I have one and only one frequency signal, a sine signal at f = 10 kHz that is sampled at Fs = 320 kHz.

I have read about applying a simple analog anti aliasing (AAA) filter then oversample, do some digital anti aliasing (DAA) and decimate to get the correct signal frequency.

I understand how this is useful if signals of interest are close to Fs/2 as it might be hard to find an AAA filter that has sharp enough roll-off, low cost, simplicity etc. And therefore DAA filter in combination with a simple AAA filter is often preferred.

But in my case I don't have a signal close to Fs/2 at all. So I was thinking maybe I can get a slower/cheaper ADC and do no oversampling, since I might be able to properly attenuate frequencies that might alias even using a simple AAA? Is what I am thinking correct?

Additional related question: How would this work if I was to apply equivalent time sampling (ETS) technique? This means that the signal is 10 kHz and the sampling frequency appears to be 320 kHz as well, but 32 samples are taken over 32 periods of the 10 kHz signal to produce 1 period. So the sampling rate is actually 10 kHz + (10 kHz / 32) = 10.3125 kHz. What I am not able to wrap my head around: will Fs = 10.3125 kHz or 320 kHz? Will the same filter work for both cases? I assume Fs = 10.3125 kHz thus giving a problem as the transition area will be relatively very small: Fs - f = 3125 Hz.

Other thoughts: I am using a DDS to generate the sine signal, it should be quite pure and maybe AAA is not "needed" / very important?


1 Answer 1


The sampling theorem states that you need at least "two samples per one Hz of bandwidth". Anti-aliasing filters are required when the bandwidth of the original analog signal is not well defined and a finite bandwidth needs to be enforced.

In your application that doesn't seem to be the case. If all you want to sample is really just an unmodulated steady-state sine wave, the bandwidth or your original signal is already limited. No anti aliasing filter is needed and you can simply sample at any rate slightly larger than twice the sine wave frequency. 22.05 kHz or 24 kHz in your case would work fine.

ETS is a bit tricky since you need some sort of a trigger mechanism and then a variable delay off that trigger. The equivalent sample rate depends on your delay steps and the whole process depends on the signal being periodic with high precision.

You actually CAN sample at a much lower rate. The theorem requires 2 samples per Hz of bandwidth. For a pures sine wave the actual bandwidth is almost zero so you can sample with a very low sample rate and still don't have any loss of information. You just need to manage the mirror spectra properly, i.e. your application needs to be aware of a frequency offset.

  • $\begingroup$ Thank you for a good and clear answer. Even though one can sample at close to 2f, only the frequency information can be properly captured, right? I am using this signal with a square reference signal to do IQ demodulation (synchronous demodulation). The sine and the square gets mixed and the sine and the square with a 90 degree offset gets mixed. I assume I need more than only close to 2 samples per Hz to be able to properly retrieve the phase information by arctan(Q/I)? $\endgroup$ Sep 6, 2013 at 6:16
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    $\begingroup$ You have to be careful to track how the processing changes the bandwidth of your signal. If you square your sin wave you have doubled the bandwidth (the signal now contains 0Hz and 20 kHz as well) so you either need to up-sample or your sample rate must be high enough to cover the maximum bandwith during processing (not just sampling) $\endgroup$
    – Hilmar
    Sep 6, 2013 at 13:34
  • $\begingroup$ So in theory for this example a sample rate slightly above 40 kHz would do? $\endgroup$ Sep 6, 2013 at 13:46
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    $\begingroup$ Yes, as far as I can tell. I don't know the details of your application though. $\endgroup$
    – Hilmar
    Sep 6, 2013 at 13:54
  • $\begingroup$ Thanks. Here are some additional information in case it would change anything: I am interested in the 0 Hz (DC) component as this would give me the amplitude of the signal by taking the root of the summed squared ( sqrt(I^2 + Q^2) ) and the phase can be calculated as arctan(Q/I). These are the parameters I want to measure. This assumes the phase is between 0 and 90 degrees. $\endgroup$ Sep 6, 2013 at 14:06

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