# IFFT function in julia

I am trying to implement the below matrix computation from matlab in julia. But the ifft function gives different output than obtained in matlab

``````a = zeros(5,10);
for i = 1:5

a(i,:) = ifft((c(i,:).*(d)));
end

``````

where ‘c’ is a non- zero 510 matrix and ‘d’ is a non- zero 110 matrix.

please provide a minimal working example to show the difference

1 Like

I can’t reproduce the problem.

Julia:

``````using FFTW
c = [-0.3  0.7  0.4  1.1 -1.3 -0.1 -1.4  0.3  0.7 -0.1;
0.1 -2.0 -1.9 -1.4 -0.6 -2.4 -0.8 -0.9 -2.4 -2.2;
0.2 -1.5 -2.5  2.2 -0.2 -1.0 -0.1  2.3  1.1 -1.9;
-1.5  0.5  0.7  2.3  1.5  2.3  1.0  1.3  1.3  2.1;
1.8  1.5 -0.6  0.4  2.0  0.4 -0.2  2.0  0.8 -1.6 ];
d = [ 0.9 -0.0 -1.0 0.6 -0.2 -0.6 0.7 -1.7 -0.9 -0.6 ]';
a = zeros(Complex, 5, 10);
for i = 1:5
a[i,:] = ifft(c[i,:] .* d);
end
round.(a, digits=4)
``````
``````5×10 Matrix{ComplexF64}:
-0.175+0.0im  -0.0064+0.2025im   0.0298-0.1789im   0.0384+0.0299im   0.0052+0.013im   -0.229+0.0im   0.0052-0.013im    0.0384-0.0299im   0.0298+0.1789im  -0.0064-0.2025im
0.716+0.0im   0.1115-0.2877im  -0.2041-0.0662im   -0.462+0.0937im   0.2286-0.3182im   0.026+0.0im   0.2286+0.3182im   -0.462-0.0937im  -0.2041+0.0662im   0.1115+0.2877im
0.081+0.0im   0.1794+0.7688im   0.1997-0.2212im  -0.4099-0.6105im  -0.0452+0.092im    0.251+0.0im  -0.0452-0.092im   -0.4099+0.6105im   0.1997+0.2212im   0.1794-0.7688im
-0.629+0.0im  -0.1634+0.4014im  -0.0811+0.0315im   0.1384-0.2139im  -0.2869+0.4296im   0.065+0.0im  -0.2869-0.4296im   0.1384+0.2139im  -0.0811-0.0315im  -0.1634-0.4014im
-0.148+0.0im   0.4013+0.4im      0.4163-0.2029im  -0.1063-0.4076im   0.0027+0.1795im    0.34+0.0im   0.0027-0.1795im  -0.1063+0.4076im   0.4163+0.2029im   0.4013-0.4im
``````

Same output in MATLAB as your code with the same input matrix/vector:

``````c = [-0.3  0.7  0.4  1.1 -1.3 -0.1 -1.4  0.3  0.7 -0.1;
0.1 -2.0 -1.9 -1.4 -0.6 -2.4 -0.8 -0.9 -2.4 -2.2;
0.2 -1.5 -2.5  2.2 -0.2 -1.0 -0.1  2.3  1.1 -1.9;
-1.5  0.5  0.7  2.3  1.5  2.3  1.0  1.3  1.3  2.1;
1.8  1.5 -0.6  0.4  2.0  0.4 -0.2  2.0  0.8 -1.6 ];
d = [ 0.9 -0.0 -1.0 0.6 -0.2 -0.6 0.7 -1.7 -0.9 -0.6 ];
a = zeros(5,10);
for i = 1:5
a(i,:) = ifft((c(i,:).*(d)));
end
a
``````
``````a =

Columns 1 through 8

-0.1750 + 0.0000i  -0.0064 + 0.2025i   0.0298 - 0.1789i   0.0384 + 0.0299i   0.0052 + 0.0130i  -0.2290 + 0.0000i   0.0052 - 0.0130i   0.0384 - 0.0299i
0.7160 + 0.0000i   0.1115 - 0.2877i  -0.2041 - 0.0662i  -0.4620 + 0.0937i   0.2286 - 0.3182i   0.0260 + 0.0000i   0.2286 + 0.3182i  -0.4620 - 0.0937i
0.0810 + 0.0000i   0.1794 + 0.7688i   0.1997 - 0.2212i  -0.4099 - 0.6105i  -0.0452 + 0.0920i   0.2510 + 0.0000i  -0.0452 - 0.0920i  -0.4099 + 0.6105i
-0.6290 + 0.0000i  -0.1634 + 0.4014i  -0.0811 + 0.0315i   0.1384 - 0.2139i  -0.2869 + 0.4296i   0.0650 + 0.0000i  -0.2869 - 0.4296i   0.1384 + 0.2139i
-0.1480 + 0.0000i   0.4013 + 0.4000i   0.4163 - 0.2029i  -0.1063 - 0.4076i   0.0027 + 0.1795i   0.3400 + 0.0000i   0.0027 - 0.1795i  -0.1063 + 0.4076i

