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stability by Routh-Hurwitz Criterion

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1 stability by Routh-Hurwitz Criterion
Nafees Ahamad AP, Department of Electrical, Electronics & Communication, DIT University, Dehradun (c) Nafees Ahamad

2 Contents Stability Why investigate stability ?
Example of unstable system Routh-Hurwitz Criterion Rules Special Cases Practical Use References (c) Nafees Ahamad

3 Stability A system is stable if every bounded input results in bounded output. A system is unstable if any bounded input results in unbounded output. (c) Nafees Ahamad

4 Why investigate stability ?
Most important consideration in design a control system. An unstable system is of NO use practically. (c) Nafees Ahamad

5 Example of unstable system
Collapse of Tacoma Narrows Bridge (US state of Washington ) in Β  It opened to traffic on July 1, 1940, and dramatically collapsed into Puget Sound on November 7 the same year. Source: Wikipedia (c) Nafees Ahamad

6 Stability Consider following closed loop system
Its closed loop transfer function 𝐢(𝑠) 𝑅(𝑠) = 𝐺(𝑠) 1+𝐺 𝑠 𝐻(𝑠) = 𝑠 𝑠 2 +31𝑠+1030 Its characteristic equation 𝑠 𝑠 2 +31𝑠+1030=0 (c) Nafees Ahamad

7 Stability… Roots of characteristic equation (=poles of transfer function) 𝑠 1 =βˆ’ , 𝑠 2 & 𝑠 3 = Β± 𝑖 All roots are on right hand side of s-plane. So, system is unstable. Re Im 𝑠 2 = 𝑖 𝑠 1 =βˆ’ , 𝑠 3 = βˆ’ 𝑖 (c) Nafees Ahamad

8 Stability… If any root lies on LHS of s-plane, then system becomes stable. If simple root(s) is on imaginary axis then system becomes marginally stable. If multiple roots are on imaginary axis then system becomes marginally unstable. (c) Nafees Ahamad

9 Stability – Practically
Stable system: If the natural response becomes zeros as time approaches infinity. Unstable system: if the natural response grows without bound as time approaches infinity. Marginally stable system : if the natural response neither decays nor grows but remains constant or oscillates as time approaches infinity. (c) Nafees Ahamad

10 Roots and Response (c) Nafees Ahamad

11 Roots and Response … (c) Nafees Ahamad

12 Routh-Hurwitz Criterion
By Routh-Hurwitz Criterion we can find out number of roots with +ve, - ve and zero real parts. That is we can find out the stability of a system. (c) Nafees Ahamad

13 Routh-Hurwitz Criterion
Consider a system with characteristic equation π‘Ž 0 𝑠 𝑛 + π‘Ž 1 𝑠 π‘›βˆ’1 + π‘Ž 2 𝑠 π‘›βˆ’2 +… π‘Ž 𝑛 =0 Necessary condition All the coefficients of the characteristic equ should have same sign. There should not be any missing term. If above two conditions are not satisfied then system will be unstable. If above two conditions are satisfied then It is not guaranteed that system is stable for that we need to apply Routh-Hurwitz Criterion. (c) Nafees Ahamad

14 Routh-Hurwitz Criterion…
𝑏 1 = π‘Ž 1 Γ— π‘Ž 2 βˆ’ π‘Ž 0 Γ— π‘Ž 3 π‘Ž 1 𝑏 2 = π‘Ž 1 Γ— π‘Ž 4 βˆ’ π‘Ž 0 Γ— π‘Ž 5 π‘Ž 1 sn a0 a2 a4 Sn-1 a1 a3 a5 Sn-2 b1 b2 Sn-3 c1 - s0 𝑐 1 = 𝑏 1 Γ— π‘Ž 3 βˆ’ π‘Ž 1 Γ— 𝑏 2 𝑏 1 If no sign change in first column β‡’ All roots are on LHS of s-plane β‡’ Stable system (c) Nafees Ahamad

15 Example Example 1: Check the stability of the system whose characteristic equation is given by 𝑠 4 + 2𝑠 3 + 6𝑠 2 +4𝑠+1=0 Necessary conditions are satisfied All the coefficients of the characteristic equation have same sign There is no missing term Make Routh array (c) Nafees Ahamad

16 Example … = 2Γ—6βˆ’1Γ—4 2 s4 1 6 s3 2 4 s2 s1 3.5 s0 = 2Γ—1βˆ’1Γ—0 2
s2 s1 3.5 s0 = 2Γ—1βˆ’1Γ—0 2 = 4Γ—4βˆ’2Γ—1 4 = 3.5Γ—1βˆ’4Γ—0 3.5 No sign change in first column β‡’ All roots are on LHS of s-plane β‡’ Stable system (c) Nafees Ahamad

