---
title: "Block diagram reduction: G1–G4, H1, H2, unity loop"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1h2-2075a"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1h2-2075a
site: BCT Engineering
---

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

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2074 Asoj, 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-h1h2-2075a
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Determine the overall transfer function C(s)/R(s) of the given system by block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → summing point S3 (+, −) → node A → G2 → node B → G3 → summing point S4 (+, +) → C(s). G4 from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (−). Unity feedback from C(s) to S1 (−).]

## Short answer

C(s)/R(s) = (G1G2G3 + G1G4) / (1 + G1G2H1 + G2G3H2 + G4H2 + G1G2G3 + G1G4).

## Answer

Name the signals: $E$ = output of $S_2$ (input of $G_1$), $A$ = output of $S_3$, $B$ = output of $G_2$.

#### Step 1: Move the take-off point of $H_1$ from $B$ back to $A$
$B = G_2A$, so the $H_1$ branch becomes $G_2H_1$ taken from $A$.

#### Step 2: Combine the paths from $A$ to $C$
$G_2$ and $G_3$ are in series, and that series pair is in parallel with $G_4$:
$$
\frac{C}{A} = G_2G_3 + G_4
$$

#### Step 3: Move the take-off point of $G_2H_1$ from $A$ forward to $C$
Since $A = \dfrac{C}{G_2G_3 + G_4}$, the branch becomes $\dfrac{G_2H_1}{G_2G_3 + G_4}$ taken from $C$.

```text
R->(S1)->(S2)-> G1 ->(S3)-> [G2G3+G4] --+--> C
    ^-     ^-          ^-               |
    |      |           +----- H2 -------+
    |      +--- G2H1/(G2G3+G4) ---------+
    +------------- 1 -------------------+
```

#### Step 4: Reduce the inner loop with $H_2$
$$
\frac{C}{G_1E} = \frac{G_2G_3 + G_4}{1 + H_2(G_2G_3 + G_4)}
$$
So the forward gain from $S_2$ is
$$
G_f = \frac{G_1(G_2G_3 + G_4)}{1 + G_2G_3H_2 + G_4H_2}
$$

#### Step 5: Combine the two outer feedbacks and close the loop
$S_1$ and $S_2$ are cascaded summers, so the feedbacks from $C$ add in parallel:
$$
H = 1 + \frac{G_2H_1}{G_2G_3 + G_4}
$$
$$
\frac{C}{R} = \frac{G_f}{1 + G_fH}
= \frac{G_1(G_2G_3 + G_4)}{1 + G_2G_3H_2 + G_4H_2 + G_1(G_2G_3 + G_4) + G_1G_2H_1}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_4H_2 + G_1G_2G_3 + G_1G_4}
$$

**Check (Mason):** paths $G_1G_2G_3$, $G_1G_4$; loops $-G_1G_2H_1$, $-G_2G_3H_2$, $-G_4H_2$, $-G_1G_2G_3$, $-G_1G_4$; all loops touch, $\Delta_1 = \Delta_2 = 1$. Same result.

## Check yourself

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

1. How is the H1 take-off moved from B back to A?
   A. B = G2A, so it becomes G2H1 taken from A
   B. B = A/G2, so it becomes H1/G2
   C. H1 is removed
   D. It becomes H1G3
   Answer: A. Since B = G2A, H1B = G2H1A. (Key point: Take-off shift for H1)

2. What is the gain from A to C?
   A. G2G3G4
   B. G2G3 + G4 (series pair in parallel with G4)
   C. G2 + G3 + G4
   D. G2G3/G4
   Answer: B. G2 and G3 are in series and G4 is parallel with that pair. (Key point: A to C)

3. What is the inner loop gain with H2?
   A. (G2G3 + G4)/(1 - H2(G2G3 + G4))
   B. G2G3/(1 + H2)
   C. (G2G3 + G4)/(1 + H2(G2G3 + G4))
   D. 1/(1 + H2)
   Answer: C. Negative feedback H2 gives G/(1 + GH) with G = G2G3 + G4. (Key point: Inner loop with H2)

4. What is the final denominator?
   A. 1 + G1G2H1 + G2G3H2 + G1G2G3
   B. 1 - G1G2H1 + G2G3H2 + G4H2
   C. 1 + G1G2H1 + G2G3H2 + G4H2 + G1G2G3H1
   D. 1 + G1G2H1 + G2G3H2 + G4H2 + G1G2G3 + G1G4
   Answer: D. Mason's check lists five loops: -G1G2H1, -G2G3H2, -G4H2, -G1G2G3, -G1G4. (Key point: Final result)

## Close variants set in other exams

- [Block diagram reduction: G1–G4 with H1, H2, H3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2h3-2073) — Control Systems, chapter 3, asked 3×
- [Block diagram reduction with positive H2 loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1-h2pos-2079b) — Control Systems, chapter 3, asked 3×
- [Block diagram reduction with G1, G2, H1, H2, feedforward](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2-h1h2-ff-unity) — Control Systems, chapter 3, asked 2×

## Related in Control Systems

- [Block diagram reduction: G1–G4, H1, H2, unity feedback](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2-unity-2076c) — Control Systems, chapter 3, asked 3×
- [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×
- [Block diagram reduction: G1–G4 with H1, H2, H3 loops](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2h3-2070r) — Control Systems, chapter 3, asked 2×
- [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×
- [Mason's gain formula: G1–G4 with H1, H2, unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Amason-g1-g4-h1h2-2075a) — Control Systems, chapter 3, asked 2×
- [Block diagram reduction: G3 feedforward, H1 and H2 loops](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2g3-h1h2-b) — Control Systems, chapter 3, asked 4×