1 / 19

IRREGULAR CASES

Learn how to handle irregular cases in perturbation method with different methods, including augmentation and coefficient listing.

josborn
Télécharger la présentation

IRREGULAR CASES

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. IRREGULAR CASES • A first entry zero • Can be handled using different methods • Entire zero row • An important case which yields additional information 5.2 Irregular Cases

  2. FIRST ENTRY ZERO(Perturbation method) + + + - + + s5 1 2 11 s4 2 4 10 s3e 6 s2(4e-12)/e 10 @-12/e 10 s1@6 s0 10 5.2 Irregular Cases

  3. REMARKS • Consider the limit as e approaches zero at the earliest convenience • Simplifies calculations • Avoids ambiguities • It may get tedious. Even a second first entry zero is possible 5.2 Irregular Cases

  4. ANOTHER METHOD(Of limited use) • D(s) and D(1/x) have the same number of RHP roots • Retry by listing coefficients in reversed order x5 10 4 2 + x4 11 2 1 + x3 24/11 12/11 + x2 -3.5 1 - x1132/7 + x0 1 + 2 sign changes, 2 RHP roots 5.2 Irregular Cases

  5. YET ANOTHER METHOD (augmentation) • Multiply D(s) by s+a with a>0 to get Q(s)=(s+a)D(s) • Q(s) & D(s) have the same number of RHP roots • Increases complexity • If it does not work, try another ‘a’ • or even a second order factor 5.2 Irregular Cases

  6. Example(previous) Q(s)=(s5+2s4+2s3+4s2+11s+10)(s+1) =s6+3s5+4s4+6s3+15s2+21s+10 Hence R-H array starts with s6 1 4 15 10 s5 3 6 21 5.2 Irregular Cases

  7. + 1 4 15 10 + 3 6 21 + - + + + s6 / / / 1 2 7 s5 / / / 1 4 5 s4 2 8 10 / / -1 1 s3 -2 2 / / 1 1 s2 5 5 2 sign changes, 2 RHP roots s1 2 s0 1 5.2 Irregular Cases

  8. 2nd irregular caseALL-ZERO ROW D(s)=s5+2s4+2s3+4s2+2s+4 s5 | 1 2 2 s4 | 2 4 4 s3 | 0 0 An all-zero row indicates a factorization of symmetrical roots D(s)=A(s).Q(s) Auxiliary polynomial 5.2 Irregular Cases

  9. Auxilliary polynomial is obtained from the last row before the zero row s5 | 1 2 2 s4 | 2 4 4 s3 | 0 0 A(s)=2s4+4s2+4 has symmetrical roots! -0.4551  j1.0987 0.4551  j1.0987 5.2 Irregular Cases

  10. Symmetrical roots(wrt real&imag axes) OR OR OR OR any combination 5.2 Irregular Cases

  11. How to continue? A’(s): derivative of A(s) Replace next (all-zero) row by the coefficients of A’(s) s5 | 1 2 2 s4 | 2 4 4 s3 | 0 0 A(s)=2s4+4s2+4 A’(s)=8s3+8s 8 8 s2 | 2 4 s1 | -8 s0 | 4 & continue! # of sign changes = # of RHP roots including those of A(s) 5.2 Irregular Cases

  12. ExampleD(s)=s5+s4-s3-2s2-2s s5 | 1 -1 -2 s4 | 1 -2 0 s3 | 1 -2 s2 | 0 0 + + + + - - A(s)=s3-2s A’(s)=3s2-2 3 -2 s1 | -4/3 s0 | -2 1 sign change 1 RHP root 5.2 Irregular Cases

  13. C(s) + K G(s) R(s) - H(s) Feedback application k K 5.2 Irregular Cases

  14. where Characteristic equation zeros Are those of i.e. zeros of 5.2 Irregular Cases

  15. Remarks Use R-H test on Q(s) with K as a parameter & investigate sign changes for different K values Note that: CL poles  Ch. Eq’n zeros Ch. Eq’n. zeros determine system stability 5.2 Irregular Cases

  16. R C + K G(s) - Example Q(s)=s3+3s2+3s+1+K(s+7) = s3+3s2+(3+K)s+1+7K 5.2 Irregular Cases

  17. -1/7 2 0 0 Q(s)=s3+3s2+(3+K)s+1+7K s3 | 1 3+K s2 | 3 1+7K s1 | (8-4K)/3 s0 | 1+7K + + 2-K 1+7K K + + - 2-K - + + 1+7K 1 RHP root No RHP roots 2 RHP roots 5.2 Irregular Cases

  18. For varying K Poles move along a trajectory! called ROOT-LOCUS 5.2 Irregular Cases

  19. End of Chapter Restart section Next chapter Restart chapter i The End General index End show 5.2 Irregular Cases

More Related