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This is a basic control theory question, since Control Theory is a part of applied mathematics but also of engineering I was unsure whether to ask this here.

The question says:

Given the transfer function of a system is $G(s)=1/(s^2+3s+2)$, consider the design of a PI closed loop control system with unit feedback using proportional gain $k_p$ and integral gain $k_i$, both of which are positive. Determine the range of gain for which the closed loop system is stable. What I did was this, I went the Routh-Hurwitz way:

My solution

After completing the Routh table, I went ahead and reasoned a little about what conditions need to be met in order to avoid sign changes on the main column, however it seems that these conditions are never met! I get that $k_i$ should be less than zero when the problem clearly specifies it will always be positive. Have I done something wrong? Is my reasoning right? Is the answer "The controller is never stable"?

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    $\begingroup$ I explained basically everything I did on the post, I think with that everyone can get every step I took, after all it's math.... $\endgroup$ – HCalderon Oct 13 '15 at 1:48
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There is a mistake in your expressions. The coefficient of $s$ is $2+k_p$.

The conditions are: $$\frac{1}{3} \left(3 \left(k_p+2\right)-k_i\right)>0$$ $$k_i>0$$ $$k_p>0$$

This simplifies to: $$k_p>0$$ $$0<k_i<3 k_p+6$$

Update:

If the transfer function is $$ \frac{1}{s^2+3 s + c} $$ where c is some positive constant, then the conditions simplify to: $$k_p>0$$ $$0<k_i<3 ( k_p+c)$$

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  • $\begingroup$ Can you be more speific about htat mistake? Because I can't seem to get to that coefficient for s^2 $\endgroup$ – HCalderon Oct 13 '15 at 15:24
  • $\begingroup$ In the first equation on the right you have $s^2+3 s+2$ as in the question, in the equation just below the figure you have $s^2+3 s+1$. $\endgroup$ – Suba Thomas Oct 13 '15 at 15:47
  • $\begingroup$ Indeed, I already corrected the error, that gives me 2+kp as the coefficient for s, not s^2 as you stated in your answer, or have I missed something else? $\endgroup$ – HCalderon Oct 13 '15 at 15:49
  • $\begingroup$ My bad, I meant coefficient for s! $\endgroup$ – Suba Thomas Oct 13 '15 at 15:50
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    $\begingroup$ @Suba I just worked through it independently and got the same result at you. +1 $\endgroup$ – Chris Mueller Oct 13 '15 at 16:53

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