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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?
nukimov
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  • 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: https://dsp.stackexchange.com/questions/63074/phase-difference-between-signals-sampled-at-different-frequencies/63079#63079 – Dan Boschen Mar 28 '20 at 19:10
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    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? – Marcus Müller Mar 29 '20 at 11:23
  • "...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. – jithin Mar 29 '20 at 11:31
  • @MarcusMüller: Alsmot perfect, thanks! I just corrected few things. – nukimov Mar 29 '20 at 15:34
  • @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 2T2, 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 2T2 or longer. We could interpolate between data points, for instance to get a period of T1, but interpolation does not create the missing information. – nukimov Mar 29 '20 at 15:37
  • @jithin That the only viable option is (C) is also what I found "empirically", my question is if such finding is in agreement with the Nyquist-Shannon theorem or, alternatively, if I can use the Nyquist-Shannon theorem as an explanation of why (C) seems to be the best. – nukimov Mar 29 '20 at 15:51
  • @DanBoschen "each accurately represent the signal at any point in continuous time for all signal content within the Nyquist bandwidth". Yes, but if I understand correctly, R1 contains information outside the Nyquist bandwidth of R2, i.e., changes on time scales of T1, which are not contained in R2. I think that matching the two signals at the period T1 would not be correct, we would be matching observations in R1 mostly to non-observed estimates (for instance interpolated values) of R2. Information is only accurate for both signals at time scales longer than 2*T2. – nukimov Mar 29 '20 at 16:04
  • @nukimov agreed and this would be consistent with Nyquist. Still there exists a continuous time waveform for each sampled waveform which does have a solution for the infinite sample locations in between each sample, and as long as we consider only spectral content within the first Nyquist Zone in frequency - there is only one solution for the trajectory when considering all previous samples. (The interpolated result based on current and past samples). – Dan Boschen Mar 29 '20 at 16:07
  • @DanBoschen: "agreed and this would be consistent with Nyquist". Great! That answers my first question, i.e., that observing the sampling theorem, the correct way of matching R1 and R2 is by downsampling of both records to 2⋅T2. It would be great to have a quote from a textbook or paper for this, some examples would be better, if there are some (this would answer question 2). – nukimov Mar 30 '20 at 10:11
  • I don't have such a quote or paper so won't be able to help you with that (so leaving the real answer blank below). However it must be emphasized that the trajectory I mention is very much dependent on the memory of the system that the signals have both gone through (for example the impulse response). Every possible sample at the output is the result of the current sample and all previous samples to the extent that the impulse response was long enough. So that is to say your result will only be equivalent if you have more samples than the time memory of the system – Dan Boschen Mar 30 '20 at 12:29
  • and the first samples will not be valid unless the original signal at the input to the system (or channel) was known to be previously 0 for all time before the signal was captured. – Dan Boschen Mar 30 '20 at 12:30
  • But short of finding a quote I recommend trying a counter-example (and I contend that you won't find one): Create a random bandlimited waveform of high time precision (many many more samples than Nyquist, emulating a continuous time waveform)- and pass that through a system with a finite impulse resposne. You will see that if you take different decimated sample sets of the output of that system that obey Nyquist (each would then be a delayed version of the same waveform); from any of these sets only the original waveform is created if resampling is done properly. – Dan Boschen Mar 30 '20 at 12:35

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