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Inverse Fourier Transform of a Double-Precision Vector
 Syntax `idbldft(`c`)` See Also idft , dbldft

Description
Returns the column by column inverse discrete transform of `c`, where `c` is a double-precision matrix with an even number of rows. The numbers 2, 3, 5, and 7 must be the only prime factors of the number of rows in `c`. This routine is faster and uses less memory than the `idft` function but it only applies to the case where the vector `c` represents a complex vector that is conjugate symmetric. If `N` is half the number of rows in `c` and for all `k` between 2 and `N`, and all `j` is between 1 and the column dimension of `c` define the complex matrix `z` by ```                       z(1, j) = c(1, j)                   z(N + 1, j) = c(2, j)      the real part of z(k, j) = c(2 k - 1, j) the imaginary part of z(k, j) = c(2 k, j)             z(2 N + 2 - k, j) = the complex conjugate of z(k, j) ```The return value of the `idbldft` function is the matrix `w` defined by ```      w(k, j) = (1 / (2 N)) * sum_{i=1}^{2 N}           z(i, j) exp[ 2 pi sqrt{-1} (i - 1) (k - 1) / (2 N)] ```Thus `w` is the column by column inverse discrete Fourier transform of `z` (`w` is real because `z` is conjugate symmetric).

Example
If you enter ```      c = double({0, 1, 0, 0}) ``` only the term with `i = 3` in the summation defining `w` is nonzero. Also note that `2 N` is 4 and element `k` of the inverse Fourier transform of `c` is given by ```      w(k) = (1/4) exp[ pi sqrt{-1} (k - 1) ] ```which is given by the following table ```      w(1) = +1/4       w(2) = -1/4      w(3) = +1/4      w(4) = -1/4 ```(Note that all of the elements of `w` are real.) If you continue this example by entering ```      idbldft(c) ``` O-Matrix will respond ```      {      .25      -.25      .25      -.25      } ```