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- 1. 26-Aug-16 1 SEMINAR PRESENTATION ON RH CRITERION Presented By: LUCKY CHAUHAN B.TECH (I.T) – 1st year (Roll No. 20190947) DEPARTMENT OF INFORMATION AND TECHNOLOGY ENGINEERING GRAPHIC ERA UNIVERSITY, DEHRADUN
- 2. CONTENTS • STABILITY • WHY INVESTIVAGATE STABILITY? • EXAMPLE OF UNSTABLE SYSTEM • ROUTH STABILITY CRITERION • RULES • SPECIAL CASES • PRACTICAL USE • REFRENCES
- 3. STABILITY A process to find out if a system is stable or not. 1. A system is stable if every bounded input yields a bounded output. 2. A system is unstable if any bounded input yields an unbounded output.
- 4. Why investigate stability? • Most important issue to design a control system. • An unstable feedback system is of no practical value. • A stable system should exhibit a bounded output if the corresponding input is bounded.
- 5. 5 Dramatic photographs showing the collapse of the Tacoma Narrows suspension bridge on November 7, 1940. (a) Photograph showing the twisting motion of the bridge’s center span just before failure. (b) A few minutes after the first piece of concrete fell, this second photograph shows a 600-ft section of the bridge breaking out of the suspension span and turning upside down as it crashed in Puget Sound, Washington. (Taken from:-1) One famous example of an unstable system:
- 6. 6 stability region: according to the roots obtained by the characteristic equation. Taken from:- 2
- 7. • Using these concepts, we can say:- 1. stable system: if the natural response approaches zero as time approaches infinity. 2. unstable system: if the natural response grows without bound as time approaches infinity. 3. marginally stable: if the natural response neither decays nor grows but remains constant or oscillates as time approaches infinity.
- 8. 8 Contributions of characteristic equation roots to closed-loop response. 𝑡𝑎𝑘𝑒𝑛 𝑓𝑟𝑜𝑚: 2
- 9. Routh-Hurwitz Stability Criterion • The characteristic equation of the nth order continuous system can be write as: • The stability criterion is applied using a Routh table which is defined as; • Where are coefficients of the characteristic equation. {𝑡aken from:3}
- 10. Routh-Hurwitz Criterion
- 11. RULES • There should be no missing term. 3 𝑠4 + 4𝑠3 + 5𝑠 + 1 = 0 𝑠2 missing. • All variables should be of same sign. 5𝑠3 − 3𝑠2 + 5𝑠 + 1 = 0 -ve sign before the variable 3
- 12. Closed-loop control system with T(s) = Y(s)/R(s) = 1/(s3 + s2+ 2s+24). EXAMPLE:
- 13. Routh array for the closed-loop control system with T(s) = Y(s)/R(s) = 1/(s3+ s2 +2s+24)
- 14. Routh-stability Criterion: Special cases • Two special cases can occur: – Routh table has zero only in the first column of a row – Routh table has an entire row that consists of zeros. s3 1 3 0 s2 3 4 0 s1 0 1 2 s3 1 3 0 s2 3 4 0 s1 0 0 0
- 15. Example of special case:1
- 16. Example of Special case:2 Determine the number of right-half-lane poles in the closed-loop transfer function 5 4 3 2 10 7 6 42 8 56 T s s s s s s 𝒔 𝟓 1 6 8 𝑠4 7 42 56 𝑠3 0 0 0 𝑠1 𝑠0
- 17. Special case:2 • To solve it we will form polynomial • P=7𝑠4+42𝑠2+56 • Then take its derivative 𝒔 𝟓 1 6 8 𝑠4 7 42 56 𝑠3 0 0 0 𝑠1 𝑠0
- 18. • Solving for the remainder of the Routh table • There are no sign changes, so there are no poles on the right half plane. The system is stable.
- 19. 19 PRACTICAL USE: Find the values of controller gain Kc that make the feedback control system of following equation stable. The characteristic equation is:- All coefficients are positive provided that 1 + Kc > 0 or Kc < -1. The Routh array is 10 8 17 1 + Kc b1 b2 c1 10𝑆3+17𝑆2+8S+(1+𝐾𝐶)=0
- 20. 20 To have a stable system, each element in the left column of the Routh array must be positive. Element b1 will be positive if Kc < 7.41/0.588 = 12.6. Similarly, c1 will be positive if Kc > -1. Thus, we conclude that the system will be stable if -1<𝐾𝐶 < 12.6
- 21. REFRENCES 1. www.calvin.edu/../chap-6 2. site.ntvc.edu.cn/..automation_7.ppt 3. www.slideshare.com 4.BOOKS:- • B.S MANKE • I.J NAGRATHA
- 22. THANK YOU