How to find the intersection a root locus plot and a line with specific angle?

There may be an analytical approach that would be better than this approach but one way to do it is to get the handle of the blue line (assuming there is only 1 blue line in the axes and the line is truely blue, [0 0 1]) and to compute the intersection of that line and the line defined by your angle from (0,0). This uses intersections.m from the file exchange.

clf
s=tf('s');
GH=(1)/(s*(s+1)^2);
rlocus(GH)
% Get handle to blue line 
ax = gca(); 
curveLine = findobj(ax, 'Type','Line','Color', 'b', 'Marker','none');
% Find intersection of the angle from x-axis at 
% origin (0,0)
ang = 60; % 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('%.1f%s (%.3f, %.3f)',ang,char(186),x0(isNot0), y0(isNot0)), ...
    'HorizontalAlignment', 'right')

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Mohamad Balawi 2021-8-27

Thanks it worked just fine!

Luke McDevitt 2023-4-26

编辑：Luke McDevitt 2023-4-26

Does the script only work with 2nd order systems? I can't seem to edit it in a way that allows for it to find the intersection for the system: GH=(s+8)/((s+6)*(s+3)*(s+10));

dashed line is between -8 and -10 on the real axis and not on the part of the rlocus that varies over the imaginary axis

The correct answer for an angle: -62.87 is

-5.415+-j10.57

EDIT: realized it just goes after the blue line, but changing the color to 'g' also does not work correctly, even if I flip the angle to a -27.13 to get the lines to intersect the script does not produce the desired result

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