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Block diagram reduction with G3 subtracted at output

Short answer

By block diagram reduction, the transfer function is C(s)/R(s) = (G1G2 - G1G3) / (1 + G1G2 - G1G3 + G1G2H1 + G2H2 + G1G2G3H1H2).

Determine the transfer function of the given system by block reduction technique. [Figure: R(s) → summing point S1 → G1 → node A → summing point S2 (+, −) → G2 → node B → summing point S3 (+) → C(s). G3 from node A to S3 (−). H2 from C(s) fed back to S2 (−). H1 from node B fed back to S1, and C(s) also fed back directly (unity) to S1 (signs at S1 not clearly printed; take as negative feedback)]

Assumption: both feedback signals at S1S_1 (H1BH_1B and CC) are negative, as stated.

Name the signals: EE = output of S1S_1, AA = output of G1G_1, BB = output of G2G_2.

E=R−H1B−C,A=G1EB=G2(A−H2C),C=B−G3A\begin{aligned} E &= R - H_1B - C, \quad A = G_1E \\ B &= G_2(A - H_2C), \quad C = B - G_3A \end{aligned}

Step 1: Move the take-off point of H1H_1 from BB past summer S3S_3

From C=B−G3AC = B - G_3A: B=C+G3AB = C + G_3A. So

H1B=H1C+G3H1A=H1C+G1G3H1EH_1B = H_1C + G_3H_1A = H_1C + G_1G_3H_1E

The H1H_1 path becomes H1H_1 from CC plus a negative self loop G1G3H1G_1G_3H_1 around G1G_1.

Step 2: Combine feedbacks and remove the self loop

Feedback from CC to S1S_1: 1+H11 + H_1 (parallel blocks). G1G_1 with negative feedback G3H1G_3H_1: G11+G1G3H1\dfrac{G_1}{1 + G_1G_3H_1}.

Step 3: Reduce from AA to CC

C=G2(A−H2C)−G3A  ⇒  CA=G2−G31+G2H2C = G_2(A - H_2C) - G_3A \;\Rightarrow\; \frac{C}{A} = \frac{G_2 - G_3}{1 + G_2H_2} 
R ->(+)-> [G1/(1+G1G3H1)] -> [(G2-G3)/(1+G2H2)] --> C
     ^-                                         |
     +--------------- (1 + H1) <----------------+

Step 4: Close the main loop

CR=G1(G2−G3)(1+G1G3H1)(1+G2H2)+G1(G2−G3)(1+H1)\frac{C}{R} = \frac{G_1(G_2 - G_3)}{(1 + G_1G_3H_1)(1 + G_2H_2) + G_1(G_2 - G_3)(1 + H_1)}

Expanding the denominator (the G1G3H1G_1G_3H_1 terms cancel):

1+G2H2+G1G3H1+G1G2G3H1H2+G1G2−G1G3+G1G2H1−G1G3H11 + G_2H_2 + G_1G_3H_1 + G_1G_2G_3H_1H_2 + G_1G_2 - G_1G_3 + G_1G_2H_1 - G_1G_3H_1

Answer:

C(s)R(s)=G1G2−G1G31+G1G2−G1G3+G1G2H1+G2H2+G1G2G3H1H2\frac{C(s)}{R(s)} = \frac{G_1G_2 - G_1G_3}{1 + G_1G_2 - G_1G_3 + G_1G_2H_1 + G_2H_2 + G_1G_2G_3H_1H_2}

Check (Mason): paths G1G2G_1G_2 and −G1G3-G_1G_3; loops −G1G2H1-G_1G_2H_1, −G2H2-G_2H_2, −G1G2-G_1G_2, +G1G3+G_1G_3, −G1G2G3H1H2-G_1G_2G_3H_1H_2; all touch, Δ1=Δ2=1\Delta_1 = \Delta_2 = 1. Same result.

✓ Answer checked by Claude Opus 5.5 (high effort)

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