FIR Averaging Filter

Frequency Response

$$H(z)=\frac{1}{2}+\frac{1}{2}{z}^{-1}$$ $$\begin{array}{rl}H(e^{j\omega })& =\frac{1}{2}+\frac{1}{2}{e}^{-j\omega }\\ & =\frac{1}{2}{e}^{-\frac{1}{2}j\omega }(e^{\frac{1}{2}j\omega }+e^{-\frac{1}{2}j\omega })\\ \text{As }& \text{ : }{e}^{\frac{1}{2}j\omega }+e^{-\frac{1}{2}j\omega }=2\cos (\omega \frac{1}{2})\\ \therefore & =e^{-\frac{1}{2}j\omega }\cos (\frac{1}{2}\omega )\end{array}$$

Real component = $\cos (\frac{1}{2}\omega )$

Amplitude Response

$$|H(e^{j\omega })|=|\cos (\frac{1}{2}\omega )|$$

In this case: $\pi$ is our nyquist, and our DC would be at 0, where we are at our maximum (1)