I need help in solving the below question. Bandwidth of the signal is 1KHz. Use a 12-bit ADC with SNR of 78dB for sampling. the sampling frequency is 12KHz. Choose suitable filter order and cut-off frequency

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    \$\begingroup\$ Welcome to EE.SE. This looks like homework with no attempt at a solution. Explain what you know and where you are stuck. Please capitalise your words properly. "KHz" would be kelvin-hertz. Use small 'k' for kilo. \$\endgroup\$ – Transistor Aug 5 '17 at 9:52
  • \$\begingroup\$ I am unable to proceed to solve this. I don't know how to select filter order based on SNR. (I ll keep in mind regarding capitalsation.) \$\endgroup\$ – Preeti Srivastava Aug 5 '17 at 10:01
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    \$\begingroup\$ You haven't mentioned anything about noise or other contaminations above 1 kHz so, on the face of it there are none and therefore no anti-alias filter is required. \$\endgroup\$ – Andy aka Aug 5 '17 at 11:52
  • \$\begingroup\$ It seems the question is suggesting that the filter attenuation must be 78 dB at the Nyquist frequency of 6 kHz. Any noise between 1kHz and 6kHz can be filtered digitally, but any noise at frequencies above 6kHz may alias into the passband. Please read this statement and study as needed until you understand it. \$\endgroup\$ – mkeith Aug 5 '17 at 17:01

Assume no crud above 1Khz to be filtered out. In that case, no filter needed.

Assume the crud above 1Khz must be filtered out. This cannot be done without infinite steepness before sampling. This cannot be done.

What can be done? Simple example is to remove all the crud above 6KHz, attenuating down to 1 quanta level or to -78dBc.

Thus need the stopband to be -78dB, above 6KHz. That ratio 6KHz/1khz, or 2.5 octaves, is key in computing the rolloff steepness.

78/2.5 ~~ 75/2.5 == 30. Thus need 30dB/octave or 5 poles of 6dB each.

In this case, the spectrum from DC to 6KHz will be properly represented, including any crud. You can use DSP methods on this properly-represented spectrum and achieve whatever rolloff you need, limited by your word-length and compute power.

  • \$\begingroup\$ I didn't really understand your Explanation. How did you come up with 6kHz? \$\endgroup\$ – Preeti Srivastava Aug 5 '17 at 10:35
  • \$\begingroup\$ The question, together with Shannon's Sampling Theorem, gives you the reasoning behind 6 kHz. \$\endgroup\$ – Brian Drummond Aug 5 '17 at 10:49
  • \$\begingroup\$ The sampling theorem talks of "folding frequency", which is 1/2 of the Sampling Frequency. \$\endgroup\$ – analogsystemsrf Aug 5 '17 at 16:19

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