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?

采纳的回答

Star Strider
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)
r =
-0.9406 + 0.0000i -0.8744 + 0.0000i -0.8685 + 0.0000i -0.8620 + 0.0000i -0.8550 + 0.0000i -0.8475 + 0.0000i -0.8394 + 0.0000i -0.8306 + 0.0000i -0.8212 + 0.0000i -0.8111 + 0.0000i -0.8003 + 0.0000i -0.7888 + 0.0000i -0.7766 + 0.0000i -0.7636 + 0.0000i -0.7500 + 0.0000i -0.7358 + 0.0000i -0.7209 + 0.0000i -0.7055 + 0.0000i -0.6896 + 0.0000i -0.6734 + 0.0000i -0.6569 + 0.0000i -0.6402 + 0.0000i -0.6236 + 0.0000i -0.6071 + 0.0000i -0.5908 + 0.0000i -0.5748 + 0.0000i -0.5593 + 0.0000i -0.5443 + 0.0000i -0.5299 + 0.0000i -0.5161 + 0.0000i -0.5030 + 0.0000i -0.4906 + 0.0000i -0.4789 + 0.0000i -0.4679 + 0.0000i -0.4576 + 0.0000i -0.4480 + 0.0000i -0.4390 + 0.0000i -0.4306 + 0.0000i -0.4229 + 0.0000i -0.4157 + 0.0000i -0.4090 + 0.0000i -0.4029 + 0.0000i -0.3972 + 0.0000i -0.3919 + 0.0000i -0.3871 + 0.0000i -0.3826 + 0.0000i -0.3785 + 0.0000i -0.3748 + 0.0000i -0.3713 + 0.0000i -0.3681 + 0.0000i -0.3652 + 0.0000i -0.3334 + 0.0000i -0.3333 + 0.0000i 0.0814 + 0.8379i 0.0483 + 0.9140i 0.0453 + 0.9212i 0.0421 + 0.9291i 0.0386 + 0.9377i 0.0349 + 0.9470i 0.0308 + 0.9573i 0.0264 + 0.9686i 0.0217 + 0.9809i 0.0167 + 0.9943i 0.0113 + 1.0090i 0.0055 + 1.0251i -0.0006 + 1.0426i -0.0071 + 1.0617i -0.0139 + 1.0826i -0.0210 + 1.1053i -0.0284 + 1.1300i -0.0362 + 1.1568i -0.0441 + 1.1859i -0.0522 + 1.2175i -0.0605 + 1.2515i -0.0688 + 1.2883i -0.0771 + 1.3278i -0.0853 + 1.3703i -0.0935 + 1.4158i -0.1015 + 1.4644i -0.1092 + 1.5162i -0.1167 + 1.5714i -0.1239 + 1.6299i -0.1308 + 1.6920i -0.1374 + 1.7578i -0.1436 + 1.8273i -0.1494 + 1.9006i -0.1549 + 1.9780i -0.1601 + 2.0594i -0.1649 + 2.1452i -0.1694 + 2.2354i -0.1736 + 2.3302i -0.1775 + 2.4299i -0.1810 + 2.5345i -0.1844 + 2.6442i -0.1875 + 2.7594i -0.1903 + 2.8802i -0.1929 + 3.0069i -0.1953 + 3.1397i -0.1976 + 3.2789i -0.1996 + 3.4247i -0.2015 + 3.5775i -0.2032 + 3.7375i -0.2048 + 3.9052i -0.2063 + 4.0807i -0.2222 +81.7209i Inf + 0.0000i 0.0814 - 0.8379i 0.0483 - 0.9140i 0.0453 - 0.9212i 0.0421 - 0.9291i 0.0386 - 0.9377i 0.0349 - 0.9470i 0.0308 - 0.9573i 0.0264 - 0.9686i 0.0217 - 0.9809i 0.0167 - 0.9943i 0.0113 - 1.0090i 0.0055 - 1.0251i -0.0006 - 1.0426i -0.0071 - 1.0617i -0.0139 - 1.0826i -0.0210 - 1.1053i -0.0284 - 1.1300i -0.0362 - 1.1568i -0.0441 - 1.1859i -0.0522 - 1.2175i -0.0605 - 1.2515i -0.0688 - 1.2883i -0.0771 - 1.3278i -0.0853 - 1.3703i -0.0935 - 1.4158i -0.1015 - 1.4644i -0.1092 - 1.5162i -0.1167 - 1.5714i -0.1239 - 1.6299i -0.1308 - 1.6920i -0.1374 - 1.7578i -0.1436 - 1.8273i -0.1494 - 1.9006i -0.1549 - 1.9780i -0.1601 - 2.0594i -0.1649 - 2.1452i -0.1694 - 2.2354i -0.1736 - 2.3302i -0.1775 - 2.4299i -0.1810 - 2.5345i -0.1844 - 2.6442i -0.1875 - 2.7594i -0.1903 - 2.8802i -0.1929 - 3.0069i -0.1953 - 3.1397i -0.1976 - 3.2789i -0.1996 - 3.4247i -0.2015 - 3.5775i -0.2032 - 3.7375i -0.2048 - 3.9052i -0.2063 - 4.0807i -0.2222 -81.7209i Inf + 0.0000i
k = 1×53
0 0.5932 0.6491 0.7103 0.7772 0.8504 0.9305 1.0182 1.1141 1.2190 1.3339 1.4595 1.5970 1.7475 1.9121 2.0922 2.2893 2.5050 2.7410 2.9992 3.2817 3.5908 3.9291 4.2993 4.7043 5.1474 5.6323 6.1629 6.7435 7.3787
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')
v2 = 1.0409
k2 = interp1(imag(r(2,fidx2)), k(fidx2), v2, 'linear','extrap')
k2 = 1.5835
v3 = interp1(real(r(3,fidx3)), imag(r(3,fidx3)), 0, 'linear','extrap')
v3 = -1.0409
k3 = interp1(imag(r(3,fidx3)), k(fidx3), v3, 'linear','extrap')
k3 = 1.5835
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
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
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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