I have a sine signal whose frequency is $f$ sampled and then reproduced by a DAC as a zero order hold signal with sampling period $T_1$. The DAC signal is then reacquired with a sampling period $T_2 = T_1 / 10 $.

In my case $f = 400 $ $Hz$, $T_1 = 1$ $ms$, $T_2 = 0.1$ $ms$.

In order to estimate the amplitude and phase of the original sine signal, I assumed I had to multiply the acquired spectrum by

$$ \frac{\exp(i\pi f T_1)}{\mbox{sinc}(fT_1) }$$

(https://en.wikipedia.org/wiki/Zero-order_hold). What I found, actually, is that I have to multiply the spectrum by

$$ \exp[i\pi f (T_1-T_2)]\frac{\mbox{sinc}(fT_2)}{\mbox{sinc}(fT_1) }$$

I tried to use also a lowpass filter after the DAC, but that did not change anything.

Could anyone please provide a formal explanation of this fact?

  • $\begingroup$ where did you get this expression? : $$ \exp[i\pi f (T_1-T_2)]\frac{\mbox{sinc}(fT_2)}{\mbox{sinc}(fT_1) }$$ $\endgroup$ Oct 14 at 3:16
  • $\begingroup$ In practice, empirically. But it works very well, I tried for different signal frequencies and sampling rates. It works too well to be a coincidence $\endgroup$ Oct 14 at 8:21
  • $\begingroup$ //The DAC signal is then reacquired with a sampling period $T_2=T_1/10$......// --- Does that mean that the output is a constant value for $T_2$ adjacent sampling periods? $\endgroup$ Oct 14 at 8:35
  • $\begingroup$ Yes, 10 samples are acquired in the period $T_1$, then the DAC changes to another value, and 10 samples are acquired again, and so on $\endgroup$ Oct 14 at 8:39
  • $\begingroup$ Then resampling imposes no new ZOH. You have the original output DAC with ZOH frequency response of $$ \exp(-i\pi f T_1) \mbox{sinc}(fT_1) $$. That is the only ZOH that is affecting your frequency response. $\endgroup$ Oct 14 at 8:53

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