---
title: "Block diagram reduction with positive H2 loop"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1-h2pos-2079b"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1-h2pos-2079b
site: BCT Engineering
---

# Block diagram reduction with positive H2 loop

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2079 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-g1-g4-h1-h2pos-2079b
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Determine the transfer function C(s)/R(s) of the block diagram given below, using block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → node A → G1 → summing point S3 (+, +) → 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+G4) / (1+G1G2H1−G2G3H2+G2G4H1H2+G1G2G3+G4).

## Answer

**Result:** $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2+G_1G_2G_3+G_4}$

The feedback $H_2$ is taken from the output, after summing point $S_4$, which is fed by both $G_3$ and $G_4$. We first shift this take-off point back over $S_4$, then reduce the loops.

```text
R->(S1)->(S2)--A--> G1 ->(S3)-> G2 --B--> G3 ->(S4)--+--> C
   -^     -^   |         +^          |         +^    |
    |      +---|----------|-- H1 ----+          |    |
    |          +----------|------- G4 ----------+    |
    |                     +---------- H2 ------------+
    +---------------------- 1 -----------------------+
```

#### Step 1: Split the $H_2$ feedback

$C = G_3B + G_4A$, so $H_2C = G_3H_2\,B + G_4H_2\,A$. The $H_2$ branch is replaced by two branches into $S_3$ (both $+$):

- $G_3H_2$ from node B (a positive feedback loop around $G_2$)
- $G_4H_2$ from node A (in parallel with $G_1$)

#### Step 2: Combine the parallel blocks from A to $S_3$

$$
G_1 + G_4H_2
$$

#### Step 3: Positive loop around $G_2$

$$
\frac{B}{E_3} = \frac{G_2}{1-G_2G_3H_2}
\quad\Rightarrow\quad
\frac{B}{A} = \frac{G_2(G_1+G_4H_2)}{1-G_2G_3H_2}
$$

#### Step 4: Negative loop through $H_1$ (B back to $S_2$)

$$
\frac{A}{E_1} = \frac{1}{1+H_1\frac{G_2(G_1+G_4H_2)}{1-G_2G_3H_2}} = \frac{1-G_2G_3H_2}{1-G_2G_3H_2+G_1G_2H_1+G_2G_4H_1H_2}
$$

#### Step 5: Output from A

$$
\begin{aligned}
\frac{C}{A} &= G_3\frac{B}{A} + G_4 = \frac{G_2G_3(G_1+G_4H_2) + G_4(1-G_2G_3H_2)}{1-G_2G_3H_2} \\
&= \frac{G_1G_2G_3+G_4}{1-G_2G_3H_2}
\end{aligned}
$$

Hence

$$
\frac{C}{E_1} = \frac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2}
$$

#### Step 6: Unity negative feedback

$$
\frac{C}{R} = \frac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2+G_1G_2G_3+G_4}
$$

**Check (Mason's rule):** forward paths $G_1G_2G_3$ and $G_4$; loops $-G_1G_2H_1$, $+G_2G_3H_2$, $-G_1G_2G_3$, $-G_4$, $-G_2G_4H_1H_2$ (A → $G_4$ → C → $H_2$ → $G_2$ → B → $H_1$ → A). All loops touch each other and both paths, so $\Delta_1=\Delta_2=1$ and the same result follows.

## Check yourself

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

1. Why is the H2 take-off point first shifted back behind summing point S4?
   A. Because C = G3*B + G4*A, so H2*C can be split into branches from B and A
   B. Because H2 is positive and positive feedback cannot be reduced directly
   C. Because G4 is in parallel with H1 and must be removed first
   D. Because the unity feedback must be reduced before any other loop
   Answer: A. The output C is the sum of G3*B and G4*A, so H2*C = G3H2*B + G4H2*A. This gives two branches into S3 that can be reduced by standard rules. (Key point: Moving the H2 take-off)

2. After splitting H2, which blocks are in parallel between node A and summing point S3?
   A. G1 and G3H2
   B. G1 and G4H2
   C. G1 and G4
   D. G2 and G4H2
   Answer: B. The branch G4H2 from node A enters S3 with a plus sign, in parallel with G1, giving G1 + G4H2. (Key point: Parallel combination A to S3)

3. The G3H2 branch from node B to S3 is positive. What is the transfer function of the loop around G2 (S3 to B)?
   A. G2/(1+G2G3H2)
   B. G2/(1-G2G3H2)
   C. G2G3H2/(1-G2)
   D. 1/(1-G2G3H2)
   Answer: B. Positive feedback gives forward gain divided by 1 minus loop gain: G2/(1-G2G3H2). (Key point: Positive loop around G2)

4. If C/E1 = N/D, with N = G1G2G3+G4, what is C/R after closing the unity negative feedback?
   A. N/(D-N)
   B. N/(1+D)
   C. N/(D+N)
   D. N/(1-D N)
   Answer: C. Closing unity negative feedback gives (N/D)/(1+N/D) = N/(D+N), which is why the denominator ends with +G1G2G3+G4. (Key point: Final unity feedback step)

## Close variants set in other exams

- [Block diagram reduction: G1–G4, H1, H2, unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2-2075a) — Control Systems, chapter 3, asked 4×
- [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×

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