# How can be Bode diagram's phase value more than 90 degree or less than -90 degree?

It doesn't have any physical meaning, i think, because phasors doesn't have more than 90 degree or less than 90 degree value ? Also negative resistance doesn't exist.

• it is up to you how you measure the transfer function; you may change the ratio from $H=V_2/V_1$ to $H'=V'_2/V_1$ where you have decided to measure $V'_2 = -V_2$ and then $phase[H'] =\pi+phase[H]$ – hyportnex Mar 10 '19 at 17:33

Consider the simple first order complex transfer $$H=\frac{-1}{1+j\omega /{{\omega }_{0}}}$$
I don't think you'll be able to find its phase in $$\left[ -\pi /2,\pi /2 \right]$$ .
But you have to remember that $$\arg (a+jb)=\left\{ \begin{matrix} \text{atan}(b/a)\text{ if }a>0 \\ \text{atan}(b/a)+\pi \text{ if }a<0 \\ \end{matrix} \right.$$