Fourier/trigonometric interpolation

Background

In a paper from Epstein (1991): On obtaining daily climatological values from monthly means, the formulation and an algorithm for calculating Fourier interpolation for periodical and even-spaced values are given.

In the paper, the goal is to obtain daily values from monthly means by interpolation.

In short, it is assumed that unknown daily values can be represented by the sum of harmonic components:
y(t)=a0+j[ajcos(2πjt/12)+bjsin(2πjt/12)]
In the paper t (time) is expressed in months.

After some derviation, it is shown that the terms can be calculated by:
a0=TYT/12aj=[(πj/12)/sin(πj/12)]×T[YTcos(2πjT/12)/6]       j=1,,5bj=[(πj/12)/sin(πj/12)]×T[YTsin(2πjT/12)/6]       j=1,,5a6=[(πj/12)/sin(πj/12)]×T[YTcos(πT)/12]b6=0
Where YT denote the monthly means and T the month.

Harzallah (1995) summarizes this aproach as follows: “The interpolation is carried out by adding zeros to the spectral coefficients of data and by performing an inverse Fourier transform to the resulting extended coefficients. The method is equivalent to applying a rectangular filter to Fourier coefficients.”


Questions

My goal is to use the above methodology for interpolation of weekly means to obtain daily data (see my previous question). In summary, I have 835 weekly means of count data (see the example dataset at the bottom of the question). There are quite a few things that I don’t understand before I can apply the approach outlined above:

  1. How would the formulas have to be changed for my situation (weekly instead of monthly values)?
  2. How could the time t be expressed? I assumed t/835 (or t/n with n data points in general), is that correct?
  3. Why does the author calculate 7 terms (i.e. 0j6)? How many terms would I have to consider?
  4. I understand that the question can probably be solved by using a regression approach and using the predictions for interpolation (thanks to Nick). Still, some things are unclear to me: How many terms of harmonics should be included in the regression? And what period should I take? How can the regression be done to ensure that the weekly means are preserved (as I don’t want an exact harmonic fit to the data)?

Using the regression approach (which is also explained in this paper), I managed to get an exact harmonic fit to the data (the j in my example would run through 1,,417, so I fitted 417 terms). How can this approach be modified – if possible – to achieve the conservation of the weekly means? Maybe by applying correction factors to each regression term?

The plot of the exact harmonic fit is:

Exact harmonic fit

EDIT

Using the signal package and the interp1 function, here’s what I’ve managed to do using the example data set from below (many thanks to @noumenal). I use q=7 as we have weekly data:

# Set up the time scale

daily.ts <- seq(from=as.Date("1995-01-01"), to=as.Date("2010-12-31"), by="day")

# Set up data frame 

ts.frame <- data.frame(daily.ts=daily.ts, wdayno=as.POSIXlt(daily.ts)$wday,
                       yearday = 1:5844,
                       no.influ.cases=NA)

# Add the data from the example dataset called "my.dat"

ts.frame$no.influ.cases[ts.frame$wdayno==3] <- my.dat$case

# Interpolation

case.interp1 <- interp1(x=ts.frame$yearday[!is.na(ts.frame$no.influ.case)],y=(ts.frame$no.influ.cases[!is.na(ts.frame$no.influ.case)]),xi=ts.frame$yearday, method = c("cubic"))

# Plot subset for better interpretation
par(bg="white", cex=1.2, las=1)
plot((ts.frame$no.influ.cases)~ts.frame$yearday, pch=20,
     col=grey(0.4),
     cex=1, las=1,xlim=c(0,400), xlab="Day", ylab="Influenza cases")
lines(case.interp1, col="steelblue", lwd=1)

Cubicinterpo

There are two issues here:

