How to interpolate a closed polar plot

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Mercedes Stringer Martin
回答： Santiago Benito 2022-9-16
Hello, I need some help interpolating a closed polar plot which currently looks like this:
figure(1)
polarplot(deg2rad(Angle),Mag)
When I interpolate it using spline it looks like this:
figure(2)
xint = linspace(0,deg2rad(360),360)';
spl = spline(deg2rad(Angle),Mag);
polarplot(xint,ppval(spl,xint))
This result is almost perfect - however, I want the angle 0 and 360 also to be interpolated/spline/curved rather than a pointed edge.
I have downloaded and tried the interpclosed function (https://uk.mathworks.com/matlabcentral/fileexchange/69055-interpclosed?s_tid=mwa_osa_a) however I get this result:
figure(3)
intc= interpclosed(deg2rad(Angle),Mag, 0:1/360:1);
polarplot(deg2rad(Angle),Mag,'o',intc(1,:),intc(2,:),':.')
Not sure if I'm missing something simple. Any advice to fix this issue would be greatly appreciated!!

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采纳的回答

DGM
DGM 2022-7-27
编辑：DGM 2022-7-27
Ran in:
EDITED:
You could enforce endslopes, but this is probably more accurate
% the same fake data
Angle = [0 90 180 270 360];
Mag = [5 1 1 1.5 5];
% replicate the data to represent multiple cycles beyond the query range
% make sure to omit duplicate endpoints if necessary
Angle = [Angle(1:end-1)-360 Angle Angle(2:end)+360];
Mag = [Mag(1:end-1) Mag Mag(2:end)];
% this is the same query range
xint = linspace(0,2*pi,360)';
spl = spline(deg2rad(Angle),Mag); % no endslopes defined
polarplot(xint,ppval(spl,xint))
It's probably not necessary to replicate the entire dataset, but you'll probably want a decent number of points to anchor the spline on either side of the query range. It all depends on how many points are in the dataset.

更多回答（1 个）

Santiago Benito
Santiago Benito 2022-9-16
Hi, I might be a little late to the party, but here's my solution.
You tried to use interpclosed to solve the issue but the function failed because it expects cartesian coordinates. To overcome this issue, just convert from polar to cartesian and then back to polar. Please see the example below. Note that, as I don't know your dataset, I made up some polar coordinates as rho and theta:
% Make up some data
rho = [14 9 6 4 2 5 6 1 1 1 3 7 4 3 5 1 2 2 14];
theta = linspace(0,360,numel(rho));
% Plot said data
figure, polarplot(deg2rad(theta),rho)
% Convert to cartesian
[x,y] = pol2cart(deg2rad(theta),rho);
% Run interpclosed
intc= interpclosed(x(1:end-1),y(1:end-1),linspace(0,1,3e2),true);
% Convert the fit to polar
[thetaint,rhoint] = cart2pol(intc(1,:),intc(2,:));
% Plot with polarplot
figure, polarplot(deg2rad(theta),rho,'o',thetaint,rhoint,':.')
The results looks like what I would expect from a spline fit.
Cheers,
Santiago

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