# 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. Commented Jun 4, 2017 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}}\$$