plz help me in this question

imnoise(I,'gaussian',0.1*5e-6); in the above line why it is 0.1*5e-6 is written.

[John D'Errico]

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]

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

Answer 1

[Star Strider]

My guess would be sloppy coding. That would evaluate to 5e-7.

[TUSHAR MURATKAR]

ya i also think that its 5e-7. and this 5e-7 represents variance ? am i right?

[Star Strider]

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.

Answer 2

[Alvi Newaz]

The 0.1*5e-6 is the mean of the Gaussian noise added.

[TUSHAR MURATKAR]

yes .....its mean not variance

Answer 3

[Image Analyst]

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

[TUSHAR MURATKAR]

thank you