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In the following example:

require(biwavelet)
t <- seq(1/24, 365, 1/24)
A <- 10
fs <- 1/24
y <- A + sin(2*pi*fs*t)

d <- cbind(t, y + rnorm(length(y))) 
wt.t1 <- wt(d)
plot(wt.t1)

If I am looking at the diurnal cycle of a given time series what is the lowest resolution possible to use? This example currently uses data measured at 1 hour intervals thus I have 24 data points in a day. Would this be as significant if I used 2,3, or even 4 hour time steps?

Also is there a method for calculating if the significance of the results varies when decreasing the resolution of the data

MichaelChirico
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KatyB
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  • Hi Kate! I'd like to help out, but need a little clarification - The fact that you are a looking at a diurnal cycle doesn't really mean much for the time resolution you'd like to choose. The actual pattern you are investigating is what determines this. For example, say your pattern was flat for 11 hrs, a hump at hour 12, then flat the rest of the day. This feature may be lost with a 4 hour time step. Therefore, some clarification on the process you are sampling would be great. Thanks! – im so confused Sep 12 '12 at 16:17
  • The time series is for air temperature, so you would expect the temperature to be higher in the day compared to the evening. So, for my example, what would be the lowest time step possible before the results become untrustworthy to detect the power at the 24 hour cycle. Thanks a lot for your help. – KatyB Sep 12 '12 at 16:27
  • Ok so here there are really no rights/wrongs because (well at least I know of none) there is no real period or frequency to temperature, but look into the Nyquist frequency and the N-S Sampling theorem (sample @ twice the maximum frequency of your signal) http://en.wikipedia.org/wiki/Nyquist%E2%80%93Shannon_sampling_theorem – im so confused Sep 12 '12 at 16:41
  • you can also determine (to a broad extend, not exact numbers i believe) from the math present there the amount of information you lose at each time sampling rate (higher frequencies of temperature fluctuations cannot be caught at lower sampling rates, etc) maybe something to start from? – im so confused Sep 12 '12 at 16:43
  • thanks for your comments. So, from what you have said and what I understand about the Nyquist frequency, would it be correct to say that according to the Nyquist theorem we require measurements at twice the frequency of the signal? – KatyB Sep 13 '12 at 08:20
  • well, I wouldn't say require for your exact situation, but yes, a bandlimited signal requires *greater than* (not >=) twice the highest frequency as the sampling frequency. Now, only you can tell the characteristics of your data, but I think that's a good place to start sampling your data. Once you gain more understanding about the specific trends you are seeing, you can further increase your sampling efficiency by making more educated guesses. – im so confused Sep 13 '12 at 14:19

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