FFT along third dimension

Hi,
I am trying to understand the fft along 3rd dimension
a = [1 2 3 4; 5 6 7 8; 9 10 11 12];
b = [1 2 3 4; 5 6 7 8; 9 10 11 12];
num_samples = 3;
num_chirps = 4;
num_of_antenna = 2;
w_range = blackman(num_samples);
w_doppler = blackman(num_chirps)';
w_angle = blackman(num_of_antenna);
window_3d = w_range.*w_doppler.*permute(w_angle,[3 2 1]);
window_2d = w_range .* w_doppler;
windowed_a = a.*window_2d;
windowed_b = b.*window_2d;
g1 = fft2(windowed_a);
g2 = fft2(windowed_b);
windowed_cat = cat(3,g1,g2).*permute(w_angle,[3 2 1]);
g3 = abs(fft(windowed_cat,[],3));
concat_3d = cat(3,a,b);
windowed_3d = concat_3d.*window_3d;
fft_2d = fft2(window_3d);
fft_3d = abs(fft(fft_2d,[],3));
ff_3d = abs(fftn(window_3d));
Shouldn't this be ture
g3==fft_3d==ff_3d
Why are they not equal?

Antworten (1)

Matt J
Matt J am 13 Aug. 2020
Bearbeitet: Matt J am 13 Aug. 2020

0 Stimmen

They are equal,
>> isequal(g3, fft_3d ), isequal(g3 , ff_3d )
ans =
logical
1
ans =
logical
1
although in general, I think you should expect they might differ by small floating point errors.
If you literally typed in g3==fft_3d==ff_3d, then this will not be true for the same reason the following is not:
>> 2==2==2
ans =
logical
0

6 Kommentare

ARN
ARN am 13 Aug. 2020
somehow in MATLAB the answer is coming in 0's while in octave its giving me valid solution.
Walter Roberson
Walter Roberson am 13 Aug. 2020
What release are you using, and which operating system? Also do you happen to have an AMD Jaguar CPU?
ARN
ARN am 13 Aug. 2020
Release 2020, os : win 10 ,cpu intel
Matt J
Matt J am 13 Aug. 2020
And what is the magnitude of the differences between the arrays?
ARN
ARN am 17 Aug. 2020
see the image
Matt J
Matt J am 18 Aug. 2020
Those look valid to me. Since w_angle contains only zeros, it makes sense that all fo the results should be approximately, if not exactly, zero.
>> w_angle = blackman(num_of_antenna)
w_angle =
0
0

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Gefragt:

ARN
am 13 Aug. 2020

Kommentiert:

am 18 Aug. 2020

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