  1. The curve seem to fit “too good”: it goes through every point
  2. The weekly means are not conserved

Example dataset

structure(list(date = structure(c(9134, 9141, 9148, 9155, 9162, 
9169, 9176, 9183, 9190, 9197, 9204, 9211, 9218, 9225, 9232, 9239, 
9246, 9253, 9260, 9267, 9274, 9281, 9288, 9295, 9302, 9309, 9316, 
9323, 9330, 9337, 9344, 9351, 9358, 9365, 9372, 9379, 9386, 9393, 
9400, 9407, 9414, 9421, 9428, 9435, 9442, 9449, 9456, 9463, 9470, 
9477, 9484, 9491, 9498, 9505, 9512, 9519, 9526, 9533, 9540, 9547, 
9554, 9561, 9568, 9575, 9582, 9589, 9596, 9603, 9610, 9617, 9624, 
9631, 9638, 9645, 9652, 9659, 9666, 9673, 9680, 9687, 9694, 9701, 
9708, 9715, 9722, 9729, 9736, 9743, 9750, 9757, 9764, 9771, 9778, 
9785, 9792, 9799, 9806, 9813, 9820, 9827, 9834, 9841, 9848, 9855, 
9862, 9869, 9876, 9883, 9890, 9897, 9904, 9911, 9918, 9925, 9932, 
9939, 9946, 9953, 9960, 9967, 9974, 9981, 9988, 9995, 10002, 
10009, 10016, 10023, 10030, 10037, 10044, 10051, 10058, 10065, 
10072, 10079, 10086, 10093, 10100, 10107, 10114, 10121, 10128, 
10135, 10142, 10149, 10156, 10163, 10170, 10177, 10184, 10191, 
10198, 10205, 10212, 10219, 10226, 10233, 10240, 10247, 10254, 
10261, 10268, 10275, 10282, 10289, 10296, 10303, 10310, 10317, 
10324, 10331, 10338, 10345, 10352, 10359, 10366, 10373, 10380, 
10387, 10394, 10401, 10408, 10415, 10422, 10429, 10436, 10443, 
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10576, 10583, 10590, 10597, 10604, 10611, 10618, 10625, 10632, 
10639, 10646, 10653, 10660, 10667, 10674, 10681, 10688, 10695, 
10702, 10709, 10716, 10723, 10730, 10737, 10744, 10751, 10758, 
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10954, 10961, 10968, 10975, 10982, 10989, 10996, 11003, 11010, 
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11080, 11087, 11094, 11101, 11108, 11115, 11122, 11129, 11136, 
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11206, 11213, 11220, 11227, 11234, 11241, 11248, 11255, 11262, 
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13411, 13418, 13425, 13432, 13439, 13446, 13453, 13460, 13467, 
13474, 13481, 13488, 13495, 13502, 13509, 13516, 13523, 13530, 
13537, 13544, 13551, 13558, 13565, 13572, 13579, 13586, 13593, 
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13663, 13670, 13677, 13684, 13691, 13698, 13705, 13712, 13719, 
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13789, 13796, 13803, 13810, 13817, 13824, 13831, 13838, 13845, 
13852, 13859, 13866, 13873, 13880, 13887, 13894, 13901, 13908, 
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14797, 14804, 14811, 14818, 14825, 14832, 14839, 14846, 14853, 
14860, 14867, 14874, 14881, 14888, 14895, 14902, 14909, 14916, 
14923, 14930, 14937, 14944, 14951, 14958, 14965, 14972), class = "Date"), 
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Answer

I am no expert on Fourier transforms, but…

Epstein’s total sample range was 24 months with a monthly sample rate: 1/12 years. Your sample range is 835 weeks. If your goal is to estimate the average for one year with data from ~16 years based on daily data you need a sample rate of 1/365 years. So substitute 52 for 12, but first standardize units and expand your 835 weeks to 835*7 = 5845 days. However, if you only have weekly data points I suggest a sample rate of 52 with a bit depth of 16 or 17 for peak analysis, alternatively 32 or 33 for even/odd comparison. So the default input options include: 1) to use the weekly means (or the median absolute deviation, MAD, or something to that extent) or 2) to use the daily values, which provide a higher resolution.

Liebman et al. chose the cut-off point jmax = 2. Hence, Fig 3. contains fewer partials and is thus more symmetrical at the top of the sine compared to Fig 2. (A single partial at the base frequency would result in a pure sine wave.) If Epstein would have selected a higher resolution (e.g. jmax = 12) the transform would presumably only yield minor fluctuations with the additional components, or perhaps he lacked the computational power.

Through visual inspection of your data you appear to have 16-17 peaks. I would suggest you set jmax or the “bit depth” to either 6, 11, 16 or 17 (see figure) and compare the outputs. The higher the peaks, the more they contribute to the original complex waveform. So assuming a 17-band resolution or bit depth the 17th partial contributes minimally to the original waveform pattern compared to the 6th peak. However, with a 34 band-resolution you would detect a difference between even and odd peaks as suggested by the fairly constant valleys. The bit depth depends on your research question, whether you are interested in the peaks only or in both peaks and valleys, but also how exactly you wish to approximate the original series.

