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Assuming a theoretical example imagine that we at first feed a 1V pk-pk noiseless sine to a 10V ADC. And in the second case we feed a 10V pk-pk noiseless sine to that 10V ADC. As far as I understand from other questions, the quantization noise decreases in the second case.

Im trying to understand why by using the above example in a step by step explanation. How can we mathematically show explicitly that in the second case the quantization noise is lower?

user1245
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  • think of it this way, the quantisation noise remains much the same regardless of signal level, so it gets relatively worse when the signal level gets smaller. – Neil_UK Sep 02 '19 at 20:47

2 Answers2

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The quantization error is typically equal to one-half the value of the LSB. That is an absolute error.

However, for a large signal the relative error (the error as a percentage of the peak signal value) is smaller.

Elliot Alderson
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  • Can you also add mathematical calculation for my example if possible? Thanks – user1245 Sep 02 '19 at 20:20
  • I don't understand what kind of math you want. You just divide 1/2 LSB by the peak signal voltage. Multiply by 100 if you want percent. – Elliot Alderson Sep 02 '19 at 20:21
  • If that ADC is 10 bit Im trying to quantity the rms values of the quantization noise in both cases. I could not find a known way:( trying to learn – user1245 Sep 02 '19 at 20:29
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For large signals ( much larger than one quanta), the RMS (standard-deviation) error is Vquanta/sqrt(12). This comes from statistical reasons.

Thus for 8 bits of deviation (256 Vquanta), the SignalNoiseRatio is

256 / [ 1/sqrt*12) ] = 256 / 0.288 = 886; the SNR in 20*log10[886] = 58.9 dB

And for 12 bits of deviation (4096 Vuanta), the SignalNoiseRatio is

4096 / [ 1/sqrt(12) ]

For a given ADC, the quantization error remains the same in volts, but becomes smaller as a fraction of the signal's voltage swing.

Good question.

analogsystemsrf
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  • Actually what you've calculated here is the PSNR (peak signal to noise ratio). The true SNR, as you point out, will depend upon the input signal level. This is typically just a little nit-picky detail, but it's important here because it's the exact nature of OP's question. – Mr. Snrub Sep 02 '19 at 21:22
  • or even more accurately, what I computed is the peak-peak signal-noise-ratio ? – analogsystemsrf Sep 05 '19 at 17:23
  • Yeah, I was just borrowing / abusing the standard terminology of [PSNR](https://en.wikipedia.org/wiki/Peak_signal-to-noise_ratio) which is a standard metric in image processing. – Mr. Snrub Sep 05 '19 at 19:55