Block diagram reduction: G1–G4 with H1, H2, H3 loops

Name the signals: $A$ is the output of $S_2$ (input of $G_1$ and of $G_4$), $B$ is the output of $G_2$.

From the figure the signal equations are:

$$\begin{aligned} A &= R - H_1C - H_2B \\ B &= G_2(G_1A - H_3C) \\ C &= G_3B + G_4A \end{aligned}$$

(The two summers $S_1$ and $S_2$ are in cascade, so they can be merged into one summer at $A$.)

Step 1: Move the take-off point of $H_2$ from $B$ past $G_3$ and $S_4$

From the last equation, $B = \dfrac{C - G_4A}{G_3}$. So the signal $H_2B$ can be produced as

$$H_2B = \frac{H_2}{G_3}C - \frac{G_4H_2}{G_3}A$$

The $H_2$ path is replaced by a block $H_2/G_3$ from $C$ (negative) and a block $G_4H_2/G_3$ from $A$ back to $A$ (positive).

Step 2: Combine the feedback blocks from $C$ and remove the self loop at $A$

Feedback from $C$ to the summer: $H_1 + \dfrac{H_2}{G_3}$ (parallel blocks).

The positive self loop at $A$ gives the block $\dfrac{1}{1 - G_4H_2/G_3}$.

Step 3: Reduce the forward part from $A$ to $C$

$$C = G_3G_2(G_1A - H_3C) + G_4A \;\Rightarrow\; \frac{C}{A} = \frac{G_1G_2G_3 + G_4}{1 + G_2G_3H_3}$$

This is the inner loop $G_2G_3$ with feedback $H_3$, plus the parallel path $G_4$.

Step 4: Close the main loop

Forward gain:

$$G_f = \frac{G_1G_2G_3 + G_4}{(1 + G_2G_3H_3)\left(1 - \frac{G_4H_2}{G_3}\right)}, \qquad H = H_1 + \frac{H_2}{G_3}$$ $$\frac{C}{R} = \frac{G_f}{1 + G_fH}$$

Multiplying out, the $G_4H_2/G_3$ terms cancel:

$$\begin{aligned} 1+G_fH &\propto (1+G_2G_3H_3)\left(1-\tfrac{G_4H_2}{G_3}\right) + (G_1G_2G_3+G_4)\left(H_1+\tfrac{H_2}{G_3}\right) \\ &= 1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3 \end{aligned}$$

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1G_2G_3 + G_4}{1 + G_1G_2H_2 + G_2G_3H_3 + G_1G_2G_3H_1 + G_4H_1 - G_2G_4H_2H_3}$$

Check by Mason's formula: forward paths $P_1 = G_1G_2G_3$, $P_2 = G_4$; loops $-G_1G_2H_2$, $-G_2G_3H_3$, $-G_1G_2G_3H_1$, $-G_4H_1$ and $+G_2G_4H_2H_3$ (path $A \to G_4 \to C \to H_3 \to G_2 \to B \to H_2 \to A$). All loops touch each other and both paths, so $\Delta_1 = \Delta_2 = 1$. This gives the same result.