Routh-Hurwitz stability criterion

Notes from this video https://www.youtube.com/watch?v=WBCZBOB3LCA and video set 5B https://vimeo.com/632436180/fdad67f3f2﻿
A transfer function 
 G(s) = \frac{X(s)}{Y(s)} G(s)=X(s)Y(s) G(s) = \frac{X(s)}{Y(s)} G(s)=Y(s)X(s)​
 is stable if all roots of  Y(s) Y(s) Y(s) Y(s)
 are in the left half of the complex plane -- that is, the real part of each root is less than 0.
This is because the inverse Laplace transform of 
 \frac{1}{s+a} 1s+a \frac{1}{s+a} s+a1​
 is  e^{-at} e−at e^{-at} e−at
, which blows up to infinity if  a < 0 a<0 a < 0 a<0
. And a product of factors like  \frac{1}{s+1}\frac{1}{s+3}\frac{1}{s-2} 1s+11s+31s−2 \frac{1}{s+1}\frac{1}{s+3}\frac{1}{s-2} s+11​s+31​s−21​
 can always be re-written as a sum of partial fractions  \frac{A}{s+1} + \frac{B}{s+3} + \frac{C}{s-2} As+1+Bs+3+Cs−2 \frac{A}{s+1} + \frac{B}{s+3} + \frac{C}{s-2} s+1A​+s+3B​+s−2C​
, which will blow up to infinity if a single RHP root is present.
We say LHP because the imaginary part of root becomes a sinusoidal factor 
 e^{ibt} eibt e^{ibt} eibt
 that doesn't contribute to stability.
Given an unfactored polynomial 
 Y(s) Y(s) Y(s) Y(s)
, it can be easy or not to determine whether all roots are in the LHP.
If the polynomial has at least one negative coefficient (except when all coefficients are negative, in which case we can factor out the negative and end up with all positive coefficients), we can immediately determine that it has at least one root in the RHP. This is because, if all roots were in the LHP, we can always write the polynomial as a product of factors 
 (s+a_n) (s+an) (s+a_n) (s+an​)
, where  Re(a_n) > 0 Re(an)>0 Re(a_n) > 0 Re(an​)>0
, resulting in a polynomial with all positive coefficients -- so if there is a single non-positive coefficient, there must be at least one positive root.
But the reverse is not necessarily true: just because a polynomial has all positive coefficients does not mean that all roots are negative and that the system is stable. Take as a counterexample 
 (s^2-s+4)(s+2)(s+1) = s^4 + 2s^3 + 3s^2 + 10s + 8 (s2−s+4)(s+2)(s+1)=s4+2s3+3s2+10s+8 (s^2-s+4)(s+2)(s+1) = s^4 + 2s^3 + 3s^2 + 10s + 8 (s2−s+4)(s+2)(s+1)=s4+2s3+3s2+10s+8
, where all coefficients of the unfactored polynomial are positive but the polynomial still has RHP roots  0.5 \pm i\sqrt{3.75} 0.5±i3.75 0.5 \pm i\sqrt{3.75} 0.5±i3.75​
.
If all coefficients are positive, we can use a Routh array to test for stability.
To construct a Routh array, make a table with as many rows as the order of your polynomial.
Start at the top left and put the coefficient of the highest order term. Move down one row and put the coefficient of the next highest order term. Move up one and right one and put the next coefficient, then down one and put the next, then up one and right one and the next, the down one...and so on in the zigzag pattern until you run out. Fill remaining spots in the first two rows with 0.
﻿
Then do some more wizardry with the following pattern to fill in the rest of the table.
﻿
The number of sign changes in the first column tell us the number of RHP roots in the polynomial. For the earlier polynomial we get that there are two RHP roots, just as expected.
﻿
And with that, yay Routh-Hurwitz criterion!
Here is the video set's statement of the same:
﻿
Routh-Hurwitz stability criterion

Comments (loading...)

E79: Engineering Systems
