Fourier theory states that we can model any time series with a series of sine curves. The only major caveat is that the series must have no overall trend (tendency to increase or decrease in value with time); if it does, we can remove the linear trend, solve for the Fourier series, and then add it back to the trend. The result will be:
Original Time Series = Trend + Mean + Series sine curves
Each sine curve has three parameters:
Because sine or cosine curves oscillate about zero, adding the mean allows the series to have an average value that is not zero.
The periods at which we get the sine curves are not random. Several choices can be made, either using powers of two or all integers. In the case of all integers, the zero-th term has infinite period and is the mean; the first term has a period equal to the length of the time series; the second term has a period equal to ½ the length of the time series (two cycles over the entire time period); the third term has a period equal to 1/3 the length of the time series (three cycles over the entire time period); and the nth term has a period equal to 1/n of the length of the time series (n cycles over the entire time period). If we only used powers of two, we would only have periods of 1, ½, ¼, 1/8, 1/16, 1/32…. of the overall period.
In the extreme case, we need the number of terms in the series to equal the number of points in the data series. In practical terms we can generally recreate any series, even a step function, with a much smaller number of terms. To look for periodicity in the series, we can order the terms by their amplitude, and pick out periods with extremely high amplitudes.
To find periodicity in the data, we need to sample over a number of periods. We can resolve very fine differences in period at short periods, but not at the long periods. If the period is long compared to overall time series, we may not be able to accurately measure it since an even number of periods will not fit into the data series.
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This graph shows a topographic profile across Mt. St. Helens in red. Superimposed in green is the fit with a series of 50 sine curves. The fit is remarkably good, except at the two ends of the profile where the Fourier curve has to repeat (note that it drops on the left side and rises on the right, which the real profile does not do). In light blue we have the sum of the two strongest terms in the series, and in dark blue is the second strongest term. |
The table below shows the data for the strongest terms in the Fourier series to represent Mt. St. Helens. Note that for each a period, phase, and amplitude appear. Each has also has a % SS for percentage of the sum of squares, to indicate how much of the variance each explains. Note that in this case a single sine curve explains almost 95% of the variance; Mt. St. Helens is almost a sine curve (the fortuitous length of the profile probably helps this).
Mean = 1668.45 Variance = 198624.668
Component Period Phase Amp % SS
1 602.000 278.3 612.845 94.54
4 150.500 332.4 75.208 1.42
5 120.400 134.0 71.785 1.30
6 100.333 352.5 64.491 1.05
3 200.667 141.8 62.420 0.98
8 75.250 359.9 28.498 0.20
2 301.000 91.1 21.209 0.11
7 86.000 146.0 19.324 0.09
11 54.727 0.9 12.635 0.04
12 50.167 271.4 12.390 0.04
9 66.889 312.5 10.960 0.03
14 43.000 292.8 9.984 0.03
10 60.200 298.2 9.586 0.02
Total Percent sum of Squares = 100.0Note that component one has a period of 602, the number of points in the data set, component 2 has a period of 301 (602/2) component 3 has a period of 150.5 (602/3), and so on. If there were any periodicity at a value of 200 or 250 it would be missed by this analysis.
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This graph shows the Annapolis tide in red, with the fit in green. The minimum resolved period does not capture the true diurnal and semidiurnal tide periods, but does capture some of the longer periods. Overall, as shown in the table below, the series only explains 30% of the variance. |
Component Period Phase Amp % SS Cum % SS
1 2880.000 171.2 0.260 18.122 18.12
4 720.000 146.8 0.117 3.667 21.79
2 1440.000 168.6 0.109 3.174 24.96
8 360.000 14.0 0.083 1.860 26.82
3 960.000 164.0 0.080 1.706 28.53
9 320.000 168.1 0.053 0.751 29.28
5 576.000 73.5 0.032 0.280 29.56
50 57.600 3.0 0.004 0.003 30.48
Total Percent sum of Squares = 30.4845With only 50 terms and a minimum resolved period of 57 hours, the series does not do a good job of capturing the tidal variation. The only strong amplitude reflects the annual cycle of the Annapolis tide, where summer water levels are significantly higher than winter levels.
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This graph shows the Annapolis tide in red, with the fit in green. This fit captures 99.7% of the variance. |
Component Period Phase Amp % SS Cum % SS 232 12.414 151.2 0.416 46.263 46.26
1 2880.000 171.2 0.260 18.122 64.39
112 25.714 295.5 0.126 4.243 68.63
4 720.000 146.8 0.117 3.667 72.30
2 1440.000 168.6 0.109 3.174 75.47
111 25.946 123.5 0.095 2.424 77.89
121 23.802 50.8 0.091 2.201 80.09
120 24.000 36.2 0.090 2.180 82.27
8 360.000 14.0 0.083 1.860 84.14
240 12.000 56.4 0.083 1.836 85.97
3 960.000 164.0 0.080 1.706 87.68
119 24.202 68.0 0.072 1.375 89.05
227 12.687 348.4 0.065 1.121 90.17
231 12.468 324.9 0.054 0.777 90.95
107 26.916 136.7 0.054 0.767 91.72
9 320.000 168.1 0.053 0.751 92.47
233 12.361 156.5 0.050 0.659 93.13
113 25.487 287.3 0.044 0.528 93.66
118 24.407 71.5 0.040 0.419 94.07
122 23.607 47.8 0.037 0.362 94.44
228 12.632 162.2 0.035 0.334 94.77
116 24.828 82.1 0.033 0.294 95.07
5 576.000 73.5 0.032 0.280 95.35
114 25.263 279.0 0.031 0.255 95.60
110 26.182 131.5 0.030 0.246 95.85
117 24.615 77.0 0.030 0.245 96.09
250 11.520 141.1 0.005 0.006 99.42
Total Percent sum of Squares = 99.7602With 250 terms in the series, note the preponderance of periods around 12 and 24 hours, and that the series almost perfectly recreates the predicted tide series.
There appears to be a bug, which can let the sum of the squares explained exceed 100%.
Last revised 3/6/2005