I'm trying to sketch the nyquist plot of $$\frac{j\omega-1}{j\omega+1}$$ but can't seem to calculate the argument correctly. I think it should be $$\arctan(-\omega) - \arctan(\omega) = -2\arctan(\omega)$$ but this doesn't give the correct nyquist plot behaviour for $\omega \to 0$ and $\omega \to \infty$ - surely $-2\arctan(\omega)$ implies that $\lim_{x\to 0} = 0^\circ$ and $\lim_{x\to \infty} = -180^\circ$?
Wolfram Alpha disagrees but I can't see where I'm going wrong. Am I making a glaring error somewhere? Any help would be greatly appreciated.
Thanks very much
$$ A(\omega) = \frac{j\omega - 1}{j\omega + 1} = \frac{(j\omega - 1)(j\omega - 1)}{(j\omega + 1)(j\omega - 1)} = -\frac{(j\omega-1)^2}{\omega^2 + 1} = -1 - j2\frac{\omega}{\omega^2 + 1} $$ so $$ |A(\omega)| = \sqrt{\frac{(1 + \omega^2)^2 + 4\omega^2}{(\omega^2 + 1)^2}} = \sqrt{1 + 4\frac{\omega^2}{(\omega^2 + 1)^2}} $$ and $$ \arg A(\omega) = \arg (-(j\omega - 1)^2) = \pi + 2 \arg(i\omega - 1) \overset{(2)} = \pi -2\arctan(\omega) \text{.} $$
I found (2) by geometric musings. The angle between $-1 + j\omega$ and the negative real axis is quite obviously $\arctan(\omega)$. Thus, the angle is $\pi - \arctan(\omega)$ when measured from the positive real axis. Doubling and adding $\pi$ yields $3\pi - 2\arctan(\omega)$, and by subtracting $2\pi$ term, we find the equivalent angle in the interval $(0,2\pi)$.
So your result looks mostly correct to me. Or I share your confusion.