# Nyquist plot Phase Question

The above graph is pulled from the book "Automatic Control Systems" by Kuo.

The subject loop transfer function is $$L(s)=\frac{K}{s(1+T_1s)}$$ or $$L(j\omega)=\frac{-jK(1-jT_1\omega)}{\omega(1+T_1^2\omega^2)}$$ or $$L(j\omega)=\frac{K(-j-T_1\omega)}{\omega(1+T_1^2\omega^2)}$$

Hence the phase equation should be $$tan(\theta)=\frac{1}{T_1\omega}$$

Now at frequency infinity, which corresponds to origin in the above plot, the value of the phase function is $$tan(\theta)= \lim_{\omega \to \infty} \frac{1}{T_1\omega}=0$$ Also $$tan(\theta)= \lim_{\omega \to 0} \frac{1}{T_1\omega}=\infty$$ At infinite frequency the phase angle can assume value of either zero degree or 180 degree. In this plot it is taken as 180 degree, I can't understand why, when zero degree is also a perfect candidate.

Also at zero frequency, the phase angle should be 90 degree but here it is taken as -90 degree. I can't understand this phase thing.

$L(j\omega)=\frac{K(-j-T_1\omega)}{\omega(1+T_1^2\omega^2)}$
$\theta=\pi+\tan^{-1}(\frac{1}{T_1\omega})$
so at $w\to0$,$\theta=180+90=270=-90$
and at $w\to\infty$,$\theta=180+0=180$