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Hi

I have a query and will appreciate your help.

If we try the code below,

>> c1=1;

>> for p=-50:50

c2=1;

for q=-50:50

pp1(c1,c2)=exp(- ( (p-10).^2 + (q-9).^2) /2) + exp(- ( (p+20).^2 + (q-30).^2) /2);

c2=c2+1;

end

c1=c1+1;

end

>> mesh(pp1)

you can see that the figure(matrix Po) has two peaks(maxima), stating that there are two targets, one at (10,9) and another at (-20,30).

If there is only one target(peak), I used [a,b]=find(Po==max(max(Po))), to find the locations (i,j) at the peak.

But in the above code, we have two peaks. Can you help me to find the

1. No of peaks(estimate no of targets)

2. Find the location of the peaks(i,j)

Cheers

[Information merged from duplicate Question]

Hello

I have a simple query, which was unanswered a few days ago. I am re-posting because I need a hint urgently. Please do not direct me to my previous post. It is not quite complete.

You can copy this code and see the result.

s1 = [5 4 0.5 -0.64]';

s2 = [0 -3.9 -0.5 -0.64]';

Sa = [-40:0.2:40];

Sb = [-40:0.2:40];

L = length(Sa);

m = zeros(L,L);

Pr = 2*ones(L,L);

Pr = Pr / sum(sum(Pr));

x1 = [s1(1);s1(2)] + sqrt(R)*randn(2,1);

x2 = [s2(1);s2(2)] + sqrt(R)*randn(2,1);

for i = 1: length(Pr)

for j = 1: length(Pr)

m(i,j) = (1/sqrt(2*pi)) * (exp(-((Sa(i)-x1(1)).^2 + (Sb(j)-x1(2)).^2)/2) + exp(-((Sa(i)-x2(1)).^2 + (Sb(j)-x2(2)).^2)/2));

m(i,j) = m(i,j) * Pr(i,j);

end

end

Po = m/sum(sum(m));

[a,b] = find(Po==max(max(Po)));

Sest = [Sa(a);Sb(b)];

meash(Sa,Sb,Po)

The matrix(surfing of the matrix) 'Po', will result in two peaks at the (s1(1),s1(2)) and (s2(1),s2(2)), show that there are two targets. If I had one peak, a,b would collect it and Sest would collect the location on the grid.

But I am unable to figure out how to find two peaks instead. So the last two statements do not hold now.

I look forward for suggestions on

1. finding the number of peaks(2 in this case)

2. finding the location on grid(Sest in this case) (Sa,Sb)

Will appreciate your help!

Cheers

Image Analyst
on 9 Mar 2012

See the code I added to yours:

c1=1;

for p=-50:50 c2=1;

for q=-50:50

pp1(c1,c2)=exp(- ( (p-10).^2 + (q-9).^2) /2) + exp(- ( (p+20).^2 + (q-30).^2) /2);

c2=c2+1;

end

c1=c1+1;

end

mesh(pp1)

Here is code I added:

[labeledImage numberOfPeaks] = bwlabel(pp1>0.01);

measurements = regionprops(labeledImage, 'Centroid');

message = sprintf('The number of peaks is %d\n', numberOfPeaks);

uiwait(msgbox(message));

for r = 1 : numberOfPeaks

message = sprintf('The centroid of peak %d is at (%.2f, %.2f)\n', ...

r, measurements(r).Centroid(1), measurements(r).Centroid(2));

uiwait(msgbox(message));

end

Yes I know the numbers aren't the same. I'm hoping you'll be able to figure out how to shift the coordinate system by subtracting 51 from x, and y.

Image Analyst
on 12 Mar 2012

Walter Roberson
on 10 Mar 2012

Image Analyst
on 10 Mar 2012

Learning how to use the debugger to step through your code is always a good course of action.

Walter Roberson
on 10 Mar 2012

Re-posting doesn't make the volunteers more interested or less busy.

If you wish to remind people about your question, you can do that by adding a comment to your question. In cases where you realized your question could be improved or did not have enough information, you can edit your question.

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