The transfer function is \$F(p)=10/(p^2+2p+1).\$
Using Wolfram Alpha, the Nichols plot shows that the argument \$\to -\pi\$ when \$\omega \to +\infty\$.
However, calculating the argument on paper gives \$-\arctan(2w/(1-w^2))\$ which \$\to 0\$ when \$w \to +\infty\$.
Where is the error, mathematically?
Probably no error with the math, but realize that the chart is not a grid. It looks like the chart below. So if charting by hand make sure the grid is a Log scale of the magnitude or \$20\log_{10}(|G(s)|)\$ on one axis and phase on the other. If the closed loop transfer function is given, then a nichols plot is the open loop version, so it will need to be converted from closed loop to open loop.
Source: http://courses.me.metu.edu.tr/courses/me442/handout.html
Look at the real and imaginary parts:
{re, im} = ComplexExpand@ReIm[10/(p^2 + 2 p + 1) /. p -> I w] // Simplify
$$\left\{-\frac{10 \left(w^2-1\right)}{\left(w^2+1\right)^2},-\frac{20 w}{\left(w^2+1\right)^2}\right\}$$
The real part starts off positive, becomes zero when \$w=1\$, and then becomes negative. The imaginary part starts off at \$0\$ and then becomes negative. Thus the phase starts at the positive real axis, goes through the fourth quadrant, then the first, and ends up on the negative real axis.
Calling ArcTan without cancelling the sign shows this.
Plot[ArcTan[re, im]/Degree, {w, 0, 20}, PlotRange -> All, Frame -> True, FrameLabel -> {"freq.", "phase"}]
If you cancel, you lose the sign information and get the result you are getting. This is not correct.
ArcTan[im/re]
$$ \tan ^{-1}\left(\frac{2 w}{w^2-1}\right)$$
Plot[%/Degree, {w, 0, 20}, PlotRange -> All, Frame -> True, FrameLabel -> {"freq.", "phase"}]
We have $$ F(j\omega)=\frac{10}{-\omega^2+2j\omega+1}=10\frac{(1-\omega^2)-2\omega j} {((1-\omega^2)+2\omega j)((1-\omega^2)-2\omega j)} $$ $$ =10\frac{(1-\omega^2)-2\omega j} {(1-\omega^2)^2+4\omega^2} =10\frac{1-\omega^2-2\omega j} {1-2\omega^2+\omega^4+4\omega^2} =10\frac{1-\omega^2-2\omega j} {1+2\omega^2+\omega^4} $$ $$ =\frac{10}{(1+\omega^2)^2}(1-\omega^2-2\omega j). $$ Since multiplication by \$\frac{10}{(1+\omega^2)^2}\$ does not change the value of an argument, consider \$\arg(1-\omega^2-2\omega j)=\alpha \$. The trigonometric form of \$1-\omega^2-2\omega j\$ is $$ |1-\omega^2-2\omega j| \left( \underbrace{\frac{1-\omega^2}{|1-\omega^2-2\omega j|}}_{\cos\alpha}+ \underbrace{\frac{-2\omega}{|1-\omega^2-2\omega j|}}_{\sin\alpha}j \right). $$ This implies that $$ \lim_{\omega\to+\infty}\cos \alpha= \lim_{\omega\to+\infty} \frac{1-\omega^2}{|1-\omega^2-2\omega j|}= -1, $$ $$ \lim_{\omega\to+\infty}\sin \alpha= \lim_{\omega\to+\infty} \frac{-2\omega}{|1-\omega^2-2\omega j|}= 0. $$ Hence, calculating the argument on paper does not give \$0\$.
