1. It is symmetrical to the real axis.
2. Number of branches of root locus is equal to number of max (poles or zero).
3. Starting point (k=0) End point(k—>∞)

           (open loop poles)       (open loop zeros)

4.     A point on the real axis lies on the root locus if number of open loop poles or zero to the right side of that point is odd in number.

5.     Value of K anywhere on the root locus is given as

             K =

6.     I)If poles > zeros then (Þ-z) branches will terminate at ∞ (where k=∞)

II)If Z > P, then (Z-P) branches will start from ∞ (K = 0)

7.     When P>Z, (P-Z) branches will terminate at ∞ (open loop zeros). But by which path. So the path is shown by asymptotes and this asymptote is given by

Asymptote =    q = o,1,2……(P-Z-1)

8.     There asymptotes intersect the real axis at a single point and this point is known as centroid.

Centroid =

9.     Break away and break in point when root locus lies between two poles it’s called break in point.

• Centroid and Breakaway points are not same
• Differentials the characteristic equation and equate to zero

10. Angle of arrival and angle of departure this print is used when the roots are complex.

Angle of departure - for complex poles

Angle of arrival – for complex zero.

 øD =(2q+1) 180° + ø

øA = (2q+1) 180° - ø               = ∠Z– ∠P.

Number of Zeros = 0

Number of polls S = (0, -1+j, -1-j) = (3).

1)     Number of Branches = max (P, Z) = max (3, 0) = 3.

2)     As there are no zeros in the system so, all branches terminate at infinity.

3)     As P>Z, branches terminate at infinity through the path shown by asymptotes

                  Asymptote = × 180°   q = 0, 1, 2………..(p-z-1)

P=3, Z=0.

q= 0, 1, 2.

For q=0

Asymptote = 1/3 × 180° = 60°

For q=1

Asymptote = × 180°

= 180°

For q=2

Asymptote = × 180° = 300°

Asymptotes = 60°,180°,300°.

4)     Asymptote intersects real axis at centroid

Centroid =

=

Centroid = -0.66

5)                                                                                                                                          

S3

1

1

0

S2

2

K

0

S1

0

S0

K

For stability   > 0. And K > 0.

0<K<2.

So, when K=2 root locus crosses imaginary axis

S3 + 2S2 + 2S + 2 =0

For k

Sn-1 = 0                   n : no. of intersection

S2-1 = 0                      at imaginary axis

S1 = 0

= 0

K<4

For Sn = 0 for valve of S at that K

S2 = 0

2S2 + K = 0

2S2 + 2 = 0

2(S2 +1) = 0

32 = -2

S = ± j

The root locus plot is shown in figure 1.

Asymptote = ×180°. q=0,1,………p-z-1

P=3, Z=0

q= 0,1,2.

For q = 0

Asymptote = × 180° = 60.

For q=1

Asymptote = × 180° = 180°

For q=2

Asymptote = × 180° = 300°

Centroid =

= = -1

6.     As root locus lies between poles S= 0, and S= -1

So, calculating breakaway point.

= 0

The characteristic equation is

1+ G(s) H (s) = 0.

1+ = 0

K = -(S3+3S2+2s)

= 3S2+6s+2 = 0

3s2+6s+2 = 0

S = -0.423, -1.577.

So, breakaway point is at S=-0.423

because root locus is between S= 0 and S= -1

G(s) =

1. Number of zeros = 0 number of poles = 4

        P = (S=0,-1,-1+j,-1-j) = 4

2.     As P>Z all the branches will terminated at infinity.

3.     As no zeros so all branches terminate at infinity.

4.     The path for branches is shown by asymptote.

Asymptote =         q = 0,1,…..(Þ-z-1)

q=0,1,2,3.             (P-Z = 4-0)

for q=0

            Asymptote = ×180° =45°

For q=1

             Asymptote = ×180° =135°

For q=2

           Asymptote = ×180° =225° 

For q=3

                            Asymptote = ×180° =315°

5.     Asymptote intersects real axis at unmarried

Centroid =  

Centroid = = = -0.75

6)     As poles are complex so angle of departure is

ØD = (2q+1) ×180 + ø                       ø = ∠Z –∠P.

7)     As the root locus lie between S=0 and S=-1

So, the breakaway point is calculated

1+ G(s)H(s) = 0

1+ = 0

(S2+S)(S2 +2S+2) + K =0

K = -[S4+S3+2s3+2s2+2s2+2s]

= 4S3+9S2+8S+2=0

S = -0.39, -0.93, -0.93.

