Need help in understanding the following FFT code (line %1 to %4)

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Siddhant Sharma
Siddhant Sharma 2020년 11월 2일
function [ y ] = FFT_DIT_R2( x )
p=nextpow2(length(x));
x=[x zeros(1,(2^p)-length(x))];
N=length(x);
S=log2(N);
Half=1;
x=bitrevorder(x);
for stage=1:S;
for index=0:(2^stage):(N-1);
for n=0:(Half-1);
pos=n+index+1;
pow=(2^(S-stage))*n; %1
w=exp((-1i)*(2*pi)*pow/N); % 2
a=x(pos)+x(pos+Half).*w; % 3
b=x(pos)-x(pos+Half).*w; % 4
x(pos)=a;
x(pos+Half)=b;
end;
end;
Half=2*Half;
end;
y=x;
end

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