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I want to find \$k_1\$ that leads to the feedbakc system having poles in the 53º line (to have overshoot \$\leq 10\%\$). From the root locus, the poles move along the vertical asymptote centered on -1. How can I find \$k_1\$ using Scilab and how can I draw diagonal lines representing the 53º line in which I want my poles to be at?

So far, I made this graph in Scilab,

Code:

s=%s;
num=2;
den=s*(s+2)
t=syslin('c',num/den);
clf;
evans(t);
mtlb_axis([-10 10 -2 2]);

Obtained graph:

enter image description here

Desired graph:

enter image description here

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  • \$\begingroup\$ To draw the asymptotes there is a formula, web.calpoly.edu/~fowen/me422/RootLocus6.html \$\endgroup\$
    – jDAQ
    Commented Jan 10, 2020 at 18:44
  • \$\begingroup\$ @jDAQ I know the formula but I don't know how to implement it in Scilab. Could you help please? \$\endgroup\$
    – Carmen González
    Commented Jan 10, 2020 at 18:51
  • \$\begingroup\$ I think that scilab is already doing it, but since the poles go along a line they go exactly over the asymptotes, try a third order system and you might see a difference in the asymptotes and pole path \$\endgroup\$
    – jDAQ
    Commented Jan 10, 2020 at 18:53
  • \$\begingroup\$ The axes must have the same scales if you're going to use geometry. \$\endgroup\$
    – Chu
    Commented Jan 10, 2020 at 19:11
  • \$\begingroup\$ @jDAQ the diagonal dashed line is given by overshoot, so the Scilab didn't draw this lines in the graph and I want to add to the graph manually. I want to know the intersection between root locus and the overshoot line to know the k of control system. \$\endgroup\$
    – Carmen González
    Commented Jan 10, 2020 at 19:20

1 Answer 1

Reset to default
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First, drawing two more lines with angles 53⁰ and -53⁰ should be easy, just do another plot while keeping the root locus. The lines should be, $$ y = \tan(53^\circ)x$$ and, $$ y = -\tan(53^\circ)x.$$

Once you have those lines, set some \$k\$ to be the maximum value of the evans function and change that value until you find the one that lies on the line you plotted. I would use a binary search to pick the values of \$k\$.

s=%s;
num=2;
den=s*(s+2)
t=syslin('c',num/den);
clf;

//test your k by changing this
k = 100;

evans(t,k);
mtlb_axis([-10 10 -2 2]);

//draw the 53 deg line
r = linspace(0,-3,10);
up_line = tan(53*%pi/180)*r;
dn_line = -tan(53*%pi/180)*r;
plot(r,[up_line; dn_line],'r-.')

//pzmap of the system with feedback
figure
plzr(1/.(t*k))

//draw the 53 deg line
r = linspace(0,-3,10);
up_line = tan(53*%pi/180)*r;
dn_line = -tan(53*%pi/180)*r;
plot(r,[up_line; dn_line],'r-.')
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  • \$\begingroup\$ Thanks for help but the code in my computer crash in the line: plzr(1/.(t*k)) \$\endgroup\$
    – Carmen González
    Commented Jan 10, 2020 at 21:48
  • 1
    \$\begingroup\$ it runs fine on my computer, try this, cloud.scilab.in \$\endgroup\$
    – jDAQ
    Commented Jan 10, 2020 at 21:54
  • 1
    \$\begingroup\$ You can also delete the code from figure downwards, that is just a pole-zero map, the root locus is done by the previous lines. \$\endgroup\$
    – jDAQ
    Commented Jan 10, 2020 at 22:12
  • \$\begingroup\$ Just to confirm: Scilab makes a 0 to infinity k scanner. If I want to make the k scanner from -infinity to 0, I need to add minus signal into the function numerator, right? \$\endgroup\$
    – Carmen González
    Commented Jan 11, 2020 at 3:20
  • \$\begingroup\$ Yes, from zero to positive infinity \$\endgroup\$
    – jDAQ
    Commented Jan 11, 2020 at 3:29

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