Increasing SQNR by Increasing the Sample Rate

February 3, 2020

The quantization noise floor is flat with power π›₯2/12 for all πœ”, as proved in the lesson Quantization and Noise.Β  However, the band-limited signal 𝖸̃(πœ”) is not. Moreover, 𝖸̃(πœ”) gets β€œtaller” and effectively more narrow as π–₯s increases (see Theorem 1 under The DTFT of a Sampled Sequence.)

As such, with proper low-pass reconstruction filtering at the output, the signal to noise ratio drops as π–₯s increases, meaning more of the quantization noise is filtered out. We refer again to the figure presented in Increasing SQNR by Increasing Bits Per Sample.

In the figure, the noise floor increases as the number of bits decreases (π›₯ will increase) but the noise floor does not increase with the sample rate.

Thus, with a sufficiently high sample rate (π–₯s) and sufficient low-pass filtering, it is possible to design a system with a high SNQR even with just one bit (𝑏 = 1). More specifically, every time the sample rate is doubled, the SQNR increases by a factor of 4 (about 6dB or 1 bit). This demonstrated by the next theorem.

Theorem 4


(1)   \begin{equation*} P'_{s}\triangleq\int_{\omega\in\mathbb{R}}[2x(2\omega)]^2d\omega \end{equation*}

(2)   \begin{equation*} =4\int_{\omega\in\mathbb{R}}x^2(2\omega)d\omega \end{equation*}

(3)   \begin{equation*} =4\int_{\omega\in\mathbb{R}}x^2(v)(1/2)dv\text{ where }v\triangleq2\omega\implies d\omega=(1/2)dv \end{equation*}

(4)   \begin{equation*} =2P_{s} \end{equation*}

As a low-pass filter eliminates half the spectral content:

(5)   \begin{equation*} P'\triangleq(1/2P)P_{n} \end{equation*}

(6)   \begin{equation*} \text{SQNR}'\triangleq(P'_{s}/P'_{n})=\left[\frac{2P}{(1/2)P_{n}}\right]=4(P_{s}/P_{n})\triangleq 4\text{SQNR} \end{equation*}

(7)   \begin{equation*} \text{SQNR}_{dB}\triangleq 10\text{log}_{10}\text{SQNR}' \end{equation*}

(8)   \begin{equation*} =10\text{log}_{10}[4\text{SQNR}]=10\text{log}_{10}(4)+10\text{log}_{10}(\text{SQNR}) \end{equation*}

(9)   \begin{equation*} \approx\text{SQNR}_{dB}+6.02 \end{equation*}


Suppose we want to sample at 𝑏 = 1 bit.

According to Theorem 3, 𝖲𝖰𝖭𝖱𝖽𝖑 β‰ˆ 6.02 Γ— 1 + 1.76 = 7.78 dB. Additionally, in an ideal system, the 𝖲𝖰𝖭𝖱𝖽𝖑 for 𝑏 = 16 bits is 𝖲𝖰𝖭𝖱𝖽𝖑 β‰ˆ 6.02 Γ— 16 + 1.76 β‰ˆ 98 dB.

To sample at 𝑏 = 1 and get 𝑏 = 16 worth of quality (98 dB), how fast do we need to sample?

According to Theorem 4, SQNR increases by 6.02 dB every time the sample rate is doubled. Therefore, the sample rate must be doubled (98 βˆ’ 7.78)/6.02 βŒ‰ = ⌈14.99βŒ‰ = 15 times. In other words, the sample rate must be increased to 215 = 32768 times the Nyquist rate.

Therefore, if our original sample rate was 44.1 kHz, our new sample rate would need to be approximately 1.45 GHz.

A high sample rate, as in the example above, tends to be impractical and sometimes undesirable. However, there are alternative techniques with more relaxed sample rates. These techniques include:

  • Differential quantization/delta modulation encodes the difference between two successive samples rather than encoding every sample as-is. If the consecutive samples are correlated⁠—as we might expect for band-limited signals⁠—the difference (delta) between the samples tends to be small and results in less quantization noise. Delta modulation is presented in the next lesson.
  • Noise shapingΒ is often implemented using a signal processing topology called delta-sigma modulation. This topology uses a feedback path that passes the desired signal through and pushes the noise energy into higher frequency bands (noise shaping) which can subsequently be filtered out. Thus, SQNR is reduced, but the quantization is only 1-bit and the sample rate may only need to be increased by 256 times the Nyquist rate. Noise shaping and delta-sigma modulation are presented in the Noise Shaping/Delta-Sigma Modulation section.