---
title: "Block diagram reduction with G4 on input and H1–H2"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1h2-2074c"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1h2-2074c
site: BCT Engineering
---

# Block diagram reduction with G4 on input and H1–H2

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2074 Chaitra, 8 marks; 2080 Bhadra, 8 marks (asked 2 times)
- 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-h1h2-2074c
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Determine the transfer function of the given system by reducing blocks. [Figure: R(s) → summing point S1 (+, −) → G1 → G3 → node A → summing point S2 (+) → C(s); G2 from node A also feeds S2 (sign printed unclear, take +). Feedback: C(s) → H1 → summing point S3 (+) → H2 → S1 (−). G4 takes the input R(s) (branch before S1) into S3 (−).]

## Short answer

By block diagram reduction, the overall transfer function is C(s)/R(s) = G1 G3 (1+G2)(1+G4 H2) / (1 + G1 G3 H1 H2 + G1 G2 G3 H1 H2).

## Answer

Reading of the figure: the forward path is $G_1$ then $G_3$ to node $A$; from $A$ the signal reaches $S_2$ directly and also through $G_2$ (sign taken as $+$). The output $C$ goes through $H_1$ to $S_3$ (+), where $G_4R$ is subtracted; the result passes through $H_2$ to $S_1$ (−).

#### Step 1: Series and parallel blocks in the forward path
$G_1$ and $G_3$ are in series; the direct path and $G_2$ from $A$ are in parallel:
$$
G(s) = G_1G_3(1 + G_2)
$$

#### Step 2: Move summing point $S_3$ past $H_2$
The signal into $S_1$ from the feedback is $H_2(H_1C - G_4R) = H_1H_2C - G_4H_2R$.
So after moving $S_3$ beyond $H_2$, the feedback path is $H_1H_2$ from $C$, and a path $G_4H_2$ from $R$ enters $S_1$ with sign $(-)(-) = +$.

#### Step 3: Combine the two input paths into $S_1$
$R$ reaches $S_1$ directly and through $G_4H_2$ (both +), a parallel combination:
$$
E = (1 + G_4H_2)R - H_1H_2C
$$
```text
 R ─►[1+G4H2]─►(Σ)─►[G1G3(1+G2)]──┬──► C
               ▲ -                │
               └─────[H1H2]◄──────┘
```

#### Step 4: Close the feedback loop
$$
\frac{C}{(1 + G_4H_2)R} = \frac{G_1G_3(1 + G_2)}{1 + G_1G_3(1 + G_2)H_1H_2}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_3(1 + G_2)(1 + G_4H_2)}{1 + G_1G_3H_1H_2 + G_1G_2G_3H_1H_2}
$$
If the sign at the $G_2$ input of $S_2$ is negative, replace $(1 + G_2)$ by $(1 - G_2)$ throughout.

## Check yourself

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

1. What is the forward path gain G(s)?
   A. G1G3(1 + G2)
   B. G1G3G2
   C. G1G3 + G2
   D. G1G3/(1 + G2)
   Answer: A. G1 and G3 are in series; the direct path and G2 from A are in parallel, giving (1+G2). (Key point: Forward path simplification)

2. After moving S3 past H2, what is the sign of the G4H2 path from R into S1?
   A. Positive
   B. Negative
   C. Zero
   D. Depends on H1
   Answer: A. G4 is subtracted at S3 and the result goes negatively into S1, so (-)(-) = +. (Key point: Moving summing point)

3. What is the equivalent feedback block from C after combining H1 and H2?
   A. H1H2
   B. H1 + H2
   C. H1/H2
   D. H2
   Answer: A. H1 and H2 are in series in the feedback path. (Key point: Equivalent feedback)

4. What is C/R?
   A. G1G3(1+G2)(1+G4H2)/(1 + G1G3H1H2 + G1G2G3H1H2)
   B. G1G3(1+G2)/(1 + G1G3H1H2)
   C. G1G3(1+G2)(1-G4H2)/(1 + G1G3H1H2)
   D. G1G3(1+G2)(1+G4H2)/(1 - G1G3H1H2)
   Answer: A. Closing the loop gives G/(1+GH1H2) multiplied by (1+G4H2) at the input. (Key point: Final result)

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