how to shift position of an index of max value to origo in matlab?

Hi, I would like to shift the position the maximum value so that the maximum value is at the center, i.e. at origo.

[Jan]

But the maximum value is at the center already?! Or do you mean vertically? What is "at origo"?

[kat001]

I meant that the x axis should be shifted so the max of y is at 0, that is at x = 0, y = 0.3

[Adam]

You haven't shown what code you are using to achieve this. You just need to give a set of x values to the plot function for which 0 matches up to 0.3 in your y data.

[kat001]

Now, this is the draft code.

clear all
clc

x = [-3:0.5:3];
norm = normpdf(x,0,1);

m = 0;
mPos = [ ];

for i = 2:length(norm)
    if(norm(i)>norm(i-1))
        m = (m + 0.065);
    else
        m = (m - 0.065);
    end
    mPos = [mPos; m];
end

plot(x, norm)
hold on
plot(mPos)
grid on

And this give me following plot:

The triangular shape should be according to the gaussian profile.

[Adam]

Well, you aren't passing x to the plot instruction when you plot mPos so it will default to 1 to 12 in your case. Give it a length 12 vector of x values too and it will plot at those values.

[kat001]

Ok. But exactly how am I gonna do that? Because x has already the length of 13. It is 'mPos' is has a length of 12. How can I get the same vector lengths?

[Adam]

Well, I haven't examined your code that closely to understand why mPos is only length 12, but presumably you know what the x values should be for it. why is it 12 instead of 13? Is it shifted by 1 or is it the middle of the other samples? (I assume the former judging by the plot). whichever it is, just give an x vector that matches - e.g. x(1:end-1) or x(2:end) or x(1:end-1) + diff( x(1:2) ) / 2 or whatever is appropriate.

As an aside, don'y use square brackets here:

x = [-3:0.5:3];

You should get an M-lint warning in the code highlighting this. The [] is un-necessary as well as being less efficient, not that efficiency matters here, but it's good to get into more efficient habits when they cost nothing.

Answer 1

[KSSV]

x = [0 30 60] ;
y = [0 0.3 0] ;
plot(x,y);

xi = x-30 ;
yi = y-0.3 ;

hold on
plot(xi,yi)

[kat001]

Thank you. Now, imagine that x axis is not only bound to 0 to 60, instead it could be any length. How would I be able to feed that into the code in such case?

[KSSV]

Let x, y be you data..

[val,idx]=max(y);

xi=x-x(idx);

yi=y-y(idx);

Answer 2

[kat001]

Now, currently the plot looks like this:

where the red line represents an position vector as the power is increasing. What I would like to do is following (used MS paint for visualization) :

In other words, as the Gaussian profile reaches its maximum, the only way to keep maximum power all time by holding the position at the same place. And I am stuck with the coding here.

So far, my code looks like this:

x = -3:0.1:3;

norm = normpdf(x,0,1);

m = 0;
mPos = zeros(length(norm),1);

for i = 2:length(norm)
    if(norm(i)>norm(i-1))
        m = m + 0.1;
    else
        m = m - 0.1;
    end
    mPos(i) =  m;

end

plot(x, mPos, 'r')
hold on
plot(x, norm, 'b')
grid on
hold off