FFT along third dimension
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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?
Risposte (1)
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 Commenti
ARN
il 13 Ago 2020
Walter Roberson
il 13 Ago 2020
What release are you using, and which operating system? Also do you happen to have an AMD Jaguar CPU?
ARN
il 13 Ago 2020
Matt J
il 13 Ago 2020
And what is the magnitude of the differences between the arrays?
ARN
il 17 Ago 2020
Matt J
il 18 Ago 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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