---
title: "Block diagram reduction: G1–G4, H1, H2, unity feedback"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1h2-unity-2076c"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1h2-unity-2076c
site: BCT Engineering
---

# Block diagram reduction: G1–G4, H1, H2, unity feedback

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2076 Chaitra, 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-h1h2-unity-2076c
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Reduce the following block diagram and find transfer function. [Figure: R → summing point S1 (+, −) → summing point S2 (+, −) → G1 → node A → summing point S3 (+, −) → G2 → node B → G3 → summing point S4 (+, +) → C. G4 from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C fed back to S3 (−). Unity feedback from C to S1 (−).]

## Short answer

Reduce by moving H1 take-off point past G3 and S4, combining feedbacks, and closing the main loop. The transfer function is (G1G2G3 + G1G4) / (1 + G1G2H1 + G2G3H2 + G1G2G3 + G1G4 - G1G2G4H1H2).

## Answer

Name the signals: $E$ = output of $S_2$ (input of $G_1$), $A$ = output of $G_1$, $B$ = output of $G_2$.
$$
\begin{aligned}
E &= R - C - H_1B, \quad A = G_1E \\
B &= G_2(A - H_2C), \quad C = G_3B + G_4A
\end{aligned}
$$
($S_1$ and $S_2$ are in cascade and are treated as one summer.)

#### Step 1: Move the take-off point of $H_1$ from $B$ past $G_3$ and $S_4$
From $C = G_3B + G_4A$: $B = \dfrac{C - G_4A}{G_3}$. Hence
$$
H_1B = \frac{H_1}{G_3}C - \frac{G_1G_4H_1}{G_3}E
$$
So the $H_1$ path becomes a block $H_1/G_3$ from $C$ (negative) and a positive self loop $G_4H_1/G_3$ around $G_1$.

#### Step 2: Combine the feedbacks and remove the self loop
Feedback from $C$: $H = 1 + \dfrac{H_1}{G_3}$.
$G_1$ with positive feedback $G_4H_1/G_3$: $\dfrac{G_1}{1 - G_1G_4H_1/G_3}$.

#### Step 3: Reduce from $A$ to $C$
$$
C = G_3G_2(A - H_2C) + G_4A
\;\Rightarrow\;
\frac{C}{A} = \frac{G_2G_3 + G_4}{1 + G_2G_3H_2}
$$

```text
R ->(+)-> [G1/(1-G1G4H1/G3)] -> [(G2G3+G4)/(1+G2G3H2)] -> C
     ^-                                               |
     +------------------ (1 + H1/G3) <----------------+
```

#### Step 4: Close the main loop
$$
\frac{C}{R} = \frac{G_1(G_2G_3+G_4)}{\left(1 - \frac{G_1G_4H_1}{G_3}\right)(1 + G_2G_3H_2) + G_1(G_2G_3 + G_4)\left(1 + \frac{H_1}{G_3}\right)}
$$
On expanding, the $G_1G_4H_1/G_3$ terms cancel.

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_1G_2G_3 + G_1G_4 - G_1G_2G_4H_1H_2}
$$

**Check (Mason):** paths $G_1G_2G_3$, $G_1G_4$; loops $-G_1G_2H_1$, $-G_2G_3H_2$, $-G_1G_2G_3$, $-G_1G_4$ and $+G_1G_2G_4H_1H_2$ (via $G_4$, $H_2$, $G_2$, $H_1$). All loops touch each other and both paths, 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. Why is the H1 take-off moved from B to C?
   A. Because H1 cannot feed back from B
   B. Since C = G3B + G4A, so B = (C - G4A)/G3 and H1B becomes (H1/G3)C minus a self loop term
   C. To remove G2 from the loop
   D. Because B equals C
   Answer: B. Moving past G3 and S4 gives a block H1/G3 from C and a loop term G4H1/G3 around G1. (Key point: Take-off shift)

2. What is the total feedback from C to the summers?
   A. 1 + H1/G3
   B. 1 - H1/G3
   C. H1/G3
   D. 1 + H1 G3
   Answer: A. The unity feedback and the H1/G3 path are in parallel and add. (Key point: Combined feedback from C)

3. What is C/A for the inner section with H2?
   A. (G2G3 + G4)/(1 - G2G3H2)
   B. G2G3/(1 + G2G3H2) x G4
   C. (G2G3 + G4)/(1 + G2G3H2)
   D. (G2 + G4)/(1 + G2H2)
   Answer: C. C = G3G2(A - H2C) + G4A gives C(1 + G2G3H2) = (G2G3 + G4)A. (Key point: A to C)

4. What is the final transfer function?
   A. (G1G2G3 + G1G4)/(1 + G1G2H1 + G2G3H2 + G1G2G3 + G1G4 + G1G2G4H1H2)
   B. G1(G2G3 + G4)/(1 + G2G3H2)
   C. (G1G2G3 + G4)/(1 + G1G2H1)
   D. (G1G2G3 + G1G4)/(1 + G1G2H1 + G2G3H2 + G1G2G3 + G1G4 - G1G2G4H1H2)
   Answer: D. Closing the loop and expanding gives a minus sign on the G1G2G4H1H2 term, confirmed by Mason's formula. (Key point: Final answer)

## Close variants set in other exams

- [Block diagram reduction of Fig. 2(a): G1–G3, H1, H2](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2g3-h1h2-2066) — Control Systems, chapter 3, asked 2×
- [Block diagram reduction: cascaded G1–G3, H1, H2, unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2g3-h1h2-bei2081) — Control Systems, chapter 3, asked 2×

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