Root locus imaginary axis intersection
显示 更早的评论
Other than using interactive data cursors, is there anyway of finding the point where the root locus intersects the imaginary axis?
采纳的回答
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.
更多回答（0 个）
类别
在 帮助中心 和 File Exchange 中查找有关 Classical Control Design 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 
