# How can I add find the gain from root locus and poles?

I try to find the P-gain from a root locus plot where I know the poles. Assume that we got a reference model:

$$G(s) = \frac{\omega_n^2}{s^2 + 2\zeta \omega_n s + \omega_n^2 }$$

Where $\zeta$ and $\omega_n$ are known.

Let's say I want to find the gain for that reference model, to implement into another controller for another model.

Then I compute the poles (s): $$0 = s^2 + 2\zeta \omega_n s + \omega_n^2$$

If I want to compute the gain, I need to use this formula:

$$K_p = abs\left [\frac{den(p)}{num(p)}\right ]$$

Question:

If I got a single row pole such as $p = a + bi$ this will work. But how would I do if I got multiple poles?

• Can you check the equation for your reference model? It is shown as a first order model, but expressed in a form typically reserved for a second order model. Note the second order model is in the form $G(s)=\frac{\omega_n^2}{s^2 + 2 \zeta \omega_n s + \omega_n^2}$, which is different from your model. – Winter Soldier Sep 12 '18 at 12:37
• @WinterSoldier Done! Have edited now. – Daniel Mårtensson Sep 12 '18 at 12:42
• Also, do you want to calculate the DC gain of the plant, or the Proportional gain, where the plant is used in a (possibly unity) feedback system with a proportional controller? – Winter Soldier Sep 12 '18 at 12:42
• @WinterSoldier I don't know if DC gain would help me. – Daniel Mårtensson Sep 12 '18 at 12:47
• In general, let's assume that the plant is $G(s) = \frac{N(s)}{D(s)}=\frac{(s+z_0)(s+z_1)...(s+z_m)}{(s+p_0)(s+p_1)...(s+p_n)}$, where $m \le n$. If we use this plant with a proportional controller of gain $K$, in a unity feedback system, then the characteristic equation is $1+KG(s)=1+K\frac{N(s)}{D(s)}=0 \Rightarrow K = -\frac{D(s)}{N(s)}$ for $K \ge 0$ since we assume negative feedback. Therefore, for a given value of $s=s_0$, $K=\left. -\frac{(s+p_0)(s+p_1)...(s+p_n)}{(s+z_0)(s+z_1)...(s+z_m)}\right|_{s=s_0}$. – Winter Soldier Sep 12 '18 at 14:20

To provide an answer: Let's consider a SISO system, where the transfer function of the plant is given by $G(s)$, and the feedback gain, is $H(s)$. Let's also assume that only a proportional control with gain $K$ is to be used. Then, the closed loop transfer function is given by: $$\frac{Y(s)}{R(s)}=\frac{KG(s)}{1+KG(s)H(s)}$$ where $Y(s)$ is the plant output, and $R(s)$ is the reference signal input. The characteristic equation is $1+KG(s)H(s)$.

As you know, the Root Locus method plots the roots of the characteristic equation - which are the closed loop poles - as a function of gain $K$, where $K \ge 0$, assuming negative feedback is required. Now for unity feedback, $H(s)=1$, and thus the characteristic equation is given by $1+KG(s)=0$.

Now, let the plant transfer function be given as $$G(s) = \frac{N(s)}{D(s)}=\frac{(s+z_0)(s+z_1)...(s+z_m)}{(s+p_0)(s+p_1)...(s+p_n)}$$ where $m \le n$.

In this case, the characteristic (or closed loop poles) equation is given by: $$\therefore 1+ KG(s)=1+ K\frac{(s+z_0)(s+z_1)...(s+z_m)}{(s+p_0)(s+p_1)...(s+p_n)}$$ $$\Rightarrow K=-\frac{(s+p_0)(s+p_1)...(s+p_n)}{(s+z_0)(s+z_1)...(s+z_m)}$$

Finally, since the characteristic equation has real coefficients, any complex poles will occur as conjugate pairs - which is why the Root locus plot is typically symmetric about the real axis.

Hence, once the desired closed loop pole, $s_0$ is calculated based on the required control system performance criteria (rise time, maximum overshoot, settling time, etc.), the desired value of the (proportional) control gain can be found as:

$$K=\left. -\frac{(s+p_0)(s+p_1)...(s+p_n)}{(s+z_0)(s+z_1)...(s+z_m)}\right|_{s=s_0}$$

Again, since negative feedback is used, we want values of $K \ge 0$.

It should be noted that for a second order realizable plant with proportional gain and unity feedback, the closed loop transfer function will also be of second order, with the characteristic equation in the form $s^2 + 2\zeta \omega_n s + \omega_n^2$.