11
$\begingroup$

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 = $0$ 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?

$\endgroup$
5
  • 1
    $\begingroup$ Is your variance, $\sigma^2$ = 63.2462 or your standard deviation $\sigma$ = 63.2462? $\endgroup$
    – Spacey
    May 16, 2012 at 20:19
  • $\begingroup$ @Mohammad sigma = standard deviation = 63.2462 $\endgroup$
    – HforHesham
    May 16, 2012 at 20:22
  • 1
    $\begingroup$ Yes, but you have written variance = 63.2462 $\endgroup$
    – Spacey
    May 16, 2012 at 20:36
  • $\begingroup$ @Mohammad I have corrected it. $\endgroup$
    – HforHesham
    May 16, 2012 at 20:48
  • $\begingroup$ to learn it, you can go to my site roesland-uwahyudi.blogspot.com $\endgroup$ Nov 18, 2014 at 5:47

1 Answer 1

10
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.