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I have two data records $R_1$ and $R_2$ with sampling periods $T_1$ and $T_2$, where $T_1$ < $T_2$. These records arise from sampling and filtering two signals to remove any noise (including aliasing noise) from signals with shorter periods. This implies that the data in $R_1$ and $R_2$ are not instantaneous observations but represent average states on time scales of $T_1$ and $T_2$, respectively.

I want to fit a model $R_2 = f(R_1)$ and, thus, I want to match the two records as if they would've been sampled simultaneously, i.e., as if they would have the same sampling period.

I tried this by

  • (A) Upsampling $R_2$ to $T_1$;
  • (B) Downsampling $R_1$ to $T_2$;
  • (C) Downsampling both, $R_1$ and $R_2$ to $2 \cdot T_2$.

Afterwards I evaluated the performance of the model with cross-validation and found that the best match was obtained with (C).

I believe this is a consequence of the Nyquist-Shannon theorem, which says that $R_2$ contains complete information only from signals with periods of $2 \cdot T_2$ or longer, while events with shorter periods (for instance, $T_1$) are not resolved by $R_2$.

This means that I cannot determine accurately the relation $R_2 = f(R_1)$ at periods shorter than $2 \cdot T_2$. Therefore, the Nyquist-Shannon theorem seem to have obvious implications on how to match two records:

For instance, to upsample $R_2$ as to match $T_1$ would not be correct, since we have inaccurate information about $R_2$ on the time scale $T_1$. On the contrary, observing the Nyquist-Shannon theorem implies matching the two signals by downsampling both of them to a period of $2 \cdot T_2$ (or longer), a time scale where both records would contain accurate information. My questions are:

  1. Is this notion correct, i.e., is the downsampling of both records to $2 \cdot T_2$ the correct way of matching the sampling period of $R_1$ and $R_2$?
  2. If yes, could somebody provide a quote for this from a text book or a paper, as well as some prominent examples in the literature where two records or signals are matched like this, for instance from signal analysis, acoustics, electronics, meteorology, etc?
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  • $\begingroup$ Assuming both signals were properly anti-alias filtered prior to sampling, then they each accurately represent the signal at any point in continuous time for all signal content within the Nyquist bandwidth. The approach taken here is similar to what you could do to deriver your specific case: dsp.stackexchange.com/questions/63074/… $\endgroup$ – Dan Boschen Mar 28 '20 at 19:10
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    $\begingroup$ puh, this was quite a wall of text. I tried to structure it a bit with line breaks, nukimov, could you please check whether this looks like you want it to? $\endgroup$ – Marcus Müller Mar 29 '20 at 11:23
  • $\begingroup$ "...R2 contains complete information only from signals with periods of 2⋅T2 or longer, while events with shorter periods (for instance, T_1) are not resolved by R2.." If the information can be resolved by $2*T_2$, why can't it be resolved by lesser time interval of $T_1$? I am feeling uneasy now thinking about it. Also, what is the ratio of $T_1/T_2$? Let us say if $T_1 = 2$, $T_2 = 3$, and Nyquist Interval is 2.5, then upsampling $R_2$ to $T_1$ wouldn't work because it is already aliased. (B) also wouldn't work. Only option is (C) here even though aliased. $\endgroup$ – jithin Mar 29 '20 at 11:31
  • $\begingroup$ @MarcusMüller: Alsmot perfect, thanks! I just corrected few things. $\endgroup$ – nukimov Mar 29 '20 at 15:34
  • $\begingroup$ @jithin : "If the information can be resolved by 2∗T2, why can't it be resolved by lesser time interval of T1": I am probably expressing it wrong in the question. Surely if we could record the signal again with period T1 we would resolve it better than 2*T2, but the record is already taken, it cannot be modified. So, what I am saying is that in that record we have only reliable information for time scales of 2*T2 or longer. We could interpolate between data points, for instance to get a period of T1, but interpolation does not create the missing information. $\endgroup$ – nukimov Mar 29 '20 at 15:37

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