How to resolve the algorithm Fast Fourier transform step by step in the Julia programming language
How to resolve the algorithm Fast Fourier transform step by step in the Julia programming language
Table of Contents
Problem Statement
Calculate the FFT (Fast Fourier Transform) of an input sequence. The most general case allows for complex numbers at the input and results in a sequence of equal length, again of complex numbers. If you need to restrict yourself to real numbers, the output should be the magnitude (i.e.: sqrt(re2 + im2)) of the complex result. The classic version is the recursive Cooley–Tukey FFT. Wikipedia has pseudo-code for that. Further optimizations are possible but not required.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Fast Fourier transform step by step in the Julia programming language
This code defines a Julia function, fft
, which computes the Fast Fourier transform (FFT) of a complex or real-valued array. To do so, it uses a recursive algorithm to compute the FFT in place.
First, we ensure that the input array is of type Array{Complex{Float64},1}
, i.e., an array of complex numbers with 64-bit floating-point precision and one dimension. If the input array is not of the correct type, we convert it using the convert
function.
Then, we initialize an array y
of the correct size to store the complex fourier coefficients of the array. We also initialize the array w
of weights used in the FFT algorithm.
Next, we iterate over the elements of the array a
, and for each element, we compute the corresponding complex fourier coefficient and store it in the array y
.
Finally, we return the array y
containing the complex fourier coefficients of the array a
.
Source code in the julia programming language
using FFTW # or using DSP
fft([1,1,1,1,0,0,0,0])
8-element Array{Complex{Float64},1}:
4.0+0.0im
1.0-2.41421im
0.0+0.0im
1.0-0.414214im
0.0+0.0im
1.0+0.414214im
0.0+0.0im
1.0+2.41421im
function fft(a)
y1 = Any[]; y2 = Any[]
n = length(a)
if n ==1 return a end
wn(n) = exp(-2*π*im/n)
y_even = fft(a[1:2:end])
y_odd = fft(a[2:2:end])
w = 1
for k in 1:Int(n/2)
push!(y1, y_even[k] + w*y_odd[k])
push!(y2, y_even[k] - w*y_odd[k])
w = w*wn(n)
end
return vcat(y1,y2)
end
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