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Block diagram reduction with G1–G4, positive feedbacks

Short answer

C(s)/R(s) = (G1 G2 G4 + G1 G3) / (1 - G1 G2 H1 - G4 H2 + G1 G2 G4 H1 H2).

Obtain the overall transfer function C(s)/R(s) using block diagram reduction technique for the system represented by the block diagram as shown in the figure below. [Figure: R(s) enters summing point S1 (+); output of S1 → G1 → node A → G2 → node B → summing point S2 (+) → G4 → summing point S3 (+) → C(s). From node A, branch → G3 → added (+) at S3. From node B, H1 feeds back (+) to S1. From C(s), H2 feeds back (+) to S2.]

Name the signals: E1E_1 is the output of S1S_1, AA the output of G1G_1, BB the output of G2G_2, E2E_2 the output of S2S_2.

The signal equations from the figure are:

E1=R+H1B,A=G1E1,B=G2AE2=B+H2C,C=G4E2+G3A\begin{aligned} E_1 &= R + H_1B, \quad A = G_1E_1, \quad B = G_2A \\ E_2 &= B + H_2C, \quad C = G_4E_2 + G_3A \end{aligned}

Step 1: Reduce the first loop (G1G_1, G2G_2 with positive feedback H1H_1)

The loop G1G2G_1G_2 with positive feedback H1H_1 gives

BR=G1G21−G1G2H1\frac{B}{R} = \frac{G_1G_2}{1 - G_1G_2H_1}

and, since A=B/G2A = B/G_2,

AR=G11−G1G2H1\frac{A}{R} = \frac{G_1}{1 - G_1G_2H_1} 
        +------- G3 (from A) -------+
        |                           v
R -> [G1/(1-G1G2H1)] -A-> G2 -B-> (S2)+-> G4 -> (S3)+-> C
                                   ^                 |
                                   +------ H2 -------+ (+)

Step 2: Reduce the second loop (G4G_4 with positive feedback H2H_2)

CC is fed back through H2H_2 to S2S_2 with a plus sign, and G3AG_3A is added after G4G_4:

C=G4(B+H2C)+G3A  ⇒  C(1−G4H2)=G4B+G3AC = G_4(B + H_2C) + G_3A \;\Rightarrow\; C(1 - G_4H_2) = G_4B + G_3A

Step 3: Substitute AA and BB

C(1−G4H2)=G1G2G4+G1G31−G1G2H1 RCR=G1(G2G4+G3)(1−G1G2H1)(1−G4H2)\begin{aligned} C(1 - G_4H_2) &= \frac{G_1G_2G_4 + G_1G_3}{1 - G_1G_2H_1}\,R \\ \frac{C}{R} &= \frac{G_1(G_2G_4 + G_3)}{(1 - G_1G_2H_1)(1 - G_4H_2)} \end{aligned}

Answer:

C(s)R(s)=G1G2G4+G1G31−G1G2H1−G4H2+G1G2G4H1H2\frac{C(s)}{R(s)} = \frac{G_1G_2G_4 + G_1G_3}{1 - G_1G_2H_1 - G_4H_2 + G_1G_2G_4H_1H_2}

Check (Mason): paths G1G2G4G_1G_2G_4 and G1G3G_1G_3; loops L1=G1G2H1L_1 = G_1G_2H_1, L2=G4H2L_2 = G_4H_2, which do not touch each other, so Δ=1−L1−L2+L1L2\Delta = 1 - L_1 - L_2 + L_1L_2. Both paths touch both loops, so Δ1=Δ2=1\Delta_1=\Delta_2=1. Same result.

✓ Answer checked by Claude Opus 5.5 (high effort)

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