Solution 3.8.8.3 C 1 C 2 R 1 R 2 V i V o Figure 1: T-network For the so called T-network of Figure 1, there are manyways arriveatthe transfer function. One method is to use a wye-delta transformation two achieveanequivalentcircuit that is a voltage divider. This process is shown in Figures 2 and 3. Then V o (s) V i (s) = Z 2 + R 1 Z 1 +Z 2 + R 1 ;; where Z 1 = R 2 C 1 s R 2 + 1 C 1 s + 1 C 2 s = R 2 C 2 s R 2 C 1 C 2 s 2 + C 1 s +C 2 s ;; and Z 2 = 1 C 1 s 1 C 2 s R 2 + 1 C 1 s + 1 C 2 s = 1 R 2 C 1 C 2 s 2 + C 1 s +C 2 s ;; Then, using these expresssions for Z 1 and Z 2 ,wehave V o (s) V i (s) = Z 2 + R 1 Z 1 + Z 2 + R 1 1 C 1 C 2 R 1 R 2 V i V o V o V i R 2 R 1 C 1 C 2 Z 1 Z 3 Z 2 Figure 2: Wye-delta transformation 2 V i V o Z 1 Z 2 Z 3 V o V i R 2 R 1 C 1 C 2 Z 1 Z 3 Z 2 R 1 Figure 3: Reduction to Voltage Divider 3 = 1 R 2 C 1 C 2 s 2 + C 1 s + C 2 s + R 1 R 2 C 2 s R 2 C 1 C 2 s 2 + C 1 s + C 2 s + 1 R 2 C 1 C 2 s 2 + C 1 s + C 2 s + R 1 = R 1 R 2 C 1 C 2 s 2 + R 1 C 1 s + R 1 C 2 s +1 R 1 R 2 C 1 C 2 s 2 + R 1 C 1 s +2R 1 C 2 s +1 = s 2 + R 1 (C 1 + C 2 )s R 1 R 2 C 1 C 2 + 1 R 1 R 2 C 1 C 2 s 2 + [(C 1 + C 2 )R 1 + R 2 C) 2 ]s R 1 R 2 C 1 C 2 + 1 R 1 R 2 C 1 C 2 = s 2 + (C 1 + C 2 )s R 2 C 1 C 2 + 1 R 1 R 2 C 1 C 2 s 2 + [(C 1 + C 2 )R 1 + R 2 C 2 ]s R 2 C 1 C 2 + 1 R 1 R 2 C 1 C 2 4