---
title: "Signal flow graph of G1–G3, H1–H3 block diagram"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sfg-g1g2g3-h1h2h3"
source: https://www.bctengineer.com/q/semester_3/control_systems/sfg-g1g2g3-h1h2h3
site: BCT Engineering
---

# Signal flow graph of G1–G3, H1–H3 block diagram

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2080 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-g1g2g3-h1h2h3
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Obtain the overall transfer function of given system by signal flow graph technique. [Figure: R(s) → summing point S1 (+, −) → G1 → summing point S2 (+, −) → summing point S3 (+, −) → node A. From node A: G2 → node B, and H1 (feedforward branch) going to summing point S4 (+). Node B → S4 (+) → G3 → C(s). From node B, H2 feeds back to S2 (−). From C(s), H3 feeds back to S3 (−). From C(s), unity feedback to S1 (−).]

## Short answer

Using Mason's gain formula with two forward paths (G1G2G3, G1H1G3) and five loops: C(s)/R(s) = G1 G3 (G2 + H1) / (1 + G2 H2 + G2 G3 H3 + G3 H1 H3 + G1 G2 G3 + G1 G3 H1).

## Answer

In the signal flow graph method, each signal becomes a node and each block a branch; then Mason's gain formula gives the transfer function.

#### Signal flow graph
Nodes: $x_1$ = output of $S_1$, $x_2$ = output of $G_1$, $x_3$ = output of $S_2$, $x_4$ = output of $S_3$ (node A), $x_5$ = node B, $x_6$ = output of $S_4$, $C$.

```text
                         H1
                   +--------------+
                   |              v
R -1-> x1 -G1-> x2 -1-> x3 -1-> x4 -G2-> x5 -1-> x6 -G3-> C
       ^                ^       ^         |                |
       |                +-(-H2)-+---------+                |
       |                        +-------(-H3)--------------+
       +--------------------(-1)---------------------------+
```
Branches: $x_4\to x_6$ ($H_1$), $x_5\to x_3$ ($-H_2$), $C\to x_4$ ($-H_3$), $C\to x_1$ ($-1$); all others as labelled.

#### Forward paths
- $P_1 = G_1G_2G_3$ (through $G_2$)
- $P_2 = G_1H_1G_3$ (through the $H_1$ branch)

#### Individual loops
- $L_1 = -G_2H_2$ ($x_3, x_4, x_5, x_3$)
- $L_2 = -G_2G_3H_3$ ($x_4, x_5, x_6, C, x_4$)
- $L_3 = -G_3H_1H_3$ ($x_4, x_6, C, x_4$)
- $L_4 = -G_1G_2G_3$ ($x_1 \dots C, x_1$)
- $L_5 = -G_1G_3H_1$ ($x_1, x_2, x_3, x_4, x_6, C, x_1$)

#### Non-touching loops
$L_1$ touches $L_2$ and $L_3$ at $x_4$, and all others pass through $x_3$/$x_4$. So there are no non-touching pairs.

$$
\Delta = 1 + G_2H_2 + G_2G_3H_3 + G_3H_1H_3 + G_1G_2G_3 + G_1G_3H_1
$$

#### Cofactors
Both forward paths pass through $x_3$, $x_4$ and $C$, so they touch every loop: $\Delta_1 = \Delta_2 = 1$.

#### Mason's gain formula
$$
T = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_3(G_2 + H_1)}{1 + G_2H_2 + G_2G_3H_3 + G_3H_1H_3 + G_1G_2G_3 + G_1G_3H_1}
$$

## Check yourself

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

1. What are the two forward paths?
   A. G1G2G3 and G1H1G3
   B. G1G2G3 and G2H1
   C. G1G3 and G2H2
   D. G1G2H1 and G3
   Answer: A. The signal reaches S4 either via G2 or via the feedforward H1 branch, then G3. (Key point: Forward paths)

2. Which pair of loops involves H3?
   A. -G2G3H3 and -G3H1H3
   B. -G2H2 and -G1G2G3
   C. -G2G3H3 and -G1G3H1
   D. -G3H3 only
   Answer: A. C feeds back through H3 to x4, which then reaches C via G2G3 or H1G3. (Key point: Loop through H3)

3. Why are there no non-touching loops?
   A. All loops pass through x3/x4
   B. There are no loops
   C. Each loop uses a different node
   D. H1 separates them
   Answer: A. Every loop shares x3 or x4 so all touch. (Key point: Non-touching)

4. What is the numerator of C/R?
   A. G1G3(G2 + H1)
   B. G1G2G3
   C. G1G3H1
   D. G1G2(G3 + H1)
   Answer: A. P1 + P2 = G1G2G3 + G1H1G3 = G1G3(G2+H1). (Key point: Result)

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