plz help me in this question

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imnoise(I,'gaussian',0.1*5e-6); in the above line why it is 0.1*5e-6 is written.
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John D'Errico
John D'Errico 2017-7-21
编辑:John D'Errico 2017-7-21
Why not just read the help for imnoise? There you will find a complete explanation of the parameters for imnoise. Instead, you want someone else to re-write the help for you here?
TUSHAR MURATKAR
TUSHAR MURATKAR 2017-7-21
John i had already read entire matlab document on imnoise .... but still i have confusion . is it bad to ask the doubt in this forum

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采纳的回答

Star Strider
Star Strider 2017-7-21
My guess would be sloppy coding. That would evaluate to 5e-7.
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TUSHAR MURATKAR
TUSHAR MURATKAR 2017-7-21
ya i also think that its 5e-7. and this 5e-7 represents variance ? am i right?
Star Strider
Star Strider 2017-7-21
As I read it, it would be the mean, since it is the first argument after 'gaussian' (at least in R2017a).
If you are using an earlier release, you need to see the documentation for the imnoise function for your MATLAB version.
Type either of these in your Command Window:
doc imnoise
or if that does not work, type:
help imnoise
to get the documentation.

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更多回答(2 个)

Alvi Newaz
Alvi Newaz 2017-7-21
The 0.1*5e-6 is the mean of the Gaussian noise added.

Image Analyst
Image Analyst 2017-7-21
Because the programmer made a mistake, in my opinion, by not being explicit. According to the help:
J = imnoise(I,'gaussian',M,V) adds Gaussian white noise of mean m and variance v to the image I. The default is zero mean noise with 0.01 variance.
so it should take 2 numbers after 'gaussian', not one like they had. Perhaps they meant:
noisyImage = imnoise(I,'gaussian',0.1, 5e-6);
where there is a comma instead of a *, but who knows. You need to ask the programmer.
If they put just one number, the number would be the mean, not the variance. The variance would take on the default value of 0.01

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