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Topic: derivative of discrete fourier transform interpolation
Replies: 8   Last Post: Jun 6, 2013 7:24 AM

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Steven G. Johnson

Posts: 413
Registered: 12/6/04
Re: derivative of discrete fourier transform interpolation
Posted: Sep 30, 2005 10:25 PM
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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_{n-k}=X_k^*, by the
Hermitian symmetry of the output. Equivalently, you are using the
aliasing property to think of the X_{n-k} output, for k > n/2, as the
X_{-k} amplitude (i.e. a negative frequency -k), and so you get
complex-conjugate 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*
mean-square 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.

Steven G. Johnson

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