---
title: "Signal flow graph and Mason's formula: G1–G4, H1–H3"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sfg-mason-g1-g4-h1-h3"
source: https://www.bctengineer.com/q/semester_3/control_systems/sfg-mason-g1-g4-h1-h3
site: BCT Engineering
---

# Signal flow graph and Mason's formula: G1–G4, H1–H3

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2079 Bhadra, 8 marks
- 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/sfg-mason-g1-g4-h1-h3
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

From the given block diagram, draw the signal flow graph and find the transfer ratio C(s)/R(s) using Mason's gain formula. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → node A → G1 → summing point S3 (+, −) → G2 → node B → G3 → summing point S4 (+, +) → C(s). G4 feeds forward from node A to S4 (+). H2 from node B fed back to S2 (−). H3 from C(s) fed back to S3 (−). H1 from C(s) fed back to S1 (−).]

## Short answer

Using Mason's formula with two forward paths (G1G2G3 and G4) and five loops, the transfer ratio is (G1G2G3 + G4) / (1 + G1G2H2 + G2G3H3 + G1G2G3H1 + G4H1 - G2G4H2H3).

## Answer

A **signal flow graph** has a node for each signal and a directed branch (with gain) for each block. Summing points become nodes; a minus sign at a summer goes into the branch gain.

#### Signal flow graph
Nodes: $x_1$ = output of $S_1$, $x_2$ = output of $S_2$ (node A), $x_3$ = output of $S_3$, $x_4$ = node B, $C$ = output.

```text
                  G4
        +--------------------------+
        |                          v
R --1--> x1 --1--> x2 --G1--> x3 --G2--> x4 --G3--> C --1--> C
         ^          ^          ^          |         |
         |          +---(-H2)--+----------+         |
         |                     +--------(-H3)-------+
         +-----------------(-H1)--------------------+
```
Branches: $R\to x_1$ (1), $x_1\to x_2$ (1), $x_2\to x_3$ ($G_1$), $x_3\to x_4$ ($G_2$), $x_4\to C$ ($G_3$), $x_2\to C$ ($G_4$), $x_4\to x_2$ ($-H_2$), $C\to x_3$ ($-H_3$), $C\to x_1$ ($-H_1$).

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

**Forward paths**
- $P_1 = G_1G_2G_3$ (R, x1, x2, x3, x4, C)
- $P_2 = G_4$ (R, x1, x2, C)

**Individual loops**
- $L_1 = -G_1G_2H_2$ (x2, x3, x4, x2)
- $L_2 = -G_2G_3H_3$ (x3, x4, C, x3)
- $L_3 = -G_1G_2G_3H_1$ (x1, x2, x3, x4, C, x1)
- $L_4 = -G_4H_1$ (x1, x2, C, x1)
- $L_5 = (G_4)(-H_3)(G_2)(-H_2) = +G_2G_4H_2H_3$ (x2, C, x3, x4, x2)

**Non-touching loops:** every pair of loops shares at least one node (all pass through x2/x3 or C), so there are none.

$$
\Delta = 1 - (L_1+L_2+L_3+L_4+L_5) = 1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3
$$

**Cofactors:** $P_1$ and $P_2$ both touch all loops (they pass through x2 and C), so $\Delta_1 = \Delta_2 = 1$.


**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_4}{1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3}
$$

## Check yourself

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

1. How many forward paths does the signal flow graph have, and what are they?
   A. One: G1G2G3 only
   B. Two: G1G2G3 and G4
   C. Two: G1G2G3 and H1
   D. Three: G1G2G3, G4 and H2
   Answer: B. R to C passes through G1, G2, G3 in one path, and G4 gives a second direct path from node A to C. (Key point: Forward paths)

2. What is the gain of the loop through G1, G2, G3 and the H1 feedback from C to the first summer?
   A. -G1G2G3H1
   B. +G1G2G3H1
   C. -G1G2H1
   D. -G2G3H1
   Answer: A. The minus sign of the summer goes into the branch, so the loop gain is -G1G2G3H1. (Key point: Loop gains)

3. The loop x2 to C (G4), C to x3 (-H3), x3 to x4 (G2), x4 to x2 (-H2) has what gain?
   A. -G2G4H2H3
   B. +G1G2G4H2H3
   C. +G2G4H2H3, because two negative signs multiply
   D. -G4H3
   Answer: C. Two negative feedback branches multiply to a positive value, giving +G2G4H2H3. (Key point: Sign of the loop through G4)

4. Why are Delta1 and Delta2 equal to 1 and no non-touching terms appear in Delta?
   A. Because there is only one loop
   B. Because the forward paths are parallel to all loops
   C. Because Mason's formula ignores feedback loops
   D. All loops share nodes with each other and with both forward paths
   Answer: D. Every loop passes through x2/x3 or C, so no pair is non-touching and both paths touch every loop. (Key point: Non-touching loops and cofactors)

## Close variants set in other exams

- [Mason gain formula with positive H2 loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asfg-mason-g1-g4-h1-h2) — Control Systems, chapter 3, asked 2×

## Related in Control Systems

- [Signal flow graph of G1–G3, H1–H3 block diagram](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asfg-g1g2g3-h1h2h3) — Control Systems, chapter 3
- [Mason's gain formula: G1–G4 with H1, H2, unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amason-g1-g4-h1h2-2075a) — Control Systems, chapter 3, asked 2×
- [Signal flow graph and Mason's formula: G3 feedforward](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asfg-g1g2g3-h1h2-2076a) — Control Systems, chapter 3, asked 2×
- [Block diagram reduction: G1–G4 with H1, H2, H3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2h3-2073) — Control Systems, chapter 3, asked 3×
- [Block diagram reduction: G1–G4 with H1, H2, H3 loops](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2h3-2070r) — Control Systems, chapter 3, asked 2×
- [Y/R of block diagram with G4 and H1–H3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2h3-2066j) — Control Systems, chapter 3