- #1
SchroedingersLion
- 211
- 56
- TL;DR
- Question to DFTs
Greetings!
I am getting started with Python's Fast Fourier Transform, and I noticed a missunderstanding on my part.
I always thought that the k's in the expression of the DFT resemble actual wavenumbers ## \frac{2\pi}{\lambda}## that form the waves the signal is composed of.
But the actual meaning seems to depend on the length of the x-interval that holds the signal.
Wiki:
$$
X_{k} = \sum_{n=0}^{N-1}x_n e^{-2\pi i k \frac n N} ~~~~(1) \\
x_n = \frac 1 N \sum_{k=0}^{N-1}X_k e^{2\pi i k \frac n N}~~~~ (2),
$$
where the ##x_n## are the ##N## signal values that are equidistantly positioned at the points ##\frac n N ##, i.e. in the interval ## [0, 1] ##.
The expressions are ##N-##periodic in ##k##, so it makes no difference whether one uses ##k\in [0,N-1]## or ##k\in [-\frac N 2 +1, \frac N 2]## (suppose ##N## is even).
In this representation, however, the ##k## are no wavenumbers. Rather, they are wavenumbers divided by ##2\pi##, i.e. ##k=\frac 1 \lambda ##.In another derivation of a book, the points ##x_n## are equidistantly spaced across ##[0, 2\pi]##.
(1) then becomes
$$ X_{k} = \sum_{n=0}^{N-1}x_n e^{-i k x_n}. $$
Here, however, it is ##k=\frac {2\pi} {\lambda}##.
Now I was wondering, is it generally true that the ##k## are given by ## \frac L \lambda ## with ##L## as the length of the ##x-##interval?
It somehow makes sense, as the sampling interval does not enter anywhere else in these formulas.SL
I am getting started with Python's Fast Fourier Transform, and I noticed a missunderstanding on my part.
I always thought that the k's in the expression of the DFT resemble actual wavenumbers ## \frac{2\pi}{\lambda}## that form the waves the signal is composed of.
But the actual meaning seems to depend on the length of the x-interval that holds the signal.
Wiki:
$$
X_{k} = \sum_{n=0}^{N-1}x_n e^{-2\pi i k \frac n N} ~~~~(1) \\
x_n = \frac 1 N \sum_{k=0}^{N-1}X_k e^{2\pi i k \frac n N}~~~~ (2),
$$
where the ##x_n## are the ##N## signal values that are equidistantly positioned at the points ##\frac n N ##, i.e. in the interval ## [0, 1] ##.
The expressions are ##N-##periodic in ##k##, so it makes no difference whether one uses ##k\in [0,N-1]## or ##k\in [-\frac N 2 +1, \frac N 2]## (suppose ##N## is even).
In this representation, however, the ##k## are no wavenumbers. Rather, they are wavenumbers divided by ##2\pi##, i.e. ##k=\frac 1 \lambda ##.In another derivation of a book, the points ##x_n## are equidistantly spaced across ##[0, 2\pi]##.
(1) then becomes
$$ X_{k} = \sum_{n=0}^{N-1}x_n e^{-i k x_n}. $$
Here, however, it is ##k=\frac {2\pi} {\lambda}##.
Now I was wondering, is it generally true that the ##k## are given by ## \frac L \lambda ## with ##L## as the length of the ##x-##interval?
It somehow makes sense, as the sampling interval does not enter anywhere else in these formulas.SL