---
title: "Y/R by block reduction with G4 from input"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:bdr-g1-g4-h1h2-2065"
source: https://www.bctengineer.com/q/semester_3/control_systems/bdr-g1-g4-h1h2-2065
site: BCT Engineering
---

# Y/R by block reduction with G4 from input

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2065 Shrawan (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-g1-g4-h1h2-2065
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Derive the overall transfer function Y(s)/R(s) for the system shown below using block reduction technique. [Figure: R(s) → node P → summing point S1 (+, −) → G1 → summing point S2 (+, −, −) → G2 → node B → G3 → node C → summing point S3 (+, +) → Y(s). G4 from node P (input R) to S3. H1 from node B fed back to both S1 (−) and S2 (−). H2 from node C fed back to S2 (−).]

## Short answer

Y(s)/R(s) = G1G2G3/(1+G1G2H1+G2H1+G2G3H2) + G4.

## Answer

**Result:** $\dfrac{Y(s)}{R(s)} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4$

The diagram is reduced step by step using block diagram algebra rules (series, parallel, feedback, moving take-off points).

#### Step 1: Separate the parallel feed-forward path

$G_4$ takes the input $R$ directly to the output summing point $S_3$ (both inputs $+$). So the system is two parallel paths:

$$
\frac{Y(s)}{R(s)} = \frac{C(s)}{R(s)} + G_4
$$

where $C(s)$ is the output of $G_3$ (node C). We now reduce the main path from $R$ to $C$.

#### Step 2: Move the take-off point of $H_2$ from C to B

Node C is after $G_3$. Moving a take-off point **before** a block multiplies the branch by that block. So the $H_2$ feedback becomes $G_3H_2$ taken from node B.

Now two negative feedbacks go from B to $S_2$: $H_1$ and $G_3H_2$. They are in parallel:

$$
H_{eq} = H_1 + G_3H_2
$$

#### Step 3: Reduce the inner loop around $G_2$

$$
\frac{B}{E_2} = \frac{G_2}{1+G_2(H_1+G_3H_2)} = \frac{G_2}{1+G_2H_1+G_2G_3H_2}
$$

#### Step 4: Series with $G_1$, then the outer loop through $H_1$ to $S_1$

Forward path from $S_1$ to B: $\frac{G_1G_2}{1+G_2H_1+G_2G_3H_2}$, with negative feedback $H_1$:

$$
\begin{aligned}
\frac{B}{R} &= \frac{\frac{G_1G_2}{1+G_2H_1+G_2G_3H_2}}{1+\frac{G_1G_2H_1}{1+G_2H_1+G_2G_3H_2}} \\
&= \frac{G_1G_2}{1+G_2H_1+G_2G_3H_2+G_1G_2H_1}
\end{aligned}
$$

#### Step 5: Series with $G_3$ and add $G_4$

$$
\frac{C}{R} = \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2}
$$

```text
        +------------ G4 -------------+
        |                             | +
 R ---->+--> [ G1G2G3 / Delta ] ----->(+)---> Y
                                        +
 Delta = 1 + G1G2H1 + G2H1 + G2G3H2
```

$$
\frac{Y(s)}{R(s)} = \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4
$$

or, over a common denominator,

$$
\frac{Y(s)}{R(s)} = \frac{G_1G_2G_3 + G_4\left(1+G_1G_2H_1+G_2H_1+G_2G_3H_2\right)}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2}
$$

**Check (Mason's rule):** loops $-G_1G_2H_1$, $-G_2H_1$, $-G_2G_3H_2$ all touch each other, so $\Delta = 1+G_1G_2H_1+G_2H_1+G_2G_3H_2$. Path $G_1G_2G_3$ touches all loops ($\Delta_1=1$); path $G_4$ touches none ($\Delta_2=\Delta$). This gives the same result.

## Check yourself

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

1. G4 runs from the input R directly to S3, so
   A. Y/R = C/R + G4
   B. Y/R = C/R * G4
   C. Y/R = C/R / G4
   D. G4 is ignored
   Answer: A. It is a parallel path, so its gain adds. (Key point: Parallel G4)

2. Moving the H2 take-off from C back to B gives
   A. H2/G3
   B. G3H2
   C. H2
   D. G3/H2
   Answer: B. Moving a take-off before a block multiplies by that block. (Key point: Moving H2)

3. The two feedbacks from B to S2 combine to
   A. H1*G3H2
   B. H1 - G3H2
   C. H1 + G3H2
   D. H1/G3H2
   Answer: C. Parallel feedback branches add. (Key point: Equivalent feedback)

4. The denominator of the C/R path is
   A. 1+G1G2H1+G2H1+G2G3H2
   B. 1+G1G2H1+G2G3H2
   C. 1+G2H1+G2G3H2
   D. 1-G1G2H1-G2H1-G2G3H2
   Answer: A. Three touching loops: -G1G2H1, -G2H1, -G2G3H2. (Key point: Result)

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