The Fourier analysis reduces your data points. If you were to inverse the function at a certain bit depth using a Fourier transform you could probably cross-check if the new mean estimates correspond to your original means. So, to answer your fourth question: the regression parameters you mentioned depend on the sensitivity and resolution that you require. If you do not wish for an exact fit, then by all means simply input the weekly means in the transform. However, beware that lower bit depth also reduces the data. For example, note how Epstein’s harmonic overlay on Lieberman and colleagues’ analysis misses the mid-point of the step function, with a skewed curve slightly to the right (i.e. temp. est. too high), in December in Figure 3.


Liebman and Colleagues’ Parameters:

  • Bit Depth: 2

Epstein’s Parameters:

  • Sample Rate: 12 [every month]
  • Sample Range: 24 months
  • Bit Depth: 6

Your Parameters:

  • Sample Rate: 365 [every day]

  • Sample Range: 5845 days

Exact Bit Depth Approach

Exact fit based on visual inspection. (If you have the power, just see what happens compared to lower bit-depths.)

  • Full Spectrum (peaks): 17
  • Full Spectrum (even/odd): 34

Variable Bit Depth Approach

This is probably what you wish to do:

  • Compare Peaks Only: 6, 11, 16, 17
  • Compare Even/Odd: 12, 22, 32, 34
  • Resynthesize and compare means

This approach would yield something similar to the comparison of Figures in Epstein if you inverse the transformation again, i.e. synthesise the partials into an approximation of the original time series. You could also compare the discrete points of the resynthesized curves to the mean values, perhaps even test for significant differences to indicate the sensitivity of your bit depth choice.


UPDATE 1:

Bit Depth

A bit – short for binary digit – is either 0 or 1. The bits 010101 would describe a square wave. The bit depth is 1 bit. To describe a saw wave you would need more bits: 0123210. The more complex a wave gets the more bits you need:

This is a somewhat simplified explanation, but the more complex a time series is, the more bits are required to model it. Actually, “1” is a sine wave component and not a square wave (a square wave is more like 3 2 1 0 – see attached figure). 0 bits would be a flat line. Information gets lost with reduction of bit depth. For example, CD-quality audio is usually 16 bit, but land-line phone quality audio is often around 8 bits.

Please read this image from left to right, focusing on the graphs:

FFT

You have actually just completed a power spectrum analysis (although at high resolution in your figure). Your next goal would be to figure out: How many components do I need in the power spectrum in order to accurately capture the means of the time series?

UPDATE 2


To Filter or not to Filter

I am not entirely sure how you would introduce the constraint in the regression as I am only familiar with interval constraints, but perhaps DSP is your solution. This is what I figured so far:

  • Step 1. Break down the series into sinus components through Fourier
    function on the complete data set (in days)

  • Step 2. Recreate the time series through an inverse Fourier
    transform, with the additional mean-constraint coupled to the
    original data: the deviations of the interpolations from the original
    means should cancel out each other (Harzallah, 1995).

My best guess is that you would have to introduce autoregression if I understand Harzallah (1995, Fig 2) correctly. So that would probably correspond to an infinite response filter (IIR)?

IIR http://paulbourke.net/miscellaneous/ar/

In summary:

  1. Derive means from Raw data
  2. Fourier Transform Raw data
  3. Fourier Inverse Transform transformed Data.
  4. Filter the result using IIR

Perhaps you could use an IIR filter without going through the Fourier analysis? The only advantage of the Fourier analysis as I see it is to isolate and determine which patterns are influential and how often they do reoccur (i.e. oscillate). You could then decide to filter out the ones that contribute less, for example using a narrow notch filter at the least contributing peak (or filter based on your own criteria). For starters, you could filter out the less contributing odd valleys that appear more like noise in the “signal”. Noise is characterized by very few cases and no pattern. A comb filter at odd frequency components could reduce the noise – unless you find a pattern there.

Here’s some arbitrary binning—for explanatory purposes only:
Can you see the noise in the valleys?

Oops – There’s an R Function for that!?

When searching for an IIR-filter I happen to discovered the R functions interpolate in the Signal package. Forget everything I said up to this point. The interpolations should work like Harzallah’s: http://cran.r-project.org/web/packages/signal/signal.pdf

Play around with the functions. Should do the trick.


UPDATE 3

interp1 not interp

case.interp1 <- interp1(x=(ts.frame$no.influ.cases[!is.na(ts.frame$no.influ.case)]),y=ts.frame$yearday[!is.na(ts.frame$no.influ.case)],xi=mean(WEEKLYMEANSTABLE),method = c("cubic"))

Set xi to the original weekly means.

Attribution
Source : Link , Question Author : COOLSerdash , Answer Author : noumenal

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