# Why RHP zero phase is not 180° to 90°

The asymptotic phase behavior of an RHP zero is from 0 degrees to -90°, the mirror of an LHP zero. Graphically, I'm confused about why this is the case and the phase is not from +180° to +90°. See the below image for my reasoning/confusion. Is the angle definition not consistent? • Occurred to me that this might be because if you define the angle in terms of the arctangent, then the range is restricted to -90 degrees to +90 degrees. Is this the case?
– knzy
May 24 at 23:12

You're right, a single (real-valued) zero in the RHP could cause a phase that moves from $$\pi$$ at $$\omega=0$$ to $$\pi/2$$ for $$\omega\to\infty$$. However, if you rewrite the corresponding contribution to the transfer function as $$s_0-s$$ (i.e., you change the sign), then you obtain a phase shift between $$0$$ at $$\omega=0$$ and $$-\pi/2$$ for $$\omega\to\infty$$.
The difference is just a sign inversion, i.e., an addition of $$\pm\pi$$ to the phase. The contribution of a single real-valued RHP zero in the form $$s_0-s$$ does not invert the sign for $$s=0$$, whereas if you write it as $$s-s_0$$ you get a sign inversion, hence the additional phase shift of $$\pi$$.