Root Locus
Simple definition – Locus of points on the splane that represents the poles of a system
as one or more parameter vary.
RL and its relation to poles of a closed loop
system
RL and its relation to transient response and
stability
RL to select a parameter (such as K) to meet
closed loop transient response specifications
a. Closed-loop system; b. equivalent
transfer function
If G ( s ) 
NG (s)
N (s)
and H ( s )  H
DG ( s )
DH ( s )
Then the CLTF T ( s ) is
T (s) 
KN G ( s ) DH ( s )
DG ( s ) DH ( s )  KN G ( s ) N H ( s )
The zeros of T ( s ) are from N G ( s ) DH ( s ) and
the poles of T ( s ) are from a contribution of a
lot factors. Also the transient response is affected
by the poles and zeros of G ( s ) and H ( s ).
The RL give a good representaion of the poles
of T ( s ) as K varies.
1
Vector
representation
of complex
numbers:
a. s =  + j;
b. (s + a);
c. alternate
representation
of (s + a);
d. (s + 7)|s5 + j2
Consider a function of the form
m
F (s) 
  s  zi 
i 1
n
 s  pj 
j 1
The parameter m is the number of zeros
and
n is the number of poles which are complex factors.
The magnitude M of F ( s ) to any point s is
m
M 
  s  zi 
i 1
n
 s  pj 
j 1
The term  s  zi  is the magnitude of the vector
from the zeros of F ( s ) at zi to the point s.
Similarly,  s  p j  is the magnitude of the vector
from the poles of F ( s ) at p j to the point s.
The angle  of F ( s ) to any point s is
 =  zero angles -  poles angles
     s  zi      s  p j 
2
The zero angle is measured in the positive sense
from the vector starting at the zero at -zi on the s plane
to the s point in question.
The pole angle is measured in the positive sense
from the vector starting at the pole - p j on the s plane
to the s point in question.
Find the vector
representation
M of F(s)
to the point -3+j4
F (s) 
M?
 s  1
s  s  2
The vector from zero at -1 to the s point is
4
22  42 (180  tan 1 )  20116.6
2
The vector from pole at 0 to the s point is
4
32  42 (180  tan 1 )  5126.9
3
The vector from pole at -2 to the s point is
4
32  12 (180  tan 1 )  17104.0
1
Using the Eq.
m
M  
  s  zi 
i 1
n
  s  pi 
    s  zi      s  pi  
i 1
M  
20
5 17
116.6  126.9  104  0.22  114.3
which is evaluating F ( s ) at the point -3+j 4
3
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b. block diagram.
c. closed-loop transfer
function.
Pole location as a function of gain for the
system
a. Pole plot from Table
b. root locus
4
Root Locus
Definition of the RL
The root locus of a closed loop TF is a
representation of a continuous path of the
closed loop poles on the s-plane as the
gain K or other parameter is varied from  to +.
For this course, the parameter K is  0
Properties of the RL
Consider the CLTF T ( s ) 
KG ( s )
1  KG ( s ) H ( s )
A pole s exists when the characteristic polynomial
in the denominator becomes zero.
Therefore,
KG ( s ) H ( s )  1
5
That pole can be represented by a vector
that has magnitude and an angle. Therefore,
a value of s is a closed loop pole if
KG ( s ) H ( s )  1
This is called the magnitude criterion.
KG ( s ) H ( s )  (2k  1)180 ; k  0, 1, 2, 3...
i.e. an odd multiple of 180. This is called the angle criterion.
A pole s exist when the char. eqn becomes zero or
KG ( s ) H ( s )  1  1(2k  1)180
The value of K can be evaluated as
K
1
G (s) H (s)
Since the magnitude of KG(s)H(s) is unity, K can be solved
as above once the pole value is substituted.
So satisfying the angle and magnitude criteria of
KG(s)H(s) indicates that the s value is a pole on the root
locus.
Example
Prove whether that the s point -2+j3 is on the RL
of a open loop system as
KG ( s ) H ( s ) 
K ( s  3)( s  4)
( s  1)( s  2)
If the s point -2+j3 is on the RL of the system, then the
magnitude and angle criteria are satisfied.
6
a. Example system;
b. pole-zero plot of G(s)
Vector representation
of G(s) from -2+ j 3
Test point -2+j3
Using the angle criterion
   s  z      s  p   (2k  1)180 ;
i
j
k  0, 1, 2, 3...
