Routh stability test question.

  • Thread starter angel23
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  • #1
21
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to find values of k for which the system is stable.

s^4+2ks^3+2s^2+(1+k)s+2=0

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??
 

Answers and Replies

  • #2
18
0
the idea is right but your numbers are wrong . it will be k > 1 .
 
  • #3
SGT
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]
 
  • #4
21
0
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???
 
  • #5
SGT
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.
 
  • #6
21
0
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????
 
  • #7
18
0
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 ..
 
  • #8
1,309
19
ha, my control systems prof basically said routh hurwitz was BS and skipped it. Can't help ya here.
 

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