--- title: "Mason's gain formula: G1–G4 with H1, H2, unity loop" subject: "Control Systems" chapter: 3 key: "semester_3/control_systems:mason-g1-g4-h1h2-2075a" source: https://www.bctengineer.com/q/semester_3/control_systems/mason-g1-g4-h1h2-2075a site: BCT Engineering --- # Mason's gain formula: G1–G4 with H1, H2, unity loop Control Systems · Chapter 3: System Transfer Function and Responses - Asked in: 2075 Asoj, 8 marks; 2081 Baisakh, 6 marks (asked 2 times) - Video: [Problem 2 on Converting Block Diagram to Signal Flow Graph](https://www.youtube.com/watch?v=uCm4n897cOY) — TutorialsPoint - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/control_systems/mason-g1-g4-h1h2-2075a - Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3 ## Question 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 (−).] ## Short answer C(s)/R(s) = (G1G2G3 + G1G4) / (1 + G1G2H1 + G2G3H2 + G4H2 + G1G2G3 + G1G4). ## Answer 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 ```text 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:** $$ \frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_4H_2 + G_1G_2G_3 + G_1G_4} $$ ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which are the forward paths? A. G1G2G3 and G1G4 B. G2G3 and G4 C. G1G2G3H1 and G1G4 D. G1 and G4 Answer: A. Both start through G1 to node A, then one goes through G2 and G3 and the other through G4. (Key point: Forward paths) 2. What are the two loops formed by the unity feedback from C to S1? A. -G1G2G3 and -G1G4 B. -G2G3H2 and -G4H2 C. -G1G2H1 and -G4H2 D. +G1G2G3 and +G1G4 Answer: A. Each forward path closed through the unity (negative) feedback gives a loop equal to minus the path gain. (Key point: Loops with unity feedback) 3. Why are there no non-touching loops? A. Every loop passes through node A B. Only one loop exists C. All loops contain H1 D. The loops are in parallel Answer: A. Since every loop passes through A, all loops touch each other. (Key point: Non-touching loops) 4. What is the denominator Δ? A. 1 + G1G2H1 + G2G3H2 + G4H2 + G1G2G3 + G1G4 B. 1 - G1G2H1 - G2G3H2 - G4H2 - G1G2G3 - G1G4 C. 1 + G1G2G3 + G1G4 D. 1 + G1G2H1 + G2G3H2 Answer: A. All five loop gains are negative, so Δ = 1 minus their sum adds all terms. 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