There is a way to stretch (scale) for a scaling factor a curve and save the stretched indexes and corresponding values ?

I stretched the value and the length of a vector for a scaling factor. Now I would like to save the new stretched indices and values? can I do that

Answer 1

[Walter Roberson]

If what you have currently is a list of x coordinates and a list of y coordinates, and you want the same density, then it would sound like you would just do

new_x = x .* x_scale;
new_y = y .* y_scale;

If you have a list of x and y coordinates, and the x is equally sampled and monotonic, and you want to increase the density, then the lazy way is to

new_length = round(length(x) * x_scale);
x_query = linspace(x(1), x(end), new_length);
y_query = interp1(x, y, x_query);
new_x = x_query .* x_scale;
new_y = y_query .* y_scale;

If the x is not equally sampled, or x is not monotonic, then I suggest you look in the File Exchange for John D'Errico's interp_arc(), scale the input x and input y first, and then interp_arc() on those.

[Paola]

Thanks Walter, the problem is that the i have to cross-correlate the new stretched vector with another vector. I don't know how to use xcorr if I have the stretched indeces and values in two separate vector. This is why I can ask if there is a way to save both on the same vector?

[Walter Roberson]

@Paola

Sure you could put them in the same vector.

combined_vector = [new_x, new_y];

However that will not be of any benefit to you.

If you need to xcorr then get the current x coordinates for the longer vector, scale down by the x stretching factor, and use those as query points in interp1(shorter_x, shorter_y, query_points) . Then take the resulting y and apply the y stretching factor to them . Now you xcorr those results against the y for the longer vector.

[If the stretched vector is to be compared to a vector that has fewer points, it is best to reverse roles of the two, so that you use the longer length as the ones being xcorr()'d )