---
title: "Block diagram reduction of Fig. 2(a): G1–G3, H1, H2"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1g2g3-h1h2-2066"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1g2g3-h1h2-2066
site: BCT Engineering
---

# Block diagram reduction of Fig. 2(a): G1–G3, H1, H2

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2066 Bhadra (old course), 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-g1g2g3-h1h2-2066
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Reduce the block diagram of fig. 2(a) and find the overall transfer function C(s)/R(s). [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → summing point S3 (+, −) → node A → G2 → node B → summing point S4 (+, +) → C(s). G3 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

Using block diagram reduction or Mason’s formula, the overall transfer function is C(s)/R(s) = (G1G2+G1G3) / (1+G1G2+G1G3+G1G2H1+G2H2+G3H2).

## 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: Combine the parallel paths from $A$ to $C$
$G_2$ and $G_3$ both go from $A$ to $S_4$:
$$
\frac{C}{A} = G_2 + G_3
$$

#### Step 2: Move the take-off point of $H_1$ from $B$ to $C$
$B = G_2A$ and $A = \dfrac{C}{G_2 + G_3}$, so the branch becomes $\dfrac{G_2H_1}{G_2 + G_3}$ taken from $C$.

#### Step 3: Reduce the inner loop with $H_2$
$$
\frac{C}{G_1E} = \frac{G_2 + G_3}{1 + H_2(G_2 + G_3)}
$$

#### Step 4: Combine the outer feedbacks
$S_1$ and $S_2$ are cascaded, so the unity feedback and the moved $H_1$ branch add in parallel:
$$
H = 1 + \frac{G_2H_1}{G_2 + G_3}
$$

```text
R ->(+)-> G1 -> [(G2+G3)/(1+G2H2+G3H2)] --+--> C
     ^-                                   |
     +------ 1 + G2H1/(G2+G3) <-----------+
```

#### Step 5: Close the loop
$$
G_f = \frac{G_1(G_2 + G_3)}{1 + G_2H_2 + G_3H_2}
$$
$$
\frac{C}{R} = \frac{G_f}{1 + G_fH} = \frac{G_1(G_2 + G_3)}{1 + G_2H_2 + G_3H_2 + G_1(G_2 + G_3) + G_1G_2H_1}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_1G_2 + G_1G_3 + G_1G_2H_1 + G_2H_2 + G_3H_2}
$$

**Check (Mason):** paths $G_1G_2$, $G_1G_3$; loops $-G_1G_2$, $-G_1G_3$, $-G_1G_2H_1$, $-G_2H_2$, $-G_3H_2$, all touching. Same result.

## Check yourself

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

1. From A to S4, the paths through G2 and G3 combine to
   A. G2G3
   B. G2+G3
   C. G2-G3
   D. G2/G3
   Answer: B. They are parallel and both enter S4 with +. (Key point: Parallel paths)

2. Moving the H1 take-off from B to C gives a branch of
   A. G2H1/(G2+G3)
   B. H1/(G2+G3)
   C. G2H1
   D. (G2+G3)H1
   Answer: A. B = G2*A, A = C/(G2+G3). (Key point: Moving H1)

3. The inner loop with H2 gives
   A. (G2+G3)/(1-H2(G2+G3))
   B. (G2+G3)H2
   C. (G2+G3)/(1+H2(G2+G3))
   D. 1/(1+H2(G2+G3))
   Answer: C. Negative feedback H2 around (G2+G3). (Key point: H2 loop)

4. Unity feedback and the moved H1 branch combine as
   A. 1 * G2H1/(G2+G3)
   B. 1 + G2H1/(G2+G3)
   C. 1 - G2H1/(G2+G3)
   D. G2H1
   Answer: B. S1 and S2 are cascaded, so both negative feedbacks add in parallel. (Key point: Combined outer feedback)

## Close variants set in other exams

- [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×

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