# Can't draw bode curve of this zero

I have a transfer function like this:

$H(s)=\frac{s^2+\frac{1}{LC}}{s^2+s\frac{R}{L}+\frac{1}{LC}}$

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?

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: $(1+\omega\sqrt{LC})(1-\omega\sqrt{LC})$

• 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. Commented Mar 9, 2018 at 3:56

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: -

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

Or

$$s = \pm j\omega_n$$

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

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