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a full scale signal of bandwidth 5 khz is sampled by an 10-bit ADC at a sampling rate of 2 Msa/sec

  1. calculate the Signal to Quantization noise of the resulting DT signal
  2. repeat for a 14-bit ADC at sampling rate 10000 sa/sec

I mean that S/N = 6.02 x b , so what is the effect if sampling rate on it

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    $\begingroup$ Hi Mohamed, what have you tried so far? The convention for these types of questions is to show some work that you have done so far. $\endgroup$ – Engineer Nov 10 '19 at 14:49
  • $\begingroup$ i triad but the equation of s/n is equal to 6.02 x no. of bits + 1.76 as mentioned in books but my professor said that is relationship also between sampling rate and bandwidth which i couldn't find or prove, i mean there is another term added to s/n $\endgroup$ – Mohamed Azab Nov 10 '19 at 16:27
  • $\begingroup$ Not a full answer to your question but this link, dsp.stackexchange.com/a/40261/31316, should be helpful for you $\endgroup$ – Engineer Nov 10 '19 at 16:58
  • $\begingroup$ @MohamedAzab The key is to know the distribution of the quantization noise in frequency. This SNR gives you the TOTAL noise relative to a full scale sinewave--- do you know how the noise is distributed in frequency? $\endgroup$ – Dan Boschen Nov 10 '19 at 18:04
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    $\begingroup$ Also to nit pick on the way the question was presented to you: The formula is 6.02 dB + 1.76 dB for the case of a full scale sinewave (which has a peak to average ratio of 3 dB). For an arbitrary waveform you must know the peak to average ratio, specifically what is the power level of the waveform at "full scale?? This can range from 3 dB to 15 dB or more--- so from the 6.02dB + 1.76 dB we know where the noise is relative to the sine wave (SNR1); if some arbitrary waveform clips with it's power 10 dB lower (13 dB peak to average ratio) then the SNR for this case would be SNR1 -10 dB. $\endgroup$ – Dan Boschen Nov 10 '19 at 18:08

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