Root locus imaginary axis intersection

Other than using interactive data cursors, is there anyway of finding the point where the root locus intersects the imaginary axis?

 Respuesta aceptada

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 comentarios

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 el 15 de Jun. de 2021
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.

Iniciar sesión para comentar.

Más respuestas (0)

Productos

Versión

R2020a

Etiquetas

Preguntada:

el 14 de Jun. de 2021

Comentada:

el 15 de Jun. de 2021

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by