Amplitude Response

We can isolate the filter amplitude response $G(\omega)$ by squaring and adding the above two equations:

$$\begin{eqnarray*} a^2(\omega) + b^2(\omega) &=& G^2(\omega)\cos^2[\Theta(\omega)] + G^2(\omega) \sin^2[\Theta(\omega)]\\ &=& G^2(\omega)\left\{\cos^2[\Theta(\omega)] + \sin^2[\Theta(\omega)]\right\}\\ &=& G^2(\omega). \end{eqnarray*}$$

This can then be simplified as follows:

$$\begin{eqnarray*} G^2(\omega) &=& a^2(\omega) + b^2(\omega)\\ &=& [1 + \cos(\omega T)]^2 + \sin^2(\omega T) \\ &=& 1 + 2 \cos(\omega T) + \cos^2(\omega T) + \sin^2(\omega T)\\ &=& 2 + 2 \cos(\omega T) \\ &=&4 \cos^2\left(\frac{\omega T}{2}\right). \end{eqnarray*}$$

So we have made it to the amplitude response of the simple lowpass filter $y(n) = x(n) + x(n - 1)$ :

$$G(\omega) = 2 \left\vert\cos\left(\frac{\omega T}{2}\right)\right\vert$$

Since $\cos(\pi fT)$ is nonnegative for $-f_s/2 \leq f \leq f_s/2$ , it is unnecessary to take the absolute value as long as $f$ is understood to lie in this range:

$$\zbox {G(\omega) = 2 \cos(\pi f T)} \qquad \left\vert f\right\vert \leq \frac{f_s}{2} \protect$$ (2.6)