---
title: "Block diagram reduction: G3 feedforward, H1 and H2 loops"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1g2g3-h1h2-b"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1g2g3-h1h2-b
site: BCT Engineering
---

# Block diagram reduction: G3 feedforward, H1 and H2 loops

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2081 Baisakh, 8 marks; 2075 Chaitra, 8 marks; 2068 Chaitra, 8 marks; 2080 Baisakh, 8 marks (asked 4 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-b
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

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

## Short answer

C(s)/R(s) = (G1G2 + G1G3) / (1 + G1G2H1 + G2H2 − G1G2G3H1H2).

## Answer

Name the signals: $E$ = output of $S_1$, $A$ = output of $G_1$, $X$ = output of $S_2$, $B = G_2X$, and $C = B + G_3A$. So $E = R - H_1B$ and $X = A - H_2C$.

#### Step 1: Move the $H_1$ take-off point behind $G_2$
Since $B = G_2X$, the feedback $H_1B$ can be taken from $X$ through $G_2H_1$.

#### Step 2: Move the $G_3$ take-off point from $A$ to after $S_2$
Because $A = X + H_2C$, the branch $G_3A$ equals $G_3X + G_3H_2C$. So $G_3$ now starts at $X$, and an extra path $G_3H_2$ goes from $C$ into $S_3$ (positive).

#### Step 3: Combine parallel blocks and the positive loop at $S_3$
$G_2$ and $G_3$ are in parallel from $X$: $(G_2 + G_3)$. The extra path forms a positive feedback loop $G_3H_2$ around $S_3$:
$$
\frac{C}{X} = \frac{G_2 + G_3}{1 - G_3H_2}
$$

#### Step 4: Move $S_1$ forward past $G_1$
$R$ becomes $G_1R$ and the $H_1$ feedback becomes $G_1G_2H_1$ from $X$, so $X = G_1R - G_1G_2H_1X - H_2C$. The loop on $X$ gives
$$
\frac{X}{U} = \frac{1}{1 + G_1G_2H_1}, \qquad U = G_1R - H_2C
$$
After Steps 3 and 4 the diagram is:
```text
 R ─►[G1]─►(Σ)──► X ──►[ (G2+G3)/(1−G3H2) ]──┬──► C
            ▲ − ▲ −                          │
            │   └──────────[H2]◄─────────────┤
            └─[G1G2H1]◄── X                  │
```

#### Step 5: Close the $H_2$ loop
Forward path from $U$: $P = \dfrac{G_2 + G_3}{(1 + G_1G_2H_1)(1 - G_3H_2)}$, feedback $H_2$:
$$
\frac{C}{G_1R} = \frac{P}{1 + PH_2} = \frac{G_2 + G_3}{(1 + G_1G_2H_1)(1 - G_3H_2) + H_2(G_2 + G_3)}
$$
Expanding the denominator: $1 + G_1G_2H_1 - G_3H_2 - G_1G_2G_3H_1H_2 + G_2H_2 + G_3H_2$; the $G_3H_2$ terms cancel.

**Answer:**
$$
\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_1G_2H_1 + G_2H_2 - G_1G_2G_3H_1H_2}
$$

#### Check by Mason's rule
Forward paths $G_1G_2$, $G_1G_3$. Loops $-G_1G_2H_1$, $-G_2H_2$, and $+G_1G_3H_2G_2H_1$ (path $A \to G_3 \to C \to H_2 \to G_2 \to B \to H_1 \to G_1 \to A$). All loops touch each other and both paths, so $\Delta_1 = \Delta_2 = 1$, giving the same result.

## Check yourself

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

1. Why is the G3 take-off from node A moved to after S2 as G3X + G3H2C?
   A. Because A = X + H2C
   B. Because A = X - H2C
   C. Because A = G2X
   D. Because A = C
   Answer: A. X = A - H2C, so A = X + H2C; hence G3A = G3X + G3H2C. (Key point: Moving a take-off point)

2. What is C/X after combining G2, G3 and the extra path G3H2 into S3?
   A. (G2 + G3)/(1 - G3H2)
   B. (G2 + G3)/(1 + G3H2)
   C. G2G3/(1 - G3H2)
   D. (G2 + G3)(1 - G3H2)
   Answer: A. G2 and G3 are parallel, and the extra path G3H2 enters S3 with a plus sign, a positive feedback giving 1 - G3H2. (Key point: Positive loop)

3. What is the final C/R?
   A. (G1G2 + G1G3)/(1 + G1G2H1 + G2H2 - G1G2G3H1H2)
   B. (G1G2 + G1G3)/(1 + G1G2H1 + G2H2 + G1G2G3H1H2)
   C. G1G2/(1 + G1G2H1)
   D. (G2 + G3)/(1 + G2H2)
   Answer: A. The G3H2 terms cancel in the expansion, leaving the given denominator. (Key point: Final result)

4. Why is Δ1 = Δ2 = 1 in the Mason's rule check?
   A. All loops touch both forward paths
   B. There are no loops
   C. Loops are all non-touching
   D. The paths have unit gain
   Answer: A. Every loop touches both paths, so no loop is left after removing a path and the cofactor is 1. (Key point: Mason check)

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