## EE G Notes Chapter Instru ctor Cheung Page Routh Array Both asymptotically stability and BIBO stability require c hecing whether Hs NsDs has all its poles on the open left half plane - Description

How to check Solution 1 Factorize Ds Re all stable This is difficult because there are NO CLOSED FORM for polynomials of degree higher than four Numerical methods are 1 Computatinally intensive 2 Solution are not exact Solution 2 Routh Array Rout ID: 27108 Download Pdf

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# EE G Notes Chapter Instru ctor Cheung Page Routh Array Both asymptotically stability and BIBO stability require c hecing whether Hs NsDs has all its poles on the open left half plane

How to check Solution 1 Factorize Ds Re all stable This is difficult because there are NO CLOSED FORM for polynomials of degree higher than four Numerical methods are 1 Computatinally intensive 2 Solution are not exact Solution 2 Routh Array Rout

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## EE G Notes Chapter Instru ctor Cheung Page Routh Array Both asymptotically stability and BIBO stability require c hecing whether Hs NsDs has all its poles on the open left half plane

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## Presentation on theme: "EE G Notes Chapter Instru ctor Cheung Page Routh Array Both asymptotically stability and BIBO stability require c hecing whether Hs NsDs has all its poles on the open left half plane"â€” Presentation transcript:

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EE 422G Notes: Chapter 6 Instru ctor: Cheung Page 6-26 6-5 Routh Array Both asymptotically stability and BIBO stability require c hecing whether H(s) = N(s)/D(s) has all its poles on the open left half plane. How to check? Solution 1: Factorize D(s): Re( )...( )( ... all stable This is difficult because there are NO CLOSED FORM for polynomials of degree higher than four. Numerical methods are: 1. Computatinally intensive 2. Solution are not exact. Solution 2: Routh Array (Routh-Hurwitz criterion). This i s EASY! Find how many poles in the right half of the s-plane? (1) Basic

Method ... The Routh Hurwitz Criterion: Number of sign changes in the first column of the array = number of poles in the (OPEN) r. h. p. Degree of the leading term PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
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EE 422G Notes: Chapter 6 Instru ctor: Cheung Page 6-27 Example 6-8 56 41 14 sign: Changed once =>one pole in the r.h.p verification: )8 )(7 )(1 56 41 14 Example 6-9 10 15 10 15 )1 15 15 10 10 Sign: changed twice => two poles in r.h.p. PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
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EE 422G Notes: Chapter 6

Instru ctor: Cheung Page 6-28 FAQ about Routh Array 1. Why do number of sign changes in Routh array has anything to do with the number of r.h.p. poles? Let say )( )( Expand it out you get, ... In this case, the Routh array looks like this: ... ... ... If all the poles are on the left half plane, must be positive and there will not be a sign change between the first and secon d row. If there is a sign change, we must have at least ONE pole on the open right half plane. 2. What about the rest of the rows? A detail explanation of the Routh-Hurwitz criterion i s beyond the scope of this course. An

elementary proof of this criterion can be found in the paper titled “Elementary proof of the Routh-Hurwitz test” by G. Meinsm a in Systems & control Letters 25 (1995) p. 237-242. 3. What about poles on the imaginary axis? It is in fact possible to also keep track of the number of pur e-imaginary poles from the Routh array as well. However, we will not co nsider such procedure here. 4. What happen when there are zeros in the first column? Excellent question! There are two cases Case 1: Only the first element of the row is 0 but the r est of the row is neither entirely zero nor empty. Procedure:

1. replace 0 by a small 2. Find # sign changes in the first column for either >0 and <0. Both cases should give you the same result. PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
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EE 422G Notes: Chapter 6 Instru ctor: Cheung Page 6-29 Example 6-10 Case 2: the whole row is zero or zero occurs in the very last row. This happens, for example, when all the odd (even) powers of the polynomial are missing: 4 2 ( ) ( 2 )( 2 )( 3 )( 3 ) 13 36 D s s j s j s j s j s s = + - + - = + + 1 13 36 0 0 0 Solution: 1. Construct an auxiliary polynomial based on the row

before the all-zero row 2. Replace the all-zero row by the DERIVATIVE of the a uxiliary polynomial. Why? As motivated by the previous example, a zero r ow implies that a polynomial D(s) has only even or odd power. It turns out in th is case, D(s) and D(s)+D’(s) have exactly the same numbers of r.h.p. poles (proo f beyond scope). As the goal is just to find the # of r.h.p. poles, we can use D’(s ) as a surrogate to continue the procedure. PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
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EE 422G Notes: Chapter 6 Instru ctor: Cheung Page 6-30 Example 6-11

Two sign change two poles on the open right half plane PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
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EE 422G Notes: Chapter 6 Instru ctor: Cheung Page 6-31 Sample Application of Routh Array: Range of system param eters. Example: 1( 3/) 8( 3/) 8( 1( 3/) 8( 1( Stable system to ensure system stable! PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com