4 Made equation use displaystyle so I can read it. :-)
source | link

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

Regarding $\sigma$, you mention in the question, simple answer is yes, longer answer is no. It refers to the variance of the noise normalized with image dimensions as follows:

$\sigma = \frac{\sigma_0\sqrt{0.5\pi}}{6(W-2)(H-2)}$$$\sigma = \frac{\sigma_0\sqrt{0.5\pi}}{6(W-2)(H-2)}$$

$-2$ is due to the convolution kernel, applied previously. However, this is not a final quantity, giving you the absolute noise level. Nevertheless, for the variance, you could use it as a percentage, just like you mentioned.

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

Regarding $\sigma$, you mention in the question, simple answer is yes, longer answer is no. It refers to the variance of the noise normalized with image dimensions as follows:

$\sigma = \frac{\sigma_0\sqrt{0.5\pi}}{6(W-2)(H-2)}$

$-2$ is due to the convolution kernel, applied previously. However, this is not a final quantity, giving you the absolute noise level. Nevertheless, for the variance, you could use it as a percentage, just like you mentioned.

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

Regarding $\sigma$, you mention in the question, simple answer is yes, longer answer is no. It refers to the variance of the noise normalized with image dimensions as follows:

$$\sigma = \frac{\sigma_0\sqrt{0.5\pi}}{6(W-2)(H-2)}$$

$-2$ is due to the convolution kernel, applied previously. However, this is not a final quantity, giving you the absolute noise level. Nevertheless, for the variance, you could use it as a percentage, just like you mentioned.

3 deleted 1 character in body
source | link

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

Regarding $\sigma$, you mention in the question, simple answer is yes, longer answer is no. It refers to the variance of the noise normalized with image dimensions as follows:

$\sigma = \sigma_0*\frac{\sqrt{0.5\pi}}{6(W-2)(H-2)}$$\sigma = \frac{\sigma_0\sqrt{0.5\pi}}{6(W-2)(H-2)}$

$-2$ is due to the convolution kernel, applied previously. However, this is not a final quantity, giving you the absolute noise level. Nevertheless, for the variance, you could use it as a percentage, just like you mentioned.

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

Regarding $\sigma$, you mention in the question, simple answer is yes, longer answer is no. It refers to the variance of the noise normalized with image dimensions as follows:

$\sigma = \sigma_0*\frac{\sqrt{0.5\pi}}{6(W-2)(H-2)}$

However, this is not a final quantity, giving you the absolute noise level. Nevertheless, for the variance, you could use it as a percentage, just like you mentioned.

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

Regarding $\sigma$, you mention in the question, simple answer is yes, longer answer is no. It refers to the variance of the noise normalized with image dimensions as follows:

$\sigma = \frac{\sigma_0\sqrt{0.5\pi}}{6(W-2)(H-2)}$

$-2$ is due to the convolution kernel, applied previously. However, this is not a final quantity, giving you the absolute noise level. Nevertheless, for the variance, you could use it as a percentage, just like you mentioned.

2 added 406 characters in body
source | link

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

Regarding $\sigma$, you mention in the question, simple answer is yes, longer answer is no. It refers to the variance of the noise normalized with image dimensions as follows:

$\sigma = \sigma_0*\frac{\sqrt{0.5\pi}}{6(W-2)(H-2)}$

However, this is not a final quantity, giving you the absolute noise level. Nevertheless, for the variance, you could use it as a percentage, just like you mentioned.

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

You could check out my MATLAB function, which estimates the variance of the noise. It is also mentioned in a Mathworks blog post, in comparison with other methods.

Regarding $\sigma$, you mention in the question, simple answer is yes, longer answer is no. It refers to the variance of the noise normalized with image dimensions as follows:

$\sigma = \sigma_0*\frac{\sqrt{0.5\pi}}{6(W-2)(H-2)}$

However, this is not a final quantity, giving you the absolute noise level. Nevertheless, for the variance, you could use it as a percentage, just like you mentioned.

1
source | link