# Equivalent compensator of pid controller?

As we know that pi controller is equivalent to lag compensator and pd controller is equivalent to lead compensator, so what will be equivalent of lead-lag compensator?

• A PID controller? Commented Nov 3, 2019 at 16:27
• A lag compensator usually means something with $H(s) = b / (s + a)$, with nonzero a. Integral control means a pole at $s = 0$ or $z = 1$. Commented Nov 3, 2019 at 17:18

Transfer fucntion lag compensator is: $$G_{lag} = \frac{Ts+1}{\alpha Ts+1}, \qquad \alpha > 1$$ The transfer funciton of PI is: $$G_{PI} = K \frac{T_Is + 1}{T_Is}$$ Lag compensator augments the open loop system magnitude by the constant gain (usually used with additional gain $$\G_{comp} = KG_{lag}\$$) whereas PI controller adds sloped gain increase starting form infinity for frequency 0. $$G_{lag}(s = 0)\to 1 \qquad G_{PI}(s=0) \to \infty$$. Therefore I would not call them equivalents.
In the case of the PD controller the same goes as before, not exactly the equivalent.Even though in PD's case there is a way around it. PD controller is not causal by itself and it is usually used in the form of PDf controller: $$G_{PDf} = K\Big(1+\frac{T_Ds}{T_fs+1}\Big) = K \frac{(T_D+T_f)s + 1}{T_f + 1} = K \frac{nT_fs + 1}{T_f + 1}, \qquad T_f = \frac{T_D}{n}, \quad n\gt 10$$ Therefore it has exaclty the form of the Lead compensator: $$G_{lead} = \frac{Ts+1}{\alpha Ts+1}, \qquad 0<\alpha < 1 \quad \to \quad\alpha = \frac{1}{n}$$
To answer your question, if you are searching to combine the PI's "lag" effect and PDf's lead compensation you should use PIDf controller in serial form: $$G_{s,PIDf} = K\Big(1+\frac{1}{T_Is}\Big)\Big(1+\frac{T_Ds}{T_fs +1}\Big) = G_{PI}G_{PDf}$$ But I would like to stress once more that it is not exactly an equivalent.