---
title: "Signal flow graph with input feedforward G4"
subject: "Control Systems"
chapter: 3
key: "semester_3/control_systems:sfg-2076c-bei"
source: https://www.bctengineer.com/q/semester_3/control_systems/sfg-2076c-bei
site: BCT Engineering
---

# Signal flow graph with input feedforward G4

Control Systems · Chapter 3: System Transfer Function and Responses

- Asked in: 2076 Chaitra, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/control_systems/sfg-2076c-bei
- Whole chapter: https://www.bctengineer.com/md/semester_3/control_systems/chapter-3

## Question

Determine the transfer function C/R for the block diagram below by signal flow graph (SFG) technique. [Figure: R → node P → summing point S1 (+, ±) → G1 → summing point S2 (+, −, −) → G2 → node B → G3 → node C → summing point S3 (+, +) → C. G4 from node P (input R) to S3 (+). H1 from node B fed back to both S1 (sign unclear in scan) and S2 (−). H2 from node C fed back to S2 (−).]

## Short answer

Convert the block diagram to a signal flow graph. Forward paths are G1G2G3 and G4. Three loops: −G1G2H1, −G2H1, −G2G3H2, all touching P1. With H1 to first summer taken negative, Mason’s formula yields C/R = G1G2G3/(1+G1G2H1+G2H1+G2G3H2) + G4.

## Answer

**Result** (taking the sign of $H_1$ at $S_1$ as negative): $\dfrac{C}{R} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4$

The block diagram is converted to a signal flow graph (SFG) and Mason's gain formula is applied:

$$
T = \frac{1}{\Delta}\sum_k P_k\Delta_k
$$

**Assumption:** the sign of $H_1$ at $S_1$ is unclear in the scan. It is taken as **negative**, as in the usual version of this question.

#### Signal flow graph

Nodes: $R$, $x_1$ (output of $S_1$), $x_2$ (output of $S_2$), $x_3$ (= B, output of $G_2$), $x_4$ (= node C, output of $G_3$), $C$.

```text
  +------------------ G4 -------------------+
  |                                         v
  R -1-> x1 -G1-> x2 -G2-> x3 -G3-> x4 -1-> C
         ^        ^^       ||       |
         |        |+(-H1)--+|       |
         +--------|--(-H1)--+       |
                  +------(-H2)------+
```

Branches: $x_3 \to x_1$ gain $-H_1$, $x_3 \to x_2$ gain $-H_1$, $x_4 \to x_2$ gain $-H_2$, $R \to C$ gain $G_4$.

#### Forward paths

| Path | Route | Gain |
|---|---|---|
| $P_1$ | $R\to x_1\to x_2\to x_3\to x_4\to C$ | $G_1G_2G_3$ |
| $P_2$ | $R\to C$ | $G_4$ |

#### Individual loops

| Loop | Route | Gain |
|---|---|---|
| $L_1$ | $x_1\to x_2\to x_3\to x_1$ | $-G_1G_2H_1$ |
| $L_2$ | $x_2\to x_3\to x_2$ | $-G_2H_1$ |
| $L_3$ | $x_2\to x_3\to x_4\to x_2$ | $-G_2G_3H_2$ |

All three loops share the branch $G_2$, so there are **no non-touching loop pairs**.

#### Determinants

$$
\begin{aligned}
\Delta &= 1-(L_1+L_2+L_3) = 1+G_1G_2H_1+G_2H_1+G_2G_3H_2 \\
\Delta_1 &= 1 \quad (P_1 \text{ touches all loops}) \\
\Delta_2 &= \Delta \quad (P_2 \text{ touches no loop})
\end{aligned}
$$

#### Transfer function

$$
\begin{aligned}
\frac{C}{R} &= \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta} \\
&= \frac{G_1G_2G_3 + G_4(1+G_1G_2H_1+G_2H_1+G_2G_3H_2)}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} \\
&= \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4
\end{aligned}
$$

If $H_1$ enters $S_1$ with a **positive** sign instead, only $L_1$ changes to $+G_1G_2H_1$, giving $\frac{C}{R} = \frac{G_1G_2G_3}{1-G_1G_2H_1+G_2H_1+G_2G_3H_2}+G_4$.

## Check yourself

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

1. Mason's gain formula states T equals
   A. Sum of loop gains over forward path gain
   B. (1/Delta) * sum of Pk*Delta_k
   C. Delta / sum of Pk
   D. Product of all path gains
   Answer: B. T = (1/Delta) * sum(Pk*Delta_k), where Delta is the graph determinant. (Key point: Mason's gain formula)

2. Which are the forward paths from R to C?
   A. G1G2G3 only
   B. G1G2G3 and G4
   C. G1G2H1 and G4
   D. G1G2G3, G4 and G2G3H2
   Answer: B. One path goes through G1, G2, G3 and the other is the direct G4 path from R to C. (Key point: Forward paths)

3. Why is Delta equal to 1 - (L1+L2+L3) with no product terms?
   A. There is only one loop
   B. The loops are all positive
   C. All loops share the branch G2, so none are non-touching
   D. Loops with H1 never touch
   Answer: C. Every loop passes through G2, so there are no pairs of non-touching loops. (Key point: Touching loops)

4. What is Delta_2 for the path G4?
   A. 1
   B. Delta
   C. 0
   D. 1 + G4
   Answer: B. G4 touches no loop, so all loops remain and Delta_2 = Delta, which gives the additive term G4. (Key point: Path G4 cofactor)

## Related in Control Systems

- [Signal flow graph of G1–G3, H1–H3 block diagram](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asfg-g1g2g3-h1h2h3) — Control Systems, chapter 3
- [Signal flow graph and Mason's formula: G1–G4, H1–H3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asfg-mason-g1-g4-h1-h3) — Control Systems, chapter 3, asked 2×
- [Block diagram reduction with G1, G2, H1, H2, feedforward](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1g2-h1h2-ff-unity) — Control Systems, chapter 3, asked 2×
- [Signal flow graph and Mason's formula: G3 feedforward](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Asfg-g1g2g3-h1h2-2076a) — Control Systems, chapter 3, asked 2×
- [Block diagram reduction: G1–G4, H1, H2, unity loop](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2-2075a) — Control Systems, chapter 3, asked 4×
- [Block diagram reduction: G1–G4 with H1, H2, H3](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abdr-g1-g4-h1h2h3-2073) — Control Systems, chapter 3, asked 3×