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9.4.3 Whittacker's interpolation formula

For time series analysis, the discrete samples in a series were represented in section 8.8 by the function
 
g(t) = f(t)I(t (9.26)
 

where

\begin{displaymath}I(t)=\sum_m\delta(t-m\Delta T) \ \ . \end{displaymath} (9.27)
 

The Fourier transform of the delta function was a series of delta functions in frequency space, and this led to mixing together of various frequency components or to aliasing. However, if the Fourier transform of f(t), $\tilde f(\nu)$, is zero for frequencies above the Nyquist frequency, no contamination of the signal by aliasing occurs, and the true Fourier transform can be recovered from

\begin{displaymath}\tilde f(\nu) = \tilde g(\nu)\tilde h(\nu)\end{displaymath} (9.28)
 

where

$\displaystyle {rl}\tilde h(\nu)$ $\textstyle = \Delta, \ \ \vert\nu\vert\le {{1}\over{2\Delta T}}$    
  $\textstyle =0\ \ {\rm otherwise}.$   (9.29)
 

The Fourier transform of $\tilde h(\nu)$ is

\begin{displaymath}h(t) = {{\sin(\pi t/\Delta)}\over{\pi t/\Delta}} \end{displaymath} (9.30)
 

so, in the case where there is no variance at frequencies higher than the Nyquist frequency, the underlying function can be recovered from the discrete series by using the formula

\begin{displaymath}f(t) = \int_{-\infty}^\infty h(t')g(t-t')dt' \ \ . \end{displaymath} (9.31)
 

This is Whittacker's interpolation formula, often used for interpolation between values of a time series for this reason. When (9.26) is substituted in (9.31) the result is

\begin{displaymath}f(t) = \sum_{n=-\infty}^\infty {{\sin[\pi(t-n\Delta T)/\DeltaT]}\over{\pi(t-n\Delta T)/\Delta T}} f(n\Delta T) \ \ . \end{displaymath} (9.32)
 

Because the terms decrease in magnitude as 1/n, and oscillate in sign, the summation converges fast enough to be practical. 


next up previous contents
Next: 9.4.4 The cubic spline Up: 9.4 Interpolation and extrapolation Previous: 9.4.2 Lagrange interpolation 


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