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The digital filter with system function $H(z)=\frac{0.5 z}{z-0.5}$ Finds the output noise power from the digital filter when input is quantized to have 8- bits.
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Solution:

$ \begin{aligned} &\sigma_e^2=\frac{2^{-2 B}}{3} \\\\ &B=8 \\\\ &\sigma_e^2=\frac{2^{-2 \times 8}}{3} \\\\ &\sigma_e^2=5.08 \times 10^{-6} \\\\ &\sigma_{e o i}^2=\left.\sigma_e^2 \sum_{i=1}^N\left[\left[z-P_i\right) H[z] H\left[z^{-1}\right]\\ z^{-1}\right]\right|_{z=P_i}\\ \end{aligned}\\ $

$ \begin{aligned}\\ H[z] \cdot H\left[z^{-1}\right] z^{-1} &=\frac{0.5 z}{z-0.5} \cdot \frac{0.5 z^{-1}}{z^{-1}-0.5} \cdot z^{-1} \\\\ &=\frac{(0.5)(0.5) z^{-1}}{(z-0.5)\left(z^{-1}-0.5\right)} \\\\ &=\frac{(0.5)(0.5) z}{(z-0.5)(1-0.5 z)} \\\\ &=\frac{0.5}{(z-0.5)(2-z)} \\\\ &=\frac{-0.5}{(z-0.5)(z-2)}\\ \end{aligned}\\ $

Here one pole $P_1=0.5$ is inside unit circle.

$ \begin{aligned}\\ \sigma_{e 0 i}^2 &=\left.\sigma_e^2(z-0.5)\left[\frac{-0.5}{(z-0.5)(z-2)}\right]\right|_{z=00} \\\\ &=\sigma_e^2 \frac{(-0.5)}{(0.5-2)} \\\\ &=5.08 \times 10^{-6}\left[\frac{-0.5}{-1.5}\right] \\\\ \sigma_{e 0 i}^2 &=1.6933 \times 10^{-6}\\ \end{aligned}\\ $

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