Block diagram reduction with positive H2 loop

Result: $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2+G_1G_2G_3+G_4}$

The feedback $H_2$ is taken from the output, after summing point $S_4$, which is fed by both $G_3$ and $G_4$. We first shift this take-off point back over $S_4$, then reduce the loops.

R->(S1)->(S2)--A--> G1 ->(S3)-> G2 --B--> G3 ->(S4)--+--> C
   -^     -^   |         +^          |         +^    |
    |      +---|----------|-- H1 ----+          |    |
    |          +----------|------- G4 ----------+    |
    |                     +---------- H2 ------------+
    +---------------------- 1 -----------------------+

Step 1: Split the $H_2$ feedback

$C = G_3B + G_4A$, so $H_2C = G_3H_2\,B + G_4H_2\,A$. The $H_2$ branch is replaced by two branches into $S_3$ (both $+$):

• $G_3H_2$ from node B (a positive feedback loop around $G_2$)
• $G_4H_2$ from node A (in parallel with $G_1$)

Step 2: Combine the parallel blocks from A to $S_3$

$$G_1 + G_4H_2$$

Step 3: Positive loop around $G_2$

$$\frac{B}{E_3} = \frac{G_2}{1-G_2G_3H_2} \quad\Rightarrow\quad \frac{B}{A} = \frac{G_2(G_1+G_4H_2)}{1-G_2G_3H_2}$$

Step 4: Negative loop through $H_1$ (B back to $S_2$)

$$\frac{A}{E_1} = \frac{1}{1+H_1\frac{G_2(G_1+G_4H_2)}{1-G_2G_3H_2}} = \frac{1-G_2G_3H_2}{1-G_2G_3H_2+G_1G_2H_1+G_2G_4H_1H_2}$$

Step 5: Output from A

$$\begin{aligned} \frac{C}{A} &= G_3\frac{B}{A} + G_4 = \frac{G_2G_3(G_1+G_4H_2) + G_4(1-G_2G_3H_2)}{1-G_2G_3H_2} \\ &= \frac{G_1G_2G_3+G_4}{1-G_2G_3H_2} \end{aligned}$$

Hence

$$\frac{C}{E_1} = \frac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2}$$

Step 6: Unity negative feedback

$$\frac{C}{R} = \frac{G_1G_2G_3+G_4}{1+G_1G_2H_1-G_2G_3H_2+G_2G_4H_1H_2+G_1G_2G_3+G_4}$$

Check (Mason's rule): forward paths $G_1G_2G_3$ and $G_4$; loops $-G_1G_2H_1$, $+G_2G_3H_2$, $-G_1G_2G_3$, $-G_4$, $-G_2G_4H_1H_2$ (A → $G_4$ → C → $H_2$ → $G_2$ → B → $H_1$ → A). All loops touch each other and both paths, so $\Delta_1=\Delta_2=1$ and the same result follows.