---
title: "Block diagram reduction with feedforward and H1, H2"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1g2g3-h1h2-ff"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1g2g3-h1h2-ff
site: BCT Engineering
---

# Block diagram reduction with feedforward and H1, H2

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2081 Bhadra, 8 marks; 2068 Baisakh (old course), 8 marks (asked 2 times)
- 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-h1h2-ff
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

From the given block diagram, find the transfer ratio C(s)/R(s) using block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → node A → G1 → summing point S2 (+) → G2 → summing point S3 (+, −) → node B → G3 → C(s). Feedforward from node A to S2 (+). H1 from node B fed back to S2 (−). Unity feedback from C(s) to S3 (−). H2 from C(s) fed back to S1 (−).]

## Short answer

C(s)/R(s) = G2 G3 (1+G1) / (1 + G3 + G2 H1 + G1 G2 G3 H2 + G2 G3 H2).

## Answer

Name the signals: $A = R - H_2C$ (output of $S_1$), $X$ = output of $S_2$, $B$ = output of $S_3$, $C = G_3B$.
Then $X = G_1A + A - H_1B$ and $B = G_2X - C$.

#### Step 1: Combine $G_1$ and the feedforward path
$G_1$ and the unity feedforward from $A$ are in parallel into $S_2$:
$$
G_a = 1 + G_1
$$

#### Step 2: Move the $H_1$ take-off point from $B$ to $C$
Since $B = C/G_3$, the feedback block becomes $\dfrac{H_1}{G_3}$, taken from $C$.

#### Step 3: Reduce the unity loop around $G_3$
$G_3$ with unity negative feedback to $S_3$:
$$
\frac{C}{G_2X} = \frac{G_3}{1 + G_3}
$$
So from $X$ to $C$: $G_b = \dfrac{G_2G_3}{1 + G_3}$.

#### Step 4: Reduce the $H_1/G_3$ loop
$$
\frac{C}{X'} = \frac{G_b}{1 + G_b\frac{H_1}{G_3}} = \frac{\frac{G_2G_3}{1+G_3}}{1 + \frac{G_2H_1}{1+G_3}} = \frac{G_2G_3}{1 + G_3 + G_2H_1}
$$
(here $X'$ is the signal $(1 + G_1)A$ entering $S_2$)

#### Step 5: Series with $(1 + G_1)$ and close the $H_2$ loop
Forward path: $G_f = \dfrac{(1 + G_1)G_2G_3}{1 + G_3 + G_2H_1}$, feedback $H_2$:
$$
\frac{C}{R} = \frac{G_f}{1 + G_fH_2}
$$
```text
 R +    ┌──────────────────────────┐
 ──►(Σ)─►│ (1+G1)G2G3/(1+G3+G2H1) ├──┬──► C
    ▲ -  └──────────────────────────┘  │
    │            ┌────┐                │
    └────────────┤ H2 │◄───────────────┘
                 └────┘
```

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_2G_3(1 + G_1)}{1 + G_3 + G_2H_1 + G_1G_2G_3H_2 + G_2G_3H_2}
$$

#### Check by Mason's rule
Forward paths: $P_1 = G_1G_2G_3$, $P_2 = G_2G_3$ (via feedforward). Loops: $-G_3$, $-G_2H_1$, $-G_1G_2G_3H_2$, $-G_2G_3H_2$. All loops touch each other and both paths ($\Delta_1 = \Delta_2 = 1$), giving the same answer.

## Check yourself

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

1. What do G1 and the unity feedforward from A into S2 combine to?
   A. 1 + G1
   B. G1
   C. 1 - G1
   D. G1/(1 + G1)
   Answer: A. They are in parallel into the same summer, so their gains add. (Key point: Feedforward combination)

2. Since B = C/G3, what happens when the H1 take-off is moved from B to C?
   A. The feedback block becomes H1/G3
   B. The block becomes H1 G3
   C. The block stays H1
   D. It becomes G3/H1
   Answer: A. Moving a take-off point forward past G3 needs the block to be divided by G3. (Key point: Take-off shifting)

3. What is C/X' after reducing the unity loop around G3 and the H1/G3 loop?
   A. G2G3/(1 + G3 + G2H1)
   B. G2G3/(1 + G3)
   C. G2G3/(1 - G3 + G2H1)
   D. G3/(1 + G2H1)
   Answer: A. G2G3/(1+G3) closed with H1/G3 gives G2G3/(1+G3+G2H1). (Key point: Inner loop)

4. What is the final C/R?
   A. G2G3(1+G1)/(1 + G3 + G2H1 + G1G2G3H2 + G2G3H2)
   B. G2G3(1+G1)/(1 + G3 + G2H1)
   C. G2G3(1+G1)/(1 + G3 + G2H1 - G2G3H2)
   D. G1G2G3/(1 + G3 + G2H1)
   Answer: A. Closing the H2 loop adds Gf H2 to the denominator, giving (1+G1)G2G3H2 terms. (Key point: Final result)

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