The Hurwitz Stability Criterion

The Hurwitz Stability Criterionbutton2

By: Song Yoong Siang (PhD Student)

 

 

2e nov15 degree polynomial, 2a nov15 of the form shown by Equation (1) is stable if all roots of this polynomial lie in the left half of the complex plane. In this situation, any solution to the linear, homogeneous differential equation will converge to zero. However, it is difficult to determine all the roots if 2b nov15 is large. The Hurwitz test provides a necessary and sufficient condition for stability without solving the Equation (2).

2c nov15(1)

    2d nov15(2)

The Hurwitz matrix for 2e nov15 degree polynomial is describe in Equation (3). 2a nov15 is stable if and only if the leading principal minors of 2f nov15 are all positive.The leading principal minors, 2g nov15 are the determinants of the upper left (1×1), (2×2), . . ., (n×n) submatrices of 2f nov15. Example of leading principal minors are shown in Equation (4) – (6).

2h nov15(3)

 

2i nov15(4)

 

2j nov15(5)

 

2k nov15

 

Reference(s):

  1. http://www.systems.caltech.edu/EE/Courses/EE32b/handouts/Hurwitz.pdf
  2. http://www.cems.uvm.edu/~tlakoba/08_fall/EE_295/Routh_Hurwitz_Criterion.pdf
  3. http://planetmath.org/sites/default/files/texpdf/35395.pdf