[[Linear Transformation|Linear]] [[Phase]] is a special case of an [[Impulse Response]] / [[Impulse Response|Finite Impulse Response]] where the [[Sequence]] of coefficients $a_{0},a_{1},\cdots,a_{n}$ is a [[Palindromic Sequence]]. $\huge \begin{align} a_{k} &= a_{n-k}\\ n &= \left| \vec a \right| \end{align} $ These types of filters will have a [[Linear Transformation|Linear]] [[Frequency Response Function|Phase Response]] with respect to $\omega$. >[!example] > $\huge \begin{align} > y_{k} &= x_{t}-2x_{t-1} - 2x_{t-2} + x_{t-3} \\ > \vec a &= \left[ 1,-2,-2,1 \right] > \end{align} $ > $\huge \begin{align} > \mathcal{H}(z) &= 1-2z^{-1} -2z^{-2}+z^{-3} \\ > H(\omega) &= 1-2e^{-i\omega} - 2e^{-i 2 \omega } + e^{-i 3 \omega} \\ > &= e^{-i \frac{3}{2}\omega} \left( e^{i \frac{3}{2}\omega} - 2 e^{i \frac{\omega}{2}}-2e^{-i \frac{\omega}{2}} + e^{-i \frac{3}{2}\omega} \right) \\ > &= e^{-i \frac{3}{2}\omega} \left( e^{i \frac{3}{2}\omega}+ e^{-i \frac{3}{2}\omega} - 2 e^{i \frac{\omega}{2}}-2e^{-i \frac{\omega}{2}} \right) \\ > &=e^{-i \frac{3}{2}\omega}\left( 2\cos \left( \frac{3}{2}\omega \right) - 4 \cos \left( \frac{\omega}{2} \right) \right) \\ > &= > \underbrace{2\left( \cos \left( \frac{3}{2}\omega \right) +\cos \left( \frac{\omega}{2} \right) \right) }_{|H(\omega)|} > \overbrace{e^{-i \frac{3}{2}\omega}}^{\theta(\omega)= -\frac{3}{2}\omega} > \end{align} $ > > $\theta(\omega)$ is a [[Linear Transformation|Linear Function]] of $\omega$. > > $\huge \begin{align} > \theta(\omega) &= -\frac{3}{2}\omega > \end{align} $ In this example, a [[Phasor]] $e^{i \omega t}$ as input will have an output $H(\omega)e^{i\omega t}$ $\huge \begin{align} H(\omega)e^{i\omega t} &= 2\left( \cos \frac{3}{2}\omega - 2\cos \frac{\omega}{2} \right) e^{i\left( t-\frac{3}{2} \right)\omega } \\ \end{align} $ This means this phasor will have a phase [[Delay Operator|delay]] by $1.5$ [[Sampling|samples]].