---
title: "Block diagram reduction with G1–G4, positive feedbacks"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1-h2"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1-h2
site: BCT Engineering
---

# Block diagram reduction with G1–G4, positive feedbacks

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2082 Baisakh, 8 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-h1-h2
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Obtain the overall transfer function C(s)/R(s) using block diagram reduction technique for the system represented by the block diagram as shown in the figure below. [Figure: R(s) enters summing point S1 (+); output of S1 → G1 → node A → G2 → node B → summing point S2 (+) → G4 → summing point S3 (+) → C(s). From node A, branch → G3 → added (+) at S3. From node B, H1 feeds back (+) to S1. From C(s), H2 feeds back (+) to S2.]

## Short answer

C(s)/R(s) = (G1 G2 G4 + G1 G3) / (1 - G1 G2 H1 - G4 H2 + G1 G2 G4 H1 H2).

## Answer

Name the signals: $E_1$ is the output of $S_1$, $A$ the output of $G_1$, $B$ the output of $G_2$, $E_2$ the output of $S_2$.

The signal equations from the figure are:
$$
\begin{aligned}
E_1 &= R + H_1B, \quad A = G_1E_1, \quad B = G_2A \\
E_2 &= B + H_2C, \quad C = G_4E_2 + G_3A
\end{aligned}
$$

#### Step 1: Reduce the first loop ($G_1$, $G_2$ with positive feedback $H_1$)
The loop $G_1G_2$ with positive feedback $H_1$ gives
$$
\frac{B}{R} = \frac{G_1G_2}{1 - G_1G_2H_1}
$$
and, since $A = B/G_2$,
$$
\frac{A}{R} = \frac{G_1}{1 - G_1G_2H_1}
$$

```text
        +------- G3 (from A) -------+
        |                           v
R -> [G1/(1-G1G2H1)] -A-> G2 -B-> (S2)+-> G4 -> (S3)+-> C
                                   ^                 |
                                   +------ H2 -------+ (+)
```

#### Step 2: Reduce the second loop ($G_4$ with positive feedback $H_2$)
$C$ is fed back through $H_2$ to $S_2$ with a plus sign, and $G_3A$ is added after $G_4$:
$$
C = G_4(B + H_2C) + G_3A
\;\Rightarrow\;
C(1 - G_4H_2) = G_4B + G_3A
$$

#### Step 3: Substitute $A$ and $B$
$$
\begin{aligned}
C(1 - G_4H_2) &= \frac{G_1G_2G_4 + G_1G_3}{1 - G_1G_2H_1}\,R \\
\frac{C}{R} &= \frac{G_1(G_2G_4 + G_3)}{(1 - G_1G_2H_1)(1 - G_4H_2)}
\end{aligned}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2G_4 + G_1G_3}{1 - G_1G_2H_1 - G_4H_2 + G_1G_2G_4H_1H_2}
$$

**Check (Mason):** paths $G_1G_2G_4$ and $G_1G_3$; loops $L_1 = G_1G_2H_1$, $L_2 = G_4H_2$, which do not touch each other, so $\Delta = 1 - L_1 - L_2 + L_1L_2$. Both paths touch both loops, so $\Delta_1=\Delta_2=1$. Same result.

## Check yourself

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

1. What is B/R for G1, G2 with positive feedback H1 from B?
   A. G1G2/(1 - G1G2H1)
   B. G1G2/(1 + G1G2H1)
   C. G1G2H1
   D. G1/(1 - G2H1)
   Answer: A. Positive feedback gives G/(1 - GH) with G = G1G2. (Key point: Positive feedback loop)

2. Which equation relates C to the other signals?
   A. C = G4(B + H2C) + G3A
   B. C = G4(B - H2C) + G3A
   C. C = G4B + G3 B
   D. C = G4(A + H2C)
   Answer: A. H2 feeds back positively to S2, and G3A is added after G4. (Key point: Output equation)

3. After solving, what factor results from the G4, H2 loop?
   A. (1 - G4H2)
   B. (1 + G4H2)
   C. G4H2
   D. (1 - G3H2)
   Answer: A. C(1 - G4H2) = G4B + G3A from the positive feedback. (Key point: Second loop)

4. Why does the denominator include +G1G2G4H1H2?
   A. The two loops are non-touching, so their product is added
   B. The loops touch
   C. It is a forward path
   D. It is a sign error
   Answer: A. Δ = 1 - L1 - L2 + L1L2 for non-touching loops. (Key point: Mason check)

## Related in Control Systems

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