---
title: "Block diagram reduction with G3 subtracted at output"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1g2g3-h1h2-2078"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1g2g3-h1h2-2078
site: BCT Engineering
---

# Block diagram reduction with G3 subtracted at output

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2078 Bhadra, 8 marks
- Video: [Block Diagram Reduction (Solved Example 1)](https://www.youtube.com/watch?v=Aabq8t9EE1M) — Neso Academy
- 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-h1h2-2078
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Determine the transfer function of the given system by block reduction technique. [Figure: R(s) → summing point S1 → G1 → node A → summing point S2 (+, −) → G2 → node B → summing point S3 (+) → C(s). G3 from node A to S3 (−). H2 from C(s) fed back to S2 (−). H1 from node B fed back to S1, and C(s) also fed back directly (unity) to S1 (signs at S1 not clearly printed; take as negative feedback)]

## Short answer

By block diagram reduction, the transfer function is C(s)/R(s) = (G1G2 - G1G3) / (1 + G1G2 - G1G3 + G1G2H1 + G2H2 + G1G2G3H1H2).

## Answer

Assumption: both feedback signals at $S_1$ ($H_1B$ and $C$) are negative, as stated.

Name the signals: $E$ = output of $S_1$, $A$ = output of $G_1$, $B$ = output of $G_2$.
$$
\begin{aligned}
E &= R - H_1B - C, \quad A = G_1E \\
B &= G_2(A - H_2C), \quad C = B - G_3A
\end{aligned}
$$

#### Step 1: Move the take-off point of $H_1$ from $B$ past summer $S_3$
From $C = B - G_3A$: $B = C + G_3A$. So
$$
H_1B = H_1C + G_3H_1A = H_1C + G_1G_3H_1E
$$
The $H_1$ path becomes $H_1$ from $C$ plus a negative self loop $G_1G_3H_1$ around $G_1$.

#### Step 2: Combine feedbacks and remove the self loop
Feedback from $C$ to $S_1$: $1 + H_1$ (parallel blocks).
$G_1$ with negative feedback $G_3H_1$: $\dfrac{G_1}{1 + G_1G_3H_1}$.

#### Step 3: Reduce from $A$ to $C$
$$
C = G_2(A - H_2C) - G_3A
\;\Rightarrow\;
\frac{C}{A} = \frac{G_2 - G_3}{1 + G_2H_2}
$$

```text
R ->(+)-> [G1/(1+G1G3H1)] -> [(G2-G3)/(1+G2H2)] --> C
     ^-                                         |
     +--------------- (1 + H1) <----------------+
```

#### Step 4: Close the main loop
$$
\frac{C}{R} = \frac{G_1(G_2 - G_3)}{(1 + G_1G_3H_1)(1 + G_2H_2) + G_1(G_2 - G_3)(1 + H_1)}
$$
Expanding the denominator (the $G_1G_3H_1$ terms cancel):
$$
1 + G_2H_2 + G_1G_3H_1 + G_1G_2G_3H_1H_2 + G_1G_2 - G_1G_3 + G_1G_2H_1 - G_1G_3H_1
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2 - G_1G_3}{1 + G_1G_2 - G_1G_3 + G_1G_2H_1 + G_2H_2 + G_1G_2G_3H_1H_2}
$$

**Check (Mason):** paths $G_1G_2$ and $-G_1G_3$; loops $-G_1G_2H_1$, $-G_2H_2$, $-G_1G_2$, $+G_1G_3$, $-G_1G_2G_3H_1H_2$; all touch, $\Delta_1 = \Delta_2 = 1$. Same result.

## Check yourself

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

1. Why is the H1 take-off point moved from B past summer S3?
   A. Because H1 cannot be connected to B
   B. Because B = C + G3A, so H1B can be written as H1C plus G1G3H1E
   C. To make G3 disappear
   D. To convert feedback into feed forward
   Answer: B. From C = B - G3A we get B = C + G3A, which separates the H1 feedback into a C part and a self loop. (Key point: Take-off shift)

2. What is the feedback from C to S1 after combining the two paths?
   A. 1 + H1 (parallel blocks)
   B. H1 only
   C. 1 - H1
   D. 1/(1 + H1)
   Answer: A. The unity feedback and the H1 feedback from C are parallel and add, giving 1 + H1. (Key point: Combined feedback)

3. What is C/A for the part after node A?
   A. (G2 + G3)/(1 + G2H2)
   B. G2/(1 + G2H2) - G3 H2
   C. (G2 - G3)/(1 + G2H2)
   D. (G2 - G3)/(1 - G2H2)
   Answer: C. C = G2(A - H2C) - G3A, so C(1 + G2H2) = (G2 - G3)A. (Key point: A to C)

4. Which is the final transfer function?
   A. (G1G2 + G1G3)/(1 + G1G2 + G1G3 + G2H2)
   B. G1(G2 - G3)/(1 + G1G2H1)
   C. (G1G2 - G1G3)/(1 - G1G2 + G1G3 - G1G2H1)
   D. (G1G2 - G1G3)/(1 + G1G2 - G1G3 + G1G2H1 + G2H2 + G1G2G3H1H2)
   Answer: D. Closing the main loop and expanding gives this result; the G1G3H1 terms cancel and Mason's formula agrees. (Key point: Final result)

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