---
title: "Block diagram reduction: G1–G4 with H1, H2, H3 loops"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1h2h3-2070r"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1h2h3-2070r
site: BCT Engineering
---

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

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2070 Chaitra, 6 marks; 2078 Bhadra, 8 marks (asked 2 times)
- 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-h1h2h3-2070r
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Reduce the following block diagram model to obtain its overall transfer function. [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 (+). H2 from node B fed back to S2 (−). H3 from C(s) fed back to S3 (−). H1 from C(s) fed back to S1 (−).]

## Short answer

Using Mason's formula: two forward paths G1G2G3 and G4, five loops. Overall transfer function: C(s)/R(s) = (G1G2G3 + G4) / (1 + G1G2H2 + G2G3H3 + G1G2G3H1 + G4H1 - G2G4H2H3).

## Answer

Name the signals: $A$ is the output of $S_2$ (input of $G_1$ and of $G_4$), $B$ is the output of $G_2$.

From the figure the signal equations are:
$$
\begin{aligned}
A &= R - H_1C - H_2B \\
B &= G_2(G_1A - H_3C) \\
C &= G_3B + G_4A
\end{aligned}
$$
(The two summers $S_1$ and $S_2$ are in cascade, so they can be merged into one summer at $A$.)

#### Step 1: Move the take-off point of $H_2$ from $B$ past $G_3$ and $S_4$
From the last equation, $B = \dfrac{C - G_4A}{G_3}$. So the signal $H_2B$ can be produced as
$$
H_2B = \frac{H_2}{G_3}C - \frac{G_4H_2}{G_3}A
$$
The $H_2$ path is replaced by a block $H_2/G_3$ from $C$ (negative) and a block $G_4H_2/G_3$ from $A$ back to $A$ (positive).

#### Step 2: Combine the feedback blocks from $C$ and remove the self loop at $A$
Feedback from $C$ to the summer: $H_1 + \dfrac{H_2}{G_3}$ (parallel blocks).

The positive self loop at $A$ gives the block $\dfrac{1}{1 - G_4H_2/G_3}$.

#### Step 3: Reduce the forward part from $A$ to $C$
$$
C = G_3G_2(G_1A - H_3C) + G_4A
\;\Rightarrow\;
\frac{C}{A} = \frac{G_1G_2G_3 + G_4}{1 + G_2G_3H_3}
$$
This is the inner loop $G_2G_3$ with feedback $H_3$, plus the parallel path $G_4$.

#### Step 4: Close the main loop
Forward gain:
$$
G_f = \frac{G_1G_2G_3 + G_4}{(1 + G_2G_3H_3)\left(1 - \frac{G_4H_2}{G_3}\right)}, \qquad H = H_1 + \frac{H_2}{G_3}
$$
$$
\frac{C}{R} = \frac{G_f}{1 + G_fH}
$$
Multiplying out, the $G_4H_2/G_3$ terms cancel:
$$
\begin{aligned}
1+G_fH &\propto (1+G_2G_3H_3)\left(1-\tfrac{G_4H_2}{G_3}\right) + (G_1G_2G_3+G_4)\left(H_1+\tfrac{H_2}{G_3}\right) \\
&= 1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3
\end{aligned}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_4}{1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3}
$$

**Check by Mason's formula:** forward paths $P_1 = G_1G_2G_3$, $P_2 = G_4$; loops $-G_1G_2H_2$, $-G_2G_3H_3$, $-G_1G_2G_3H_1$, $-G_4H_1$ and $+G_2G_4H_2H_3$ (path $A \to G_4 \to C \to H_3 \to G_2 \to B \to H_2 \to A$). All loops touch each other and both paths, so $\Delta_1 = \Delta_2 = 1$. This gives the same result.

## Check yourself

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

1. How are the cascaded summers S1 and S2 treated?
   A. Merged into one summer at A
   B. Kept separate with extra blocks
   C. Replaced by a multiplication
   D. Removed together with H1
   Answer: A. Summers in cascade can be combined into a single summer. (Key point: Cascade summers)

2. Why does moving the H2 take-off from B to C create a block G4H2/G3 from A?
   A. Because B = (C - G4A)/G3
   B. Because B = G3C
   C. Because B = C + G4A
   D. Because B = G1A
   Answer: A. From C = G3B + G4A, B = (C - G4A)/G3, which introduces the G4 term. (Key point: Moving H2 take-off)

3. What is C/A for the inner part (G1, G2, G3, G4, H3)?
   A. (G1G2G3 + G4)/(1 + G2G3H3)
   B. (G1G2G3 + G4)/(1 - G2G3H3)
   C. G1G2G3/(1 + G2G3H3)
   D. G4/(1 + G2G3H3)
   Answer: A. C = G3G2(G1A - H3C) + G4A rearranges to this. (Key point: Inner forward part)

4. Which denominator is the final answer's?
   A. 1 + G1G2H2 + G2G3H3 + G1G2G3H1 + G4H1 - G2G4H2H3
   B. 1 + G1G2H2 + G2G3H3 + G1G2G3H1 + G4H1 + G2G4H2H3
   C. 1 + G1G2H1 + G2G3H3
   D. 1 - G1G2H2 - G2G3H3
   Answer: A. Loop +G2G4H2H3 (positive) enters as a minus term in Δ; the other four are negative loops. (Key point: Final denominator)

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