled middle nodes $A$ (after $G_{1}$ ) and $B$ (before the gain $2$ ). $G_{2}$ and $G_{3}$ both run forward from $A$ to $B$ . The arrowhead on the $- 1$ arc points up into $A$ , so it is a feedback branch from $B$ to $A$ .
Forward paths from $X$ to $Y$ : $P_{1} = G_{1} \cdot G_{2} \cdot 2 = 2 G_{1} G_{2}$ and $P_{2} = G_{1} \cdot G_{3} \cdot 2 = 2 G_{1} G_{3}$ .
Individual loops (each goes $A \to B$ forward and closes through the $- 1$ branch): $L_{1} = G_{2} (- 1) = - G_{2}$ and $L_{2} = G_{3} (- 1) = - G_{3}$ .
Non-touching loops: none, because $L_{1}$ and $L_{2}$ share nodes $A$ and $B$ .
Graph determinant: $Δ = 1 - (L_{1} + L_{2}) = 1 + G_{2} + G_{3}$ .
Both forward paths pass through $A$ and $B$ , so each touches both loops: $Δ_{1} = Δ_{2} = 1$ .
Mason's gain formula: $\frac{Y ( s )}{X ( s )} = \frac{P _{1} Δ _{1} + P _{2} Δ _{2}}{Δ} = \frac{2 G _{1} ( s ) G _{2} ( s ) + 2 G _{1} ( s ) G _{3} ( s )}{1 + G _{2} ( s ) + G _{3} ( s )}$ This is option (A).
Quick check by block reduction: $G_{2}$ and $G_{3}$ are in parallel between $A$ and $B$ , giving $G_{2} + G_{3}$ , and the $- 1$ branch closes a unity negative feedback loop around it, giving $\frac{G _{2} + G _{3}}{1 + G _{2} + G _{3}}$ . Multiplying by $G_{1}$ in front and $2$ behind gives the same result.

KEY IDEA

$G_{2}$ and $G_{3}$ are two parallel forward branches from the same node to the same node, and the $- 1$ branch feeds back around both of them. That makes two forward paths and two loops that touch each other and both paths. Mason's formula, or block reduction (parallel, then unity negative feedback), gives $\frac{2 G _{1} ( G _{2} + G _{3} )}{1 + G _{2} + G _{3}}$ .

WATCH OUT FOR

Reading the $- 1$ arc as a third forward branch from $A$ to $B$ . It looks symmetric with the $G_{3}$ arc, but its arrowhead points into the left node, so it is feedback. Treating it as forward removes both loops and gives the wrong result $2 G_{1} (G_{2} + G_{3} - 1)$ . A second slip is to count only the $G_{2}$ loop and forget that $G_{3}$ also closes a loop through the $- 1$ branch, which gives a denominator of $1 + G_{2}$ .

From the GATE 2022 IN paper. Every question of this paper · Download the full paper · All 10 Block diagrams & SFG questions, with notes

GATE 2022 IN Q.37: solution and answer

Answer and solution

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