``````
1 Like

Thank you, the c and d input matrices I am using are complex and am I getting little difference

``````c =

Columns 1 through 5

97.9713 + 0.0000i   3.7743 -90.6967i -70.9956 - 6.0008i  -5.8363 +44.4675i  18.1929 + 3.5097i
97.9619 + 0.0000i   3.7759 -90.6841i -70.9751 - 6.0029i  -5.8372 +44.4393i  18.1620 + 3.5090i
97.9337 + 0.0000i   3.7791 -90.6464i -70.9149 - 6.0057i  -5.8356 +44.3569i  18.0727 + 3.5015i
97.8861 + 0.0000i   3.7812 -90.5838i -70.8161 - 6.0049i  -5.8257 +44.2231i  17.9287 + 3.4810i
97.8153 + 0.0000i   3.7835 -90.4930i -70.6775 - 6.0029i  -5.8118 +44.0387i  17.7319 + 3.4534i

Columns 6 through 10

0.1821 + 1.6917i  11.9049 + 2.6249i   3.7467 -12.7256i  -7.3258 - 2.8909i  -0.7121 + 0.1283i
0.1806 + 1.7176i  11.9192 + 2.6249i   3.7430 -12.7262i  -7.3161 - 2.8828i  -0.7013 + 0.1158i
0.1706 + 1.7907i  11.9568 + 2.6305i   3.7372 -12.7221i  -7.2808 - 2.8632i  -0.6726 + 0.0715i
0.1455 + 1.9077i  12.0153 + 2.6483i   3.7360 -12.7122i  -7.2198 - 2.8388i  -0.6323 - 0.0040i
0.1124 + 2.0667i  12.0940 + 2.6710i   3.7329 -12.6976i  -7.1353 - 2.8048i  -0.5774 - 0.1089i

``````

d=

``````  Columns 1 through 5

3.1922 + 0.0000i   0.3826 - 6.7552i -15.0138 - 1.4812i  -3.4435 +23.5588i  30.5882 + 6.0793i

Columns 6 through 10

9.3782 -37.2328i -44.3985 -13.4759i -18.3235 +51.3590i  57.3788 +23.7564i  29.7857 -62.8610i
``````
``````a = zeros(5,10);

for i = 1:5
a(i,:) = ifft(c(i,:) .* d);
end
``````

a in MATLAB:
a =

``````   1.0e+02 *

Columns 1 through 5

0.0578 - 0.0712i   0.0089 + 0.0020i  -0.1765 + 0.2240i  -0.9484 + 0.7314i   2.4748 - 3.7660i
0.0576 - 0.0718i   0.0085 + 0.0015i  -0.1773 + 0.2235i  -0.9504 + 0.7310i   2.4760 - 3.7627i
0.0567 - 0.0735i   0.0070 - 0.0001i  -0.1800 + 0.2221i  -0.9567 + 0.7301i   2.4797 - 3.7510i
0.0549 - 0.0762i   0.0043 - 0.0023i  -0.1845 + 0.2200i  -0.9670 + 0.7292i   2.4863 - 3.7306i
0.0525 - 0.0799i   0.0005 - 0.0054i  -0.1908 + 0.2173i  -0.9809 + 0.7281i   2.4951 - 3.7025i

Columns 6 through 10

2.0623 - 0.3361i  -0.6152 + 3.8925i  -0.0011 - 0.3264i   0.1463 - 0.2339i   0.1185 - 0.1163i
2.0594 - 0.3350i  -0.6109 + 3.8924i  -0.0006 - 0.3275i   0.1464 - 0.2346i   0.1184 - 0.1169i
2.0529 - 0.3325i  -0.5984 + 3.8916i   0.0005 - 0.3309i   0.1464 - 0.2369i   0.1180 - 0.1189i
2.0440 - 0.3306i  -0.5787 + 3.8899i   0.0020 - 0.3367i   0.1463 - 0.2408i   0.1171 - 0.1220i
2.0318 - 0.3277i  -0.5516 + 3.8872i   0.0040 - 0.3446i   0.1460 - 0.2461i   0.1158 - 0.1264i
``````

In Julia,

a =

``````5×10 Array{Complex,2}:
6.10126-7.5036im    1.08646-0.29492im   …  12.2904-11.8982im
6.0691-7.54846im   1.03751-0.332524im     12.2716-11.9514im
5.95686-7.69857im  0.870022-0.454864im      12.205-12.1323im
5.76391-7.94203im  0.587843-0.649582im     12.0858-12.4299im
5.49364-8.27828im  0.194955-0.91658im      11.9176-12.8427im

``````

Do these examples have to be so big? Can you not show the problem with e.g. just a 2x4 matrix? And can you please include both the full MATLAB and the full Julia code that you used (including the assignments to c and d, as well as the using statement), such that we have nothing other to do than to copy&paste to reproduce exactly what you are getting? Otherwise trying to reproduce your matrices in Julia from your MATLAB outputs is quite labour intensive.

1 Like

Julia always calculates the fft over all dimensions.

Maybe Matlab does that differently and you maybe need so specify over which dimensions you want to transform?

3 Likes

@Sukera already answered almost the same question for `fft`, which boiled down to the fact that Matlab by default computes the `fft` for each column, whereas Julia computes a multi-dimensional one.

The same applies to `ifft`. I’d like to gently suggest looking at the documentation, `?ifft` in Julia and `help ifft` in Matlab.

1 Like