Root locus imaginary axis intersection
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Other than using interactive data cursors, is there anyway of finding the point where the root locus intersects the imaginary axis?
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Star Strider
2021-6-14
Try something like this —
sys = tf([3 1],[9 7 5 6]); % Example From The Documentation
[r,k] = rlocus(sys)
fre2 = isfinite(real(r(2,:)));
fim2 = isfinite(imag(r(2,:)));
fidx2 = fre2 & fim2;
fre3 = isfinite(real(r(3,:)));
fim3 = isfinite(imag(r(3,:)));
fidx3 = fre3 & fim3;
v2 = interp1(real(r(2,fidx2)), imag(r(2,fidx2)), 0, 'linear','extrap')
k2 = interp1(imag(r(2,fidx2)), k(fidx2), v2, 'linear','extrap')
v3 = interp1(real(r(3,fidx3)), imag(r(3,fidx3)), 0, 'linear','extrap')
k3 = interp1(imag(r(3,fidx3)), k(fidx3), v3, 'linear','extrap')
figure
plot(real(r(1,:)),imag(r(1,:)), '-g')
hold on
plot(real(r(2,:)),imag(r(2,:)), '-b')
plot(real(r(3,:)),imag(r(3,:)), '-r')
plot(0, v2, 'sr')
plot(0, v3, 'sb')
hold off
grid
ylim([-6 6])
.
2 个评论
Paul
2021-6-15
Will this solution work if a branch of the root locus crosses the imaginary axis twice? For example if
sys = tf([3 1],[9 7 5 6]) * tf(20,[1 20])
Can this solution be generalized to loop over all of the rows of r?
As I understand it, this solution assumes that the rows of r are, in some sense, smooth. I think that rlocus() tries to ensure this, but I'm not sure it's guaranteed.
Star Strider
2021-6-15
This is prototype code.
It simply shows the correct approach, and would likely have to be adapted to specific situations that did not follow the same sort of loci.
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