Block diagram reduction: G1–G4, H1, H2, unity loop

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

Step 1: Move the take-off point of $H_1$ from $B$ back to $A$

$B = G_2A$, so the $H_1$ branch becomes $G_2H_1$ taken from $A$.

Step 2: Combine the paths from $A$ to $C$

$G_2$ and $G_3$ are in series, and that series pair is in parallel with $G_4$:

$$\frac{C}{A} = G_2G_3 + G_4$$

Step 3: Move the take-off point of $G_2H_1$ from $A$ forward to $C$

Since $A = \dfrac{C}{G_2G_3 + G_4}$, the branch becomes $\dfrac{G_2H_1}{G_2G_3 + G_4}$ taken from $C$.

R->(S1)->(S2)-> G1 ->(S3)-> [G2G3+G4] --+--> C
    ^-     ^-          ^-               |
    |      |           +----- H2 -------+
    |      +--- G2H1/(G2G3+G4) ---------+
    +------------- 1 -------------------+

Step 4: Reduce the inner loop with $H_2$

$$\frac{C}{G_1E} = \frac{G_2G_3 + G_4}{1 + H_2(G_2G_3 + G_4)}$$

So the forward gain from $S_2$ is

$$G_f = \frac{G_1(G_2G_3 + G_4)}{1 + G_2G_3H_2 + G_4H_2}$$

Step 5: Combine the two outer feedbacks and close the loop

$S_1$ and $S_2$ are cascaded summers, so the feedbacks from $C$ add in parallel:

$$H = 1 + \frac{G_2H_1}{G_2G_3 + G_4}$$ $$\frac{C}{R} = \frac{G_f}{1 + G_fH} = \frac{G_1(G_2G_3 + G_4)}{1 + G_2G_3H_2 + G_4H_2 + G_1(G_2G_3 + G_4) + G_1G_2H_1}$$

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_4H_2 + G_1G_2G_3 + G_1G_4}$$

Check (Mason): paths $G_1G_2G_3$, $G_1G_4$; loops $-G_1G_2H_1$, $-G_2G_3H_2$, $-G_4H_2$, $-G_1G_2G_3$, $-G_1G_4$; all loops touch, $\Delta_1 = \Delta_2 = 1$. Same result.