I am using a 12-bit DAC to produce signals between 10 kHz and 400 kHz, produced at a rate of 3.33 MSamples/s (Nyquist frequency of 1.665 MHz).

I would like to create a "reconstruction filter" that will be placed after the DAC. My question is, how much attenuation should the reconstruction filter achieve at the Nyquist frequency of 1.665 MHz to avoid alias images?

I am aware of the ~6 dB per bit heuristic that is commonly used, but I was wondering if there is a better approach of calculating this.

  • $\begingroup$ how much attenuation can you tolerate at your highest frequency of content: 400 kHz? you have a nice 2 octave transition band between your highest frequency of content and the Nyquist frequency. also, you might need to consider the attenuation due to the inherent zero-order hold. there won't be much effect at 400 kHz, but there will be an extra 4 dB of attenuation at 1.665 MHz. you can put that into your attenuation budget. $\endgroup$ Sep 6 at 13:16
  • $\begingroup$ Near 400 kHz, up to 10 dB of attenuation can be tolerated. $\endgroup$
    – Rhodes
    Sep 6 at 15:42
  • $\begingroup$ okay, so now how much attenuation do you need at your first image at 2.93 MHz (which is 3.33 MHz - 0.4 MHz)? nailing down that number will help you decide how sharp your anti-imaging filter needs to be which will decide the order of the filter. $\endgroup$ Sep 6 at 16:41
  • $\begingroup$ But that is my original question. I am building a measurement system where I need to create an anti-aliasing filter for the ADC and reconstruction filter for the DAC. I am uncertain of the attenuation levels that the reconstuction filter needs to achieve. $\endgroup$
    – Rhodes
    Sep 6 at 16:46
  • $\begingroup$ then, what you can do is guess. -60 dB attenuation means that the attenuated signal is $\frac{1}{1000}$ of the amplitude of the unattenuated signal. is -60 dB enough to sufficiently kill the images? $\endgroup$ Sep 6 at 17:22

Well, you cannot completely avoid images, you can only suppress them sufficiently. How much "sufficiently" is is completely up to your application! So, we can't tell you this.

However, thanks to the fact that you're comfortably oversampling your 400 kHz, the first aliases would appear at 2.93 MHz (=lowest frequency in signal, i.e. -400 kHz, plus sampling rate). How sensitive is whatever will get that signal to energy at 2.93 MHz or higher?

Assuming 50 dB attenuation totally suffice at 2.9 MHz, a simple split-into-two-stages, fourth order Butterworth filter would do:

Magnitude response of a two-stage Butterworth filter Designed with the analog devices filter design wizard

Implemented as Sallen-Key active filter, that'd amount to four capacitors, four resistors and two relatively cheap opamps. Plus one capacitor and a high-valued resistive voltage divider to AC-couple in the signal, which, since it's got no DC component, can be shifted to an arbitrary voltage, so that you can work with a single supply voltage here.


So the answer here is: well, no matter what you need, it's not very hard, might as well go for a pretty good filter. Assuming you want your passband to be very flat, you want a Butterworth filter, and if you choose that, you'll want an even order one, and second-order isn't going to be steep enough, so fourth order it is. You can do better, but unless you know why, you won't need to.

  • $\begingroup$ Do the number of DAC bits affect the attenuation levels at all? $\endgroup$
    – Rhodes
    Sep 6 at 14:50
  • $\begingroup$ no. you are conflating two different things. $\endgroup$ Sep 6 at 16:45
  • $\begingroup$ Marcus, if passband flatness is needed but you don't care about the stopband, a Tchebyshev Type 2 filter will work and might be lower order. but it might not be a lower parts count because the Butterworth or Tchebyshev Type 1 have only poles in the $s$-plane (and that Sallen-Key circuit can do it) while the Tchebyshev Type 2 (or the elliptical) require zeros and a different circuit. $\endgroup$ Sep 6 at 16:48
  • $\begingroup$ @robertbristow-johnson ah right, tcheby II can have strictly limited passband ripple, and then it beats Butterworth in the transition region $\endgroup$ Sep 7 at 10:16
  • 1
    $\begingroup$ tcheby II is strictly monotonic in the passband. no ripple at all. $\endgroup$ Sep 8 at 16:57

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