Tutorial Solutions - Block Diagram Reductions 1. (Exam Q1, 2007) + U(s) + _ 1/s _ Y(s) 4 + + 7 Figure Q1 a) Use block diagram reduction techniques to show that the system shown in Figure Q1 can be represented by Y( s) K  U ( s) s 2  4  7K s  K Solution: a) Firstly, move the take-off point ahead of block '1/s'. U(s) + + _ 1/s _ Y(s) 4 + + s 7 Then the two parallel feedback loops can be combined as 7s+1 U(s) + + _ 1/s _ Y(s) 4 7s+1 The inner loop in the forward path can then be removed. 1 U(s) + 1 𝑠+4 _ Y(s) 7s+1 The final closed-loop transfer function is therefore: K 1 K K Y ( s) s s4    U ( s) 1 K 1 7 s 1 s 2  4s  7 Ks  K s 2  4  7 K s  K     s s4 2. G3(s) U(s) +_ G1(s) ++ _ Y(s) G2(s) H1(s) H2(s) (a) Figure Q2 Use block diagram reduction techniques to show that the system shown in Figure Q1 can be represented by Y  s G1  s G2  s  G2  s G3  s  U  s 1  G2  s  H1  s  G1  s G2  s  H2  s  [ 10 Marks] Solution: (a) Firstly, remove the inner feedback loop as shown below: G3(s) U(s) +_ G1(s) + G2  s  Y(s) 1  G2  s H1  s H2(s) [2 marks] 2 then we can move the summing point in the middle ahead of block G 1(s) as follows: G 3 s  G1  s  U(s) + +_ G2  s G1(s) Y(s) 1 G 2  s  H 1  s  H2(s) [3 marks] This can be further reduced to: U(s) 1 G3  s  G1  s  G1 sG2 s 1G2 sH1 sG1sG2 sH2 s Y(s) [3 marks] The closed-loop transfer function is therefore:  G3  s   1  G1  s  G2  s  G1  s   Y  s G1 s G 2  s  G 2  s G3 s     U  s 1  G2  s  H1  s   G1  s  G2  s  H 2  s  1  G2  s  H1  s   G1  s  G2  s  H 2  s  The final block diagram with the closed-loop transfer function is: U(s) G1  s G2  s  G2  s G3  s  Y(s) 1  G 2 s H 1 s   G1 s G 2 s H 2 s  [2 marks] 3 3. Use block diagram reduction techniques to achieve the overall transfer functions for the following systems Y(s)/R(s). (a). Solution for (a): First, eliminate loop 1 by combining the feedback loop H2 with G2, so the block diagram becomes: Then, move pickoff point A behind the new block to B: Next, combine the bottom two branches (note that this is not a feedback loop, but the two branches are in parallel): 4 In the end, eliminate the outer loop by using the feedback loop formula to obtain the overall transfer function: Y ( s) G1G2 .  R(s ) 1  G2 H 2  G1G2 H 3  G1H1  G1G2 H1H 2 (b). Solution for (b): First, move a pickoff point as follows: Then using the cascade and parallel rules to combine branches on the left hand side of the main forward path, and using the positive feedback rule (add a ‘-’ sign to H1) to combine the inner feedback loop on the right hand side of the main forward path, give: 5 6

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