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Mason gain formula with positive H2 loop

Short answer

Draw signal flow graph; two forward paths P1=G1G2G3 and P2=G1G4; five loops, none non-touching; all cofactors are 1. By Mason's formula, C/R = (G1G2G3 + G1G4) / (1 + G1G2H1 - G2G3H2 + G1G2G3 + G1G4 + G1G2G4H1H2).

From the given block diagram, draw the signal flow graph and find the transfer ratio C(s)/R(s) using Mason's gain formula. [Figure: R(s) → summing point S1 (+, −) → summing point S2 (+, −) → G1 → node A → summing point S3 (+, +) → G2 → node B → G3 → summing point S4 (+, +) → C(s). G4 from node A to S4 (+). H1 from node B fed back to S2 (−). H2 from C(s) fed back to S3 (+). Unity feedback from C(s) to S1 (−).]

Each signal becomes a node and each block a branch; summer signs go into branch gains. Then Mason's gain formula is applied.

Signal flow graph

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

                         G4
               +------------------------+
               |                        v
R -1-> x1 -1-> x2 -G1-> x3 -1-> x4 -G2-> x5 -G3-> C -1-> C
       ^       ^                ^        |        |
       |       +-----(-H1)------+--------+        |
       |                        +-----(+H2)-------+
       +-------------------(-1)-------------------+

Branches: x3→Cx_3\to C (G4G_4), x5→x2x_5\to x_2 (−H1-H_1), C→x4C\to x_4 (+H2+H_2), C→x1C\to x_1 (−1-1).

Forward paths

  • P1=G1G2G3P_1 = G_1G_2G_3
  • P2=G1G4P_2 = G_1G_4

Individual loops

  • L1=−G1G2H1L_1 = -G_1G_2H_1 (x2,x3,x4,x5,x2x_2, x_3, x_4, x_5, x_2)
  • L2=+G2G3H2L_2 = +G_2G_3H_2 (x4,x5,C,x4x_4, x_5, C, x_4)
  • L3=−G1G2G3L_3 = -G_1G_2G_3 (x1,…,C,x1x_1, \dots, C, x_1)
  • L4=−G1G4L_4 = -G_1G_4 (x1,x2,x3,C,x1x_1, x_2, x_3, C, x_1)
  • L5=G1G4(H2)(G2)(−H1)=−G1G2G4H1H2L_5 = G_1G_4(H_2)(G_2)(-H_1) = -G_1G_2G_4H_1H_2 (x2,x3,C,x4,x5,x2x_2, x_3, C, x_4, x_5, x_2)

Non-touching loops

Every pair of loops shares at least one node (for example, L2L_2 and L4L_4 share CC), so there are no non-touching loops.

Δ=1+G1G2H1−G2G3H2+G1G2G3+G1G4+G1G2G4H1H2\Delta = 1 + G_1G_2H_1 - G_2G_3H_2 + G_1G_2G_3 + G_1G_4 + G_1G_2G_4H_1H_2

Cofactors

P1P_1 touches all loops. P2P_2 passes through x2x_2, x3x_3 and CC, and every loop contains at least one of these nodes. So Δ1=Δ2=1\Delta_1 = \Delta_2 = 1.

Mason's gain formula

CR=P1Δ1+P2Δ2Δ\frac{C}{R} = \frac{P_1\Delta_1 + P_2\Delta_2}{\Delta}

Answer:

C(s)R(s)=G1G2G3+G1G41+G1G2H1−G2G3H2+G1G2G3+G1G4+G1G2G4H1H2\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 - G_2G_3H_2 + G_1G_2G_3 + G_1G_4 + G_1G_2G_4H_1H_2}

✓ Answer checked by Claude Opus 5.5 (high effort)

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