Mason's gain formula states T equals

Sum of loop gains over forward path gain

Why is Delta equal to 1 - (L1+L2+L3) with no product terms?

All loops share the branch G2, so none are non-touching

2076 Chaitra · 8 marks

Signal flow graph with input feedforward G4

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.

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 (−).]

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 $G_3$ ), $C$ .

  +------------------ 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 $\frac{C}{R} = \frac{G_1G_2G_3}{1-G_1G_2H_1+G_2H_1+G_2G_3H_2}+G_4$ .

✓ Answer checked by Claude Opus 5.5 (high effort)

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