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Routh stability test question.

  1. Dec 20, 2006 #1
    to find values of k for which the system is stable.


    first (1+k)must be >0 and 2K must be >0 then i construct routh array
    to get 3k-1/2k as a coefficent of s^2 and
    (3k-1)/2k *(1+k) - 4k as a coefficent of s .
    then k must be>1/3 and K>2.15 and K>-0.154 then k must be >2.15 ????
    is this right or there is something wrong??
  2. jcsd
  3. Dec 21, 2006 #2
    the idea is right but your numbers are wrong . it will be k > 1 .
  4. Dec 24, 2006 #3


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    For the line [tex]s^1[/tex] I got
    [tex]-\frac{1}{3} < k < 1[/tex]
    And for the line [tex]s^2[/tex]
    [tex]\frac{1}{3} < k [/tex]
    So, you should have
    [tex]\frac{1}{3} < k < 1[/tex]
  5. Dec 25, 2006 #4
    how did you get that for s^1 k<1
    the coeff is (3k+1)(k-1)(2k) so K must be >1 not <1 !!!!!!
    i just want to know how did you get this as i substituted by a value greater than 1 and found system to be unstable.is there anything wrong with my rules or numbers???
  6. Dec 25, 2006 #5


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    I did the calculations wrong. The coeff for s is [tex](3k-1)(1+k) - 8k^2 = -5k^2 + 2k -1[/tex], whose roots are complex. Since the higher power of k has a negative coeff, the parabola has the concavity down. For all values of k the polynomial has negative value.
    Your system is unstable for every k.
  7. Dec 25, 2006 #6
    i am really too confused everytime i solve this problem i get different solution !!!!!!!!!!!!! i solved now and got the same result as yours.

    but plz tell me if the coeff of s^1 is(3k+1)(k-1) then how to get the range????
  8. Dec 26, 2006 #7
    i think you reached this for S^1 :

    [ (3k-1)(1+k)-8k^2 ] / 3k-1 > 0

    you cant multiply by 3k-1 , so :

    (1+k) - (8k^2 / 3k-1) > 0

    simplify to get : 5k^2 - 2K +1 < 0

    and since your first condition from S^3 is K>0 , then :

    system is unstable for all values of k ..
  9. Dec 28, 2006 #8
    ha, my control systems prof basically said routh hurwitz was BS and skipped it. Can't help ya here.
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