circuit analysiscircuit-designcompensationpole-zeroplot
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.
I got the position of the pole to be \$8.873\$ and the gain to be \$8.873\$.
First,
yields \$z = 0.8873\$. Next plug in \$z\$ into the transfer function above and you get \$-0.1127\$. The gain is then \$\frac{1}{0.1127} = 8.873\$.
Step response is as follows
I also had issues with the overshoot (about \$33 \%\$) but at least it meets the pole-zero relationship. I'm guessing it's an issue with the two poles at \$0\$.
Given that your first equation has the correct signs, \$ a\$ must be negative otherwise it's an unstable pole and \$ s\rightarrow j\omega\$ is not a valid operation, i.e. there is no steady state, therefore a frequency response does not exist.
Hence it would be better (less confusing) to let the complex pole be: $$ H(s)=\dfrac{1}{s+(a-jb)}\:, \:a>0$$
Now, determining the amplitude Bode plot for this isolated complex pole is mathematically simple, but in practice it's meaningless. Rather, you need to consider the frequency response of the complex conjugate pair.
To illustrate the problem, the DC gain of the single pole (obtained by taking \$\small s=0)\$ would be \$\small \dfrac{1}{(a-jb)}\$. A complex gain is not something that is physically realisable.
However, including the complex conjugate pole in the formulation would give a real DC gain of \$\small \dfrac{1}{(a^2+b^2)}\$, which complies with the 2nd order TF: \$\small \dfrac{1}{s+(a-jb)}.\dfrac{1}{s+(a+jb)}=\dfrac{1}{s^2+2as+(a^2+b^2)}\$
Best Answer
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:
\$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}}\$
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:
http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf
http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf