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Block diagram reduction techniques
- 1. Block Diagram Reduction Techniques Prepared by, A.Parimala Gandhi, AP(SS)/ECE Department, KIT/CBE CONTROL SYSTEM ENGINEERING
- 2. Block diagram Transfer Function: Ratio between transformation of output to the transformation of input when all the initial conditions are zero. A Block diagram is basically modelling of any simple or complex system. It Consists of multiple Blocks connected together to represent a system to explain how it is functioning )(sR 2G 3G1G 4G 1H 2H )(sY )(sG )(sC)(sR G(s)=C(s)/R(s) Simple system: Complex System:
- 3. • It is normally required to reduce multiple blocks into single block or for convenient understanding it may sometimes required to rearrange the blocks from its original order. • For the calculation of Transfer function its required to be reduced. Need for block diagram reduction
- 4. Block Diagram Reduction techniques 2G1G 21GG 2. Moving a summing point behind a block G G G 1G 2G 21 GG + 1. Combining blocks which are in cascade or in parallel
- 5. 5. Moving a pickoff point ahead of a block G G G G G 1 G 3. Moving a summing point ahead of a block G G G 1 4. Moving a pickoff point behind a block
- 6. 6. Eliminating a feedback loop G H GH G 1 7. Swapping with two adjacent summing points A B AB G 1=H G G 1
- 7. Example 1 Find the transfer function of the following block diagrams 2G 3G1G 4G 1H 2H )(sY)(sR (a)
- 8. 1. Apply the rule that Moving pickoff/takeoff point ahead of block 2. Eliminate loop I & simplify as 324 GGG + B 1G 2H )(sY 4G 2G 1H AB 3G 2G )(sR I Solution: 2G
- 9. 3. Moving pickoff point B behind block 324 GGG + 1G B )(sR 21GH 2H )(sY )/(1 324 GGG + II 1G B )(sR C 324 GGG + 2H )(sY 21GH 4G 2G A 3G 324 GGG +
- 10. 4. Eliminate loop III )(sR )(1 )( 3242121 3241 GGGHHGG GGGG +++ + )(sY )()(1 )( )( )( )( 32413242121 3241 GGGGGGGHHGG GGGG sR sY sT +++++ + == )(sR 1G C 324 12 GGG HG + )(sY 324 GGG + 2H C )(1 3242 324 GGGH GGG ++ + Using rule 6
- 12. Solution: 1. Eliminate loop I 2. Moving pickoff point A behind block 22 2 1 HG G + 1G 1H )(sR )(sY 3H BA 22 2 1 HG G + 2 221 G HG+ 1G 1H )(sR )(sY 3H 2G 2H BA II I 22 2 1 HG G + Not a feedback loop ) 1 ( 2 22 13 G HG HH + +
- 13. 3. Eliminate loop II )(sR )(sY 22 21 1 HG GG + 2 221 3 )1( G HGH H + + 21211132122 21 1)( )( )( HHGGHGHGGHG GG sR sY sT ++++ == Using rule 6
- 15. Solution: 2G 4G1G 4H )(sY 3G 1H 2H )(sR A B 3H 4 1 G 4 1 G I 1. Moving pickoff point A behind block 4G 4 3 G H 4 2 G H
- 16. Solution: 2G 4G1G 4H )(sY 3G 1H 2H )(sR A B 3H 4 1 G 4 1 G I 1. Moving pickoff point A behind block 4G 4 3 G H 4 2 G H
- 17. Solution: 2G1G 4H )(sY 43GG 1H 2H )(sR B 3H 4 1 G 4 1 G I 1. Moving pickoff point A behind block 4G 4 3 G H 4 2 G H
- 18. Solution: 2G1G )(sY 1H 2H )(sR B 3H 4 1 G 4 1 G I 1. Moving pickoff point A behind block 4G 4 3 G H 4 2 G H 2 443 43 1 HGG GG +
- 19. 2. Eliminate loop I and Simplify II III 443 432 1 HGG GGG +1G )(sY 1H B 4 2 G H )(sR 4 3 G H II 332443 432 1 HGGHGG GGG ++ III 4 142 G HGH − Not feedbackfeedback
- 20. )(sR )(sY 4 142 G HGH − 332443 4321 1 HGGHGG GGGG ++ 3. Eliminate loop II & III 143212321443332 4321 1)( )( )( HGGGGHGGGHGGHGG GGGG sR sY sT −+++ == Using rule 6
- 22. Solution: 1. Moving pickoff point A behind block 3G I 1H 3 1 G )(sY 1G 1H 2H )(sR 4G 2G A B 3 1 G 3G
- 23. 2. Eliminate loop I & Simplify 3G 1H 2G B 3 1 G 2H 32GG B 2 3 1 H G H + 1G )(sR )(sY 4G 3 1 G H 23212 32 1 HGGHG GG ++ II
- 24. )(sR )(sY 12123212 321 1 HGGHGGHG GGG +++ 3. Eliminate loop II 12123212 321 4 1)( )( )( HGGHGGHG GGG G sR sY sT +++ +== 4G
- 25. End