$\huge \begin{align} w_{t} &= x_{t}+R^{L}y_{t-L} \\ y_{t} &= \frac{1}{2}\left( w_{t}+w_{t-1} \right) \end{align} $ ![[Plucked String Filter .excalidraw.svg]] %%[[Plucked String Filter .excalidraw.md|🖋 Edit in Excalidraw]]%% Where $y_{t}$ is the [[Low Pass Filter|Low Pass Delay]]. $ \huge \begin{align} \mathcal{H}(z) &= \frac{1}{2}\left( 1+z^{-1} \right) \\ H(\omega) &= \frac{1}{2}\left( 1+e^{-i\omega} \right) \\ &= \frac{1}{2}e^{-\frac{i\omega}{2}}\left( e^{\frac{i\omega}{2}}+e^{- \frac{i\omega}{2}} \right) \\ &= \frac{1}{2}e^{-i \frac{\omega}{2}} \cos \frac{\omega}{2} \\ &= e^{{-i \frac{\omega}{2}}} \cos \frac{\omega}{2} \\ \left| H(\omega) \right| &= \cos \frac{\omega}{2} \\ \theta \left( \omega \right) &= -\frac{1}{2}\omega \end{align} $ $\huge \begin{align} e^{i \omega t} &\mapsto_{L.P} H(\omega)e^{i \omega t} \\ &=e^{-i \frac{\omega}{2}} e^{i \omega t} \cos \frac{\omega}{2} \\ &= e^{i\omega \left( t-\frac{1}{2} \right) } \cos \frac{\omega}{2} \end{align} $ Therefore there is a [[Frequency Response Function|phase delay]] of $\frac{1}{2}$ [[Sampling|samples]] for all [[Frequency|frequencies]] $\omega$. This gives "coherent phase" [[Delay Operator|delay]]. Recall that if $\vec x$ is the [[Kronecker Delta]] $\vec\delta$ impulse, then $\vec y = \vec h$ [[Impulse Response|impulse response]] has [[Fundamental Frequency]] $f= \frac{f_{s}}{L}$, due to the [[Inverse Comb Filters|comb filter]] delay of $L$ samples. ## [[Resonant Frequency|Resonance]] $\huge \begin{align} W &= X + R^{L}z^{-L}Y \\ zy &= \frac{1}{2}\left( 1+z^{-1} \right) W \\ \mathcal{H}(z) &= \frac{Y}{X} \\ \end{align} $ $\huge \begin{align} Y &= \frac{1}{2}\left( 1+z^{-1} \right) \left( X+R^{L} z^{-{L}}Y\right) \\ &=\frac{1}{2}\left( 1+z^{-1} \right) X + \frac{1}{2}\left( 1+z^{-1} \right) R^{L}z^{-L}Y \\ \end{align} $ $\huge \begin{align} Y- \frac{1}{2}\left( 1+z^{-1} \right) R^{L}z^{-L}Y &= \frac{1}{2}\left( 1+z^{-1} \right) X \\ Y &= \frac{ \frac{1}{2}\left( 1+z^{-1} \right) }{1- \frac{1}{2}\left( 1+z^{-1} \right)R^{L} z^{-L} } \end{align} $ $\huge \begin{align} \mathcal{H}(z) &= \frac{z^{L+1}+z^{L}}{2z^{L+1}-R^{L}z-R^{L}} \end{align} $ ## All-Pass Filter All-Pass [[Signal Filter|Filter]] with any delay $\delta$ and $|H(\omega)|=1$. This will give: $\huge \begin{align} f &= \frac{f_{s}}{L+\frac{1}{2}+\delta} \\ \delta &: 0\leq \delta < 1 \end{align} $ ![[Plucked String Filter _0.excalidraw.svg]] %%[[Plucked String Filter _0.excalidraw.md|🖋 Edit in Excalidraw]]%% $\huge \begin{align} \mathcal{H}(z) &= \frac{z-\frac{1}{p}}{z-p}C_{0} \\ H(\omega) &= \frac{e^{i\omega}- \frac{1}{p}}{e^{i\omega}-p}C_{0} \\ \left| H(\omega) \right| &= \frac{C}{B} \end{align} $ $\huge \begin{align} B^{2} &= p^{2} + 1^{2} - 2p_{1} \cos \theta \\ C^{2} &= \frac{1}{p^{2}} + 1^{2} - 2 \frac{1}{p} 1 \cos \theta \\ p^{2} C^{2} &= 1 + p^{2} - 2 p \cos \theta \\ \frac{C^{2}}{B^{2}} &= \frac{1}{p^{2}} \\ \left| \frac{C}{B} \right| &= \frac{1}{p} \end{align} $ $\huge \begin{align} \mathcal{H}(z) &= K \frac{z+ \frac{1}{a}}{z+a} \\ &= K \frac{1+ \frac{1}{a}z^{-1}}{1+az^{-1}} \end{align} $ Has a pole at $z=-a$ and a zero at $z=- \frac{1}{a}$. Normalize: $\mathcal{H}(1)=1$, $H(0)=1$. $\huge \begin{align} H(\omega) &= \mathcal{H}\left( e^{i\omega} \right) \\ \mathcal{H}(1) &= K \frac{a+1}{1+a} \implies K=1 \end{align} $ $\huge \begin{align} \mathcal{H}(z) &= \frac{a+z^{-1}}{1+az^{-1}} \end{align} $ Filter Equation: $\huge \begin{align} Y &= \frac{a+z^{-1}}{1+az^{-1}}X \\ Y + az^{-1}Y &= aX + z^{-1}X \\ y_{t} &= ax_{t} + x_{t-1} - ay_{t-1} \end{align} $ ![[Plucked String Filter _1.excalidraw.svg]] %%[[Plucked String Filter _1.excalidraw.md|🖋 Edit in Excalidraw]]%% $\huge \begin{align} H(\omega) &= \frac{a+ e^{-i\omega}}{1+ae^{-i\omega}} \\ &= \frac{ae^{i \frac{\omega}{2}}+e^{-i \frac{\omega}{2}}}{e^{i \frac{\omega}{2}}+ae^{-i \frac{\omega}{2} }}\\ &= \frac{z_{1}}{z_{2}} \end{align} $ $\huge \begin{align} z_{1}^{*} &= \left( ae^{i \frac{\omega}{2}}+e^{-i \frac{\omega}{2}} \right) ^{*} \\ &= \left( ae^{i \frac{\omega}{2}} \right)^{*} + \left( e^{-i \frac{\omega}{2}} \right) ^{*} \\ &= e^{i \frac{\omega}{2}}+ ae^{-i \frac{\omega}{2}} \\ &= z_{2} \end{align} $ $\huge \begin{align} \frac{z_{1}}{z_{2}} = \frac{R e^{ i\theta }}{R e^{ -i \theta }} =e^{i{2} \theta} \end{align} $ Where $\theta$ is the [[Phase]] of $\left( ae^{i\frac{\omega}{2}} +e^{-i \frac{\omega}{2}} \right)$. $\huge \begin{align} z_{1} &= a\left( \cos \frac{\omega}{2} + i \sin \frac{\omega}{2} \right) + \cos -\frac{\omega}{2} + i\sin -\frac{\omega}{2} \\ &= a\left( \cos \frac{\omega}{2}+i\sin \frac{\omega}{2} \right) + \cos \frac{\omega}{2} - i\sin \frac{\omega}{2} \\ &=\left( a+1 \right) \cos \frac{\omega}{2} + i(a-1)\sin \frac{\omega}{2} \\ \arg\left( z_{1} \right) &= \arctan \left( \frac{(a-1)\sin \frac{\omega}{2}}{(a+1) \cos \frac{\omega}{2}} \right) \\ &= \arctan \left( \frac{a-1}{a+1}\tan \left( \frac{\omega}{2} \right) \right) \end{align} $ So, $\huge \begin{align} H(\omega) &= 1 e^{i \theta(\omega)} \\ &= e^{i 2\arctan \left( \frac{a-1}{a+1}\tan \frac{\omega}{2} \right) } \end{align} $ For $\theta(\omega)$, there is an approximation that for small values of $x$, $\huge \begin{align} \arctan(x) \approx x \\ \tan(x) \approx x \end{align} $ $\huge \begin{align} \theta(\omega) &\approx \frac{a-1}{a+1} \frac{\omega}{2} \end{align} $