
Re: derivative of discrete fourier transform interpolation
Posted:
Sep 30, 2005 10:25 PM


Peter Spellucci wrote: > the FFT gives a real sin/cos series only for special N and in this case your sum > is running from N/2 to N/2 . the real sin/cos (Fourier ) sum can then be > differentiated in the usual manner with no trouble.
Special N?? The DFT of real inputs x_n, for any N, can be interpreted as the amplitudes of a real sin/cos series (with real derivatives).
To see this, you pair the output X_k with X_{nk}=X_k^*, by the Hermitian symmetry of the output. Equivalently, you are using the aliasing property to think of the X_{nk} output, for k > n/2, as the X_{k} amplitude (i.e. a negative frequency k), and so you get complexconjugate pairs of sinusoids. (The k=n/2 Nyquist element, for even n, must be treated specially. Since it is purely real, it can be thought of as 1/2 k=n/2 and 1/2 k=+n/2.)
In order to take the derivative, you need to realize that the interpolation implied by the DFT, and hence the slope, is not unique because of aliasing. Normally, however, you want the interpolation corresponding to exactly the aliasing described above: the k > n/2 outputs are *negative* frequency amplitudes.
This choice means that your frequencies run from N/2+1 to N/2 (for even N). Not only does it guarantee real derivatives from real inputs, but it also corresponds to the interpolation with the *minimal* meansquare slope.
Hence, it is the interpretation of the DFT that you want to use for solving differential equations by spectral methods, regardless of whether your data are real.
Cordially, Steven G. Johnson