The breakaway point is at S = -0.39 as root locus exists between S= 0 and S=-1

I + G(s) H(s) = 0

K+S4+3S3+4S2+2S=0

1)     Number of zero = 0 number of poles = 4 located at S=0, -2, -1+j, -1-j.

2)     As no zeros are present so all branches terminated at infinity.

3)     As P>Z, the path for branches is shown by asymptote

Asymptote =                 

q = 0,1,2……p-z-1

For q = 0

Asymptote = 45°

q=1

Asymptote = 135°

q=2

Asymptote = 225°

q=3

Asymptote = 315°

4)     Asymptote intersects real axis at centroid.

Centroid =

=

Centroid = -1.

5)     As poles are complex so angle of departure is

ØD=(2q+1)180° + Ø

             ø = ∠Z –∠P

             = 0-[135°+45°+90°]

= 180°- 270°

ØD = -90°

6)     As root locus lies between two poles so calculating point. The characteristic equation is

1+ G(s)H(s) = 0

1+ = 0.

K = -[S4+2S3+2S2+2S3+4S2+4S]

K = -[S4+4S3+6S2+4S]

= 0

= 4s3+12s2+12s+4=0

S = -1

So, breakaway point is at S = -1

7)     Intersection of root locus with imaginary axis is given by Routh Hurwitz.

S4+4S3+6S2+4s+K = 0

            ≤ 0

K≤5.

For K=5 valve of S will be.

5S2+K = 0

5S2+5 = 0

S2 +1 = 0

S2 = -1

S = ±j.

1. Number of zeros = 0. Number of poles = 4 located at S=0, -3, -1+j, -1-j.
2. As no. zero so all branches terminate at infinity.
3.  The asymptote shows the both to the branches terminating at infinity.

Asymptote =                    q=0,1,….(p-z).

For q = 0

Asymptote = 45

For q = 1

Asymptote = 135

For q = 2

Asymptote = 225

For q = 3

Asymptote = 315

(4). The asymptote intersects real axis at centroid.

Centroid = ∑Real part of poles - ∑Real part of zero / P – Z

= [-3-1-1] – 0 / 4 – 0

Centroid = -1.25

(5). As poles are complex so angle of departure

φD = (29 + 1)180 + φ

ø = ∠Z –∠P.

= 0 – [ 135 + 26.5 + 90 ]

= -251.56

For q = 0

φD = (29 + 1)180 + φ

= 180 – 215.5

φD = - 71.56

(6). Break away point  dk / ds = 0 is at   S = -2.28.

(7). The intersection of root locus on imaginary axis is given by Routh Hurwitz.

1 + G(S)H(S) = 0

K + S4 + 3S3 + 2S3 + 6S2 + 2S2 + 6S = 0

S4 1 8 K

S3 5 6

S2 34/5 K

S1 40.8 – 5K/6.8

K ≤ 8.16

For K = 8.16 value of S will be

6.8 S2 + K = 0

 6.8 S2 + 8.16 = 0

S2 = - 1.2

S = ± j1.09

G(S) = K (S + 6)/S (S + 4)

1. Number of zeros = 1(S = -6)

Number of poles = 2(S = 0, -4)

2.     As P > Z one branch will terminate at infinity and the other at S = -6.

3.     For Break away and breaking point

1 + G(S)H(S) = 0

1 + K (S + 6)/S (S + 4) = 0

dk/ds = 0

S2 + 12S + 24 = 0

S = -9.5, -2.5

Breakaway point is at -2.5 and Break in point is at -9.5.

4.     Root locus will be in the form of a circle. So finding the center and radius. Let S =  + jw.

G( + jw) =  K(  + jw + 6)/(   + jw)(  + jw + 4 ) = +- π 

tan-1  w/  + 6 - tan-1 w/  – tan-1 w /  + 4 = - π

taking tan of both sides.

w/  + w/  + 4 / 1 – w/  w/  + 4 = tan π + w /  + 6 / 1 - tan π w/  + 6

w/  + w/  + 4 = w/  + 6[ 1 – w2 /  ( + 4) ]

(2 + 4)(  + 6) = (2 + 4  – w2)

2 2 + 12 + 4 + 24 = 2 + 4 – w2

22 + 12 + 24 = 2 – w2

2 + 12 – w2 + 24 = 0

Adding 36 on both sides

( + 6)2 + (w + 0)2 = 12