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

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

$$\begin{aligned} E &= R - C - H_1B, \quad A = G_1E \\ B &= G_2(A - H_2C), \quad C = G_3B + G_4A \end{aligned}$$

($S_1$ and $S_2$ are in cascade and are treated as one summer.)

Step 1: Move the take-off point of $H_1$ from $B$ past $G_3$ and $S_4$

From $C = G_3B + G_4A$: $B = \dfrac{C - G_4A}{G_3}$. Hence

$$H_1B = \frac{H_1}{G_3}C - \frac{G_1G_4H_1}{G_3}E$$

So the $H_1$ path becomes a block $H_1/G_3$ from $C$ (negative) and a positive self loop $G_4H_1/G_3$ around $G_1$.

Step 2: Combine the feedbacks and remove the self loop

Feedback from $C$: $H = 1 + \dfrac{H_1}{G_3}$. $G_1$ with positive feedback $G_4H_1/G_3$: $\dfrac{G_1}{1 - G_1G_4H_1/G_3}$.

Step 3: Reduce from $A$ to $C$

$$C = G_3G_2(A - H_2C) + G_4A \;\Rightarrow\; \frac{C}{A} = \frac{G_2G_3 + G_4}{1 + G_2G_3H_2}$$

R ->(+)-> [G1/(1-G1G4H1/G3)] -> [(G2G3+G4)/(1+G2G3H2)] -> C
     ^-                                               |
     +------------------ (1 + H1/G3) <----------------+

Step 4: Close the main loop

$$\frac{C}{R} = \frac{G_1(G_2G_3+G_4)}{\left(1 - \frac{G_1G_4H_1}{G_3}\right)(1 + G_2G_3H_2) + G_1(G_2G_3 + G_4)\left(1 + \frac{H_1}{G_3}\right)}$$

On expanding, the $G_1G_4H_1/G_3$ terms cancel.

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_1G_4}{1 + G_1G_2H_1 + G_2G_3H_2 + G_1G_2G_3 + G_1G_4 - G_1G_2G_4H_1H_2}$$

Check (Mason): paths $G_1G_2G_3$, $G_1G_4$; loops $-G_1G_2H_1$, $-G_2G_3H_2$, $-G_1G_2G_3$, $-G_1G_4$ and $+G_1G_2G_4H_1H_2$ (via $G_4$, $H_2$, $G_2$, $H_1$). All loops touch each other and both paths, so $\Delta_1 = \Delta_2 = 1$. Same result.