Question For a 12 bit ADC with analog input voltage ranging from \( -2 \mathrm{~V} \) to \( 2 \mathrm{~V} \). Determine the following ? i.) Number of Quantization levels ii.) Step Size iii.) Quantization level, when the analog votage is \( 1.33 \mathrm{~V} \) iv.) Quantization error and dynamic range? The Nyquist sampling rate for the signal: \[ For a 12 bit ADC with analog input voltage ranging from −2 V to 2 V. Determine the following ? i.) Number of Quantization levels ii.) Step Size iii.) Quantization level, when the analog votage is 1.33 V iv.) Quantization error and dynamic range? The Nyquist sampling rate for the signal: s(t)=πtsin(500πt)×πtsin(700πt)

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Transcribed Image Text: For a 12 bit ADC with analog input voltage ranging from −2 V to 2 V. Determine the following ? i.) Number of Quantization levels ii.) Step Size iii.) Quantization level, when the analog votage is 1.33 V iv.) Quantization error and dynamic range? The Nyquist sampling rate for the signal: s(t)=πtsin(500πt)×πtsin(700πt)

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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 ...