---
title: "Signal flow graph and Mason's formula: G3 feedforward"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sfg-g1g2g3-h1h2-2076a"
source: https://www.bctengineer.com/q/semester_3/control_systems/sfg-g1g2g3-h1h2-2076a
site: BCT Engineering
---

# Signal flow graph and Mason's formula: G3 feedforward

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2076 Asoj, 2+6 marks; 2067 Asar (old course), 8 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/sfg-g1g2g3-h1h2-2076a
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Develop signal flow graph for the block diagram model below and find transfer function using Mason's gain formula. [Figure: R → summing point S1 (+, −) → G1 → node A → summing point S2 (+, −) → G2 → node B → summing point S3 (+, +) → C. G3 from node A to S3 (+). H1 from node B fed back to S1 (−). H2 from C fed back to S2 (−).]

## Short answer

C(s)/R(s) = (G1G2 + G1G3) / (1 + G1G2H1 + G2H2 − G1G2G3H1H2).

## Answer

#### Signal flow graph
Nodes: $R$, $e_1$ (after $S_1$), $A$ (after $G_1$), $e_2$ (after $S_2$), $B$ (after $G_2$), $C$. A dummy output node $C$ is drawn with unity gain.
```text
       1      G1      1       G2       1
 R ───►e1 ───►A ────►e2 ────► B ─────►C ──1──► C
       ▲      │       ▲       │        │
       │      └───────┼──G3───┼───────►│
       │              │       │        │
       │              └──────(−H2)─────┘
       └────────(−H1)─────────┘
```
Branches: $R\to e_1$ (1), $e_1\to A$ ($G_1$), $A\to e_2$ (1), $e_2\to B$ ($G_2$), $B\to C$ (1), $A\to C$ ($G_3$), $B\to e_1$ ($-H_1$), $C\to e_2$ ($-H_2$).

#### Mason's gain formula
$$
T = \frac{1}{\Delta}\sum_k P_k\Delta_k
$$

**Forward paths**
- $P_1 = G_1G_2$ ($R\to e_1\to A\to e_2\to B\to C$)
- $P_2 = G_1G_3$ ($R\to e_1\to A\to C$)

**Individual loops**
- $L_1 = -G_1G_2H_1$ ($e_1\to A\to e_2\to B\to e_1$)
- $L_2 = -G_2H_2$ ($e_2\to B\to C\to e_2$)
- $L_3 = +G_1G_3G_2H_1H_2$ ($e_1\to A\to C\to e_2\to B\to e_1$; two negative branches give a positive gain)

**Non-touching loops:** none (all share node $e_2$ or $B$).

**Determinant**
$$
\Delta = 1 - (L_1 + L_2 + L_3) = 1 + G_1G_2H_1 + G_2H_2 - G_1G_2G_3H_1H_2
$$

**Cofactors:** both forward paths touch all loops, so $\Delta_1 = \Delta_2 = 1$.

#### Transfer function
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_1G_2H_1 + G_2H_2 - G_1G_2G_3H_1H_2}
$$
This agrees with block diagram reduction of the same figure.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What are the forward paths from R to C?
   A. G1G2 and G1G3
   B. G1G2G3 only
   C. G2 and G3
   D. G1G2H1 and G1G3H2
   Answer: A. One path goes through G1 and G2, the other through G1 and the G3 branch from A to C. (Key point: Forward paths)

2. Which set gives the three individual loops?
   A. -G1G2H1, -G2H2, +G1G2G3H1H2
   B. -G1G2H1, -G2H2, -G1G2G3H1H2
   C. +G1G2H1, +G2H2, +G1G3
   D. -G1H1, -G3H2, -G2
   Answer: A. The third loop passes through two negative branches (H1 and H2), so its gain is positive. (Key point: Loop gains)

3. What are Δ1 and Δ2 and why?
   A. Both 1, because every loop touches both forward paths
   B. Both 0, no loops
   C. Both equal to Δ
   D. 1 - L1 for both
   Answer: A. All loops share node e2 or B with both paths, so no loops remain after removing a path. (Key point: Cofactors)

4. What is Δ?
   A. 1 + G1G2H1 + G2H2 - G1G2G3H1H2
   B. 1 - G1G2H1 - G2H2 + G1G2G3H1H2
   C. 1 + G1G2H1 + G2H2 + G1G2G3H1H2
   D. 1 + G1G2 + G1G3
   Answer: A. Δ = 1 - (L1+L2+L3) with L1 and L2 negative and L3 positive. (Key point: Determinant)

## Related in Control Systems

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