Stability part 2

Stability part 2 - Special Cases Special Cases Auxiliary...

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pecial Cases Special Cases • Auxiliary polynomial with reciprocal roots: Original Polynomial: 1 11 0 ... 0 nn n sa s a s a  Replacing s by 1/d: 1 0 1 ... 0 n aa a dd d    1( 1 ) 1 1 n a a  0 1 ... n n n d d  1 0 1 1. . . 0 n ad a d a d d   
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pecial Cases Special Cases s it could be seen, the Auxiliary polynomial is a As it could be seen, the Auxiliary polynomial is a polynomial with the coefficients written in inverse order. • Recalling the previous example: 0 5432 10 () 2365 3 Ts sssss  () 3 5 6 3 2 1 Ds s s s s s  (c) 2010 Farrokh Sharifi
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pecial Cases Special Cases lgorithm: Algorithm: 2- Entire row is zero: There is an even polynomial that is a factor of the original polynomial. gp y - In this case an auxiliary polynomial should be used. - Example: Number of RHP poles for the following losed-l op transfer function? cosed oop s e u c o ? 543 2 10 () 64 28 5 6 Ts ss  (c) 2010 Farrokh Sharifi 764 sss 
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pecial Cases Special Cases - tep 1: Forming the table: 10 ) s Step 1: Forming the table: 7 543 2 () 764 28 5 6 Ts sss ss  tep 2: Return to the row immediately above the row of - Step 2: Return to the row immediately above the row of zeros and form an auxiliary polynomial: 42 ) 6 8 s s s - Step 3: Differentiate the polynomial with respect to s: Ps  ) P s (c) 2010 Farrokh Sharifi 3 41 2 0 dP s ds
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pecial Cases Special Cases - Step 4: Replace the obtained coefficients with ements of the row of zeros: elements of the row of zeros: 3 () 41 2 0 dP s s s ds  4 - Step 5: The rest of the table is formed following the standard form. (c) 2010 Farrokh Sharifi
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pecial Cases Special Cases All entries in the first column are positive, therefore there are no RHP poles.
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This note was uploaded on 02/20/2011 for the course MEC 709 taught by Professor Sharifi during the Fall '11 term at Ryerson.

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Stability part 2 - Special Cases Special Cases Auxiliary...

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