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
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
Does the above answer your question?
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.
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?
@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.