1-d gradient

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Yostena Wassef
Yostena Wassef le 17 Déc 2018
Modifié(e) : Jan le 17 Déc 2018
can i use covlution to get the same result of 1-d gradient
Screenshot (58).png
Screenshot (59).png
  1 commentaire
KSSV
KSSV le 17 Déc 2018
Read about Taylor Series.

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Jan
Jan le 17 Déc 2018
Your question contains the answer already. If the data is a vector, conv is sufficient instead of conv2:
x = rand(1, 10)
dx_1 = gradient(x)
dx_2 = conv(x, [0.5, 0, -0.5], 'same')
The first and last element of dx_2 differ from dx_1, but you can adjust it with the one-sided quotient of differences easily.
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Yostena Wassef
Yostena Wassef le 17 Déc 2018
input is matrix
so i can't get the same result for 1dgradient
Jan
Jan le 17 Déc 2018
Modifié(e) : Jan le 17 Déc 2018
Then the screenshot of the text "When F is a vector" is simply confusing. Please ask the questions as clear as possible.
dx_2 = conv2(x, [0.5, 0, -0.5], 'same')
This replies the same as gradient, except for the first and last column. Then:
dx_2(:, 1) = diff(x(:, 1:2), 1, 2);
dx_2(:, end) = diff(x(:, end-1:end), 1, 2);
Or if you love conv2:
d = conv2(x, [1, -1], 'valid');
dx_2(:, 1) = d(:, 1);
dx_2(:, end) = d(:, end);

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