---
title: "Transfer function with G3 feedforward, H1 and unity loop"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1g2g3-h1-unity-2070o"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1g2g3-h1-unity-2070o
site: BCT Engineering
---

# Transfer function with G3 feedforward, H1 and unity loop

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2070 Chaitra (old course), 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-g1g2g3-h1-unity-2070o
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Determine the transfer function C(s)/R(s) for the following system. [Figure: R(s) → summing point S1 (+, −) → G1 → summing point S2 (+, −) → node A → G2 → node B → summing point S3 (+, +) → C(s). G3 from node A to S3 (+). H1 from node B fed back to S1 (−). Unity feedback from C(s) to S2 (−).]

## Short answer

Combine parallel G2 and G3 to obtain C/A = G2+G3. The inner unity feedback reduces this to (G2+G3)/(1+G2+G3). Closing the outer loop with H1 feedback yields C(s)/R(s) = G1(G2+G3)/(1+G2+G3+G1G2H1).

## Answer

Name the signals: $E$ = output of $S_1$, $A$ = output of $S_2$, $B$ = output of $G_2$.

#### Step 1: Combine the parallel paths from $A$ to $C$
$G_2$ (through $B$) and $G_3$ both go from $A$ to $S_3$:
$$
\frac{C}{A} = G_2 + G_3
$$

#### Step 2: Reduce the inner unity-feedback loop
$C$ is fed back to $S_2$ with unity gain:
$$
\frac{C}{G_1E} = \frac{G_2 + G_3}{1 + G_2 + G_3}
$$

#### Step 3: Move the take-off point of $H_1$ from $B$ to $C$
$B = G_2A$ and $A = \dfrac{C}{G_2 + G_3}$, so
$$
H_1B = \frac{G_2H_1}{G_2 + G_3}\,C
$$

```text
R ->(S1)-> G1 -> [(G2+G3)/(1+G2+G3)] --+--> C
     ^-                                |
     +----- G2H1/(G2+G3) <-------------+
```

#### Step 4: Close the outer loop
$$
\begin{aligned}
G_f &= \frac{G_1(G_2 + G_3)}{1 + G_2 + G_3}, \qquad H = \frac{G_2H_1}{G_2 + G_3} \\
\frac{C}{R} &= \frac{G_f}{1 + G_fH} = \frac{G_1(G_2 + G_3)}{1 + G_2 + G_3 + G_1G_2H_1}
\end{aligned}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_2 + G_3 + G_1G_2H_1}
$$

**Check (Mason):** paths $G_1G_2$, $G_1G_3$; loops $-G_2$, $-G_3$, $-G_1G_2H_1$, all touching each other and both paths. Same result.

## Check yourself

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

1. G2 and G3 both go from node A to summing point S3. Their combined gain is
   A. G2G3
   B. G2 + G3
   C. G2/G3
   D. 1/(G2+G3)
   Answer: B. Parallel blocks with the same input and summed outputs add. (Key point: Parallel paths)

2. The unity feedback from C to S2 around G1 output gives the loop
   A. (G2+G3)/(1+G2+G3)
   B. (G2+G3)/(1-G2-G3)
   C. (G2+G3)
   D. 1/(1+G2+G3)
   Answer: A. Negative unity feedback around (G2+G3) gives (G2+G3)/(1+G2+G3). (Key point: Inner loop)

3. Moving the H1 take-off from B to C changes the feedback branch to
   A. H1/(G2+G3)
   B. G2H1
   C. G2H1/(G2+G3)
   D. G3H1/(G2+G3)
   Answer: C. B = G2*A and A = C/(G2+G3), so H1*B = G2H1/(G2+G3) * C. (Key point: Moving take-off)

4. The overall C/R is
   A. G1(G2+G3)/(1+G2+G3)
   B. G1(G2+G3)/(1+G2+G3+G1G2H1)
   C. G1(G2+G3)/(1+G1G2H1)
   D. G1G2/(1+G2+G3+G1G2H1)
   Answer: B. Closing the outer loop with Gf and H gives this result, confirmed by Mason's formula. (Key point: Overall TF)

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