Block Reduction Method
- 1. Name :- Smit Shah -140410109096 T.Y Electrical 2 Sem 5 Subject:-Control System Engineering.. Topic :- 1
- 2. Block diagram Transfer Function Consists of Blocks Can be reduced )(sR 2G 3G1G 4G 1H 2H )(sY G )(sY)(sR 2
- 3. Reduction techniques 2G1G 21GG 2. Moving a summing point behind a block G G G 1G 2G 21 GG 1. Combining blocks in cascade or in parallel 3
- 4. 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 4
- 5. 6. Eliminating a feedback loop G H GH G 1 7. Swap with two neighboring summing points A B AB G 1H G G 1 5
- 6. Example 1 Find the transfer function of the following block diagrams 2G 3G1G 4G 1H 2H )(sY)(sR (A) 6
- 7. 1. Moving pickoff point A ahead of block 2G 2. Eliminate loop I & simplify 324 GGG B 1G 2H )(sY 4G 2G 1H AB 3G 2G )(sR I Solution: 7
- 8. 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 8
- 9. 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 9
- 11. 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 11
- 12. 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 12
- 14. 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 14
- 15. 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 15
- 16. )(sR )(sY 4 142 G HGH 332443 4321 1 HGGHGG GGGG 3. Eliminate loop II & IIII 143212321443332 4321 1)( )( )( HGGGGHGGGHGGHGG GGGG sR sY sT Using rule 6 16
- 18. 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 18
- 19. 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 19
- 20. )(sR )(sY 12123212 321 1 HGGHGGHG GGG 3. Eliminate loop II 12123212 321 4 1)( )( )( HGGHGGHG GGG G sR sY sT 4G 20
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