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.

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  • 1
    \$\begingroup\$ 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. \$\endgroup\$
    – Andy aka
    Commented Jun 4, 2017 at 9:40

1 Answer 1


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}\$.

Voilà! The complete transfer function is therefore:


It is difficult to do simpler and faster. With some habit, you can visualize in your head the various time constants and instantaneously infer the transfer function. Just as I did. Checkout the seminar I taught at APEC in 2016 and the examples solved in the book:




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