Using the previous figure, we need to
evaluate all the angles from the zeros and poles
to the point in question (  2  j3) and observe
whether the result is an odd multiple of 180.
The result of 1   2   3   4 has to be evaluated.
7
3
1  tan 1 ( )  56.31
2
3
 2  tan 1 ( )  71.57
1
3  90
-2+j3
This image cannot currently be display ed.
3
 4  180  tan 1 ( )  108.43
1
1   2  3   4  70.55  (2k  1)180 ;
k  0, 1, 2, 3...
Therefore, the point  2  j3 is not on the root locus of
K ( s  3)( s  4)
.
( s  1)( s  2)
2
is a point on the RL so the
2
angle add up to a 180 (check).
The point  2  j
Now the gain K has to be evaluated at this
point of  2  j
2
using the magnitude criterion
2
m
K
pole vector lengths
1
1 
=
 i n1
G (s) H (s)
M  zero vector lengths
j1
2
 2
L1  22  
  2.121
 2 
K
2
 2
L2  12  
  1.223
 2 
L3 
2
 0.707
2
2
 2
L4  12  
  1.223
 2 
LL
K  3 4  0.33
L1 L2
2
K ( s  3)( s  4)
is a point on the RL of
2
( s  1)( s  2)
with a gain of 0.33.
The point  2  j
8
Sketching the root locus
• Number of branches  the number of branches of the
root locus equals the number of closed loop poles
• Symmetry  root locus is symmetrical about the real axis
• Real axis segment  on the real axis, for K>0 the root
locus exists to the left of an odd number of real axis, finite
open-loop poles and/or finite open loop zeroes
example
Complete root locus for the system
The number of paths
to infinity =
none
n (poles) - m (zeros)
9
From Matlab
System: sys
Root Locus
Gain: 0.333
Pole: -2 + 0.705i
Damping: 0.943
Overshoot (%): 0.013
Frequency (rad/sec): 2.12
0.8
0.6
0.4
Imag Axis
0.2
0
-0.2
-0.4
-0.6
-0.8
-4
-3.5
-3
-2.5
-2
-1.5
-1
-0.5
0
Real Axis
Stable for all K
• Start and end points of the RL
T (s) 
KN G ( s ) D H ( s )
DG ( s ) D H ( s )  KN G ( s ) N H ( s )
The RL begins at K  0 and ends a t K  
As K  0 ; T ( s ) 
KN G ( s ) D H ( s )
DG ( s ) D H ( s )  
As  approaches zero due to K  0, the closed loop poles of T ( s )
becomes the poles of DG ( s ) D H ( s ).
This implies that the RL commences at the poles of DG ( s ) D H ( s ).
The RL therefore begins at the poles of the loop transfer function
at K  0.
Analytically, this can also be seen from
DG (s) DH (s)  KNG (s) N H ( s)  0
As K  0 ; DG (s) DH (s)  0
RL end
T (s) 
KN G ( s ) DH ( s )
DG ( s ) DH ( s )  KN G ( s ) N H ( s )
T (s) 
KN G ( s ) DH ( s )
  KN G ( s ) N H ( s )
As K   ; The poles of T ( s ) is therefore the zeros of N G ( s ) N H ( s )
10
T (s) 
KN G ( s ) DH ( s )
DG ( s ) DH ( s )  KN G ( s ) N H ( s )
DG ( s ) DH ( s )  KN G ( s ) N H ( s )  0
Dividing by K  K   
DG ( s ) DH ( s )
 NG (s) N H (s)  0
K
NG (s) N H (s)  0
The RL ends at K  , at the zeros of the open loop
transfer function N G ( s ) N H ( s ).
Therefore, the RL starts at the poles of G ( s ) H ( s ) and ends at
the zeros of G ( s ) H ( s ) (the open loop TF).
• Behavior at infinity  If there are n poles of P(s) and m
finite zeros of P(s), the number of loci that approaches
infinity as K approaches infinity is n-m. They will
approach infinity along asymptotes with angles of 1800
(n-m=1); +900 (n-m=2); 1800 and +600 (n-m=3), or +450
and +1350 (n-m=4).
