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8.11 Removing the trend and tapering the ends of a time series

Because analysis in terms of the Fourier series is based on the assumption that the time series is periodic, analysis of a finite segment of data results in characterization of a series that matches that segment but then repeats the series periodically beyond the measured segment. This can cause a problem when there is a trend (i.e., a mean slope) in the series, because at the end of the series there is an artificial jump back to the beginning point, as illustrated in Fig. 8.6.  The Fourier decomposition of this sawtooth pattern is a broad frequency spectrum that contaminates the desired spectrum. For this reason, it is conventional to remove a trend such as that shown from the time series before calculating the variance spectrum. (The mean is usually also removed, but the reason for this is that it improves the numerical precision of computer routines in cases where the mean is large compared to the variance.) If there is evidence for a higher-order tendency, removal of that tendency can be justified in the same way.




 
Figure 8.6: Periodic repetition of a time series with a trend, showing the effective sawtooth pattern that results.


Removing the trend does not fully solve the problem, however, because even with trend removal there are still artificial jumps or correlations at the ends of the time series that arise from the assumed periodicity. For example, the jump from the last point in the series to the first point in the assumed periodic extension of the series is a jump that probably does not occur in the real time series, so it introduces an artificial high-frequency contribution to the spectrum. For this reason, the series is often tapered slowly to zero at the ends to minimize these wrap-around effects.
 


Exercise 8.4: To illustrate further the effect of the assumption that the series in periodic, and to show the effects of aliasing, consider the following time series which has an analytic Fourier transform:
f(t) $\textstyle =1-\vert t\vert, \ \vert t\vert\le 1$    
  $\textstyle =0, \ \ \vert t\vert>1.$   (8.60)
 

The Fourier transform of this function is

\begin{displaymath}\hat f(\nu) = {{\sin^2(\pi\nu)}\over{(\pi\nu)^2}} \ .\end{displaymath} (8.61)
 

Suppose that this function is sampled over the time interval from -1 to 1, in 16 samples so that the Fourier coefficients can be determined at the frequencies $n\delta\nu$ where $\delta\nu=0.5$ and $0\le n\le 4$. (a.) Evaluate the Fourier coefficients, using (8.11), that represent the repeating periodic series matching f(t) in the interval from -1 to 1, and compare to the exact coefficients obtained from the Fourier transform. (b). Evaluate the effects of aliasing in this case, and show that aliasing accounts for the difference obtained in (a).



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