Figure 1: Magnitude and phase of the frequency response of the 5pt-averager filter. Below is a derivation of the analytical expression for the frequency response:

\( \begin{array}{rcl} H(e^{j\hat\omega})&=&\sum\limits_{k=0}^{4}{b_k e^{-j\hat\omega}} \\&=&\frac15 (\frac{1-e^{-j5\hat\omega}}{1-e^{-j\hat\omega}}) \\&=&\frac15 (\frac{e^{-j\hat\omega5/2}(e^{j\hat\omega5/2}- e^{-j\hat\omega5/2}} {e^{-j\hat\omega/2}(e^{j\hat\omega/2}-e^{-j\hat\omega/2})}) \\&=&(\frac{sin(\hat\omega5/2)}{5sin(\hat\omega/2}) e^{-j2\hat\omega} \end{array} \)