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An audio tone \$Acos(\omega t + \theta)\$ undergoes uniform quantization by a quantizer that has the average quantization noise power \$q^2/6\$ joules, where \$q\$ is the quantization step size. If the dynamic range of the quantizier is adjusted to 15dB and the signal-to-quantization-noise ratio (SQNR) is targeted to be at least 40dB, how many bits per sample are needed to code each sample?

This is an past-exam exercise from my university. I tried to solve it, but I only got a solution as a function of \$A\$. Anyone can give me a hint? I don't know what I'm missing. Thanks!

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  • \$\begingroup\$ Is this for an over-sampling ADC? \$\endgroup\$ – analogsystemsrf May 4 at 4:40
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I got to this solution:

Starting from SQNR formula: \begin{equation} \text{SQNR} = 10log(\frac{A^2/2}{q^2/6}) \end{equation} simplified: \begin{equation} \text{SQNR} = 4.77 + 20log(\frac{A}{q}) \end{equation} As, \begin{equation} q = \frac{2*V_{max}}{2^N} \end{equation} \begin{equation} \text{SQNR} = 4.77 + 20log(\frac{2^NA}{2V_{max}}) \end{equation} \begin{equation} \text{SQNR} = -1.25 + 6.02N + 20log(\frac{A}{V_{max}}) \end{equation} Dynamic range in this case: \begin{equation} \text{DR} = 20log(\frac{V_{max}}{A}) \end{equation} Substitution: \begin{equation} 40 = -1.25 + 6.02N - 15 \end{equation} \begin{equation} N = 9 \text{ bits (ceiling function)} \end{equation}

I would appreciate if someone can give his approval!

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