Block diagram reduction with G4 on input and H1–H2

Reading of the figure: the forward path is $G_1$ then $G_3$ to node $A$; from $A$ the signal reaches $S_2$ directly and also through $G_2$ (sign taken as $+$). The output $C$ goes through $H_1$ to $S_3$ (+), where $G_4R$ is subtracted; the result passes through $H_2$ to $S_1$ (−).

Step 1: Series and parallel blocks in the forward path

$G_1$ and $G_3$ are in series; the direct path and $G_2$ from $A$ are in parallel:

$$G(s) = G_1G_3(1 + G_2)$$

Step 2: Move summing point $S_3$ past $H_2$

The signal into $S_1$ from the feedback is $H_2(H_1C - G_4R) = H_1H_2C - G_4H_2R$. So after moving $S_3$ beyond $H_2$, the feedback path is $H_1H_2$ from $C$, and a path $G_4H_2$ from $R$ enters $S_1$ with sign $(-)(-) = +$.

Step 3: Combine the two input paths into $S_1$

$R$ reaches $S_1$ directly and through $G_4H_2$ (both +), a parallel combination:

$$E = (1 + G_4H_2)R - H_1H_2C$$

 R ─►[1+G4H2]─►(Σ)─►[G1G3(1+G2)]──┬──► C
               ▲ -                │
               └─────[H1H2]◄──────┘

Step 4: Close the feedback loop

$$\frac{C}{(1 + G_4H_2)R} = \frac{G_1G_3(1 + G_2)}{1 + G_1G_3(1 + G_2)H_1H_2}$$

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1G_3(1 + G_2)(1 + G_4H_2)}{1 + G_1G_3H_1H_2 + G_1G_2G_3H_1H_2}$$

If the sign at the $G_2$ input of $S_2$ is negative, replace $(1 + G_2)$ by $(1 - G_2)$ throughout.