I have the following Plant function: $$ P(s) = \frac{50(s-1)}{(s-5)(s+5)} $$

And a controller $ C(s) = K_p + sK_d$

I want to Check if I can stabilize the closed loop system with this controller.

My try:

so what I did was to look at the transfer function:

$$ S(s) = \frac{1}{1 + P(s)\cdot C(s)} $$

Calling for now $ P(s) := \frac{n(s)}{d(s)} $

I want to look at the denominator of that transfer function:

$$ d(s) + C(s)\cdot n(s) $$

If this polynomial is stable, then our closed loop system is stable (?)

Then I did some algebra and gotten the following polynomial:

$$ s^2(1+ 50Kd) + 50s(Kp - Kd) - 25 - 50Kp $$

Then, With a stability criterion I got the following equations:

$$ 1+ 50Kd > 0 $$ $$ Kp - kd > 0$$ $$ -25 - 50Kp > 0$$

And I couldn't find any $Kp$ values that satisfy those equations.

I don't know is this way is wrong, and I don't know if that is the way of going about it. I would greatly appreciate help on the subject, Thank you!

  • 1
    $\begingroup$ It is very likely that a PD controller might not be able to stabilize this system. Namely, rules of thumb are that your bandwidth should be below the RHP zeros and your bandwidth should be above the RHP poles. But those contradict each other due to the locations of the RHP pole and zero of your system. $\endgroup$ Commented May 11, 2022 at 9:52
  • 1
    $\begingroup$ What you did is completely correct and you just proved the fact that one cannot stabilize that system using a PD controller. So, you will have to consider another controller. $\endgroup$
    – KBS
    Commented May 12, 2022 at 13:42

1 Answer 1


You made one incorrect assumption. Namely, that the coefficient of the quadratic term also has to be positive. However, if all coefficients are negative one can factor out the minus sign and the resulting characteristic equation can still have all roots in the left half plane.

So the relevant inequalities would become:

\begin{align} 1+50 K_d &< 0 \\ K_p - K_d &< 0 \\ -25 - 50 K_p &< 0 \end{align}

for which solutions can be found. But I will leave finding those solutions up to you.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .