Question

A Unity feedback system has an open loop transfer function of

G(s) = K / s (s+1) (s+5)

Draw the root locus plot and determine the value of K to give a damping ratio of 0.3 A network having a transfer function of 10(1 +10s) /(1 +100s) is now introduced in tandem. Find the new value of K, which gives the same damping ratio for the closed -loop response. Compare the velocity error constant and settling time of the original and the compensated systems

Answer 1

The plot of the rootlocus and the zeta line on the same graphs is drawn as below in matlab for accuracy

by zooming in at the intersection part of the plots we get as below

Hence we get from the above plots the value of K is 7.05

The poles are -0.34 + 1.1 j , -0.34 - 1.1 j

for the compensated system

the root locus of the compensated system is

by zooming in at the intersection we get as below

The value of K for the required zeta value is K = 6.23

Kv and settling time values comparison

We have already compared the Kv values and we got the Kv value improved and hence we can say error at steady state is reduced

for settling time

we know considering the second-order approximation of the dominant poles we get

where the denominator is the magnitude of the real part of the poles comparing we get as below

for uncompensated system: real part of the poles is 0.346 in magnitude. so settling time is 4 / 0.346 = 11.56 s

for the compensated system, the real part of the poles is 0.315 so settling time is 4 / 0.315 = 12.698 s

The Matlab code used for all parts is

g1 = tf([1],[1 1 0])
g2 = tf([1],[1 5])          
c = 10*tf([10 1],[100 1])  % compensator

x = 0:-0.1:-5;
y = -3.17979*x;

g = g1*g2;              % given transfer function 


hold on
rlocus(g*c)          % rootlocus of system 
plot(x,y);            % zeta line 
hold off

in summary

we have found the root locus and zeta line in each case and found the k value

then we found the error constant and then found the settling time