Why the angle is $-180°$ at $\omega = 0$ for this system

Question

I'm trying to plot the nyquist from the analytical expression of the system but the bode plot generated by matlab yields an angle -180 whereas the analytical expression yields zero when $\omega=0$. The open-loop is $$ G(s) = \frac{s+4}{(s+2)(s-3)} $$ The analytical expression for this system is $$ G(j\omega) = \frac{-(5\omega^2+24)-j(\omega^3+2\omega)}{\omega^4+13\omega^2+36} $$ The angle is defined as follows: $$ \angle G(j\omega) = \tan^{-1}\left(\frac{\omega^3+2\omega}{5\omega^2+24}\right) $$

the standard arctan function is defined as follows:

In my case, $x$ is always positive. Now if $\omega = 0$, the angle is zero. According to the Bode plot generated by Matlab, the angle is -180 shown below

Am I missing something in here?

Answer 1

while $\text{atan2} \frac {x}{y}$ may not be defined when $\omega = 0$

$\lim_\limits{\omega\to 0} \text{atan2} \frac {y}{x}$ exists.

$G(j\omega) = \frac{-(5\omega^2+24)-j(\omega^3+2\omega)}{\omega^4+13\omega^2+36}$

$x = \frac {-(5\omega^2+24)}{\omega^4+13\omega^2+36}$

Which means that $x$ is always negative or $0$.

And where does that put your angle?