For some [[Frequency Response Function]] $|H(\omega)|$, the frequency $\omega_{m}$ at the [[Maxima|maxima]] of the function is called the Resonance in the [[Frequency Response Function|Magnitude Response]], or the Resonant Frequency of the system. $\huge \begin{align} \omega_{m}: \max_{\omega_{m}} |H(\omega_{m})| \end{align} $ This resonance frequency can be thought of as the 'natural' frequency of a system. ### Half Point Bandwidth $\huge \begin{align} \hat{H} (\omega) &= \left| \frac{H(\omega)}{H(\omega_{m})} \right| \end{align} $ The bandwidth is the range of frequencies containing the [[Resonant Frequency]] starting at $b_{1}$ where the normalized function $\hat{H}(b_{1})= \frac{1}{\sqrt{ 2 }}$ *before* $\omega_{m}$ and the frequency $b_{2}$ where the normalized function equals $\frac{1}{\sqrt{ 2 }}$ *after* $\omega_{m}$. $\huge \begin{align} b_{1} &: \hat{H}(b_{1} )= \frac{1}{\sqrt{ 2 }}, b_{1} < \omega_{m}\\ b_{2} &: \hat{H}(b_{2} )= \frac{1}{\sqrt{ 2 }}, b_{2} > \omega_{m} \\ B &= b_{2}-b_{1} \end{align} $ ## Constructing a filter Goal: Construct a [[Signal Filter]] with [[Transfer Function]] $\mathcal{H}(z)$ and magnitude response with resonance having center frequency $\omega_{m}$ and bandwidth $B$. $\huge \begin{align} y_{t}&=x_{t}+a_{1}y_{t-1} \\ Y&=X+a_{1}z^{-1}Y \\ Y &= \frac{1}{1-a_{1}z^{-1}}X \\ \mathcal{H}(z) &= \frac{1}{1-a_{1}z^{-1}} \\ &= \frac{z}{z-a_{1}} \end{align} $ This has a single [[Pole]] at $z=a_{1}$ and a single zero at $z=0$. $z=0$ is not in the [[Unit Circle]] however $a_{1}$ gives flexibility. $\huge \begin{align} H(\omega) &= \frac{e^{i\omega}}{e^{i\omega}-a_{1}} \\ |H(\omega)| &= \frac{e^{i\omega}}{|e^{i\omega}-a_{1}|} \\ \end{align} $ If we choose values of $a_{1}$, we can control where the peak of the magnitude response peaks. We use this filter with $a_{1}\in\mathbb{R}, a_{1}\approx 1$, and let $a_{1}=R$. To normalize $|H|$, we use scale $\frac{1}{|H(\omega)|}$, but it is easier to work with $\frac{1}{|H(\omega)|^{2}}$. $\huge \begin{align} \frac{1}{\left| H(\omega) \right| ^{2}} &= \left| e^{i\omega}-R \right| ^{2}\\ \end{align} $ Set $\omega_{m}=\phi$ and solve for the maxima. $\huge \begin{align} \left| e^{i\phi}-R \right| ^{2} &= \left| \cos \phi+i\sin \phi-R \right| ^{2}\\ &=\left( \cos \phi-R \right) ^{2}+\sin ^{2}\phi \\ &= \cos ^{2}\phi+\sin ^{2}+\phi-2R\cos \phi+R^{2} \\ &= 1-2R\cos \phi + R^{2} \end{align} $ At center of $\phi=0$, we get: $\huge \begin{align} \frac{1}{\left| H(0) \right|^{2} } &= 1-2R+R^{2} \\ &= (1-R)^{2} \end{align} $ To get the half power points for the bandwidth we look for frequencies: $\huge \begin{align} 1-2R\cos \phi+R^{2} &= 2(1-R)^{2} \end{align} $ Solve for $\cos \phi$, $ \huge \begin{align} \cos \phi&= \frac{{2(1-R)^{2}-1-R^{2}}}{-2R} \\ &= \frac{2-4R+2R^{2}-1-R^{2}}{-2R} \\ &= \frac{{R^{2}-4R+1}}{-2R} \\ &= -\frac{1}{2}R + 2 - \frac{1}{2 }\frac{1}{R} \\ \cos \phi&= 2 - \frac{1}{2}\left( R+\frac{1}{R} \right) \end{align} $ So, $ \huge \begin{align} B &= 2\phi \\ &= 2\arccos \left( 2-\frac{1}{2}\left( R+\frac{1}{R} \right) \right) \\ \end{align} $ For an approximation, $ \huge \begin{align} \let \epsilon &\approx 0 \\ R &= 1-\epsilon \\ \cos \phi&= 2 - \frac{1}{2}\left( 1-\epsilon+ \frac{1}{1-\epsilon} \right) \\ &= 2-\frac{1}{2}\left( 1-\epsilon + \sum_{n=0}^{\infty}\epsilon^{n} \right) \\ &= 2-\frac{1}{2}\left( 2 + \sum_{n=2}^{\infty}\epsilon^{n} \right) \\ &=2-1- \frac{1}{2} \sum_{n=2}^{\infty}\epsilon^{n} \\ &= 1-\frac{1}{2}\epsilon^{2} + \epsilon^{3} \underbrace{ \sum_{n=1}^{\infty} \frac{1}{2}\epsilon^{n}}_{\text{Higher Order Terms}} \end{align} $ $ \huge \begin{align} \cos(x) &= 1- \frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\cdots \\ &= 1 - \frac{1}{2}x^{2} + \underbrace{x^{4}\left( \cdots \right) }_{\text{Higher erms}} \end{align} $ $ \huge \begin{align} \cos\epsilon &= 1- \frac{1}{2} \epsilon^{2} + \cdots \\ \cos \phi &\approx \cos\epsilon \\ \phi &\approx \epsilon = 1-R \end{align} $ Using bandwidth $B=2\phi$, we get: $ \huge \begin{align} B \approx 2\epsilon \approx 2(1-R) \end{align} $ >[!example] For ex. $f_{s}=441000\pu{ Hz }$, bandwidth $B=20\pu{ Hz }$ yields $R=1-\pi \left( \frac{20}{44100} \right)\approx 0.9985$ >[!example] However for this to be flexible enough, we would need $a_{1}\in\C$ - which may break our original signal. To keep the original signal $Y\in\R$, we can introduce *two* poles of the transfer function that are [[Complex Conjugate|Conjugates]] of one another.