Engineering Discover the Stability of the System through a Visualization | Learn More

AI Thread Summary
The discussion centers on determining the stability of a polynomial using the Routh-Hurwitz criterion. The critical value for k is identified as -1/2, indicating that for stability, k must be greater than -1/2 to avoid sign changes in the Routh array. However, it is noted that the correct answer for stability is k > -2. The polynomial's coefficients are analyzed, revealing that for the roots to have negative real parts, the conditions require k to be greater than -1. Overall, the conversation emphasizes the importance of applying the Routh-Hurwitz criterion correctly to assess system stability.
engnrshyckh
Messages
51
Reaction score
2
Homework Statement
For what value of k does the polynomial shown in picture have roots with negative real parts
Relevant Equations
Rooth array criteria
See the picture
I am stuck at 12(1+2k)=0
So k=-1/2 for stability k must have value greater the - 1/2 which means there will no sign changes in rooth array and equation represents a stable system
 

Attachments

  • IMG_20200409_131514.jpg
    IMG_20200409_131514.jpg
    46.4 KB · Views: 193
  • IMG_20200409_131402.jpg
    IMG_20200409_131402.jpg
    34.1 KB · Views: 190
Physics news on Phys.org
engnrshyckh said:
Homework Statement:: For what value of k does the polynomial shown in picture have roots with negative real parts
Relevant Equations:: Rooth array criteria

See the picture
I am stuck at 12(1+2k)=0
So k=-1/2 for stability k must have value greater the - 1/2 which means there will no sign changes in rooth array and equation represents a stable system
But correct ans is k>-2
 
engnrshyckh said:
Homework Statement:: For what value of k does the polynomial shown in picture have roots with negative real parts
Relevant Equations:: Rooth array criteria

See the picture
I am stuck at 12(1+2k)=0
So k=-1/2 for stability k must have value greater the - 1/2 which means there will no sign changes in rooth array and equation represents a stable system
First off, the name is Routh. I've never heard of this algorithm, but I found something about Routh-Hurwitz stability at this wiki page - https://en.wikipedia.org/wiki/Routh–Hurwitz_stability_criterion

In the section titled Routh–Hurwitz criterion for second and third order polynomials, it says,
The third-order polynomial
##P ( s ) = s^3 + a_2 s^2 + a_1 s + a_0## has all roots in the open left half plane if and only if
##a_2 , a_0## are positive and ##a_2 a_1 > a_0## .
In your problem, ##a_2 = 4 + 4k, a_1 = 6, a_0 = 12##
The solution to both inequalities is k > -1, so it seems to me that the closest of the given answers is k > -2.
 
Back
Top