Can SQRT be used on an RGB image with double precision?

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FJ
FJ 2018-4-6
评论： Rik 2018-4-6
I am trying to square root my RGB image but it keeps showing me a white output image.

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Rik
Rik 2018-4-6
That is due to the way you display the image. You probably have a uint8 input with values from 0-255. sqrt then does its thing and computes the square root for each element, and converts it to a double. Then you try to use imshow, but now your image is a double, so imshow assumes your data is scaled from 0-1, instead of 0-255 (or 0-15.9687).
You should read the doc for instructions on how you can specify the range, or just scale the image yourself.
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FJ
FJ 2018-4-6
编辑：FJ 2018-4-6
Thank you. I'll look into that. So you are saying that once an uint8 is converted into a double (and applied with operations), the values change and I need to scale back to see the original image(with operations applied)?
Rik
Rik 2018-4-6
You could scale the image to 0-1 doubles with the line below before you calculate the root.
I2 = im2double(I);
Then you don't have to worry about imshow, but your absolute values will be a factor sqrt(255) lower.

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