Obtain the transfer function of the control system represented by block below, by the signal flow graph method

We define the following nodes:

• $R$: Input
• $N_1$: Output of the first summing junction (SJ1)
• $N_2$: Output of $G_1$
• $N_3$: Output of the second summing junction (SJ2)
• $N_4$: Output of $G_2$
• $N_5$: Output of $G_3$
• $C$: Output of the third summing junction (SJ3)

The branches and their gains are:

• $R \xrightarrow{1} N_1$
• $N_1 \xrightarrow{G_1} N_2$
• $N_2 \xrightarrow{1} N_3$
• $N_3 \xrightarrow{G_2} N_4$
• $N_4 \xrightarrow{G_3} N_5$
• $N_5 \xrightarrow{1} C$
• $N_4 \xrightarrow{H_1} N_1$ (positive feedback)
• $N_4 \xrightarrow{-1} N_3$ (negative feedback)
• $N_5 \xrightarrow{-H_2} N_3$ (negative feedback)
• $R \xrightarrow{G_4} C$ (parallel path)

• $P_1$: $R \to N_1 \to N_2 \to N_3 \to N_4 \to N_5 \to C$
  • Gain $P_1 = 1 \cdot G_1 \cdot 1 \cdot G_2 \cdot G_3 \cdot 1 = G_1 G_2 G_3$
• $P_2$: $R \to C$ (via $G_4$)
  • Gain $P_2 = G_4$

• $L_1$: $N_3 \to N_4 \to N_3$ (via $G_2$ and $-1$)
  • Gain $L_1 = G_2 \cdot (-1) = -G_2$
• $L_2$: $N_3 \to N_4 \to N_5 \to N_3$ (via $G_2, G_3$ and $-H_2$)
  • Gain $L_2 = G_2 \cdot G_3 \cdot (-H_2) = -G_2 G_3 H_2$
• $L_3$: $N_1 \to N_2 \to N_3 \to N_4 \to N_1$ (via $G_1, 1, G_2$ and $H_1$)
  • Gain $L_3 = G_1 \cdot 1 \cdot G_2 \cdot H_1 = G_1 G_2 H_1$

• $L_1$ involves nodes $N_3, N_4$.
• $L_2$ involves nodes $N_3, N_4, N_5$.
• $L_3$ involves nodes $N_1, N_2, N_3, N_4$. All three loops share common nodes ($N_3$ and $N_4$). Therefore, there are no non-touching loops. Thus, $\sum L_i L_j = 0$, $\sum L_i L_j L_k = 0$, etc.

• For $P_1 = G_1 G_2 G_3$:

  • $P_1$ passes through nodes $N_1, N_2, N_3, N_4, N_5$.
  • $L_1$ (nodes $N_3, N_4$) touches $P_1$.
  • $L_2$ (nodes $N_3, N_4, N_5$) touches $P_1$.
  • $L_3$ (nodes $N_1, N_2, N_3, N_4$) touches $P_1$.
  • Since all loops touch $P_1$, $\Delta_1 = 1$.

• For $P_2 = G_4$:

  • $P_2$ is the path $R \to C$. It does not pass through any of the intermediate nodes $N_1, N_2, N_3, N_4, N_5$.
  • $L_1$ (nodes $N_3, N_4$) does not touch $P_2$.
  • $L_2$ (nodes $N_3, N_4, N_5$) does not touch $P_2$.
  • $L_3$ (nodes $N_1, N_2, N_3, N_4$) does not touch $P_2$.
  • Since no loops touch $P_2$, $\Delta_2 = \Delta = 1 + G_2 + G_2 G_3 H_2 - G_1 G_2 H_1$.

The final transfer function is: