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Consider a D/A converter for audio signals consisiting of a zero-order-hold interpolator followed by a continuous-time lowpass filter with positive passband between 0 and 20KHz and stopband starting at fa= 40KHz.

Assume we want to convert a digital signal originally sampled at 16KHz. What is the minimum oversampling factor that we need to use?

for getting 16 khz i used 2 as my oversampling factor which seems not right. can you elaborate on top of it ?? i used 2 because (40khz/20khz)

dsp
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    exact same problem as your previous question: this is homework, dumped on us, without showing any own attempt. We don't do that here. – Marcus Müller Sep 13 '20 at 09:37
  • for getting 16 khz i used 2 as my oversampling factor which seems not right. can you elaborate on top of it ?? i used 2 because (40khz/20khz) – dsp Sep 13 '20 at 09:54
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    ... edit your question to include that, instead of posting it as a comment. – Marcus Müller Sep 13 '20 at 09:54
  • done sir. Hope to get reply soon:) – dsp Sep 13 '20 at 10:05
  • Does this answer your question? D/A converter with – Dan Boschen Sep 13 '20 at 21:16
  • please tell me methodology used to solve this question – dsp Sep 14 '20 at 05:05
  • what have you tried so far ? do you know the implications of sampling? what about DAC sampling frequency? The zero-order hold method? What's the effect of interpolation filter on the output spectrum of a DAC? What's the procedure to playback a sequence $x[n]$ at a different rate then it was sampled? You have to put your effords. Your answer is very easy. Just one divison of two numbers suffice, but the problem is actually more deeper. You should consider the whole system from sampling to interpolation stages... – Fat32 Sep 14 '20 at 10:03
  • for getting 16 khz i used 2 as my oversampling factor which seems not right. can you elaborate on top of it ?? i used 2 because (40khz/20khz) – dsp Sep 14 '20 at 10:06
  • @DanBoschen for 16 khz sampling the bandwidth is 8khz and applying it as ((x*16)/2 ) -(8), i get 6 as my oversampling factor, is it a right approach for this question.. – dsp Sep 17 '20 at 16:46
  • @dsp ask me the questions in the linked question that answers this to keep those together-- but let me know what you don't understand in the answer there as it has the background needed to solve the problem so good if you could work through completely understanding that. We should close this one since it is an exact duplicate. We're happy to help you - this comes across as a classical homework problem so we want to help you get the learning part of it without doing out the whole solution. Ask questions there and we'll help you get there! – Dan Boschen Sep 17 '20 at 18:55
  • @DanBoschen From this question, it has been clearly told that at 16 khz we need to find out the oversampling factor and acc to nyquist rate it should be 32 khz but now my question is that how can it be used to sample 40khz stop band and how come minimum oversampling be cal. – dsp Sep 18 '20 at 09:36
  • @dsp right that is all answered with the same details of the other question; it requires the same understanding so if you get the full details in the other answer you will also get this- please work with me to get through your other question and it will all be clearer for you! – Dan Boschen Sep 18 '20 at 11:20
  • thank you sir for helping me out! i did it in this way.. as per the nyquist theorem the range of x[n] will be (-16/2)<x[n]<(16/2), i.e., -8<x[n]<8, the bits of quantization is given in the form of 2^n, which is equal to 8, so 2^n=8, i.e., n=3. – dsp Sep 18 '20 at 16:20

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