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Question:


Sketch the root locus for the open loop transfer function of a unity feedback control sytem given below and determine the value of K for \$\xi=0.5\$

\$G(s)=\frac{K}{s(s+1)(s+3)}\$

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I have drawn the root locus diagram like this

enter image description here

Now i tried all possible way to get the intersection point of rootlocus diagram with the 60 degree line.If i get that point i can easily find the value of K for \$\xi=0.5\$.

Is is possible to find that point without any software?

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The characteristic equation is:

$$s(s+1)(s+3)=-K$$

Hence, plug in any point on the locus, and that will give the corresponding value of K.

It appears that your selected point on the locus is approximately: \$s=-0.4+j0.7\$, which will give \$K\approx 1.8\$

For an accurate answer to this particular problem (\$\small \zeta=0.5\$), write the CE as: $$s(s+1)(s+3)+K= (s+\alpha)(s^2+\omega _n s+ \omega_n^2)$$

solve for \$\alpha\$ and \$\omega_n\$, and hence find K. Thus:

$$ s^3+(\alpha +\omega_n)s^2+(\alpha\omega_n+\omega_n^2)s+\alpha\omega_n^2=s^3 +4s^2+3s+K$$ Giving \$\small K=1.83\$.

Systems higher than 3rd order will need a root solver.

But this approach begs the question, why bother to sketch the root locus in the first place?

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  • \$\begingroup\$ I want that \$"point"\$ to plug into that equation.I think you did't get what i am asking...i am asking that point coordinates where \$\xi\$ becomes 0.5 \$\endgroup\$ – Rohit Oct 8 '17 at 8:48
  • \$\begingroup\$ how did you get that point coordinates please tell me?thats my doubt only \$\endgroup\$ – Rohit Oct 8 '17 at 8:53
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    \$\begingroup\$ From the graph, assuming the axes are to the same scale. \$\endgroup\$ – Chu Oct 8 '17 at 8:55
  • \$\begingroup\$ Really i am not getting what you are trying to say sir \$\endgroup\$ – Rohit Oct 8 '17 at 8:56
  • \$\begingroup\$ The point on the graph is at (-0.4, 0.7) if the 'y-axis' scale is the same as the 'x-axis' scale. \$\endgroup\$ – Chu Oct 8 '17 at 8:58
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We search a given line for a point that yields a summation of angles (for both zero and pole angles) equal to an odd multiple of \$180^\circ\$. Take a look at the following picture

enter image description here

The line in which a point that yields a summation of angles equal to an odd multiple of 180 deg is shown as the Green line in the above picture. The angle of this line is \$ \beta = \cos^{-1} (\zeta) \$ and the radius that satisfies the aforementioned condition is searched by a software that we code. In this example, this is Matlab code that searches for the target point.

r=.75;
while r < .76
zeta=.5; 
a=acos(zeta);
x=r*cos(a);
y=r*sin(a);

%poles' angles
p1=0;
Th1=pi-atan( y/x );

p2=-1;
Th2=atan(y/(abs(p2)-x));

p3=-3;
Th3=atan(y/(abs(p3)-x));

angle=-(Th1+Th2+Th3);
fprintf('Th1=%.3f : Th2=%.3f : Th3=%.3f  :  angle=%.3f  : r=%.3f\n', rad2deg(Th1),rad2deg(Th2), rad2deg(Th3),rad2deg(angle),r);

r = r + .00001;
end 

If you run the code, you notice the following result

Th1=120.000 : Th2=46.102 : Th3=13.898  :  angle=-180.000  : r=0.750

The aforementioned condition is satisfied if \$\theta_1 = 120^\circ, \theta_2=46.102^\circ \$ and \$\theta_3 = 13.898^\circ\$ which yields a radius \$r=0.75\$. The point \$P\$ is then \$ (r\cos(\beta), r\sin(\beta))\$, or \$P=-0.3750+j0.6495\$. Now, to compute the gain K, we substitute P in the following formula:

$$ \begin{align} K &= \frac{1}{|G(s)||H(s)|} \\ &= |s(s+1)(s+3)| \Big|_{s=P} \\ &= 1.8281. \end{align} $$

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