A 2-Pole Reson filter is a filter with a [[Transfer Function]] of the form: $\huge \begin{align} c &= Re^{i\theta} \\ \mathcal{H}(z) &= \frac{1}{ \left( 1-cz^{-1} \right) \left( 1-\bar{c}z^{-1} \right) } \\ H(\omega) &= \frac{1}{ \left( 1-ce^{-i\omega} \right) \left( 1-\bar{c}e^{-i\omega } \right) } \end{align} $ Which can be simplified into, $\huge \begin{align} \mathcal{H}(z) &= \frac{1}{ 1-(c+\bar{c})z^{-1} + R^{2}z^{-2} } \\ &= \frac{1}{ 1-2R\cos(\theta)z^{-1}+R^{2}z^{-2} } \end{align} $ $\huge \begin{align} Y\left( 1-2R\cos(\theta)z^{-1}+R^{2}z^{-2} \right) &= X \\ \end{align} $ $\huge \begin{align} y_{t} &= x_{t}+2R\cos(\theta )y_{t-1}-R^{2}y_{t-2} \\ a_{1} &= 2R\cos\theta \\ a_{2} &= R^{2} \\ y_t &= x_{t}+a_{1}y_{t-1}+a_{2}y_{t-2} \end{align} $ We expect a [[Resonant Frequency|Resonance]] in the [[Frequency Response Function|Magnitude Response]] $|H(\omega)|$ where $\omega =\theta$. To find the [[Resonant Frequency]] $\omega_{m}$, where $\max_{\omega_{m}}|H(\omega_{m})|$, we will minimize $\frac{1}{|H(\omega)|^{2}}=f(\omega )$ $\begin{align} f(\omega) &= \frac{1}{ \left| H(\omega) \right| ^{2} } \\ &= \left| e^{i\omega}-R{e}^{i\theta} \right|^{2} \left| e^{i\omega }-R e^{-i\theta} \right| ^{2} \\ &= \left| \left( e^{i\omega}-R e^{i\theta} \right) \left( e^{i\omega}-R e^{i\theta} \right) \right| ^{2} \\ &=\left| e^{2i\omega } -R e^{i\left( \omega +\theta \right) }-R e^{i\left( \omega -\theta \right) }+R^{2} \right|^{2} \\ &= \left| \begin{align} \left( \cos 2\omega + i\sin 2\omega \right) - R\left( \cos \left( \omega+\theta \right) + i\sin \left( \omega+\theta \right) \right) - R\left( \cos \left( \omega-\theta \right) +i\sin \left( \omega-\theta \right) \right) +R^{2} \end{align} \right|^{2} \\ &= \left| \begin{align} &\left| \left( \cos{2}\omega - R\left( \cos \left( \omega+\theta \right) + \cos \left( \omega-\theta \right) \right) +R^{2}jk \right) \right| + i\left( \sin{2}\omega - R\left( \sin \left( \omega+\theta \right) +\sin \left( \omega-\theta \right) \right) \right) \end{align} \right|^{2} \\ &=\left( \cos 2\omega - 2R \cos\omega\cos\theta + R^{2} \right)^{2} +\left( \sin 2\omega - 2R\sin\omega\cos\theta \right) ^{2} \\ &= \left( \begin{align} &\cos ^{2} 2\omega + 4R^{2} \sin ^{2} \omega \cos ^{2}\theta + R^{4} + 2R^{2}\cos 2\omega -4R\cos{2}\omega \cos \omega \cos \theta -R^{3} \cos \omega \cos \theta \\ &+ \sin ^{2} 2 \omega + 4R^{2} \sin ^{2} \omega \cos ^{2} \omega - 4R \sin 2\omega \sin \omega \cos \theta \end{align} \right)\\ &=\left( \begin{align} &1+R^{4}+ 4R^{2} \cos ^{2}\theta \\ - &4R(1-2\sin ^{2} \omega) \cos \omega \cos \theta \\ + &2R^{2} \cos 2\omega \\ - &4R^{3} \cos \omega \cos \theta \\ -& 4R 2\sin \omega \cos \omega \sin \omega \cos \theta \end{align} \right) \\ &= \left( \begin{align} & 1+R^{4}+4R^{2}\cos ^{2}\theta \\ -& 4R \cos \omega \cos \theta \\ +& 2R^{2} \cos 2\omega \\ -& 4R^{3}\cos \omega \cos \theta \end{align} \right) \\ &=1+R^{4}+4R^{2}\cos ^{2}\theta - 4\left( R+R^{3}\right) \cos \omega \cos \theta + 2R^{2} \cos 2\omega \\ &= 1-2R^{2} + R^{4} + 4R^{2} \cos ^{2} \theta -4\left( R+R^{3} \right) \cos \theta\cos \omega + 4R^{2} \cos ^{2} \omega \\ &=1+R^{4}+4R^{2}\cos ^{2}\theta - 4\left( R+R^{3}\right) \cos \omega \cos \theta + 2R^{2} \cos 2\omega \\ &= 1-2R^{2} + R^{4} + 4R^{2} \cos ^{2} \theta -4R\left( 1+R^{2} \right) \cos \theta\cos \omega + 4R^{2} \cos ^{2} \omega \\ \end{align} $ $\large \begin{align} f(\omega ) &= \left( 1-R^{2} \right) sr + 4R^{2} \cos ^{2} \theta -4R\left( 1+R^{2} \right) \cos \theta\cos \omega + 4R^{2} \cos ^{2} \omega \\ \end{align}$ We find critical points by finding $f'$. $ \begin{align} f'(\omega ) &= 0 \\ 0&= 0 - 4 R \left( 1+R^{2} \right) \cos \theta \left( -\sin \omega g \right) + 4R^{2} \cdot 2 \cos \omega \left( -\sin \omega \right) \\ 0 &= 4R\left( 1+R^{2} \right) \cos\theta \sin \omega - 8R^{2}\cos \omega \sin \omega \\ 4R\left( 1+R^{2} \right) \cos \theta \sin \omega &= 8R ^{2} \cos \omega \sin \omega \\ \left( 1+R^{2} \right) \cos \theta &= 2R ^{2} \cos \omega \\ \frac{1+R^{2}}{2R} \cos \theta &=\cos \omega \\ \let \omega _{m} &: f(\omega _{m}) =0 , \omega _{m} \in (0,\pi) \\ \end{align} $ $\theta$ is the pole angle for the [[Transfer Function]] $\mathcal{H}(z)$. $ \begin{align} f''(\omega) &= 4R \left( 1+R^{2} \right) \cos \theta \cos \omega - 8R^{2} \left( \cos 2\omega - 1 \right) \\ f''(\omega_{m}) &= 4R\left( 1+R^{2} \right) \cos \theta + \frac{1+R^{2}}{2R}\cos \theta - 8R^{2}\left( 2\cos ^{2} \theta \left( \frac{1+R^{2}}{2R} \right) ^{2} - 1 \right) \\ &= 2 \left( 1+R^{2} \right) ^{2} \cos ^{2} \theta - 2 \cdot 2 \cos ^{2} \theta \left( 1+R^{2} \right)^{2} + 8R^{2} \\ &=8R^{2} - 2\left( 1+R^{2} \right) ^{2} \cos ^{2} \theta \\ &= 8R^{2} \left( 1- \frac{2\left( 1+R^{2} \right) ^{2}}{8R^{2}} \cos ^{2} \theta \right) \\ &=8R^{2} \left( 1 - \left( \frac{1+R^{2}}{2R} \cos \theta \right) ^{2} \right) \\ &= 8R^{2} \left( 1- \cos \omega_{m} \right) ^{2} \\ &= 8R^{2} \sin ^{2} \omega_{m} \end{align} $ Design steps: 1. Closest bandwidth $B$ and [[Resonant Frequency]] $\omega _m$ 2. Calculate pole radius $R$ with $R\approx 1 - \frac{B}{2}$ 3. $\theta = \arccos \left( \frac{2R}{1+R^{2}} \cos \omega_{m}\right)$ 4. Set $A_{0}=(1-R^{2})\sin \theta$ And you get the final filter: $\huge \begin{align} y_{t} &= A_{0}x_{t} + \left( 2R\cos \theta \right) y_{t-1} - R^{2} y_{t-2} \end{align} $