4 Made equation use displaystyle so I can read it. :-) edited Sep 4 '15 at 13:15 Peter K.♦ 17.8k88 gold badges3232 silver badges6565 bronze badges 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 edited Sep 4 '15 at 12:37 Tolga Birdal 4,69311 gold badge99 silver badges3232 bronze badges 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 edited Sep 4 '15 at 11:44 Tolga Birdal 4,69311 gold badge99 silver badges3232 bronze badges 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 answered Sep 3 '15 at 17:38 Tolga Birdal 4,69311 gold badge99 silver badges3232 bronze badges