We will learn how to draw root locus using a solved
example.
Example :- A feedback
control system has open-loop transfer function
(OLTF).
Draw root locus as K is varied from 0 to ∞
.
Solution :- Let NP be representing number of
open-loop (OL) poles and NZ the number of open-loop (OL) zeros.
The characteristic equation will be S(S+3)(S2+2S+2)+Ka
=0
Step-1 :- Find OL
poles and OL zeros from the OLTF
OL poles are S=0,-3,(-1+j1) and (-1-j1)
There are no finite OL zeros.
Mark OL pole with cross and OL zero with circle in
S-plane as shown.
Step-2 :- Find number
of asymptotes
Number of asymptotes = No of root locus branches
tending to infinity = | NP-NZ | = 4
Step-3 :- Find
centroid (common point on real axis at which asymptotes cut the real axis)
as
Step-4 :- Find
asymptote angles (angles that asymptotes make with the real axis in the counter
clock wise direction)
So, from steps 2,3 and 4 , four asymptotes cut the
real axis at -1.25 and make angles 450,
1350, 2250 and 3150 , as shown below.
Step-5 :- Find the
parts of the real axis at which root locus lies.
A point on real axis lies on root locus if the number of OL poles+OL zeros on the real axis to the right of the point is odd.
Therefore, root-locus lies between 0 and -3 on the
real axis.
Step-6 :- Find the
breakaway points (points at which two or more root locus branches meet)
Breakaway points are the solutions of (dKa/ds)=0
From the characteristic equation,
Ka = -S(S+3)(S2+2S+2) = -(S4+5S3+8S2+6S)
We get, S = -2.3 , (-0.725±j0.365)
Not all values obtained
as solutions of (dKa/ds)=0 need to be necessarily the breakaway points. Out
of the obtained s values only those values of S which satisfy angle condition are the actual breakaway points.
On checking angle condition we find that (-0.725±j0.365) do not satisfy it. Therefore, only S= -2.3 is the only breakaway point.
So, from step 5 and 6 , the real axis from 0 to -3 contains root
locus which breakdown at -2.3 as shown.
Step-7 :- The number of
branches approaching breakdown point (r) = number of branches leaving the
breakdown point (r) = 2
The leaving branches make ±1800/r
= ±900 angles with the approaching branches as
shown above in step-6 diagram.
Step-8 :- Find angles of departure (angle which a root locus
branch starting from an open loop pole, makes with a line parallel to the real
axis, in the counter clockwise direction).
It is calculated as
ΦA = 1800+ ϴ
Where,
ϴ = Σangles subtended by OL zeros - Σangles subtended by OL poles
From
above figure, Ï´
= 0 – ( 900+1350+26.60 ) = -251.60
Therefore,
angle of departure = 1800+ (-251.60) = -71.60
So, root locus branch starts from (-1+j1) at an angle
-71.60
Since root locus is always mirror image about real axis , therefore, root locus starts from (-1-j1) at +71.60
Step-9 :- Find the points
at which root locus branches intersect jw axis.
By simply replacing S with jw in characteristic
equation and solving it. The equation will have two variables, gain Ka
and frequency w. Take only those values of Ka which are greater than
zero because Ka cannot be -ve.
The points of intersection comes out to be +j1.1 and
–j1.1
The complete
root locus is shown below.
Related :-
(1) Bode plot
(1) Bode plot
