I have a transfer function like this:


I can separate the denominator into an expression of the type: \$(1+s/a)(1+s/b)\$ and then plot these two curves.

However, my textbook doesn't tell me how to plot the numerator: \$s^2+\frac{1}{LC}\$

It teaches how to draw curves for quadratic poles but they have a diferente form.

How do I manipulate it to make it look like one of these?

enter image description here

All these types of poles and zeros have \$j\omega\$ instead of \$\omega\$ which is what I would get if I separate in two terms: \$(1+\omega\sqrt{LC})(1-\omega\sqrt{LC})\$

Appreciate your attention.

  • 2
    \$\begingroup\$ Factor \$\frac{1}{LC}\$ in the numerator and the denominator. In the denominator, you should rework your expression so that it fits a second-order canonical such as \$D(s)=1 + \frac{s}{Q\omega_0} + (\frac{s}{\omega_0})^2\$ and identify \$Q\$ and \$\omega_0\$. In the numerator, you will have \$N(s)=1+(\frac{s}{\omega_0})^2\$ which also is the correct canonical form. \$\endgroup\$ Commented Mar 9, 2018 at 3:56

1 Answer 1


Rewritten in standard form, \$s^2 +\frac{1}{LC}\$ is: -

$$s^2 +\omega_n^2$$

Where \$\omega_n\$ is the natural resonant frequency. So equating to zero we find that: -

$$s^2 = -\omega_n^2$$


$$s = \pm j\omega_n$$

Because the square root of -1 is either +j or -j

You have a notch filter with zeroes at \$\pm j\omega_n\$


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