Community Answer

Honor CodeSolved 1 Answer

See More Answers for FREE

Enhance your learning with StudyX

Receive support from our dedicated community users and experts

See up to 20 answers per week for free

Experience reliable customer service

Get Started

n=12 \text { bit }Voltage range =-2 v to 2 \mathrm{v}\begin{aligned}\text { Number of quantization levels } & =2^{n} \\(q) & =2^{12} \\q & =4096 .\end{aligned}(b)\text { Step size } \begin{aligned}(\Delta) & =\frac{\text { Voltage range }}{q} \\\Delta & =\frac{-2}{4096}+\frac{2}{4096} \\D & =0.0004882 \text { volt. } \\D & =4.88 \text { mvolt }\end{aligned}(c) Quantization level, when antog voltage is 1.33 \mathrm{~N}\begin{array}{l}10^{-3} \times 0.00048 \times L=1.33 \Rightarrow 10^{-3} \times 4.88 \times L=1.33 \\L=638\end{array}(d) (638)_{10}=(1001111110)_{2}(e) Quantization error\begin{array}{l}\left.=11.33-\frac{63842}{4096}\right] \\=0.55 \times 10^{-4} \mathrm{~V}\end{array}PLEASE UPVOTE AND POST YOUR DOUBTS IN COMMENTS ...