180 0  k .360 0
, k  0,1,2,  , n  m  1
nm

p1  p2   pn    z1  z 2   z m 
Real axis intercept  c 
,n  m  2
nm
Angles
k 
11
Example
The real axis intercept
 1  2  4   3  4
a 
3
4 1
The angle of the lines that intersect at -4/3 are
θa = π/3 for k = 0
θa = π
for k = 1
θa = 5π/3 for k = 2
Root locus and asymptotes for the system
G ( s) 
K ( s  3)
s 4  7 s 3  14s 2  8s
# of Paths to infinity
n–m=3
# of zeros
# of poles
Root Locus
4
3
Imag Axis
2
System: sys
Gain: 9.53
Pole: 0.000296 + 1.58i
Damping: -0.000187
Overshoot (%): 100
Frequency (rad/sec): 1.58
1
0
-1
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
Real Axis
Stable up to a limiting K value
12
Root Locus
3
2
Imag Axis
1
System: sys
Gain: 0.534
Pole: -0.45
Damping: 1
Overshoot (%): 0
Frequency (rad/sec): 0.45
0
-1
-2
-3
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
Real Axis
K at the breakaway point
• Root locus example showing real- axis breakaway (-1)
and break-in points (2)
Variation of gain along the real
axis for the previous root locus
13
The previous plot show that the gain reaches a maximum
between the poles (where K starts off at 0). This occurs at
the breakaway point.
The gain is a minimum as the RL plot comes back on the
real axis and goes towards the zeros (K becomes infinite).
This occurs at the break-in point.
Therefore, we can use basic calculus to find the
breakaway and break-in points  first method.
Repeating part (d), Recall
1
Recall K  
G (s) H (s)
Subst s   in the above
1
K 
G ( ) H ( )
dK ( )
0
d
we can solve for the values of  (or s values) where
the RL leaves and arrives on the real axis (the breakaway point
and break-in points).
Differentiating with respect to  , with
Example
KG ( s ) H ( s ) 
K ( s  3)( s  5) K ( s 2  8s  15)

 1
( s  1)( s  2)
s 2  3s  2
Subst. s  
K ( 2  8  15)
 1
 2  3  2
Making K the subject of the fromula
K 
 2  3  2
 2  8  15
Diff. with respect to  and equating to 0
dK 11 2  26  61

0
d  ( 2  8  15) 2
11 2  26  61  0
Solving for  gives   1.45 and  =3.82
the breakaway and break-in ponts
14
Second method  breakaway and break-in point without
differentiation (transition method). These points satisfy
the relationship:
m
n
1
1
1   z  1   p
i
i
where zi and pi are the negative of the zero and pole
values
From the example
1
1
1
1



  3   5  1   2
11 2  26  61  0
  1.45 and   3.82
Data for breakaway and break-in points for the root locus
• j crossing
CLTF T ( s ) 
K ( s  3)
s 4  7 s 3  14s 2  (8  K ) s  3K
Completing the Routh array
15
Only s1 can from a row of zeros
Solving
- K 2  65 K  720  0
K  9.65
Using the auxillary equation of the s 2 term and K  9.65
(90  K )  21K  80.35s 2  202.7  0
s   j1.59
The RL crosses the imaginary axis at s   j1.59 at a
gain K  9.65. At this gain, marginal stability occurs.
Also, the system is stable for 0  K  9.65
• Angles of departure from poles of P(s) and angles of
arrival at finite zeros of P(s) can determined by
application of the angle criterion to a point selected
arbitrarily close to the departure or arrival point
Example
Given a unity feedback system that has a forward TF
K ( s  2)
, do the following
( s 2  4 s  13)
(a) Obtain a RL using Matlab  Roots: 2+3i
G (s) 
(b) Find the complex poles that crosses on the imaginary axis
(c) Determine the gain at the j crossing
(d) Determine the break-in point
Discuss plant stability
16
Repeating part (d), Recall
1
G (s) H (s)
Subst s   in the above
Recall K  
K 
1
G ( ) H ( )
dK ( )
0
d
we can solve for the values of  (or s values) where
Differentiating with respect to  , with
the RL leaves and arrives on the real axis (the breakaway point
and break-in points).
17
KG ( s ) H ( s ) 
K ( s  2)
 1
( s 2  4 s  13)
+ 4 - 13
 2
dK (  2)(2  4)  ( 2  4  13)

0
d
(  2) 2
Subst s   ; K 
-
2
 2  4  21  0
  3,  7
The break-in point occurs at s  7.
