106 Gain and phase

so the gain is

$$\mathrm{\Gamma }(\omega )=|T(i\omega )|=\frac{1}{\sqrt{{(1-a{\omega }^2)}^2+b^2{\omega }^2}};$$

while the phase $\varphi (\omega )$ satisfies

$$\tan \varphi =\frac{b\omega }{{\omega }^{2}a-1},\varphi ={\tan }^{-1}\frac{b\omega }{{\omega }^{2}a-1}$$

which has sign depending on the value of $\omega$. Note that

$$\varphi (\omega )=\arg (1-a{\omega }^2-ib\omega )$$

and so $T(0)=1,$ hence $\varphi (0)=0$; while $T(i\omega )$ is in the third quadrant as $\omega \to \mathrm{\infty }$, whereas $T(i\omega )$ is in the second quadrant as $\omega \to -\mathrm{\infty }$; at $1-a{\omega }^2=0$, $T(i\omega )$ is on the imaginary axis so $\varphi (\pm 1/\sqrt{a})=\mp \pi /2$.