Phase margin

If i consider the open loop gain of an opamp with negative feedback
composed of 2 poles (with negative real part) and one zero (with
positive real part) and if i want a phase margin of 45°, from its
definition
360°-tan^-1[-A(jwo)F(jwo)]=45°
i have
+180°-tan^-1[A(jwo)F(jwo)]=+180°-[-tan^-1(w/p1)-tan^-1(w/p2) -tan^-1(w/
z)]=45°
where +180 is obtained from 360°-tan^-1[negative constant]=360°-
tan^-1[-1]=360°-180°
Instead i've found as solution
+-180°-tan^-1(w/p1)-tan^-1(w/p2) -tan^-1(w/z) =45°
what's wrong?
thanks

 

[Eeyore]

silusilusilu@gmail.com wrote:

If i consider the open loop gain of an opamp with negative feedback
composed of 2 poles (with negative real part) and one zero

Are you trying to do one of those cute pole-zero compensations with the
extra pole inside the op-amp itself.

Fun isn't it ? I just hack it with simulations. I know why the lecturers at
Uni said they didn't fancy getting into it !

Graham