MATLAB Answers

Root locus imaginary axis intersection

18 views (last 30 days)
Other than using interactive data cursors, is there anyway of finding the point where the root locus intersects the imaginary axis?

Accepted Answer

Star Strider
Star Strider on 14 Jun 2021
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 Comments
Star Strider
Star Strider on 15 Jun 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.

Sign in to comment.

More Answers (0)

Tags

Products


Release

R2020a

Community Treasure Hunt

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

Start Hunting!