Y/R by block reduction with G4 from input

Result: $\dfrac{Y(s)}{R(s)} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4$

The diagram is reduced step by step using block diagram algebra rules (series, parallel, feedback, moving take-off points).

Step 1: Separate the parallel feed-forward path

$G_4$ takes the input $R$ directly to the output summing point $S_3$ (both inputs $+$). So the system is two parallel paths:

$$\frac{Y(s)}{R(s)} = \frac{C(s)}{R(s)} + G_4$$

where $C(s)$ is the output of $G_3$ (node C). We now reduce the main path from $R$ to $C$.

Step 2: Move the take-off point of $H_2$ from C to B

Node C is after $G_3$. Moving a take-off point before a block multiplies the branch by that block. So the $H_2$ feedback becomes $G_3H_2$ taken from node B.

Now two negative feedbacks go from B to $S_2$: $H_1$ and $G_3H_2$. They are in parallel:

$$H_{eq} = H_1 + G_3H_2$$

Step 3: Reduce the inner loop around $G_2$

$$\frac{B}{E_2} = \frac{G_2}{1+G_2(H_1+G_3H_2)} = \frac{G_2}{1+G_2H_1+G_2G_3H_2}$$

Step 4: Series with $G_1$, then the outer loop through $H_1$ to $S_1$

Forward path from $S_1$ to B: $\frac{G_1G_2}{1+G_2H_1+G_2G_3H_2}$, with negative feedback $H_1$:

$$\begin{aligned} \frac{B}{R} &= \frac{\frac{G_1G_2}{1+G_2H_1+G_2G_3H_2}}{1+\frac{G_1G_2H_1}{1+G_2H_1+G_2G_3H_2}} \\ &= \frac{G_1G_2}{1+G_2H_1+G_2G_3H_2+G_1G_2H_1} \end{aligned}$$

Step 5: Series with $G_3$ and add $G_4$

$$\frac{C}{R} = \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2}$$

        +------------ G4 -------------+
        |                             | +
 R ---->+--> [ G1G2G3 / Delta ] ----->(+)---> Y
                                        +
 Delta = 1 + G1G2H1 + G2H1 + G2G3H2

$$\frac{Y(s)}{R(s)} = \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2} + G_4$$

or, over a common denominator,

$$\frac{Y(s)}{R(s)} = \frac{G_1G_2G_3 + G_4\left(1+G_1G_2H_1+G_2H_1+G_2G_3H_2\right)}{1+G_1G_2H_1+G_2H_1+G_2G_3H_2}$$

Check (Mason's rule): loops $-G_1G_2H_1$, $-G_2H_1$, $-G_2G_3H_2$ all touch each other, so $\Delta = 1+G_1G_2H_1+G_2H_1+G_2G_3H_2$. Path $G_1G_2G_3$ touches all loops ($\Delta_1=1$); path $G_4$ touches none ($\Delta_2=\Delta$). This gives the same result.