User:DukeEgr93/BPF

\( \begin{align} \mathbb{H}(j\omega)&=\frac{K}{1+jQ\left(\frac{\omega}{\omega_n}-\frac{\omega_n}{\omega}\right)}\\ |\mathbb{H}(j\omega)|^2&=\frac{K^2}{1^2+Q^2\left(\frac{\omega}{\omega_n}-\frac{\omega_n}{\omega}\right)^2} \end{align} \) Maximum magnitude when $$\omega=\omega_n$$ of $$K$$; to get half-power frequency\[ \begin{align} |\mathbb{H}(j\omega_{hp})|^2=\frac{K^2}{1^2+Q^2\left(\frac{\omega_{hp}}{\omega_n}-\frac{\omega_n}{\omega_{hp}}\right)^2}&=\frac{K^2}{2}\\ 1^2+Q^2\left(\frac{\omega_{hp}}{\omega_n}-\frac{\omega_n}{\omega_{hp}}\right)^2 &= 2\\ Q^2\left(\frac{\omega_{hp}}{\omega_n}-\frac{\omega_n}{\omega_{hp}}\right)^2 &= 1\\ Q\left(\frac{\omega_{hp}^2-\omega_n^2}{\omega_{hp}\omega_n}\right)&=\pm 1\\ Q\omega_{hp}^2-Q\omega_n^2&=\pm \omega_{hp}\omega_n\\ Q\omega_{hp}^2\mp \omega_{hp}\omega_n-Q\omega_n^2&=0\\ \omega_{hp}&=\frac{\pm\omega_n\pm\sqrt{\omega_n^2+4Q^2\omega_n^2}}{2Q} \end{align} \] Since $$\sqrt{\omega_n^2+4Q^2\omega_n^2}>\omega_n$$, \( \begin{align} \omega_{hp}&=\frac{\pm\omega_n+\sqrt{\omega_n^2+4Q^2\omega_n^2}}{2Q} \end{align} \)