So I asked this question previously in the control theory section and got no responses. Ill try it here. Basically I need to find the ranges of k and \$\beta\$ such that the steady state error will be less than 10% for unit step input.

I'm currently at the point where I know that the steady state error is given by:

$$ 9 < \beta k$$

Now I'm using the closed loop characteristic equation:

$$2s^4 +2s^3+ (2 + \beta)s^2 + (\beta - 10 + 2k)s + 1 +\beta k = 0$$

$$ \begin{matrix} s^4 & 2 & (2+\beta) & (1+\beta k) \\ s^3 & 2 & \beta-10+2k \\ s^2 &12-2k & 1+\beta k \\ s^1 & \frac{(12-2k)(\beta-10+2k)-(2)(1+\beta k)}{12-2k} \\ s^0 & 1+\beta k \end{matrix} $$

So at this point I'm not sure how to find the ranges for k and \$\beta\$ and apply it to the steady state error? Can I say that:

$$ 12 - 2k > 0$$ $$ - 2k > -12$$ $$ 2k < 12$$ $$ k < 6$$

I'm just stuck as this point. Any thoughts? Thanks.

  • \$\begingroup\$ FYI inline MathJax is [backslash][dollar sign] Math here! [backslash][dollarsign] Refresher here: meta.electronics.stackexchange.com/questions/5565/… \$\endgroup\$
    – Daniel
    Mar 28, 2016 at 6:11
  • 1
    \$\begingroup\$ Thanks for the tips! I didn't even know this sub existed and I'm very excited I found it. \$\endgroup\$
    – user108698
    Mar 28, 2016 at 6:14

1 Answer 1


First, your Routh table is not correct. Third element in first column is wrong. When you get correct table, you need to find intersection of all inequalities in first column so that all elements are higher than zero including your condition for steady state error.

  • \$\begingroup\$ ok I think I fixed the column. I understand that the first column must contain no sign changes but I'm just confused on the ranges and applying them to the steady state error \$\endgroup\$
    – user108698
    Mar 28, 2016 at 10:16
  • \$\begingroup\$ When you calculate all permissible intersections (for your coefficients), you can use any one combination that satisfy your condition 9<kb, as long as you don't know a real system and what those coefficients represent. \$\endgroup\$
    – Haris778
    Mar 28, 2016 at 10:28
  • \$\begingroup\$ Yeah I'm not to sure I guess Im mostly confused on what exactly the question wants. \$\endgroup\$
    – user108698
    Mar 28, 2016 at 10:41

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