problem with arg(z)

Question

I have second order system $$\ F(j\omega)=\frac{1 }{(1-\frac{\omega^2}{\omega_f^2})+j\omega\frac{2.ξ}{\omega_f}}$$ with $\ ξ<1$ and i'm trying to find his phase Margin $\phi$=$\ Arg(F(j\omega))$ for $\omega=2\omega_f$

I know that $\ Arg(F(j\omega))$= $\ Arg(1)-Arg((1-\frac{\omega^2}{\omega_f^2})+j\omega\frac{2.ξ}{\omega_f})$

I consider a= $\ (1-\frac{\omega^2}{\omega_f^2})<0 $ and b= $\frac{2.ξ}{\omega_f^2}>0$ then $$\phi=-(\pi+atan(\frac{2.ξ.\frac{\omega}{\omega_f} }{1-\frac{\omega^2}{\omega_f^2}}))$$

when i replace with the value $\omega=2\omega_f$ and $\ ξ=0,25$ i find: $$\phi=-(\pi+atan(\frac{4.ξ }{1-4}))$$

$$\phi=-180°+18,44°=-161°$$

This is my result but in the book they have found $$\phi=-173°$$ could someone please help me to find my error

Answer 1

Well, we have:

$$\text{F}\left(\omega\cdot\text{j}\right):=\frac{1}{\text{n}_1+\text{n}_2\cdot\text{j}}=\frac{\overline{\text{n}_1+\text{n}_2\cdot\text{j}}}{\left(\text{n}_1+\text{n}_2\cdot\text{j}\right)\cdot\overline{\text{n}_1+\text{n}_2\cdot\text{j}}}=$$ $$\frac{\text{n}_1-\text{n}_2\cdot\text{j}}{\text{n}_1^2+\text{n}_2^2}=\frac{\text{n}_1}{\text{n}_1^2+\text{n}_2^2}-\frac{\text{n}_2}{\text{n}_1^2+\text{n}_2^2}\cdot\text{j}\tag1$$

Now, for $\text{n}_1$ we have:

$$\text{n}_1:=1-\frac{\omega^2}{\omega_\text{f}^2}=1-\frac{\left(2\cdot\omega_\text{f}\right)^2}{\omega_\text{f}^2}=1-\frac{4\omega_\text{f}^2}{\omega_\text{f}^2}=1-4=-3\tag2$$

And for $\text{n}_2$:

$$\text{n}_2:=\omega\cdot\frac{2\cdotξ}{\omega_\text{f}}=2\cdot\omega_\text{f}\cdot\frac{2\cdotξ}{\omega_\text{f}}=4\cdotξ\tag3$$

Now, when $ξ\in\mathbb{R}$ and $0\leξ<1$, we can write:

$$\arg\left(\text{F}\left(\omega\cdot\text{j}\right)\right)=\arg\left(\frac{-3}{\left(-3\right)^2+\left(4\cdotξ\right)^2}-\frac{4\cdotξ}{\left(-3\right)^2+\left(4\cdotξ\right)^2}\cdot\text{j}\right)=$$ $$\arg\left(-\frac{3}{9+16\cdotξ^2}-\frac{4\cdotξ}{9+16\cdotξ^2}\cdot\text{j}\right)=\pi+\arctan\left(\frac{\left(\frac{4\cdotξ}{9+16\cdotξ^2}\right)}{\left(\frac{3}{9+16\cdotξ^2}\right)}\right)=$$ $$\pi+\arctan\left(\frac{4\cdotξ}{3}\right)\tag4$$

So, when $ξ=0.25$, we get:

$$\arg\left(\text{F}\left(\omega\cdot\text{j}\right)\right)=\pi+\arctan\left(\frac{4\cdot0.25}{3}\right)=\pi+\arctan\left(\frac{1}{3}\right)\approx198.4349488^\circ=$$ $$198.4349488^\circ-360^\circ\approx-161.5650512^\circ\tag5$$

So, you're right!