Block diagram reduction: G3 feedforward, H1 and H2 loops

Name the signals: $E$ = output of $S_1$, $A$ = output of $G_1$, $X$ = output of $S_2$, $B = G_2X$, and $C = B + G_3A$. So $E = R - H_1B$ and $X = A - H_2C$.

Step 1: Move the $H_1$ take-off point behind $G_2$

Since $B = G_2X$, the feedback $H_1B$ can be taken from $X$ through $G_2H_1$.

Step 2: Move the $G_3$ take-off point from $A$ to after $S_2$

Because $A = X + H_2C$, the branch $G_3A$ equals $G_3X + G_3H_2C$. So $G_3$ now starts at $X$, and an extra path $G_3H_2$ goes from $C$ into $S_3$ (positive).

Step 3: Combine parallel blocks and the positive loop at $S_3$

$G_2$ and $G_3$ are in parallel from $X$: $(G_2 + G_3)$. The extra path forms a positive feedback loop $G_3H_2$ around $S_3$:

$$\frac{C}{X} = \frac{G_2 + G_3}{1 - G_3H_2}$$

Step 4: Move $S_1$ forward past $G_1$

$R$ becomes $G_1R$ and the $H_1$ feedback becomes $G_1G_2H_1$ from $X$, so $X = G_1R - G_1G_2H_1X - H_2C$. The loop on $X$ gives

$$\frac{X}{U} = \frac{1}{1 + G_1G_2H_1}, \qquad U = G_1R - H_2C$$

After Steps 3 and 4 the diagram is:

 R ─►[G1]─►(Σ)──► X ──►[ (G2+G3)/(1−G3H2) ]──┬──► C
            ▲ − ▲ −                          │
            │   └──────────[H2]◄─────────────┤
            └─[G1G2H1]◄── X                  │

Step 5: Close the $H_2$ loop

Forward path from $U$: $P = \dfrac{G_2 + G_3}{(1 + G_1G_2H_1)(1 - G_3H_2)}$, feedback $H_2$:

$$\frac{C}{G_1R} = \frac{P}{1 + PH_2} = \frac{G_2 + G_3}{(1 + G_1G_2H_1)(1 - G_3H_2) + H_2(G_2 + G_3)}$$

Expanding the denominator: $1 + G_1G_2H_1 - G_3H_2 - G_1G_2G_3H_1H_2 + G_2H_2 + G_3H_2$; the $G_3H_2$ terms cancel.

Answer:

$$\frac{C(s)}{R(s)} = \frac{G_1G_2 + G_1G_3}{1 + G_1G_2H_1 + G_2H_2 - G_1G_2G_3H_1H_2}$$

Check by Mason's rule

Forward paths $G_1G_2$, $G_1G_3$. Loops $-G_1G_2H_1$, $-G_2H_2$, and $+G_1G_3H_2G_2H_1$ (path $A \to G_3 \to C \to H_2 \to G_2 \to B \to H_1 \to G_1 \to A$). All loops touch each other and both paths, so $\Delta_1 = \Delta_2 = 1$, giving the same result.