How to draw Root Locus | Solved Example

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 N_P be representing number of open-loop (OL) poles and N_Z the number of open-loop (OL) zeros.

The characteristic equation will be       S(S+3)(S²+2S+2)+K_a =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 = | N_P-N_Z | = 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  (dK_a/ds)=0

From the characteristic equation, 

K_a = -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 (dK_a/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

Φ_A = 180⁰+ ϴ

Where, ϴ = Σangles subtended by OL zeros - Σangles subtended by OL poles

From above figure,    ϴ = 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 K_a and frequency w. Take only those values of K_a which are greater than zero because K_a 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