# SQNR and Dynamic Range

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!

• Is this for an over-sampling ADC? – analogsystemsrf May 4 at 4:40

## 1 Answer

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!