FFT along third dimension
이전 댓글 표시
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?
답변 (1개)
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
댓글 수: 6
ARN
2020년 8월 13일
Walter Roberson
2020년 8월 13일
What release are you using, and which operating system? Also do you happen to have an AMD Jaguar CPU?
ARN
2020년 8월 13일
Matt J
2020년 8월 13일
And what is the magnitude of the differences between the arrays?
ARN
2020년 8월 17일
Matt J
2020년 8월 18일
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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