Block diagram reduction: cascaded G1–G3, H1, H2, unity loop

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

The diagram has three nested negative feedback loops. We remove them from the inside out.

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

Step 1: Move the take-off point of $H_2$ from C to B

C $= G_3 \times$ B. Moving the take-off point to before the block $G_3$ means the feedback branch becomes $G_3H_2$, from node B to $S_3$.

Step 2: Reduce the innermost loop ($G_2$ with feedback $G_3H_2$)

$$\frac{B}{E_3} = \frac{G_2}{1+G_2G_3H_2}$$

Step 3: Series with $G_1$, then the $H_1$ loop

The forward gain from $S_2$ to B is $\frac{G_1G_2}{1+G_2G_3H_2}$, with negative feedback $H_1$:

$$\begin{aligned} \frac{B}{E_2} &= \frac{\frac{G_1G_2}{1+G_2G_3H_2}}{1+\frac{G_1G_2H_1}{1+G_2G_3H_2}} = \frac{G_1G_2}{1+G_2G_3H_2+G_1G_2H_1} \end{aligned}$$

Step 4: Series with $G_3$

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

Step 5: Unity negative feedback loop

$$\begin{aligned} \frac{C}{R} &= \frac{\frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2}}{1+\frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2}} \\ &= \frac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3} \end{aligned}$$

Check (Mason's rule): one forward path $P_1=G_1G_2G_3$; loops $-G_1G_2H_1$, $-G_2G_3H_2$, $-G_1G_2G_3$, all touching each other and the path. So $\Delta = 1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3$ and $\Delta_1=1$, giving the same answer.

Answer: $\dfrac{C(s)}{R(s)} = \dfrac{G_1G_2G_3}{1+G_1G_2H_1+G_2G_3H_2+G_1G_2G_3}$