---
title: "Block diagram reduction: cascaded G1–G3, H1, H2, unity loop"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1g2g3-h1h2-bei2081"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1g2g3-h1h2-bei2081
site: BCT Engineering
---

# Block diagram reduction: cascaded G1–G3, H1, H2, unity loop

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2081 Bhadra, 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-bei2081
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Reduce the blocks and find transfer function for the following model. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → summing point S3 (+, −) → G2 → node B → G3 → C(s). 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 / (1 + G1G2H1 + G2G3H2 + G1G2G3).

## Answer

**Result:** $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3}$

The diagram has three nested negative feedback loops. We remove them from the inside out.

```text
R -->(S1)-->(S2)--> G1 -->(S3)--> G2 --+--> G3 --+--> C
      -^     -^            -^          |B        |
       |      +--- H1 ------|----------+         |
       |                    +------- H2 ---------+
       +------------------- 1 -------------------+
```

#### Step 1: Move the take-off point of $H_2$ from C to B

C $= G_3 \times$ B. Moving the take-off point to before the block $G_3$ means the feedback branch becomes $G_3H_2$, from node B to $S_3$.

#### Step 2: Reduce the innermost loop ($G_2$ with feedback $G_3H_2$)

$$
\frac{B}{E_3} = \frac{G_2}{1+G_2G_3H_2}
$$

#### Step 3: Series with $G_1$, then the $H_1$ loop

The forward gain from $S_2$ to B is $\frac{G_1G_2}{1+G_2G_3H_2}$, with negative feedback $H_1$:

$$
\begin{aligned}
\frac{B}{E_2} &= \frac{\frac{G_1G_2}{1+G_2G_3H_2}}{1+\frac{G_1G_2H_1}{1+G_2G_3H_2}} = \frac{G_1G_2}{1+G_2G_3H_2+G_1G_2H_1}
\end{aligned}
$$

#### Step 4: Series with $G_3$

$$
\frac{C}{E_1} = \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2}
$$

#### Step 5: Unity negative feedback loop

$$
\begin{aligned}
\frac{C}{R} &= \frac{\frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2}}{1+\frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2}} \\
&= \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3}
\end{aligned}
$$

**Check (Mason's rule):** one forward path $P_1=G_1G_2G_3$; loops $-G_1G_2H_1$, $-G_2G_3H_2$, $-G_1G_2G_3$, all touching each other and the path. So $\Delta = 1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3$ and $\Delta_1=1$, giving the same answer.

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

## Check yourself

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

1. H2 from C to S3: moving the take-off to node B makes it
   A. H2/G3
   B. G3H2
   C. H2
   D. G2H2
   Answer: B. C = G3*B, so the branch is G3H2 from B. (Key point: Moving H2)

2. The loop around G2 with G3H2 gives
   A. G2/(1+G2G3H2)
   B. G2/(1-G2G3H2)
   C. G2G3H2
   D. G2/(1+G2)
   Answer: A. Negative feedback. (Key point: Innermost loop)

3. After the H1 loop, B/E2 is
   A. G1G2/(1+G1G2H1)
   B. G1G2/(1+G2G3H2+G1G2H1)
   C. G1G2/(1+G2G3H2)
   D. G1G2H1/(1+G2G3H2)
   Answer: B. Both loops add to the denominator. (Key point: H1 loop)

4. After unity feedback, the denominator is
   A. 1+G1G2H1+G2G3H2
   B. 1+G1G2G3
   C. 1+G1G2H1+G2G3H2+G1G2G3
   D. 1+G1G2H1+G2G3H2-G1G2G3
   Answer: C. Unity feedback adds G1G2G3 to the denominator. (Key point: Final TF)

## 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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