# Determine the gain K from a root locus

I'm studying Root Locus method and I still confused. The question below exemplifiques my doubt.

Determine the $$K$$ gain so that the dominant roots have a damping factor equal to $$0.5$$, where

$$G(s) = \frac{K(s^2+5s+9)}{s^2(s+3)}$$

I have successfully draw the root locus from this $$G(s)$$, but I can't figure out how I could calculate the gain $$K$$.

An analytical solution.

The closed-loop system is $$G(s)/(1+G(s))$$ and its poles are those of $$1+G(s)=0$$.

In this case that is $$k \left(s^2+5 s+9\right)+(s+3) s^2=0 \ \ \ \ (1)$$.

For general third-order system with a pair of complex dominant poles, the poles are the roots of $$(\alpha +s) \left(s^2 + 2 \zeta s \omega _n+\omega _n^2\right)=0$$. Here $$\alpha$$ is the real pole, $$\zeta$$ is the damping factor, and $$\omega _n$$ is the natural frequency. In this case $$\zeta=0.5$$ and hence the equation becomes

$$(\alpha +s) \left(s^2 + s \omega _n+\omega _n^2\right)=0 \ \ \ \ (2)$$

If we equate the coefficients of (1) and (2) we will have 3 equations and 3 unknowns of which $$k$$ is one, and hence can be computed.

Expand[s^2*(s + 3) + k*(s^2 + 5*s + 9) - (s^2 + 2*(1/2)*wn *s + wn^2)*(s + a)]
eqns = Thread[CoefficientList[%, s] == 0]
sols = Solve[eqns, {a, k, wn}, Reals]
k /. sols


9 k + 5 k s + 3 s^2 - a s^2 + k s^2 - a s wn - s^2 wn - a wn^2 - s wn^2

{9 k - a wn^2 == 0, 5 k - a wn - wn^2 == 0, 3 - a + k - wn == 0}

{{a -> 3, k -> 0, wn -> 0}, {a -> 9/2, k -> 9/2, wn -> 3}}

{0, 9/2}

$$k=0$$ is an artifact of the calculations. The answer turns out to be $$k=\frac{9}{2}$$.

• Is a quite complex to solve this equations by hand. There are non-linear, right? – Alessandro Melo Apr 25 at 18:55
• Yes, they are nonlinear. 'Quite complex' is a subjective term. Personally, it has been ages since I have solved even a quadratic equation manually. :) – Suba Thomas Apr 26 at 3:06

The closed loop system should have three poles, whose characteristic polynomial can be factored into

$$(s - a)(s^2 + 2\,\zeta\,\omega\,s + \omega^2),$$

where $$\zeta$$ refers to the damping factor. The poles are the roots of that polynomial, which for the second factor can be shown to be $$s=\omega(-\zeta \pm i \sqrt{1 - \zeta^2})$$. It can be noted that for $$|\zeta|\leq 1$$ those poles are each others complex conjugate with their magnitude only a function of $$\omega$$ and their phase only a function of $$\zeta$$.

The given damping factor therefore defines what the phase of complex conjugate poles should be. The corresponding value for $$K$$ can thus be found by looking when the root locus crosses the line with the appropriate angle.

Adding some graphical information to what fibonatic included in his own answer. The root locus of the trasfer function $$G$$ with some random value for $$K$$, let's say $$K=1$$ (it is almost always good to start with the value $$1$$ as starting point since it is like adding no controller to the closed loop and be able to see the behaviour of your system just by closing the loop), is the following where the poles are noted with an x and the zeros with an o:

Next step is to add the lines which correspond to the design requirement of the damping ratio $$ζ = 0.5$$ and the following root locus graph is produced:

Now, your goal is to pick the gain $$K$$ of your compensator which moves the poles of your closed loop system inside the white area which is defined under the two diagonal lines. This area generally defines a regio where the damping ratio of the system belongs the interval $$ζ<0.5$$. So if you want your damping ratio to be exactly $$ζ=0.5$$ you need to place your poles exactly on the diagonal lines where the lines cross the root locus graph (the poles of the closed loop system are noted with a pink o). Perfoming these modifications lead to the following root locus graph of the closed loop (see that the poles are now placed exactly on the two lines):

After the design procedure, it is always useful to obtain the step response of your closed loop tuned system to see if it achieves the required behaviour. So, the step response for this system gives an overshoot of $$\%OS = 25.5\%$$ which is not what the damping ratio of $$0.5$$ explicitly dictates by working out the math of the equation:

$$\%PO = 100\cdot e^{({\frac{-ζ\pi}{\sqrt{1-ζ^2}}})}$$

$$ζ=0.5 \rightarrow \%PO = 16.3\%$$

This happens because you have a third pole which is not that far away of the dominant poles of the system and as a consequence this third pole influences the behaviour of the system (the further away the third pole is compared to the dominant poles the least it influences the behaviour of the system):