---
title: "Block diagram reduction: G1–G4 with H1, H2, H3"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1h2h3-2073"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1h2h3-2073
site: BCT Engineering
---

# Block diagram reduction: G1–G4 with H1, H2, H3

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2073 Shrawan, 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-h1h2h3-2073
- 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 (+, −) → node A → G1 → summing point S2 (+, −) → G2 → summing point S3 (+, −) → G3 → node B → summing point S4 (+, +) → C(s). G4 from node A to S4 (+). H2 from node B fed back to S2 (−). H1 from C(s) fed back to S3 (−). H3 from C(s) fed back to S1 (−).]

## Short answer

C(s)/R(s) = (G1G2G3 + G4 + G2G3G4H2) / (1 + G3H1 + G2G3H2 + G1G2G3H3 + G4H3 + G2G3G4H2H3).

## Answer

Name the signals: $A$ = output of $S_1$, $B$ = output of $G_3$. From the figure, $C = B + G_4A$.

#### Step 1: Move the take-off point of $H_1$ from $C$ back to $B$
Since $C = B + G_4A$:
$$
H_1C = H_1B + G_4H_1A
$$
So $H_1$ is now taken from $B$ (a loop around $G_3$), and an extra branch $G_4H_1$ goes from $A$ to $S_3$ with a minus sign.

#### Step 2: Move the extra input at $S_3$ back before $G_2$
A signal entering after $G_2$ can be moved before it by dividing by $G_2$: the branch becomes $G_4H_1/G_2$ entering $S_2$ (minus).

#### Step 3: Reduce the loop $G_3$, $H_1$
$$
\frac{G_3}{1 + G_3H_1}
$$

#### Step 4: Reduce the loop with $H_2$
Forward: $G_2\cdot\dfrac{G_3}{1+G_3H_1}$, feedback $H_2$:
$$
\frac{B}{E_2} = \frac{G_2G_3}{1 + G_3H_1 + G_2G_3H_2}
$$
where $E_2 = G_1A - \dfrac{G_4H_1}{G_2}A$ is the input of $G_2$. Hence
$$
\frac{B}{A} = \frac{G_3(G_1G_2 - G_4H_1)}{1 + G_3H_1 + G_2G_3H_2}
$$

#### Step 5: Add the parallel path $G_4$
$$
\begin{aligned}
\frac{C}{A} &= \frac{G_3(G_1G_2 - G_4H_1) + G_4(1 + G_3H_1 + G_2G_3H_2)}{1 + G_3H_1 + G_2G_3H_2} \\
&= \frac{G_1G_2G_3 + G_4 + G_2G_3G_4H_2}{1 + G_3H_1 + G_2G_3H_2}
\end{aligned}
$$

```text
R ->(S1)--A--> [ N / D' ] -----+---> C
     ^-                        |
     +--------- H3 ------------+
 N  = G1G2G3 + G4 + G2G3G4H2
 D' = 1 + G3H1 + G2G3H2
```

#### Step 6: Close the outer loop with $H_3$
$$
\frac{C}{R} = \frac{N/D'}{1 + H_3N/D'} = \frac{N}{D' + H_3N}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_4 + G_2G_3G_4H_2}{1 + G_3H_1 + G_2G_3H_2 + G_1G_2G_3H_3 + G_4H_3 + G_2G_3G_4H_2H_3}
$$

**Check (Mason):** paths $P_1 = G_1G_2G_3$ ($\Delta_1 = 1$), $P_2 = G_4$ ($\Delta_2 = 1 + G_2G_3H_2$, since the $H_2$ loop does not touch it). Loops $-G_2G_3H_2$, $-G_3H_1$, $-G_1G_2G_3H_3$, $-G_4H_3$; the pair ($-G_2G_3H_2$, $-G_4H_3$) is non-touching. Same result.

## Check yourself

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

1. Since C = B + G4A, how is H1C rewritten?
   A. H1C = H1B - G4H1A
   B. H1C = H1B + G4H1A
   C. H1C = H1G4
   D. H1C = H1B/G4
   Answer: B. Substituting C gives a term from B and an extra branch G4H1 from A. (Key point: Moving H1 take-off)

2. How is the extra branch G4H1 at S3 moved before G2?
   A. Multiply by G2
   B. Multiply by G1
   C. Divide by G2, giving G4H1/G2 into S2
   D. Leave it unchanged
   Answer: C. A signal entering after a block moves before it by dividing by that block. (Key point: Moving a branch before G2)

3. What is C/A after the inner loops?
   A. (G1G2G3 + G4 + G2G3G4H2)/(1 + G3H1 + G2G3H2)
   B. (G1G2G3 + G4)/(1 + G3H1)
   C. (G1G2G3 - G4)/(1 + G2G3H2)
   D. G1G2G3/(1 + G3H1 + G2G3H2)
   Answer: A. Adding G4 to B/A and simplifying gives this numerator and denominator. (Key point: Parallel path G4)

4. With N/D' forward and H3 feedback, what is C/R?
   A. N/(D' - H3N)
   B. N D'/(1 + H3)
   C. N/(1 + H3)
   D. N/(D' + H3N)
   Answer: D. G/(1 + GH) with G = N/D' gives N/(D' + H3N). (Key point: Closing the outer loop)

## 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 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×

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