Block diagram reduction with G1, G2, H1, H2, feedforward

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

From the figure:

$$\begin{aligned} E &= R - C - H_1B, \quad A = G_1E \\ B &= G_2(A - H_2C), \quad C = B + A \end{aligned}$$

($S_1$ and $S_2$ are in cascade, so they act as one summer.)

Step 1: Move the take-off point of $H_1$ from $B$ past summer $S_4$

Since $C = B + A$, we have $B = C - A$. So

$$H_1B = H_1C - H_1A = H_1C - G_1H_1E$$

The $H_1$ path becomes $H_1$ from $C$ (negative) plus a positive self loop $G_1H_1$ around $G_1$.

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

Feedback from $C$: $1 + H_1$ (parallel). Loop $G_1$ with positive feedback $H_1$: $\dfrac{G_1}{1 - G_1H_1}$.

Step 3: Reduce from $A$ to $C$

$$C = G_2(A - H_2C) + A \;\Rightarrow\; \frac{C}{A} = \frac{1 + G_2}{1 + G_2H_2}$$

(Inner loop $G_2$, $H_2$ plus the parallel unity feedforward path.)

R ->(+)-> [G1/(1-G1H1)] -> [(1+G2)/(1+G2H2)] --> C
     ^-                                      |
     +------------- (1 + H1) <---------------+

Step 4: Close the main loop

$$G_f = \frac{G_1(1+G_2)}{(1 - G_1H_1)(1 + G_2H_2)}, \qquad H = 1 + H_1$$ $$\begin{aligned} \frac{C}{R} &= \frac{G_f}{1+G_fH} = \frac{G_1(1+G_2)}{(1-G_1H_1)(1+G_2H_2) + G_1(1+G_2)(1+H_1)} \\ \text{Denominator} &= 1 + G_2H_2 - G_1H_1 - G_1G_2H_1H_2 + G_1 + G_1G_2 + G_1H_1 + G_1G_2H_1 \end{aligned}$$

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1 + G_1G_2}{1 + G_1 + G_1G_2 + G_2H_2 + G_1G_2H_1 - G_1G_2H_1H_2}$$

Check (Mason): paths $G_1G_2$, $G_1$; loops $-G_1G_2H_1$, $-G_2H_2$, $-G_1G_2$, $-G_1$, and $+G_1G_2H_1H_2$ (through the feedforward, $H_2$, $G_2$, $H_1$). All loops touch, $\Delta_1=\Delta_2=1$. Same result.