Bandpass filter

The resonant circuit has an complex electrical impedance $Z = \frac{U_\text{in}}{I} = i \omega L + R + \frac{1}{i \omega C} = R + i \omega L \left(1 - \frac{\omega_0^2}{\omega^2} \right)$ The voltage drop at the resistor is taken as output signal $\begin{aligned} & & U_\text{out} &= RI = \frac{R}{Z} U_\text{in} \\ \Rightarrow & & A(\omega) &= \frac{U_\text{out}}{U_\text{in}} = \frac{1}{ 1 + i \omega \dfrac{L}{R} \left(1 - \dfrac{\omega_0^2}{\omega^2} \right) } \\ \Rightarrow & & |A(\omega)| &= \frac{1}{ \sqrt{1 + \omega^2 \dfrac{L^2}{R^2} \left(1 - \dfrac{\omega_0^2}{\omega^2} \right)^2} } \end{aligned}$ To estimate the bandwidth $\Delta \omega$ , we put $|A(\omega)| = 1/\sqrt{2}$ , so that $\begin{aligned} & & \omega \frac{L}{R} \left(1 - \frac{\omega_0^2}{\omega^2} \right) &= \pm 1 \\ \Rightarrow & & \omega^2 \pm \frac{R}{L} \omega - \omega_0^2 &= 0 \\ \Rightarrow & & \omega_{1,2} &= \pm \frac{R}{2 L} + \sqrt{\frac{R^2}{4 L^2} + \omega_0^2} \\ \Rightarrow & & \Delta \omega &= \omega_1 - \omega_2 = \frac{R}{L} \\ \Rightarrow & & Q &= \frac{\omega_0}{\Delta \omega} = \frac{1}{R} \sqrt{\frac{L}{C}} = 20 \end{aligned}$