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Quantization Noise and Signal-Noise Ratio (SNR)

What do you need to know to understand this topic?

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 $\pm Q/2$.

Any value of the error is equally likely, so it has a uniform distribution ranging from $-Q/2$ to $+Q/2$. Then, this error can be considered a quantization noise with RMS: $$ v_{qn} = \sqrt{\frac{1}{Q}\int_{-Q/2}^{+Q/2}x^2dx}=\sqrt{\frac{1}{Q}\left[\frac{x^3}{3}\right]_{-Q/2}^{+Q/2}} = \sqrt{\frac{Q^2}{2^3 3} + \frac{Q^2}{2^3 3}} = \frac{Q}{\sqrt{12}}$$

What is the frequency spectrum of the quantization noise?

We know the quantization noise power is $v_{qn}^2$, but where is it concentrated or spread in the frequency domain? The quantization error creates harmonics in the signal that extend well above the Nyquist frequency. Due to the sampling step of an ADC, these harmonics get folded to the Nyquist band, pushing the total noise power into the Nyquist band and with an approximately white spectrum (equally spread across all frequencies in the band).

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 $$V_{rms} = \frac{V_{ref}}{2\sqrt{2}} = \frac{2^NQ}{2\sqrt{2}}.$$ where $N$ is the number of bits available for discretization. The relation $V_{ref} = 2^NQ$ comes from the fact that the range $V_{ref}$ is divided among $2^N$ steps, each with quantum $Q$. 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 equation: $$SNR = 6.02N + 1.76 (dB)$$ 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.

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