Matlab interp1 function x-points for given y-points
显示 更早的评论
Dear Community Members,
Hopefully this finds you safe and well. Using interp1 to acquire x-points for given y-points does not seem to work. Data in this particular instance generated via a function however in reality it is experimental data.
%**main function
x=[3:0.1:6];
y=sin(x);
%***y-array for new x-values
y_array = [-0.9:0.005:0.1];
x_new = interp1(y,x,y_array);
figure
plot(x,y,'k+-',x_new,y_array,'rx-')
grid on
Resutant image see attachment. 
Any help on how to extrapolate between current and next data opints would be appreciated.
采纳的回答
Ameer Hamza
2020-10-8
Inverting nonlinear function using interp1() like this is not correct. Check the following code
%**main function
x=[3:0.1:6];
y=sin(x);
%***y-array for new x-values
y_array = -0.9:0.005:0.1;
interp_x = @(xq) interp1(x, y, xq);
x_new = fsolve(@(xq) interp_x(xq) - y_array, 3+rand(size(y_array)));
figure
plot(x,y,'k+-',x_new,y_array,'rx-')
grid on
4 个评论
Marathon_Mike
2020-10-8
Ameer Hamza
2020-10-8
编辑：Ameer Hamza
2020-10-8
Yes, for functions with two x values for the same y-value, it can be difficult to find a solution. This is equivalent to finding multiple roots of a system of equations. There is no universal solver that can find all the roots. However, particular problems can have some specific features which can help in solving them. For example, in this case, we can see that there are exactly two y-values. fsolve() converges to the value closest to the initial point. If we give want two solutions, we need to provide two different initial points. For example
%**main function
x=[3:0.1:6];
y=sin(x);
%***y-array for new x-values
y_array = -0.9:0.005:0.1;
interp_x = @(xq) interp1(x, y, xq, 'linear', 'extrap');
x_new1 = fsolve(@(xq) interp_x(xq) - y_array, 3*ones(size(y_array)));
x_new2 = fsolve(@(xq) interp_x(xq) - y_array, 6*ones(size(y_array)));
figure
plot(x,y,'k+-',x_new1,y_array,'rx-',x_new2,y_array,'rx-')
grid on
But here you will notice that the solutions are going beyond the range of 
. The y-value at x=3 and x=6 are not equal. To only get solution in this range, following code can be used
. The y-value at x=3 and x=6 are not equal. To only get solution in this range, following code can be used%**main function
x=[3:0.1:6];
y=sin(x);
%***y-array for new x-values
y_array = -0.9:0.005:0.1;
interp_x = @(xq) interp1(x, y, xq);
x_new1 = fsolve(@(xq) interp_x(xq) - y_array, 3*ones(size(y_array)));
x_new2 = zeros(size(y_array));
valid_sols = zeros(size(y_array));
for i = 1:numel(x_new2)
[x_new2(i), ~, valid_sols(i)] = fsolve(@(xq) interp_x(xq) - y_array(i), 6);
end
idx = valid_sols > 0; % only get the valid solutions
figure
plot(x,y,'k+-',x_new1,y_array,'rx-',x_new2(idx),y_array(idx),'rx-')
grid on
Marathon_Mike
2020-10-11
Ameer Hamza
2020-10-12
I am glad to be of help!
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Graphics Object Properties 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
