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Block diagram reduction with G3 from before G1

Short answer

Shift the H2 take-off point before summing point S3, then combine parallel paths G1 and G3H2 into G1−G3H2. Reduce the inner loop with G2 and H2, then close the outer loop through H1. The overall transfer function is (G1G2+G3) / (1+G2H2+G1G2H1−G2G3H1H2).

Find overall transfer function of the following diagram using block diagram reduction technique. [Figure: R(s) → summing point S1 (+, −) → node A → G1 → summing point S2 (+, −) → G2 → node B → summing point S3 (+, +) → C(s). G3 from node A (before G1) to S3 (+). H2 from C(s) fed back to S2 (−). H1 from node B fed back to S1 (−).]

Result: C(s)R(s)=G1G2+G31+G2H2+G1G2H1−G2G3H1H2\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2+G_3}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2}

The feedback H2H_2 is taken from C, after summing point S3S_3 where G3G_3 joins. We shift this take-off point back over S3S_3, then reduce.

R ->(S1)--A--> G1 ->(S2)-> G2 --B-->(S3)---+-> C
    -^    |         -^          |     +^   |
     |    |          +-----H2---|------|---+
     +----|------ H1 -----------+      |
          +----------- G3 -------------+

Step 1: Move the H2H_2 take-off point before S3S_3

C=B+G3AC = B + G_3A, so H2C=H2B+G3H2AH_2C = H_2B + G_3H_2A. The single H2H_2 branch is replaced by:

  • H2H_2 from node B to S2S_2 (negative)
  • G3H2G_3H_2 from node A to S2S_2 (negative)

Step 2: Combine blocks from A into S2S_2

G1G_1 (+) and G3H2G_3H_2 (−) both start at A and end at S2S_2, so they are in parallel: G1−G3H2G_1 - G_3H_2.

Step 3: Inner loop around G2G_2

BE2=G21+G2H2BA=G2(G1−G3H2)1+G2H2\frac{B}{E_2} = \frac{G_2}{1+G_2H_2} \qquad \frac{B}{A} = \frac{G_2(G_1-G_3H_2)}{1+G_2H_2}

Step 4: Loop through H1H_1 back to S1S_1

AR=11+G2H1(G1−G3H2)1+G2H2=1+G2H21+G2H2+G1G2H1−G2G3H1H2\frac{A}{R} = \frac{1}{1+\frac{G_2H_1(G_1-G_3H_2)}{1+G_2H_2}} = \frac{1+G_2H_2}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2}

Step 5: Output

CA=BA+G3=G1G2−G2G3H2+G3+G2G3H21+G2H2=G1G2+G31+G2H2CR=CA⋅AR=G1G2+G31+G2H2+G1G2H1−G2G3H1H2\begin{aligned} \frac{C}{A} &= \frac{B}{A} + G_3 = \frac{G_1G_2 - G_2G_3H_2 + G_3 + G_2G_3H_2}{1+G_2H_2} = \frac{G_1G_2+G_3}{1+G_2H_2} \\ \frac{C}{R} &= \frac{C}{A}\cdot\frac{A}{R} = \frac{G_1G_2+G_3}{1+G_2H_2+G_1G_2H_1-G_2G_3H_1H_2} \end{aligned}

Check (Mason's rule): paths G1G2G_1G_2, G3G_3; loops −G1G2H1-G_1G_2H_1, −G2H2-G_2H_2, and +G2G3H1H2+G_2G_3H_1H_2 (A → G3G_3 → C → −H2-H_2 → G2G_2 → B → −H1-H_1 → A). All loops touch both paths, so Δ=1+G1G2H1+G2H2−G2G3H1H2\Delta = 1+G_1G_2H_1+G_2H_2-G_2G_3H_1H_2, matching the result.

✓ Answer checked by Claude Opus 5.5 (high effort)

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