I have two signal (1.5Mhz) on one degree phase. I want to use two AD9642 to convert these signals
my question is : what sampling frequency should I use to get a significant result.


  • \$\begingroup\$ Your question is very difficult to understand. Please edit it and be clear about your question. \$\endgroup\$
    – jwh20
    Feb 17 at 13:36
  • \$\begingroup\$ sure I'll edited \$\endgroup\$
    – Marouane
    Feb 17 at 13:41
  • \$\begingroup\$ What does "on one degree phase" mean? When you say "convert" what EXACTLY do you mean? What do you mean by "significant result"? \$\endgroup\$
    – jwh20
    Feb 17 at 13:43
  • \$\begingroup\$ Triggering those 2 ADC simultaneously should be fairly easy. What will be difficult is maintaining both analog paths equal in phase and amplitude. \$\endgroup\$
    – glen_geek
    Feb 17 at 13:56

Simple calculation: 1.5MHz is a period of about 666ns You need 1° which is 1/360 of a period, so about 1.8ns

The absolute minimum to theorically discriminate them (for example by zero crossing) would be the 540Msps. That is minimun in a Nyquist sense of the term; you'll probably want to use 5-10 times that sample rate to account for noise, slight deviations from a perfect sine and so one.

I don't known exactly what you need that for but I wouldn't use less than 2Gsps (hey, you'll need an heck of an ADC, I didn't think you'll need so much)

  • 2
    \$\begingroup\$ Just three orders of magnitude off. It's a period of 666 ns. \$\endgroup\$
    – Arsenal
    Feb 17 at 13:50
  • \$\begingroup\$ 1.5Mhz is period of about 0.66µs. but yes that was I was looking for. Thank you for the answer. \$\endgroup\$
    – Marouane
    Feb 17 at 14:12
  • \$\begingroup\$ Yup horrible fault. I'll fix the answer \$\endgroup\$ Feb 17 at 14:23
  • 2
    \$\begingroup\$ The minimum in the Nyquist sense is >3 MHz since the completely captures all information, which you can then feed into fft, interpolation, etc to extract the phase difference. Zero crossing is a non-Nyquist approach since you do not try to reconstruct the original waveform, only measure difference in zero crossing. Either approach could work, but using more dsp may be more cost effective then gigasample ADCs. \$\endgroup\$ Feb 17 at 14:40
  • \$\begingroup\$ This answer is wrong (and user1850... is right). You don't need to sample at the time difference's Nyquist rate to determine the phase of an oscillation. That claim makes no sense: You can determine the phase of a sufficiently sampled oscillation (meaning f_sample > 2·bandwidth, nothing more, nothing less, so anything f_sample > 3 MHz works out of the box) with arbitrary precision, given sufficient SNR. \$\endgroup\$ Feb 17 at 16:21

If you have two signals of 1.5 MHz and want to discriminate their phase shift to 1°, then you have two options

The hardware obvious way, sample at 360*1.5 MHz = 540 MHz. Each sample will be 1° of phase. Count the samples between zero crossings. You can't do that with an AD9642, the fastest grade is 250 MHz, and cheaper ones are 170 MHz.

If you are prepared to do some processing, then a bit more than Nyquist on your highest expected frequency, say 5 to 10 MHz, though with a 9642 you can sample rather faster. Do a DFT of the two sampled waveforms, and extract the phases. I've discriminated 0.01° between two 10 kHz signals sampled at 44.1 kHz on a PC soundcard

As an alternative, if those 1.5 MHz +/- are not going to change in frequency between measurements, then sample at pretty much any frequency you like, except an exact multiple or sub-multiple of 1.5 MHz. The input signals will alias, but the phase relationship between the two input signals will be preserved in their aliased versions. If you sampled at 501 kHz, then you'd get 3 kHz aliased signals. If the 1.5 MHz signals were 1° apart in phase, then so would the 3 kHz signals.

  • \$\begingroup\$ thank you for replying. you gave a lot of useful information, that I'll definitely use later. really appreciate. \$\endgroup\$
    – Marouane
    Feb 17 at 14:17

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