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Unit-III - Root Locus

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IFC course

Feedback Control System (FCS)


[ICE(IF)-20001]

Unit:
Stability Analysis using Root Locus
Dr. S. D. Hanwate
Faculty, Dept. of Instrumentation and Control
College of Engineering Pune
Introduction
• Primary aim of a control engineer: To design a control system that
meets the desired specifications.
• Stability of the system is also a necessary condition.
• Analytical approach can be followed till second order systems while
Routh-Hurwitz criterion can be followed for higher order systems.
• Drawbacks of Routh-Hurwitz criterion -
(a) does not provide sufficient information about relative stability, which
may make the system unstable,
(b) does not help in designing a control system where tuning of parameters
is important.
Example:
Root Locus Technique
Root Locus: Example
Root Locus: Example (contd.)
Root Locus: Example (contd.)
Advantages
• The advantages of root locus technique are:
• Clearly showing the contributions of each loop poles or zeros to the
location of the closed loop poles.
• Indicating the manner in which the loop poles and zeros should be
modified so that the response meets system performance
specifications.
• Knowledge of the open loop system is sufficient to analyse the
behaviour of the system, detailed study of the closed loop system is
not required.
System Parameters and Pole Locations
System Parameters and Pole Locations
System Parameters and Pole Locations
Evans’ Conditions
Evans’ Conditions (contd.)
Points on Root locus
Example
Rules for constructing Root Loci
• Segment of root loci can exist in the right half of s-plane. This signifies instability, the points at
which the root loci cross the imaginary axis define the stability limits.
• Basic application of the Routh array determines the gains at the satability limits. By substituting
this value of gain in the auxiliary equation, the value of s=jw at the stability limit is evaluated.
Illustrative example:
Design problem:
Over all Closed loop transfer function:
Effects of Pole and Zeros on Root locus
Example 2:
Control system Design Problem:
Effect of addition of zero (derivative) term
Example:
• For the unity feedback system in Figure,

• Design a controller for following specification:


a) Design rate feedback to yield a step response with no more than
25% overshoot and settling time < 2 seconds.
b) Plot the states, control signal u without and with controller.
Solution:
• The desired operating point is found from the desired specifications,

• Thus,
• Hence the design point is –2 +j4.5324.
• Now, add a pole at the origin to increase system type and drive error
to zero for step inputs.
• Now design a PD controller. The angular contribution to the design
point of the system poles and pole at the origin is 101.9 ⁰. Thus, the
compensator zero must contribute 180⁰ – 101.9 ⁰ =78.1 ⁰. Using
thegeometry below,
• Using Hence, zc = 2.955.
• The compensated open-loop transfer function with PD compensation
is

• Adding the compensator zero to the system and evaluating the gain
for this at the point –2 + j4.5324,
• yields K = 294.51.
Matlab code:
K= 294.51 %root locus gain for new compensated system
num = [-2.995]
den = [0 -4 -6 -10]
sys = zpk(num,den,K)
rlocus(sys)
grid
v = [-15 1 -10 10];axis(v)
hold on
plot(-2,0,'*')
hold on
plot(0,4.5324,'*')
T=feedback(sys,1);
figure
step(T)
stepinfo(T)
• Since we have able to obtain the both the desired performance
• Mp < 25% = 19.3%
• Ts < 2 sec = 1.78 sec.
References:
1) K. Ogata, Modern Control Engineering. Prentice Hall, 2010.
2) Norman Nise, “Control System Engineering”, Wiley India, Fifth ed., 2009
3) G. C. Goodwin, S. F. Graebe, M. E. Salgado; “Control System Design”, PHI, First ed. 2002
4) Friedland, “Control System Design”, Dover Publication, First ed., 2005.
5) NPTEL Video lecture:
➢ Lec1
➢ Lec2
➢ Lec3

FCS (AY 2021-22) / COEP Dr. S. D. Hanwate

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