This computer science problem involves algorithmic thinking and programming concepts. The solution below explains the approach, logic, and implementation step by step.
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To obtain the transfer function (C)/(R) using the signal flow graph method, we will apply Mason's Gain Formula: T = (C)/(R) = (1)/() _k=1^N P_k _k where P_k is the gain of the k-th forward path, is the system determinant, and _k is the determinant of the graph when all loops touching the k-th forward path are removed. Step 1: Convert the block diagram into a signal flow graph. 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 1 N_1 N_1 G_1 N_2 N_2 1 N_3 N_3 G_2 N_4 N_4 G_3 N_5 N_5 1 C N_4 H_1 N_1 (positive feedback) N_4 -1 N_3 (negative feedback) N_5 -H_2 N_3 (negative feedback) R G_4 C (parallel path) Step 2: Identify all forward paths (P_k) and their gains. P_1: R N_1 N_2 N_3 N_4 N_5 C Gain P_1 = 1 · G_1 · 1 · G_2 · G_3 · 1 = G_1 G_2 G_3 P_2: R C (via G_4) Gain P_2 = G_4 Step 3: Identify all individual loops (L_i) and their gains. L_1: N_3 N_4 N_3 (via G_2 and -1) Gain L_1 = G_2 · (-1) = -G_2 L_2: N_3 N_4 N_5 N_3 (via G_2, G_3 and -H_2) Gain L_2 = G_2 · G_3 · (-H_2) = -G_2 G_3 H_2 L_3: N_1 N_2 N_3 N_4 N_1 (via G_1, 1, G_2 and H_1) Gain L_3 = G_1 · 1 · G_2 · H_1 = G_1 G_2 H_1 Step 4: Identify all non-touching loops. 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, L_i L_j = 0, L_i L_j L_k = 0, etc. Step 5: Calculate the system determinant . = 1 - L_i = 1 - (L_1 + L_2 + L_3) = 1 - (-G_2 - G_2 G_3 H_2 + G_1 G_2 H_1) = 1 + G_2 + G_2 G_3 H_2 - G_1 G_2 H_1 Step 6: Calculate _k for each forward path. 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, _1 = 1. For P_2 = G_4: P_2 is the path R 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, _2 = = 1 + G_2 + G_2 G_3 H_2 - G_1 G_2 H_1. Step 7: Apply Mason's Gain Formula. (C)/(R) = (P_1 _1 + P_2 _2)/() (C)/(R) = ((G_1 G_2 G_3)(1) + G_4 (1 + G_2 + G_2 G_3 H_2 - G_1 G_2 H_1))/(1 + G_2 + G_2 G_3 H_2 - G_1 G_2 H_1) (C)/(R) = (G_1 G_2 G_3 + G_4 + G_2 G_4 + G_2 G_3 G_4 H_2 - G_1 G_2 G_4 H_1)/(1 + G_2 + G_2 G_3 H_2 - G_1 G_2 H_1) The final transfer function is: (C)/(R) = (G_1 G_2 G_3 + G_4 + G_2 G_4 + G_2 G_3 G_4 H_2 - G_1 G_2 G_4 H_1)/(1 + G_2 + G_2 G_3 H_2 - G_1 G_2 H_1) Drop the next question.