Finding point of zeta line & rootlocus

I have a transfer function, constant zeta, I need easy way to find the intersection point between root locus & zeta line. Now, after plotting root locus & zeta using sgrid function, I can find approximately correct intersection point by mouse draging. But I need exact point. You know, mouse draging is too difficult. Besides, zeta line is so narrow linewidth which is too hard to see in plot. Zeta line should increase default linewidth. Can anyone help?

Answer 1

[Star Strider]

I am not certain what you are starting with, so I cannot provide an exact response. However one approach may simply be to use interp1 to interpolate the existing ‘r’ and ‘k’ values that correspond to the ζ value you want.

Using an example from the rlocus documentation —

H = tf([2 5 1],[1 2 3]);

[r,kout] = rlocus(H);

zetav = cos(angle(r(1,:)));

zetaq = [0.64 0.76 0.80 0.86 0.90 0.985];                           % Desired ‘zeta’ Values

for k = 1:numel(zetaq)
    zidx = find(diff(sign(zetav + zetaq(k))));
    idxrng = max(1,zidx-1) : min(numel(zetav),zidx+1);
    rq(k,:) = interp1(zetav(idxrng), r(1,idxrng), -zetaq(k));
    kq(k,:) = interp1(zetav(idxrng), kout(idxrng), -zetaq(k));
end

Result = table(zetaq(:),rq,kq, VariableNames={'zeta','r','k'})

Result = 6x3 table
    zeta            r               k    
    _____    ________________    ________

     0.64    -1.0373+1.2454i     0.087697
     0.76    -1.0931+0.93478i     0.29697
      0.8    -1.1081+0.83101i     0.38138
     0.86    -1.1279+0.66911i     0.52458
      0.9    -1.1396+0.55131i      0.6333
    0.985     -1.161+0.18852i     0.90624

    
figure
rlocus(H)
sgrid
hold on
plot(real(rq), imag(rq), 'sr')                                        % Plot ‘zeta’ Value As Red Squares
hold off

[min_zeta,max_zeta] = bounds(zetav)

min_zeta = -1

max_zeta = -0.5774

% disp(zetav)

The values corresponding to the plotted ζ l;ines plot at their intersections with the root locus.

The ‘Result’ table shows only one value for ‘r’ for convenience. (It likely has a complex conjugate that you can easily calculate.)

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