---
title: "Block diagram reduction with G1, G2, H1, H2, feedforward"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1g2-h1h2-ff-unity"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1g2-h1h2-ff-unity
site: BCT Engineering
---

# Block diagram reduction with G1, G2, H1, H2, feedforward

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2080 Baisakh, 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-g1g2-h1h2-ff-unity
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Obtain the overall transfer function of given system by block diagram reduction techniques. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → node A → summing point S3 (+, −) → G2 → node B → summing point S4 (+, +) → C(s). Feedforward from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (−). Unity feedback from C(s) to S1 (−).]

## Short answer

The overall transfer function is C(s)/R(s) = (G1 + G1 G2) / (1 + G1 + G1 G2 + G2 H2 + G1 G2 H1 - G1 G2 H1 H2).

## Answer

Name the signals: $E$ = output of $S_2$ (input of $G_1$), $A$ = output of $G_1$, $B$ = output of $G_2$.

From the figure:
$$
\begin{aligned}
E &= R - C - H_1B, \quad A = G_1E \\
B &= G_2(A - H_2C), \quad C = B + A
\end{aligned}
$$
($S_1$ and $S_2$ are in cascade, so they act as one summer.)

#### Step 1: Move the take-off point of $H_1$ from $B$ past summer $S_4$
Since $C = B + A$, we have $B = C - A$. So
$$
H_1B = H_1C - H_1A = H_1C - G_1H_1E
$$
The $H_1$ path becomes $H_1$ from $C$ (negative) plus a positive self loop $G_1H_1$ around $G_1$.

#### Step 2: Combine the feedback from $C$ and remove the self loop
Feedback from $C$: $1 + H_1$ (parallel).
Loop $G_1$ with positive feedback $H_1$: $\dfrac{G_1}{1 - G_1H_1}$.

#### Step 3: Reduce from $A$ to $C$
$$
C = G_2(A - H_2C) + A
\;\Rightarrow\;
\frac{C}{A} = \frac{1 + G_2}{1 + G_2H_2}
$$
(Inner loop $G_2$, $H_2$ plus the parallel unity feedforward path.)

```text
R ->(+)-> [G1/(1-G1H1)] -> [(1+G2)/(1+G2H2)] --> C
     ^-                                      |
     +------------- (1 + H1) <---------------+
```

#### Step 4: Close the main loop
$$
G_f = \frac{G_1(1+G_2)}{(1 - G_1H_1)(1 + G_2H_2)}, \qquad H = 1 + H_1
$$
$$
\begin{aligned}
\frac{C}{R} &= \frac{G_f}{1+G_fH} = \frac{G_1(1+G_2)}{(1-G_1H_1)(1+G_2H_2) + G_1(1+G_2)(1+H_1)} \\
\text{Denominator} &= 1 + G_2H_2 - G_1H_1 - G_1G_2H_1H_2 + G_1 + G_1G_2 + G_1H_1 + G_1G_2H_1
\end{aligned}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1 + G_1G_2}{1 + G_1 + G_1G_2 + G_2H_2 + G_1G_2H_1 - G_1G_2H_1H_2}
$$

**Check (Mason):** paths $G_1G_2$, $G_1$; loops $-G_1G_2H_1$, $-G_2H_2$, $-G_1G_2$, $-G_1$, and $+G_1G_2H_1H_2$ (through the feedforward, $H_2$, $G_2$, $H_1$). All loops 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. Using C = B + A, how is H1B rewritten?
   A. H1C - H1A = H1C - G1H1E
   B. H1C + H1A
   C. H1A - H1C
   D. H1 G2 C
   Answer: A. B = C - A, so H1B = H1C - H1A, with A = G1E. (Key point: Moving H1 take-off)

2. What is the combined feedback from C to the summer?
   A. 1 + H1
   B. H1
   C. 1 - H1
   D. H1/(1+H1)
   Answer: A. Unity feedback and H1 from C are in parallel. (Key point: Parallel feedback)

3. What is C/A for G2 with H2 and the unity feedforward?
   A. (1 + G2)/(1 + G2H2)
   B. (1 + G2)/(1 - G2H2)
   C. G2/(1 + G2H2)
   D. 1 + G2/(1 + H2)
   Answer: A. C = G2(A - H2C) + A gives (1+G2)A = (1 + G2H2)C. (Key point: Inner loop)

4. What is the final C/R?
   A. (G1 + G1G2)/(1 + G1 + G1G2 + G2H2 + G1G2H1 - G1G2H1H2)
   B. (G1 + G1G2)/(1 + G1 + G1G2 + G2H2)
   C. (G1 + G1G2)/(1 + G1G2H1 + G2H2)
   D. G1G2/(1 + G1G2H1)
   Answer: A. Expanding the loop closure, the G1H1 terms cancel, leaving the stated denominator. (Key point: Final result)

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