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Mason's gain formula: G1–G4 with H1, H2, unity loop

Short answer

C(s)/R(s) = (G1G2G3 + G1G4) / (1 + G1G2H1 + G2G3H2 + G4H2 + G1G2G3 + G1G4).

Using Mason's gain formula, find the transfer function C(s)/R(s) of the fig given below. [Figure: R → summing point S1 (+, −) → summing point S2 (+, −) → G1 → summing point S3 (+, −) → node A → G2 → node B → G3 → summing point S4 (+, +) → C. G4 from node A to S4 (+). H2 from C fed back to S3 (−). H1 from node B fed back to S2 (−). Unity feedback from C to S1 (−).]

Name the nodes: $R$ , $x_1$ (after $S_1$ ), $x_2$ (after $S_2$ ), $x_3$ (after $G_1$ ), $A$ (after $S_3$ ), $B$ (after $G_2$ ), $x_4$ (after $G_3$ ), $C$ (after $S_4$ ).

Signal flow graph

     1     1     G1     1     G2     G3     1
 R ─►x1 ─►x2 ──►x3 ──►A ───►B ───►x4 ───►C
     ▲     ▲          ▲│     │            ▲│
     │     │          ││     │            ││
     │     └───(−H1)──┼┼─────┘            ││
     │                │└──────G4──────────┘│
     │                └───────(−H2)────────┤
     └──────────────────(−1)───────────────┘

Forward paths

$P_1 = G_1G_2G_3$ (through $G_2$ , $G_3$ )
$P_2 = G_1G_4$ (through $G_4$ )

Individual loops

Loop	Path	Gain
$L_1$	$x_2\to x_3\to A\to B\to x_2$	$-G_1G_2H_1$
$L_2$	$A\to B\to x_4\to C\to A$	$-G_2G_3H_2$
$L_3$	$A\to C$ (via $G_4$ ) $\to A$	$-G_4H_2$
$L_4$	$x_1\to\dots\to B\to x_4\to C\to x_1$	$-G_1G_2G_3$
$L_5$	$x_1\to x_2\to x_3\to A\to C\to x_1$ (via $G_4$ )	$-G_1G_4$

Non-touching loops: none; every loop passes through node $A$ .

Determinant

\Delta = 1 - \sum L_i = 1 + G_1G_2H_1 + G_2G_3H_2 + G_4H_2 + G_1G_2G_3 + G_1G_4

Cofactors

Both forward paths pass through $A$ , so they touch all loops: $\Delta_1 = \Delta_2 = 1$ .

Transfer function

\frac{C(s)}{R(s)} = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}

Answer:

\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_4H_2 + G_1G_2G_3 + G_1G_4}

✓ Answer checked by Claude Opus 5.5 (high effort)

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