# Question about the derive of filter roll-off

I don't understand the follow step in my notes:

It is about the filter roll-off.

$H(w)={1\over 1+j(w/w_c)}$

$\Rightarrow$ $|H(w)|= {1\over \sqrt(1+(w/w_c)^2)}$

where $H(w) = {v_o \over v_i}$

$w_c$ = cut-off frequency

• What's your question? Oct 23 '12 at 14:06
• why the first step can jump to the second step? Oct 23 '12 at 14:09
• cross-posted: electronics.stackexchange.com/questions/45313/… Oct 23 '12 at 22:01

For any arbitrary rational complex function $H(\omega) = \frac{N(\omega)}{D(\omega)}$,

$$|H(\omega)| = \frac{|N(\omega)|}{|D(\omega)|}$$

This follows straightforwardly from the properties of the polar (i.e. magnitude/phase) representation of complex numbers. So, for your example:

$$|H(\omega)| = \frac{|1|}{|1 + j \frac{\omega}{\omega_c}|}$$

The magnitude of an arbitrary complex number is:

$$|a + jb| = \sqrt{a^2+b^2}$$

\begin{align} |H(\omega)| &= \frac{1}{\sqrt{1^2 + \left(\frac{\omega}{\omega_c}\right)^2}} \\ &= \frac{1}{\sqrt{1 + \left(\frac{\omega}{\omega_c}\right)^2}} \end{align}