Block diagram reduction with H4 input feedforward

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

Step 1: Combine the cascaded summers $S_1$ and $S_2$

$R$ enters $S_1$ directly and $S_2$ through $H_4$. Since the two summers are in series:

$$E = R + H_4R - H_3B - H_1C = (1 + H_4)R - H_3B - H_1C$$

So the input becomes a block $(1 + H_4)$ in front of one summer.

Step 2: Move the take-off point of $H_3$ from $B$ back to $A$

$B = G_2A$, so $H_3$ becomes $G_2H_3$ taken from $A$.

Step 3: Reduce the loop $G_1$, $G_2H_3$

$$\frac{A}{E'} = \frac{G_1}{1 + G_1G_2H_3}$$

where $E'$ is the summer output before this loop.

Step 4: Parallel paths from $A$ to $C$

$G_2$ and $H_2$ both go from $A$ to $S_3$:

$$\frac{C}{A} = G_2 + H_2$$

R -> [1+H4] ->(+)-> [G1/(1+G1G2H3)] -> [G2+H2] --+-> C
               ^-                                |
               +------------- H1 ----------------+

Step 5: Close the loop with $H_1$

$$G_f = \frac{G_1(G_2 + H_2)}{1 + G_1G_2H_3}$$ $$\frac{C}{R} = (1 + H_4)\frac{G_f}{1 + G_fH_1} = \frac{(1 + H_4)G_1(G_2 + H_2)}{1 + G_1G_2H_3 + G_1G_2H_1 + G_1H_1H_2}$$

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1(G_2 + H_2)(1 + H_4)}{1 + G_1G_2H_1 + G_1G_2H_3 + G_1H_1H_2}$$

Check (Mason): four forward paths $G_1G_2$, $G_1H_2$, $H_4G_1G_2$, $H_4G_1H_2$; loops $-G_1G_2H_3$, $-G_1G_2H_1$, $-G_1H_2H_1$, all touching each other and all paths. Same result.