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Signal flow graph of G1–G3, H1–H3 block diagram

Short answer

Using Mason's gain formula with two forward paths (G1G2G3, G1H1G3) and five loops: C(s)/R(s) = G1 G3 (G2 + H1) / (1 + G2 H2 + G2 G3 H3 + G3 H1 H3 + G1 G2 G3 + G1 G3 H1).

Obtain the overall transfer function of given system by signal flow graph technique. [Figure: R(s) → summing point S1 (+, −) → G1 → summing point S2 (+, −) → summing point S3 (+, −) → node A. From node A: G2 → node B, and H1 (feedforward branch) going to summing point S4 (+). Node B → S4 (+) → G3 → C(s). From node B, H2 feeds back to S2 (−). From C(s), H3 feeds back to S3 (−). From C(s), unity feedback to S1 (−).]

In the signal flow graph method, each signal becomes a node and each block a branch; then Mason's gain formula gives the transfer function.

Signal flow graph

Nodes: x1x_1 = output of S1S_1, x2x_2 = output of G1G_1, x3x_3 = output of S2S_2, x4x_4 = output of S3S_3 (node A), x5x_5 = node B, x6x_6 = output of S4S_4, CC.

                         H1
                   +--------------+
                   |              v
R -1-> x1 -G1-> x2 -1-> x3 -1-> x4 -G2-> x5 -1-> x6 -G3-> C
       ^                ^       ^         |                |
       |                +-(-H2)-+---------+                |
       |                        +-------(-H3)--------------+
       +--------------------(-1)---------------------------+

Branches: x4→x6x_4\to x_6 (H1H_1), x5→x3x_5\to x_3 (−H2-H_2), C→x4C\to x_4 (−H3-H_3), C→x1C\to x_1 (−1-1); all others as labelled.

Forward paths

  • P1=G1G2G3P_1 = G_1G_2G_3 (through G2G_2)
  • P2=G1H1G3P_2 = G_1H_1G_3 (through the H1H_1 branch)

Individual loops

  • L1=−G2H2L_1 = -G_2H_2 (x3,x4,x5,x3x_3, x_4, x_5, x_3)
  • L2=−G2G3H3L_2 = -G_2G_3H_3 (x4,x5,x6,C,x4x_4, x_5, x_6, C, x_4)
  • L3=−G3H1H3L_3 = -G_3H_1H_3 (x4,x6,C,x4x_4, x_6, C, x_4)
  • L4=−G1G2G3L_4 = -G_1G_2G_3 (x1…C,x1x_1 \dots C, x_1)
  • L5=−G1G3H1L_5 = -G_1G_3H_1 (x1,x2,x3,x4,x6,C,x1x_1, x_2, x_3, x_4, x_6, C, x_1)

Non-touching loops

L1L_1 touches L2L_2 and L3L_3 at x4x_4, and all others pass through x3x_3/x4x_4. So there are no non-touching pairs.

Δ=1+G2H2+G2G3H3+G3H1H3+G1G2G3+G1G3H1\Delta = 1 + G_2H_2 + G_2G_3H_3 + G_3H_1H_3 + G_1G_2G_3 + G_1G_3H_1

Cofactors

Both forward paths pass through x3x_3, x4x_4 and CC, so they touch every loop: Δ1=Δ2=1\Delta_1 = \Delta_2 = 1.

Mason's gain formula

T=P1Δ1+P2Δ2ΔT = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}

Answer:

C(s)R(s)=G1G3(G2+H1)1+G2H2+G2G3H3+G3H1H3+G1G2G3+G1G3H1\frac{C(s)}{R(s)} = \frac{G_1G_3(G_2 + H_1)}{1 + G_2H_2 + G_2G_3H_3 + G_3H_1H_3 + G_1G_2G_3 + G_1G_3H_1}

✓ Answer checked by Claude Opus 5.5 (high effort)

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