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hotpaw2
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A standard even length DFT has to be symmetric, not around the middle, but circularly symmetric around the first element x[0], to have a zero imaginary component, as that is the point of symmetry of all the (circular) basis vectors of the matrix transform.

By that definition of symmetry, your 20 point example is not symmetric.

For symmetry (matching the symmetry of the basis vectors), x[i] must = x[N-i], for all i modulo N

A standard even length DFT has to be symmetric, not around the middle, but circularly symmetric around the first element x[0], to have a zero imaginary component, as that is the point of symmetry of all the (circular) basis vectors of the matrix transform.

By that definition of symmetry, your 20 point example is not symmetric.

A standard even length DFT has to be symmetric, not around the middle, but circularly symmetric around the first element x[0], to have a zero imaginary component, as that is the point of symmetry of all the (circular) basis vectors of the matrix transform.

By that definition of symmetry, your 20 point example is not symmetric.

For symmetry (matching the symmetry of the basis vectors), x[i] must = x[N-i], for all i modulo N

Source Link
hotpaw2
  • 35.7k
  • 9
  • 48
  • 92

A standard even length DFT has to be symmetric, not around the middle, but circularly symmetric around the first element x[0], to have a zero imaginary component, as that is the point of symmetry of all the (circular) basis vectors of the matrix transform.

By that definition of symmetry, your 20 point example is not symmetric.