Block diagram reduction of Fig. 2(a): G1–G3, H1, H2

Name the signals: $E$ = output of $S_2$ (input of $G_1$), $A$ = output of $S_3$, $B$ = output of $G_2$.

Step 1: Combine the parallel paths from $A$ to $C$

$G_2$ and $G_3$ both go from $A$ to $S_4$:

$$\frac{C}{A} = G_2 + G_3$$

Step 2: Move the take-off point of $H_1$ from $B$ to $C$

$B = G_2A$ and $A = \dfrac{C}{G_2 + G_3}$, so the branch becomes $\dfrac{G_2H_1}{G_2 + G_3}$ taken from $C$.

Step 3: Reduce the inner loop with $H_2$

$$\frac{C}{G_1E} = \frac{G_2 + G_3}{1 + H_2(G_2 + G_3)}$$

Step 4: Combine the outer feedbacks

$S_1$ and $S_2$ are cascaded, so the unity feedback and the moved $H_1$ branch add in parallel:

$$H = 1 + \frac{G_2H_1}{G_2 + G_3}$$

R ->(+)-> G1 -> [(G2+G3)/(1+G2H2+G3H2)] --+--> C
     ^-                                   |
     +------ 1 + G2H1/(G2+G3) <-----------+

Step 5: Close the loop

$$G_f = \frac{G_1(G_2 + G_3)}{1 + G_2H_2 + G_3H_2}$$ $$\frac{C}{R} = \frac{G_f}{1 + G_fH} = \frac{G_1(G_2 + G_3)}{1 + G_2H_2 + G_3H_2 + G_1(G_2 + G_3) + G_1G_2H_1}$$

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_1G_2 + G_1G_3 + G_1G_2H_1 + G_2H_2 + G_3H_2}$$

Check (Mason): paths $G_1G_2$, $G_1G_3$; loops $-G_1G_2$, $-G_1G_3$, $-G_1G_2H_1$, $-G_2H_2$, $-G_3H_2$, all touching. Same result.