Finding intersection of rlocus branch and a line at an angle

I have the same question as: How to find the intersection a root locus plot and a line with specific angle? - MATLAB Answers - MATLAB Central (mathworks.com)

but the solution posed in the answer https://www.mathworks.com/matlabcentral/answers/1442099-how-to-find-the-intersection-a-root-locus-plot-and-a-line-with-specific-angle#answer_775729 does not seem to work in my use case.

Modified code to suit my use case:

clf

s=tf('s');
GH=(s+8)/((s+6)*(s+3)*(s+10));
rlocus(GH)

% Get handle to red line 
ax = gca(); 
curveLine = findobj(ax, 'Type','Line','Color', 'r', 'Marker','none');

% Find intersection of the angle from x-axis at 
% origin (0,0)
ang = -62.87; % deg from negative x-axis into quadrant 3
xLine = [0, ax.XLim(1)]; 
yLine = [0, tand(-ang)*xLine(2)];
[x0,y0,~,~] = intersections(curveLine.XData, curveLine.YData, xLine, yLine);

% Plot the lines and intersection
hold(ax,'on')
plot(ax, curveLine.XData, curveLine.YData, 'k--', 'LineWidth', 2)
plot(ax, xLine, yLine, 'k--', 'LineWidth', 2)

% remove (0,0) intersection and label intersection
isNot0 = x0~=0 & y0~=0; 
plot(ax, x0(isNot0), y0(isNot0), 'mo','MarkerSize', 12)
text(ax, x0(isNot0)*2, y0(isNot0), ...
    sprintf('(%.3f, %.3f)',x0(isNot0), y0(isNot0)), ...
    'HorizontalAlignment', 'right')

Expected results:

The correct answer for an angle: +-62.87 is

-5.415+-j10.57

The issue that I am getting is the code outputs 2 different sets of intersections instead of 1 x,y group. I suspect that this is because it has a "phantom" intersection.

Answer I got (-5.415, -2.750)(-10.569, -11.710)

Expected (-5.415, -10.569)

Bonus points for outputting K :p

Answer 1

[Paul]

One way would be to use the output arguments from rlocus to get the closed loop pole locations and the associated gain.

Here's the root locus plot.

s=tf('s');
GH=(s+8)/((s+6)*(s+3)*(s+10));
rlocus(GH),sgrid

Get closed loop pole locations and associated gains.

[r,k] = rlocus(GH);

Compute the angles from the -x-axis of the pole locations.

angles = 180/pi*atan2(imag(r),-real(r));

At this point, I'll leave the rest to you to figure out the value of the gain k that corresponds to the desired angle. A couple of things to keep in mind. Some problems might have no value of k that satisfies the criteria and others might have more than one, so you'll have to consider all possbilities. The other is that rlocus does not gurantee (at least I don't think it does) that each column of r is a continous branch of the root locus (though I do think rlocus tries very hard to make that so). So there's a possbility that two or more columns of r can jump from one branch to another, and if that happens at the critical point where the angle is what you're looking for there will be complications.