Quantization Noise and Signal-Noise Ratio (SNR)

What do you need to know to understand this topic?

• Root-Mean Square (RMS)

What is Quantization Noise?

When an Analog-Digital Converter (ADC) converts a continuous signal into a discrete digital representation, there is a range of input values that produces the same output. That range is called quantum (Q) and is equivalent to the Least Significant Bit (LSB). The difference between input and output is called the quantization error. Therefore, the quantization error can be between $\pm 1/2Q$.

This error can be considered a quantization noise with RMS: $$ v_{qn} = \sqrt{\frac{1}{Q}\int_{-Q/2}^{+Q/2}x^2dx}=\frac{Q}{\sqrt{12}}$$

How does the Signal-Noise Ratio (SNR) relates to the number of bits in the digital representation?

Assuming an input sinusoidal with peak-to-peak amplitude $V_{ref}$, where $V_{ref}$ is the reference voltage of an N-bit ADC (therefore, occupying the full-scale of the ADC), its RMS value is $$V_{rms} = \frac{V_{ref}}{2\sqrt{2}} = \frac{2^NQ}{2\sqrt{2}}.$$ To calculate the Signal-Noise Ratio, we divide the RMS of the input signal by the RMS of the quantization noise: $$SNR = 20\log\left(\frac{V_{rms}}{v_{qn}}\right) = 20\log\left(\frac{\frac{2^NQ}{2\sqrt{2}}}{\frac{Q}{\sqrt{12}}}\right) = 20\log\left(\frac{2^N\sqrt{12}}{2\sqrt{2}}\right)$$ $$ = 20\log\left(2^N\right) + 20\log\left(\frac{\sqrt{6}}{2}\right) = 6.02N + 1.76 (dB).$$

In fact, the term: $$SNR = 6.02N + 1.76 (dB)$$ generalizes to any system using a digital representation. So, a microprocessor representing values with N bits will have a SNR defined by the above formula.