I am currently reading the 'Coding' chapter on Rabiner & Schafer's Book on speech processing.

In one of its exercises, the reader is given a simple A/D converter using 16-bit uniform quantization. The converter uses a sampling rate of $Fs = 8000 Hz$. Then it asks what is the effect on $SNR_Q$ (quantization signal-to-noise ration) if we double the sampling rate.

From what I understand, doubling the sampling rate has obviously no effect on the quantizer's range $X_{max}$ as well as the quantization step $\Delta$. Furthermore, the theoretical model for $SNR_Q$ on uniform quantizers is built on top of the hypothesis that the quantization error $e(n)$ follows a uniform probability density, so:

$$ SNR_Q = \frac{\mathcal{E}(\sum_n x^2(n))}{\mathcal{E}(\sum_n e^2(n))} $$

should remain constant, since both its numerator and denominator are calculated off expected values.

Is my thinking correct or does the change of the sampling rate break any of my assumptions?


1 Answer 1


You are correct. When using uniform sampling, the sampling rate is unrelated to the quantization noise. Only the quantization step and the signal's dynamic range are relevant for calculating the quantization SNR.

  • $\begingroup$ It should be noted though that if you downsample using a perfect lowpass filter with a cut-off of half the signal band (4000 Hz), you will gain 3.01 dB as only half of the original quantization noise power is left in the remaining signal. So plainly oversampling is one feasible way to increase the effective resolution of an A/D-converter. (Which may quite likely be a forthcoming exercise...) $\endgroup$
    – Oscar
    Feb 5, 2015 at 17:28
  • $\begingroup$ @Oscar I'm not sure I understand; the signal's band is 4000 Hz. Are you referring to dithering? If so, that doesn't fundamentally contradict what I said, but it's a good point and something to keep in mind. $\endgroup$
    – MBaz
    Feb 5, 2015 at 19:36
  • $\begingroup$ No, just plain oversampling. If you sample at 16 kHz and then downsample to 8 kHz (using a lowpass filter with cutoff 4 kHz), you will only have half of the original quantization noise power in your signal, and, hence, have gained 3.01 dB SNR. Your answer is fully correct, I just speculated on the next question in the book, sort of. The comment was more of a follow-up to VHarisop than a comment on your answer, although related. $\endgroup$
    – Oscar
    Feb 6, 2015 at 9:11
  • $\begingroup$ @Oscar won't that only work if the signal's bandwidth is below 4kHz? I'm not quite sure I follow. $\endgroup$
    – VHarisop
    Feb 7, 2015 at 14:30
  • $\begingroup$ @VHarisop Yes, that is correct. However, based on the question (fs = 8000 Hz) I got the idea that it was assumed to be below 4kHz. $\endgroup$
    – Oscar
    Feb 8, 2015 at 15:14

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