python - What is numpy.fft.rfft and numpy.fft.irfft and its equivalent code in MATLAB -

i converting python code matlab , 1 of code uses numpy rfft. in documentation of numpy, says real input.

compute one-dimensional discrete fourier transform real input.

so did in matlab using abs results different.

python code

ffta = np.fft.rfft(a)  

matlab code

ffta = abs(fft(a)); 

what have misunderstood?

the real fft in numpy uses fact fourier transform of real valued function "skew-symmetric", value @ frequency k complex conjugate of value @ frequency n-k k=1..n-1 (the correct term hermitian). therefore rfft returns part of result corresponds nonpositive frequences.

for input of size n rfft function returns part of fft output corresponding frequences @ or below n/2. therefore output of rfft of size n/2+1 if n (all frequences 0 n/2), or (n+1)/2 if n odd (all frequences 0 (n-1)/2). observe function floor(n/2+1) returns correct output size both , odd input sizes.

so reproduce rfft in matlab

function rfft = rfft(a)      ffta = fft(a);      rfft = ffta(1:(floor(length(ffta)/2)+1)); end 

for example

a = [1,1,1,1,-1,-1,-1,-1]; rffta = rfft(a) 

would produce

rffta =   columns 1 through 3:     0.00000 + 0.00000i   2.00000 - 4.82843i   0.00000 + 0.00000i      columns 4 through 5:     2.00000 - 0.82843i   0.00000 + 0.00000i 

now compare python

>>> np.fft.rfft(a) array([ 0.+0.j        ,  2.-4.82842712j,  0.-0.j        ,           2.-0.82842712j,  0.+0.j        ]) 

reproducing irfft

to reproduce basic functionality of irfft need recover missing frequences rfft output. if desired output length even, output length can computed input length 2 (m - 1). otherwise should 2 (m - 1) + 1.

the following code work.

function irfft = irfft(x,even=true)      n = 0; % output length      s = 0; % variable hold index of highest             % frequency below n/2, s = floor((n+1)/2)      if (even)         n = 2 * (length(x) - 1 );         s = length(x) - 1;      else         n = 2 * (length(x) - 1 )+1;         s = length(x);      endif      xn = zeros(1,n);      xn(1:length(x)) = x;      xn(length(x)+1:n) = conj(x(s:-1:2));      irfft  = ifft(xn); end 

now should have

>> irfft(rfft(a)) ans =     1.00000   1.00000   1.00000   1.00000  -1.00000  -1.00000  -1.00000  -1.00000 

and also

abs( irfft(rfft(a)) - ) < 1e-15 

for odd output length get

>> irfft(rfft(a(1:7)),even=false) ans =     1.0000   1.0000   1.0000   1.0000  -1.0000  -1.0000  -1.0000