Fourier Transformation

Goal: to expand a function $f(x)$

  • i.e. write it as a linear combination
  • $f(x) = a_0 + a_1 \cos x + b_1 \sin x + a_2 \cos 2x + b_2 \sin 2x + \ ...$


Basis

These functions $\cos nx$ and $\sin nx$ are orthogonal

  • they form an orthogonal basis
  • so basis is $\big[ 1, \cos x, \sin x, \cos 2x, \sin 2x, \ ... \big]$
  • inner product in functions space is $\langle f, g \rangle = \int\limits_0^{2\pi} f(x) g(x) \, dx$
    • because these functions are all periodic and analytical, we take the integral only over $[0, 2 \pi]$
  • e.g. $\int \sin x \, \cos x \, dx = 0.5 (\sin x)^2 \mathop{|}\limits_0^{2\pi} = 0$
  • so we have orthogonal $\infty$-dimensional basis for this functional space
  • and we want to express some function $f(x)$ in this basis


Coefficients

Let's start with $a_0$

  • $f(x) = a_0 1 + a_1 \cos x + b_1 \sin x + a_2 \cos 2x + b_2 \sin 2x + \ ...$
  • let's multiply by \cos x and integrate
  • $\int f(x) \cos x \, dx = 0 + a_1 \underbrace{\int \cos x \, \cos x \, dx}_{\pi} + 0 + 0 + \ ...$
  • $\int f(x) \cos x \, dx = a_1 \pi$
  • so, $a_1 = \cfrac{1}{\pi} \int f(x) \cos x \, dx$

We can do it for all the coefficients

  • this is called "Euler's formula"


Discrete Fourier Transform

Fourier Matrix

Let $F_n$ be a Fourier matrix:

  • [math]F_n = \begin{bmatrix} 1 & 1 & 1 & \cdots & 1 \\ 1 & w^2 & w^2 & \cdots & w^{n - 1} \\ 1 & w^3 & w^4 & \cdots & w^{2(n-1)} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & w^{n-1} & w^{2(n-1)} & \cdots & w^{(n-1)^2} \\ \end{bmatrix}[/math]
  • each element is $(F_n)_{ij} = w^{ij}$ for all $i,j$ (indexes of $F_n$)
  • matrix $F_n$ is a symmetric

where $w \in \mathbb C$:

  • $w^n = 1$, so $w = \sqrt[n]{1}$
  • $w = \exp \left( i \ \cfrac{2\pi}{n} \right) = \cos \cfrac{2\pi}{n} + i \, \sin \cfrac{2\pi}{n}$
  • e.g. $w^2 = \exp \left( 2 \ \cfrac{2\pi}{n} \right)$
  • $w$ is $n$th root of 1 ("roots of unity")


Example

  • $n = 6, w = \exp \left( 2 \ \cfrac{2\pi}{6} \right) = \exp \left( \cfrac{2\pi}{3} \right)$
  • 6f050622c0eb47e4bd5eeb8c3dfcd463.png

$n = 4$

  • $w = \exp \left( i \ \cfrac{2\pi}{4} \right) = i$
  • so we have $1, i, i^2 = -1, i^3 = 1$
  • thus, [math]F_4 = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & w^2 & w^3 & w^4 \\ 1 & w^3 & w^4 & w^5 \\ 1 & w^5 & w^5 & w^6 \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & i^2 & i^3 & i^4 \\ 1 & i^3 & i^4 & i^5 \\ 1 & i^5 & i^5 & i^6 \\ \end{bmatrix} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & i & -1 & -i \\ 1 & -1 & 1 & -1 \\ 1 & -i & -1 & i \\ \end{bmatrix}[/math]
  • for 4-point Fourier transform: for a vector with 4 components


Columns of $F_n$ are orthogonal

  • Let's check it for $n=4$:
  • [math]\overline{\Big[ 1 \ i \ -1 \ -i \Big]} \begin{bmatrix} 1 \\ -i \\ -1 \\ i \end{bmatrix} = 1 + 1 - 1 - 1 = 0[/math]
  • but they are not orthonormal
  • e.g. for $F_4$ all columns have length 2
  • so let's define a Fourier Matrix $W_4$ as $W_4 = \cfrac{1}{2} F_4$
  • now $W_4^H W_4 = I$
  • so let $W_n = \cfrac{1}{\sqrt{n}} F_n$
  • we call this $W_n$ a Fourier Matrix


Discrete Fourier Transform

So, given a matrix $W_n$ and a vector $\mathbf u \in \mathbb C^{n}$ (or in $\mathbb R^{n}$)

  • $\mathbf u \cdot W_n$ is the direct transformation
  • $\mathbf u \cdot W_n^{-1}$ is the inverse transformation


Fast Fourier Transform

The idea of FFT

  • There's a connection between $W_6$ and $W_3$, $W_8$ and $W_4$, $W_{2n}$ and $W_n$

Example:

  • suppose we have $W_{64}$, it's a $64 \times 64$ matrix
  • $w$ is 64th root of 1
  • in $W_{32}$, $w$ is 32th root of 1
  • so $w_{64}^2 = w_{32}$
  • 96f2c3858129488290280b709be08893.png

How can we use this fact?

  • we want to go from $W_64$ to a matrix [math]\left[ \begin{array}{c|c} W_{32} & 0 \\ \hline 0 & W_{32} \\ \end{array} \right][/math]
  • i.e. factorize $W_{64}$ in terms of $W_{32}$
  • can factorize it as [math]W_{64} = \begin{bmatrix} I_{32} & D_{32} \\ I_{32} & -D_{32} \end{bmatrix} \begin{bmatrix} W_{32} & 0 \\ 0 & W_{32} \end{bmatrix} P_{64}[/math]
    • where $P_n$ is a $n \times n$ permutation matrix [math]P_n = \begin{bmatrix} 1 & & & & & & \\ & & 1 & & & & \\ & & & & \ddots & & \\ & & & & & 1 & \\ \hline & 1 & & & & & \\ & & & 1 & & & \\ & & & & \ddots & & \\ & & & & & & 1 \\ \end{bmatrix}[/math]
    • first, in $P$ we have rows with even columns containing $1$, and then, in the second half, rows that contain $1$ in odd columns (here we start indexing columns from 0)
    • $P_n$ takes even-numbered components first, and then odd-numbered
    • $D_n$ is a diagonal matrix,
    • [math]D_n = \begin{bmatrix} 1 & & & & \\ & w & & & \\ & & w^2 & & \\ & & & \ddots & \\ & & & & w^{n/2 - 1} \end{bmatrix}[/math]
  • now we can break $W_{32}$ down in the same way!
    • [math]W_{32} = \begin{bmatrix} I_{16} & D_{16} \\ I_{16} & -D_{16} \end{bmatrix} \begin{bmatrix} W_{16} & 0 \\ 0 & W_{16} \end{bmatrix} P_{32}[/math]
  • use recursion


This way we reduce computation from $O(n^2)$ to $\cfrac{n}{2} \, \log_2 n$


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