---
title: "Block diagram reduction with G3 from before G1"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g3-before-g1-2078k"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g3-before-g1-2078k
site: BCT Engineering
---

# Block diagram reduction with G3 from before G1

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2078 Kartik, 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-g3-before-g1-2078k
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Find overall transfer function of the following diagram using block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → node A → G1 → summing point S2 (+, −) → G2 → node B → summing point S3 (+, +) → C(s). G3 from node A (before G1) to S3 (+). H2 from C(s) fed back to S2 (−). H1 from node B fed back to S1 (−).]

## Short answer

Shift the H2 take-off point before summing point S3, then combine parallel paths G1 and G3H2 into G1−G3H2. Reduce the inner loop with G2 and H2, then close the outer loop through H1. The overall transfer function is (G1G2+G3) / (1+G2H2+G1G2H1−G2G3H1H2).

## Answer

**Result:** $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2+G_3}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2}$

The feedback $H_2$ is taken from C, after summing point $S_3$ where $G_3$ joins. We shift this take-off point back over $S_3$, then reduce.

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

#### Step 1: Move the $H_2$ take-off point before $S_3$

$C = B + G_3A$, so $H_2C = H_2B + G_3H_2A$. The single $H_2$ branch is replaced by:

- $H_2$ from node B to $S_2$ (negative)
- $G_3H_2$ from node A to $S_2$ (negative)

#### Step 2: Combine blocks from A into $S_2$

$G_1$ (+) and $G_3H_2$ (−) both start at A and end at $S_2$, so they are in parallel: $G_1 - G_3H_2$.

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

$$
\frac{B}{E_2} = \frac{G_2}{1+G_2H_2}
\qquad
\frac{B}{A} = \frac{G_2(G_1-G_3H_2)}{1+G_2H_2}
$$

#### Step 4: Loop through $H_1$ back to $S_1$

$$
\frac{A}{R} = \frac{1}{1+\frac{G_2H_1(G_1-G_3H_2)}{1+G_2H_2}} = \frac{1+G_2H_2}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2}
$$

#### Step 5: Output

$$
\begin{aligned}
\frac{C}{A} &= \frac{B}{A} + G_3 = \frac{G_1G_2 - G_2G_3H_2 + G_3 + G_2G_3H_2}{1+G_2H_2} = \frac{G_1G_2+G_3}{1+G_2H_2} \\
\frac{C}{R} &= \frac{C}{A}\cdot\frac{A}{R} = \frac{G_1G_2+G_3}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2}
\end{aligned}
$$

**Check (Mason's rule):** paths $G_1G_2$, $G_3$; loops $-G_1G_2H_1$, $-G_2H_2$, and $+G_2G_3H_1H_2$ (A → $G_3$ → C → $-H_2$ → $G_2$ → B → $-H_1$ → A). All loops touch both paths, so $\Delta = 1+G_1G_2H_1+G_2H_2-G_2G_3H_1H_2$, matching the result.

## Check yourself

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

1. Why is the H2 take-off shifted back before S3?
   A. C = B + G3*A, so H2*C = H2*B + G3H2*A
   B. To make H2 positive
   C. Because G3 and G1 are in series
   D. To remove the H1 loop
   Answer: A. Since C = B + G3A, the single feedback splits into H2 from B and G3H2 from A, both into S2. (Key point: Shifting take-off over S3)

2. Between A and S2, G1 (+) and G3H2 (-) combine to
   A. G1 + G3H2
   B. G1 - G3H2
   C. G1*G3H2
   D. G1/(G3H2)
   Answer: B. They are in parallel with opposite signs, giving G1 - G3H2. (Key point: Parallel blocks from A)

3. The inner negative loop around G2 with H2 has transfer function
   A. G2/(1-G2H2)
   B. G2/(1+G2H2)
   C. G2H2/(1+G2)
   D. 1/(1+G2H2)
   Answer: B. Negative feedback: forward gain over 1 plus loop gain, G2/(1+G2H2). (Key point: Inner loop)

4. The overall transfer function C/R is
   A. (G1G2+G3)/(1+G2H2+G1G2H1+G2G3H1H2)
   B. G1G2/(1+G2H2+G1G2H1)
   C. (G1G2+G3)/(1+G1G2H1)
   D. (G1G2+G3)/(1+G2H2+G1G2H1-G2G3H1H2)
   Answer: D. C/A = (G1G2+G3)/(1+G2H2) and A/R = (1+G2H2)/(1+G2H2+G1G2H1-G2G3H1H2); the product gives option A. (Key point: Final result)

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