The output signal of an A/D converter is passed through a lowpass filter with transfer function is given by H(z)=(1-a)z/(z-a) for 0<a<1.Find the steady state output noise power due to quantization at the output of the digital filter.</p>

Solution:

$ \begin{aligned}\\ & \sigma_{\varepsilon}^2=\sigma_e^2 \frac{1}{2 \pi j} \oint_c H(z) H\left(z^{-1}\right) z^{-1} d z-(r) \\\\ H(z) &=\frac{(1-a) z}{z-a} ; H\left(z^{-1}\right)=\frac{(1-a) z^{-1}}{z^{-1}-a} \\\\ \sigma_{\varepsilon}^2 &=\sigma_c^2 \frac{1}{2 \pi j} \oint_c \frac{(1-a){ }^2\left(z^{-1}\right) d z}{(z-a)\left(z^{-1}-a\right)} \\\\ &=\sigma_e^2\left[\text { residue of } H(z) H\left(z^{-1}\right) z^{-1} \text { at } z=a\right.\\\\ +& \text { residue of } H(z) H\left(z^{-1}\right) z^{-1} \text { at } z=1 / a\\ \end{aligned}\\ $

$ \begin{aligned}\\ &=\sigma_e^2\left[(z-a) \frac{(1-a)^2 z^{-1}}{(z-a)\left(z^{-1}-a\right)}+0\right] \\\\ &=\sigma_e^2\left[\frac{1-a)^2 \cdot 1 / a}{(1 / a-a)}\right] \\\\ &=\sigma^2\left[\frac{11-a)^2}{a\left(\frac{1-a}{a}\right)^2}\right] \\\\ &=\sigma_e^2\left[\frac{(1-a)^2}{(1+a)(1-a)}\right] \\\\ &=\sigma_c^2\left[\frac{1-a}{1+a}\right]\\ \end{aligned}\\ $

$ \begin{aligned}\\ &=\sigma_c^2\left[\frac{1-a}{1+a}\right] \\\\ &=\frac{2^{-2 b}}{12}\left[\frac{1-a}{1+a}\right]\\ \end{aligned}\\ $