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Let Y be a measured (noisy) image Y= X+ noise, where X is an image contains 0(Background) and 200(object). I need to create a decision rule that determines whether the true pixel value was 0 or 200 given the image Y.

the noise is gaussian with mean=0 and standard deviation=sigma

I_true = [zeros(50,140);zeros(60,40),(ones(60,60)*200),zeros(60,40);zeros(50,140)];
[nrows ncolumns] = size(I_true);
sigma = 63.246;
gaussian_noise = sigma*randn(size(I_true));
I_noisy = I_true + gaussian_noise;

After adding the Gaussian noise to the true image the PDF of the intensity of a background pixel will be Gaussian with mean = zero and variance= $63.2462^2$ and the PDF of the intensity of an object pixel will be Gaussian with mean = 200 and variance= $63.2462^2$

I used MAP rule and assumed that P(Y=0)=p(Y=200)

Likelihood ratio

$(P(Y=j|X=200))/(P(Y=j|X=0))≥P(X=0)/(P(X=200))=1$

$exp((400Y−(200)^2)/(2σ^2))≥1$

$Y≥100$

so if $Y≥100$ the pixel will be considered as object.

my questions are :.

1) is my solution is right?

2) in the case of two objects with gray levels 150 and 200 what will be the steps of Map decision rule?

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Is your variance, $\sigma^2$ = 63.2462 or your standard deviation $\sigma$ = 63.2462? –  Mohammad May 16 '12 at 20:19
    
@Mohammad sigma = standard deviation = 63.2462 –  HforHesham May 16 '12 at 20:22
    
Yes, but you have written variance = 63.2462 –  Mohammad May 16 '12 at 20:36
    
@Mohammad I have corrected it. –  HforHesham May 16 '12 at 20:48

1 Answer 1

up vote 7 down vote accepted

1) Yes, your solution is correct.

2) If you assume that all a priori probabilities are equal, then the boundaries for AWGN is always the middle points between the possible values of X. In this case, then, the decision boundaries are at 75 and 175.

I believe that this rule (decision boundary at the middle points) can be generalised to applying to any noise probability distribution that is symmetric and monotonically decreases as the distance from zero increases.

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