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Signal flow graph with G3 subtracted at output

Short answer

Forward paths: P1 = G1G2, P2 = −G3; loop L1 = −G1H gives Δ = 1+G1H, and all Δ_k = 1. Using Mason's rule, C(s)/R(s) = (G1G2 − G3)/(1+G1H).

Convert the given block diagram to signal flow graph and determine the overall transfer function using Mason's Gain Formula. [Figure: R(s) → summing point S1 (+, −) → node A → G1(s) → node B → G2(s) → summing point S2 (+, −) → C(s). G3(s) from node A to S2 (−). H(s) from node B fed back to S1 (−).]

Signal flow graph

Nodes: RR, AA (output of S1S_1), BB (output of G1G_1), CC.

From the diagram: A=R−HBA = R - HB, B=G1AB = G_1A, C=G2B−G3AC = G_2B - G_3A.

              -G3
        +----------------+
        |                v
R --1--> A --G1--> B --G2--> C
         ^         |
         +---(-H)--+

Branches: R→AR\to A (1), A→BA\to B (G1G_1), B→CB\to C (G2G_2), A→CA\to C (−G3-G_3), B→AB\to A (−H-H).

Mason's gain formula

T=∑PkΔkΔT = \frac{\sum P_k\Delta_k}{\Delta}

Forward paths

  • P1=G1G2P_1 = G_1G_2 (R, A, B, C)
  • P2=−G3P_2 = -G_3 (R, A, C)

Loops

  • L1=−G1HL_1 = -G_1H (A, B, A)

There is only one loop, so no non-touching loops.

Δ=1−L1=1+G1H\Delta = 1 - L_1 = 1 + G_1H

Cofactors: both paths pass through node AA, which is on L1L_1, so Δ1=Δ2=1\Delta_1 = \Delta_2 = 1.

Transfer function

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

Answer:

C(s)R(s)=G1G2−G31+G1H\frac{C(s)}{R(s)} = \frac{G_1G_2 - G_3}{1 + G_1H}

✓ Answer checked by Claude Opus 5.5 (high effort)

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