---
title: "Block diagram reduction with H4 input feedforward"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-h4-g1g2-h1h2h3-2072"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-h4-g1g2-h1h2h3-2072
site: BCT Engineering
---

# Block diagram reduction with H4 input feedforward

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2072 Chaitra, 8 marks
- Video: [Finding Transfer Function of a Block Diagram Example (Block Diagram Reduction Method)](https://www.youtube.com/watch?v=dXUAJtWNuCU) — DMD Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/bdr-h4-g1g2-h1h2h3-2072
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Find transfer function of the following system. [Figure: R(s) → node P → summing point S1 (+, −) → summing point S2 (+, +, −) → G1 → node A → G2 → node B → summing point S3 (+, +) → C(s). H4 from node P (input R) to S2 (+). H2 from node A to S3 (+). H1 from C(s) fed back to S2 (−). H3 from node B fed back to S1 (−).]

## Short answer

C(s)/R(s) = G1(G2+H2)(1+H4) / (1 + G1G2H1 + G1G2H3 + G1H1H2).

## Answer

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

#### Step 1: Combine the cascaded summers $S_1$ and $S_2$
$R$ enters $S_1$ directly and $S_2$ through $H_4$. Since the two summers are in series:
$$
E = R + H_4R - H_3B - H_1C = (1 + H_4)R - H_3B - H_1C
$$
So the input becomes a block $(1 + H_4)$ in front of one summer.

#### Step 2: Move the take-off point of $H_3$ from $B$ back to $A$
$B = G_2A$, so $H_3$ becomes $G_2H_3$ taken from $A$.

#### Step 3: Reduce the loop $G_1$, $G_2H_3$
$$
\frac{A}{E'} = \frac{G_1}{1 + G_1G_2H_3}
$$
where $E'$ is the summer output before this loop.

#### Step 4: Parallel paths from $A$ to $C$
$G_2$ and $H_2$ both go from $A$ to $S_3$:
$$
\frac{C}{A} = G_2 + H_2
$$

```text
R -> [1+H4] ->(+)-> [G1/(1+G1G2H3)] -> [G2+H2] --+-> C
               ^-                                |
               +------------- H1 ----------------+
```

#### Step 5: Close the loop with $H_1$
$$
G_f = \frac{G_1(G_2 + H_2)}{1 + G_1G_2H_3}
$$
$$
\frac{C}{R} = (1 + H_4)\frac{G_f}{1 + G_fH_1} = \frac{(1 + H_4)G_1(G_2 + H_2)}{1 + G_1G_2H_3 + G_1G_2H_1 + G_1H_1H_2}
$$

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1(G_2 + H_2)(1 + H_4)}{1 + G_1G_2H_1 + G_1G_2H_3 + G_1H_1H_2}
$$

**Check (Mason):** four forward paths $G_1G_2$, $G_1H_2$, $H_4G_1G_2$, $H_4G_1H_2$; loops $-G_1G_2H_3$, $-G_1G_2H_1$, $-G_1H_2H_1$, all touching each other and all paths. Same result.

## Check yourself

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

1. How are S1 and S2 with the H4 path combined?
   A. E = (1 - H4)R - H3B - H1C
   B. E = (1 + H4)R - H3B - H1C, so a block (1 + H4) sits in front of one summer
   C. E = H4R - H3B - H1C
   D. E = R/(1 + H4) - H3B - H1C
   Answer: B. R enters S1 directly and S2 through H4, both positive, so the R terms add. (Key point: Cascaded summers with H4)

2. How is the H3 take-off at B handled?
   A. B = G2A, so H3 becomes G2H3 from A, giving the loop G1/(1 + G1G2H3)
   B. It becomes H3/G2 from A
   C. It is removed
   D. It becomes G1H3 from E
   Answer: A. Moving a take-off backward past G2 multiplies the feedback by G2. (Key point: Moving H3)

3. What is the gain from A to C?
   A. G2H2
   B. G2 - H2
   C. G2 + H2, since both branches go from A to S3
   D. G2/(1 + H2)
   Answer: C. H2 and G2 are parallel branches feeding S3 with positive signs. (Key point: Parallel A to C)

4. What is the final transfer function?
   A. G1(G2 + H2)(1 + H4)/(1 + G1G2H1 + G1G2H3 + G1H1H2)
   B. G1(G2 + H2)/(1 + G1G2H1 + G1G2H3)
   C. G1(G2 + H2)(1 - H4)/(1 + G1G2H1 + G1H1H2)
   D. G1G2(1 + H4)/(1 + G1G2H3)
   Answer: A. Closing the H1 loop with Gf = G1(G2+H2)/(1+G1G2H3) and multiplying by (1+H4) gives this. (Key point: Final result)

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