How do I use a root locus to find a value of K such that the damping ratio of dominant closed loop poles is a specified value?

How do I use a root locus to find a value of K such that the damping ratio of dominant closed loop poles is 0.7071?

Using the following TF and associated root locus.

Many thanks

Answer 1

[Paul]

Click on branch of the root locus and drag along the branch. The values in the datatip, which include the damping ratio, will correspond to the datamarker point on the locus.

[Luke McDevitt]

How would you find that programmatically?

[Paul]

Here's my attempt. Can't say I've thorougly tested it, but maybe it's a start. I know for sure it will break down if the desired damping ratio is 1.

Transfer function for testing and its root locus

h = tf([2 5 1],[1 2 3]);
s = tf('s');
h = h/(s+4)/(s+5);
figure
rlocus(h);
hold on

Desired damping ratios

zetad = [0.2, 0.6, 0.99];

Loop through the zetad, compute the gains and corresponding root locations, which are added to plot as filled, black dots

for ii = 1:numel(zetad)
[r,k{ii}] = zetagain(h,zetad(ii));
plot(real(r),imag(r),'k.','MarkerSize',15);
end

Warning: The numerator or denominator of this transfer function has complex-valued coefficients.

Warning: The numerator or denominator of this transfer function has complex-valued coefficients.

Warning: The numerator or denominator of this transfer function has complex-valued coefficients.

sgrid(zetad,1:5:25)

k

k = 1×3 cell array
    {[214.6052]}    {[15.2534 0.3113]}    {[0.4787 9.2402]}

Cases with no solution

[r,k] = zetagain(tf(1,[1 1]),0.2)

Warning: The numerator or denominator of this transfer function has complex-valued coefficients.

r =

  0×1 empty double column vector


k =

  1×0 empty double row vector

[r,k] = zetagain(tf(1,[1 2*.5*1 1]),0.7)

Warning: The numerator or denominator of this transfer function has complex-valued coefficients.

r =

  0×1 empty double column vector


k =

  1×0 empty double row vector

Other corner cases would be where one or more branches of the root locus either lie exactly on a line of constant zetad or if the line of constant zetad is a root locus asymptote.

function [r,k] = zetagain(h,zetad)

% compute the rotation angle
theta = asin(zetad);
m = numel(h.num{:})-1;
n = numel(h.den{:})-1;
h1 = tf(h.num{:}.*exp(-1j*theta).^(m:-1:0),h.den{:}.*exp(-1j*theta).^(n:-1:0));

% gain margin in rotated complex plane
s = allmargin(h1);
k = s.GainMargin;

% roots at values of k
r = rlocus(h,k);
r = r(:);

% eliminate extraneous roots that aren't at the the desired zeta. use 0.02 as a tolerance,
% maybe this should be an input parameter. maybe this test could could be made more robust.
zeta = sin(atan2(-real(r),abs(imag(r))));
r(abs(zeta-zetad)>0.02) = [];

end

Answer 2

[Sam Chak]

Hi @Thomas and @Luke McDevitt

The steps are generally described in the following documentation, as well as in most undergrad control textbooks.

https://www.mathworks.com/help/control/ug/design-a-yaw-damper-for-a-747-jet-transport.html#f0-1000201

Demo:

% Plant
Gp  = tf(32.5, [1 5 40])

Gp =
 
       32.5
  --------------
  s^2 + 5 s + 40
 
Continuous-time transfer function.

% rlocus(Gp), sgrid

From the Root Locus of uncompensated system Gp with the s-plane grid of constant damping factors, it is clear that no matter how we move the 'square' data marker (■) to track the gain and damping values, it cannot achieve the desired 0.7071 damping ratio. Thus, a feedback controller is needed. In this example, a PID controller is designed and its gains are tuned.

% PID Controller
kp  = -0.5066;
ki  =  0.8703;
kd  =  0.38;
Tf  =  0.7071;
Gc  = pid(kp, ki, kd, Tf)

Gc =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = -0.507, Ki = 0.87, Kd = 0.38, Tf = 0.707
 
Continuous-time PIDF controller in parallel form.

rlocus(Gc*Gp), sgrid

From the Root Locus of the compensated system Gc*Gp, we clearly have more rooms to move the data marker (■). Now, if we zoom into the locus and move the data marker (■) until the desired 0.707 damping ratio is observed, then the corresponding gain is found to be .