Block diagram reduction with feedforward and H1, H2

Name the signals: $A = R - H_2C$ (output of $S_1$), $X$ = output of $S_2$, $B$ = output of $S_3$, $C = G_3B$. Then $X = G_1A + A - H_1B$ and $B = G_2X - C$.

Step 1: Combine $G_1$ and the feedforward path

$G_1$ and the unity feedforward from $A$ are in parallel into $S_2$:

$$G_a = 1 + G_1$$

Step 2: Move the $H_1$ take-off point from $B$ to $C$

Since $B = C/G_3$, the feedback block becomes $\dfrac{H_1}{G_3}$, taken from $C$.

Step 3: Reduce the unity loop around $G_3$

$G_3$ with unity negative feedback to $S_3$:

$$\frac{C}{G_2X} = \frac{G_3}{1 + G_3}$$

So from $X$ to $C$: $G_b = \dfrac{G_2G_3}{1 + G_3}$.

Step 4: Reduce the $H_1/G_3$ loop

$$\frac{C}{X'} = \frac{G_b}{1 + G_b\frac{H_1}{G_3}} = \frac{\frac{G_2G_3}{1+G_3}}{1 + \frac{G_2H_1}{1+G_3}} = \frac{G_2G_3}{1 + G_3 + G_2H_1}$$

(here $X'$ is the signal $(1 + G_1)A$ entering $S_2$)

Step 5: Series with $(1 + G_1)$ and close the $H_2$ loop

Forward path: $G_f = \dfrac{(1 + G_1)G_2G_3}{1 + G_3 + G_2H_1}$, feedback $H_2$:

$$\frac{C}{R} = \frac{G_f}{1 + G_fH_2}$$

 R +    ┌──────────────────────────┐
 ──►(Σ)─►│ (1+G1)G2G3/(1+G3+G2H1) ├──┬──► C
    ▲ -  └──────────────────────────┘  │
    │            ┌────┐                │
    └────────────┤ H2 │◄───────────────┘
                 └────┘

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_2G_3(1 + G_1)}{1 + G_3 + G_2H_1 + G_1G_2G_3H_2 + G_2G_3H_2}$$

Check by Mason's rule

Forward paths: $P_1 = G_1G_2G_3$, $P_2 = G_2G_3$ (via feedforward). Loops: $-G_3$, $-G_2H_1$, $-G_1G_2G_3H_2$, $-G_2G_3H_2$. All loops touch each other and both paths ($\Delta_1 = \Delta_2 = 1$), giving the same answer.