17 Example… Example 2: Check the stability of the system whose characteristic equation is given by 2𝑠 4 + 2𝑠 3 + 𝑠 2 +3s+2=0 Necessary conditions are satisfied All the coefficients of the characteristic equation have same sign There is no missing term Make Routh array (c) Nafees Ahamad

18 Example … = 2Γ—1βˆ’2Γ—3 2 s4 2 1 s3 3 s2 -2 s1 5 s0 = 2Γ—2βˆ’2Γ—0 2
s2 -2 s1 5 s0 = 2Γ—2βˆ’2Γ—0 2 = βˆ’2Γ—3βˆ’2Γ—2 βˆ’2 = 5Γ—2βˆ’(βˆ’2)Γ—0 5 Two sign change in first column β‡’ Two roots are on LHS of s-plane β‡’ Unstable system (c) Nafees Ahamad

19 Example: Special case 1 ( First element in a row of Routh array is zero)
Example 3: Check the stability of the system whose characteristic equation is given by 𝑠 5 +𝑠 4 + 2𝑠 3 + 2𝑠 2 +3s+5=0 Necessary conditions are satisfied All the coefficients of the characteristic equation have same sign There is no missing term Make Routh array (c) Nafees Ahamad

20 Example: Special case 1 …
= 1Γ—2βˆ’1Γ—2 1 s5 1 2 3 s4 5 S3 -2 s2 s1 s0 = 1Γ—3βˆ’1Γ—5 1 This is the special case where first element of a row is zero. Replace zero with a very small +ve number say πœ€ and continue the procedure. Finally put πœ€β†’+0 and note the number of sign changes (c) Nafees Ahamad

21 Example: Special case 1 …
2 3 s4 5 S3 πœ€ -2 s2 2πœ€+2 πœ€ β†’+𝑣𝑒 s1 βˆ’4πœ€βˆ’9 2πœ€+2 β†’βˆ’π‘£π‘’ s0 Note: Alternate method Multiply characteristic equ with (s+a). Where β€˜a’ is any +ve real number. The simplest value of β€˜a’ is 1. So, multiply cha equ with (s+1) and apply normal procedure. Two sign change in first column β‡’ Two roots are on RHS of s-plane β‡’ Unstable system (c) Nafees Ahamad

22 Example: Special case 2 ( Any one row of Routh array is zero)
Example 4: Check the stability and position of roots of the system whose characteristic equation is given by 𝑠 𝑠 5 +8𝑠 𝑠 𝑠 2 +16s+16=0 Necessary conditions are satisfied All the coefficients of the characteristic equation have same sign There is no missing term Make Routh array (c) Nafees Ahamad

23 Example: Special case 2 …
1 8 20 16 s5 2 12 s4 S3 s2 s1 s0 Divide each by 2 Note: We can also divide any row by a common factor (c) Nafees Ahamad

24 Example: Special case 2 …
1 8 20 16 s5 6 s4 S3 s2 s1 s0 Auxiliary row Vanishing row A(s)=1s4+6s2+8 (c) Nafees Ahamad

25 Example: Special case 2 …
Auxiliary equation A(s)=1s4+6s2+8 Take the first derivative of above equation 𝑑𝐴 𝑠 𝑑𝑠 =4 𝑠 𝑠 1 Put the coefficients of 𝑑𝐴 𝑠 𝑑𝑠 in vanishing row and proceed (c) Nafees Ahamad

26 Example: Special case 2 …
1 8 20 16 s5 6 s4 S3 4 12 s2 s1 s0 Divide each by 4 (c) Nafees Ahamad

27 Example: Special case 2 …
1 8 20 16 s5 6 s4 s3 3 s2 s1 1/3 s0 No sign change in first column β‡’ No roots on RHS of s-plane. Now system may be marginally stable. (c) Nafees Ahamad

28 Example: Special case 2 …
Now, system may be marginally stable. Find out the roots of auxiliary equation A(s)=1s4+6s2+8 = 0 Put s2 = x in above equ. β‡’ π‘₯ 2 +6π‘₯ β‡’π‘₯=βˆ’2, βˆ’4 ⇒𝑠=±𝑗 & ±𝑗2 These two pairs of the roots are also the roots of the original characteristic equation. So, 4 roots are on Imaginary axis, 2 roots are on LHS of s-plane. Hence, system is marginally stable. (c) Nafees Ahamad

29 Application of Routh – Hurwitz criterion
Example 5: The open loop transfer function of a feedback control system is given by 𝐺 𝑠 𝐻(𝑠)= 𝐾 𝑠(𝑠+4)( 𝑠 2 +2𝑠+2) Find the range of the values of K for stability, also determine the stability of the system when K = 12. Answer: 0<K<11.56 and at K = 12 system is Unstable. (c) Nafees Ahamad

30 Thanks ? (c) Nafees Ahamad


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