# Intuitive to get pole and zero points for lead compensator

The circuit below is a lead compensator. I can easily know exact the pole and zero of the network by deriving the transfer function.
However, I am wondering if there is an intuitive method to get the pole and zero of the network (even approximation). Thanks. • For this network yes, one can learn to recognize intuitely the poles and zeroes but it comes through learning and reinforcement through reuse. But, if suddenly presented a lag compensator for the first time and asked to use intuition then no, it's not intuitive. Also, what might be intuitve for one is not necessarily intuitive for someone else. – Andy aka Jun 4 '17 at 9:40

Yes, there is an intuitive approach and it with the fast analytical techniques (FACTs). Reduce the excitation to 0 V ($$\V_1=0\$$ and look at the resistance driving the capacitor (you temporarily remove the cap and "see" what resistance is offered between its connecting terminals). You "see" $$\R_1||R_2\$$. This is it, you have the time constant of this circuit $$\\tau_1=C_1(R_1||R_2)\$$. The pole for a 1st-order circuit is the inverse of the time constant. Therefore $$\\omega_p=\frac{1}{C_1(R_1||R_2)}\$$. For the zero, what condition in this circuit would prevent the excitation $$\V_1\$$ from producing a response? In other words, what condition creates a null in $$\V_2\$$ despite a signal in $$\V_1\$$? When the parallel association of $$\R_1\$$ and $$\C_1\$$ become a transformed open-circuit. The pole of the impedance $$\R_1||C_1\$$ is our zero: $$\\omega_z=\frac{1}{R_1C_1}\$$. Then, for $$\s=0\$$, we have $$\H_0=\frac{R_2}{R_2+R_1}\$$.
$$\H(s)=\frac{R_2}{R_2+R_1}\frac{1+sR_1C_1}{1+sC_1(R_1||R_2)}=H_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\$$