Routh stability test question.

In summary, to find values of k for which the system is stable, the conditions are that (1+k) must be greater than 0 and 2k must be greater than 0. The Routh array should have a coefficient of s^2 as (3k-1)/2k and a coefficient of s as (3k-1)(1+k)-8k^2. The range for k for s^2 is (1/3, +infinity) and for s^1 is (-infinity, 1). However, after further calculations, it is determined that the system is actually unstable for all values of k.
  • #1
angel23
21
0
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??
 
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  • #2
the idea is right but your numbers are wrong . it will be k > 1 .
 
  • #3
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
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
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
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 please tell me if the coeff of s^1 is(3k+1)(k-1) then how to get the range?
 
  • #7
i think you reached this for S^1 :

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

you can't 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
ha, my control systems prof basically said routh hurwitz was BS and skipped it. Can't help you here.
 

Related to Routh stability test question.

1. What is the Routh stability test?

The Routh stability test is a mathematical method used to determine the stability of a system. It is commonly used in control engineering to analyze the stability of a control system.

2. How does the Routh stability test work?

The Routh stability test works by creating a table of coefficients from the characteristic equation of a system. The table is then used to determine the number of roots with positive and negative real parts, which can indicate the stability of the system.

3. What are the conditions for stability in the Routh stability test?

In order for a system to be stable according to the Routh stability test, all the coefficients in the first column of the table must be positive. If there are any negative or zero coefficients, the system is unstable.

4. What are the advantages of using the Routh stability test?

The Routh stability test is a quick and efficient method for determining the stability of a system. It also provides a clear graphical representation of the system's stability, making it easy to interpret and analyze.

5. Are there any limitations to the Routh stability test?

Yes, the Routh stability test can only be used for systems with real coefficients. It also cannot be used for systems with repeated roots or systems with time delays.

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