lecture10 - 10.RouthHurwitzCriterion

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
  1 10.  Routh-Hurwitz Criterion Solving the characteristic equation for roots is  DIFFICULT. Question: Can we conclude stability without solving  the characteristic equation? Answer: Yes, use the Routh-Hurwitz Criterion It will tell if the characteristic equation has unstable  poles or not. Furthermore, if it is unstable, it will also tell how many. This involves constructing an array and observe the  sign changes in the first column.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
  2 Routh-Hurwitz Array Let the characteristic equation be         a 1 s n +a 2 s n-1  +a 3 s n-2  +a 4 s n-3  +…+a n s+a n+1  = 0 We form the array: s n : a 1 a 3 a 5 s n-1  : a 2 a 4 a 6 Then we construct the s n-2  row as follows:  s n : a 1 a 3 a 5 s n-1  : a 2 a 4 a 6 s n-2  : b 1 b 2 where ... etc. a a a a a b ; a a a a a b 2 6 2 5 1 2 2 4 2 3 1 1 - = - =
Background image of page 2
  3 Routh-Hurwitz Array (cont.) Then we construct the s
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/09/2009 for the course ENG 91301 taught by Professor Lui during the Spring '08 term at Hong Kong Institute of Vocational Education.

Page1 / 6

lecture10 - 10.RouthHurwitzCriterion

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online