# Resampling a Zero Order Hold signal

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?

• where did you get this expression? : $$\exp[i\pi f (T_1-T_2)]\frac{\mbox{sinc}(fT_2)}{\mbox{sinc}(fT_1) }$$ Oct 14 at 3:16
• 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 Oct 14 at 8:21
• //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? Oct 14 at 8:35
• 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 Oct 14 at 8:39
• 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. Oct 14 at 8:53