0
\$\begingroup\$

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?

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.

\$\endgroup\$
1
  • 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

0
\$\begingroup\$

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\$

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.