Gradient of a 2D plot

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Prince Alex
Prince Alex 2020 年 9 月 30 日
コメント済み: Star Strider 2020 年 9 月 30 日
I have a 2D plot of potential (V) (circular contour plot). I have to find the electric field (E) by taking the gradient. How is it possible to see the gradient of V. [U,W]=gradient(V) returns the gradient in X and Y direction and quiver(X,Y,U,W) helps to visualize the vector field. How is it possible to view effective gradient. Contour (X,Y,U) will only help me to see gradient in x direction. How can I see the effect of gradient which is a vector sum of both gradient in X and Y.?
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Ameer Hamza
Ameer Hamza 2020 年 9 月 30 日
What is wrong with quiver. It generates arrows in the gradient direction.
Prince Alex
Prince Alex 2020 年 9 月 30 日
I am looking for contourf plot of the gradient. Thank you

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採用された回答

Star Strider
Star Strider 2020 年 9 月 30 日
See the gradient documentation section on Contour Plot of Vector Field . It demonstrates exactly what you are describing.
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Prince Alex
Prince Alex 2020 年 9 月 30 日
Thank you. I think it will work. But the above script is showing matrix dimension error. Thank you so much
Star Strider
Star Strider 2020 年 9 月 30 日
My pleasure!
It should not be showing any error. It is an example from the documentation (that I linked to in my oridginal Answer), and ran for me without error in R2020b. The ‘px’ and ‘py’ matrices are both (21x21), so the addition should work. If it is not working with your code, please post it.

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その他の回答 (1 件)

KSSV
KSSV 2020 年 9 月 30 日
M = sqrt(U.^2+W.^2) ;
contour(X,Y,M)
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Prince Alex
Prince Alex 2020 年 9 月 30 日
編集済み: Prince Alex 2020 年 9 月 30 日
Its a nice way. But since it is electric field it will have both positive and negative values. Squaring results in absolute value. @ KSSV a simple U + W will give the answer?

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