---
title: "Y/R of block diagram with G4 and H1–H3"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1h2h3-2066j"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1h2h3-2066j
site: BCT Engineering
---

# Y/R of block diagram with G4 and H1–H3

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2066 Jestha (old course), 6 marks
- Video: [Problem 1 on Block Diagram Reduction](https://www.youtube.com/watch?v=7a2pyG1JkHQ) — TutorialsPoint
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1h2h3-2066j
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Evaluate the transfer function Y(s)/R(s) for the system represented by following block diagram. [Figure: R(s) → summing point S1 (+, −) → node A → G1 → summing point S2 (+, −) → node B → G2 → summing point S3 (+, −) → node C → G3 → summing point S4 (+, +) → Y(s). G4 from node A to S4 (+). H1 from node B fed back to S1 (−). H2 from node C fed back to S2 (−). H3 from Y(s) fed back to S3 (−).]

## Short answer

By Mason’s gain formula, Y(s)/R(s) = [G1G2G3 + G4(1+G2H2)] / [1+G1H1+G2H2+G3H3+G1G3H1H3+G4H1H2H3].

## Answer

Mason's gain formula is used (the block diagram reduction gives the same result). Name the signals: $A$ = output of $S_1$, $B$ = output of $S_2$, $N$ = output of $S_3$ (node C in the figure), $Y$ = output.

#### Signal flow graph
```text
                     G4
        +-----------------------------+
        |                             v
R -1--> A --G1--> B --G2--> N --G3--> Y
        ^         |^        |^        |
        +--(-H1)--+|        ||        |
                   +-(-H2)--+|        |
                             +-(-H3)--+
```
Branches: $B\to A$ ($-H_1$), $N\to B$ ($-H_2$), $Y\to N$ ($-H_3$), $A\to Y$ ($G_4$).

#### Forward paths
- $P_1 = G_1G_2G_3$ (A, B, N, Y)
- $P_2 = G_4$ (A, Y)

#### Individual loops
- $L_1 = -G_1H_1$ (A, B, A)
- $L_2 = -G_2H_2$ (B, N, B)
- $L_3 = -G_3H_3$ (N, Y, N)
- $L_4 = G_4(-H_3)(-H_2)(-H_1) = -G_4H_1H_2H_3$ (A, Y, N, B, A)

#### Non-touching loops
$L_1$ (A, B) and $L_3$ (N, Y) do not touch: $L_1L_3 = G_1G_3H_1H_3$. No three loops are mutually non-touching.

$$
\begin{aligned}
\Delta &= 1 - (L_1 + L_2 + L_3 + L_4) + L_1L_3 \\
&= 1 + G_1H_1 + G_2H_2 + G_3H_3 + G_4H_1H_2H_3 + G_1G_3H_1H_3
\end{aligned}
$$

#### Cofactors
- $P_1$ touches all loops: $\Delta_1 = 1$.
- $P_2$ (A, Y) does not touch $L_2$ (B, N): $\Delta_2 = 1 - L_2 = 1 + G_2H_2$.

#### Transfer function
$$
\frac{Y}{R} = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}
$$

**Answer:**
$$
\frac{Y(s)}{R(s)} = \frac{G_1G_2G_3 + G_4(1 + G_2H_2)}{1 + G_1H_1 + G_2H_2 + G_3H_3 + G_1G_3H_1H_3 + G_4H_1H_2H_3}
$$

## Check yourself

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

1. The forward paths from A to Y are
   A. G1G2G3 and G4
   B. G1G2G3 only
   C. G4H1
   D. G1 and G4
   Answer: A. One path through all three blocks and one direct path G4. (Key point: Forward paths)

2. Which pair of loops does not touch?
   A. G1H1 and G2H2
   B. G1H1 and G3H3
   C. G2H2 and G3H3
   D. G1H1 and G4H1H2H3
   Answer: B. G1H1 uses nodes A,B and G3H3 uses N,Y, so they share no node. (Key point: Non-touching loops)

3. Delta2 for path G4 is 1+G2H2 because
   A. G4 touches all loops
   B. G4 has no loops
   C. the path avoids loop G2H2 (B,N)
   D. G2H2 is positive
   Answer: C. Path A-Y does not touch the loop B-N, so Delta2 = 1 - L2. (Key point: Cofactor of G4)

4. The numerator of Y/R is
   A. G1G2G3 + G4
   B. G1G2G3 + G4(1+G2H2)
   C. (G1G2G3+G4)(1+G2H2)
   D. G1G2G3(1+G2H2)+G4
   Answer: B. P1*Delta1 + P2*Delta2 = G1G2G3 + G4(1+G2H2). (Key point: Numerator)

## Related in Control Systems

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