We define the [[Transfer Function]] to be $\mathcal{H}(z)$, which has the property: $\huge \begin{align} \mathcal{H}(e^{i\omega}) &= H(\omega) \\ \left( H(e^{i\omega})\right) &= \left| H(\omega) \right| \end{align} $ Where $H$ is the [[Frequency Response Function]] of some [[Finite Impulse Response Filters|filter]]. This extends the [[Domain]] of $H$ to be $S^{1}$. $\left| \mathcal{H}(z) \right|$ can be thought of as a [[Surface]] above the [[Complex Plane]]. With the previous example, $\huge \begin{align} H(\omega) &= 1+e^{-i\omega}\\ &= 1 + \left( e^{i\omega}\right)^{-1}\\ \mathcal{H}(e^{i\omega}) &=H(\omega) \\ \mathcal{H}(z) &=1+z^{-1} \end{align}$ To go directly from the filter equation to [[Transfer Function]], we can think of $z^{-1}$ as a "delay operation", ie a [[Signal]] $X$ or $\vec x$ is delayed by 1 sample when multiplied by $z^{-1}$. So for a filter, if we are adding values of $X,x_{t}$ and $z^{-1}X,x_{t-1}$ at time $t$ to get the radius of $Y$ at time $t$ ($y_{t}$). Then, $\huge \begin{align} y_{t} &= x_{t} + x_{t-1} \\ Y &= X + z^{-1}X \\ &= (1+z^{-1})X \\ &= \mathcal{H}(z)X \end{align} $ So $\mathcal{H}(z)$ is a [[Linear Combination]] of delay operators that is applied to $X$ to get $Y$. >[!example] (3.1) > Where $t$ is in [[Sampling|samples]]. >$\huge > y_{t} = x_{k} + a_{i}x_{t-1} > $ > > Find $H(\omega)$ & $\mathcal{H}(z)$. > > $\huge \begin{align} > x_t &= e^{i\omega t} \\ > y_{t} &= e^{i\omega t} + a_{1} e^{i\omega(t-1)} \\ > &= e^{i\omega t}\left( 1+ a_{1}e^{-i\omega} \right) \\ > H(\omega) &= 1+a_{1}e^{-i\omega} \\ > &= 1+ a_{1}\left( e^{i\omega} \right) ^{-1} \\ > \mathcal{H}(z) &= 1+a_{1}z^{-1} > \end{align} $ > > So, > $\huge \begin{align} > Y &= X+ a_{1}z^{-1}X \\ > &=(1+a_{1}z^{-1})X \\ > &= \mathcal{H}(z)X > \end{align} $ >