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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?

enter image description here

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  • $\begingroup$ 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? $\endgroup$ – knzy May 24 at 23:12
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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 $\pi$.

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