Example
Given a unity feedback system that has a forward TF
K ( s  2)( s  4)
G (s)  2
, do the following
( s  6 s  25)
(a) Obtain a RL using Matlab
(b) Find the complex poles that crosses on the imaginary axis
(c) Determine the gain K at the j crossing
(d) Determine the break-in point
(e) Find the point where the RL crosses the 0.5 damping ratio line
(f) Find the gain at the point where the RL crosses the 0.5 damping ratio line
(g) Find the range of gain K , for which the system is stable
18
Root Locus
4
System: sys
Gain: 1
Pole: 0.00264 + 4.05i
Damping: -0.000652
Overshoot (%): 100
Frequency (rad/sec): 4.05
3
2
Imag Axis
1
0
-1
-2
-3
-4
-3
-2
-1
0
1
2
3
4
Real Axis
Root Locus
4
3
2
System: sys
Gain: 51.3
Pole: 2.89 - 3.67e-008i
Damping: -1
Overshoot (%): Inf
Frequency (rad/sec): 2.89
Imag Axis
1
0
-1
-2
-3
-4
-3
-2
-1
0
1
2
3
4
Real Axis
19
Root Locus
4
System: sys
Gain: 0.108
Pole: -2.42 + 4.18i
Damping: 0.5
Overshoot (%): 16.3
Frequency (rad/sec): 4.83
3
2
Imag Axis
1
0
-1
-2
-3
-4
-3
-2
-1
0
1
2
3
4
Real Axis
Repeat part (f) using the magnitude criterion.
This is to be done at the point  2.42  j 4.18
First from the poles (3  j 4)
L1  (3  2.42) 2  (4.18  4) 2  0.607
L2  (3  2.42) 2  (4  4.18) 2  8.2
Now from the zeros (2, 4)
L3  (2  2.42) 2  4.182  6.08
System: sys
Gain: 0.108
Pole: -2.42 + 4.18i
Damping: 0.5
Overshoot (%): 16.3
Frequency (rad/sec): 4.83
3
2
1
Imag Axis
L4  (4  2.42) 2  4.182  7.66
Root Locus
4
0
-1
-2
-3
K
L1 L2
 0.107
L3 L4
-4
-3
-2
-1
0
1
2
3
4
Real Axis
Transient Response via Gain Adjustment
2nd order approximation must be upheld since the RL
provides various damping ratios, settling time, peak time
etc.
 Higher order poles are much farther from the dominant
second-order pair.
 Closed loop zeros near the closed loop second poles are
canceled or nearly cancelled by the close proximity of other
higher order closed loop poles.
20
 Closed loop zeros not cancelled by the close proximity
of other higher order poles are far removed from the
closed loop second order dominant pair.
Design procedure for higher order systems
 Sketch the root locus for the given system
 Assume the system is second order with no zeroes.
Find the gain to meet the desired spec
 Justify second order approximation
 If it is not justified, then perform constrol simulations to
ensure that the specs are met
Second-order approximations
Example: For the system as shown, determine the value of K
to give a 1.52% overshoot. Evaluate the settling time, peak
time and SSE.
The system is third order with a zero at s = -1.5
21
 First get the Root locus
Assuming the system can be 2nd order approx draw the
damping ratio line
Searching for the closed loop poles at  = 0.8
Root Locus
3
2.5
2
System: sys
Gain: 12.7
System: sys
Pole: -1.19 + 0.893i
Damping: 0.8 Gain: 7.36
Overshoot (%): Pole:
1.52 -0.874 + 0.655i
Damping:
Frequency (rad/sec):
1.490.8
Overshoot (%): 1.51
Frequency (rad/sec): 1.09
1
0.5
0
-0.5
-1
-1.5
-2
-1.5
-1
-0.5
0
Real Axis
Root Locus
10
8
System: sys
Gain: 39.4
Pole: -4.56 + 3.42i
Damping: 0.8
Overshoot (%): 1.51
Frequency (rad/sec): 5.7
6
4
2
Imag Axis
Imag Axis
1.5
0
-2
-4
-6
-8
-10
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
Real Axis
22
The point where the RL crosses the  =0.8 is at 3
points yielding 3 sets of closed loop poles at
0.87  j 0.66,  1.19  j 0.9 and  4.6  j 3.45.
The respective gains from the RL plot are 7.36, 12.79 and 39.64.
For each point the settling time, time to first peak can be evaluated from
4

Ts 
; Tp 
n
d
The third closed loop pole must be obtained for each dominant set
having the same corresponding gain.
Searching for the third pole on the RL at each of the
corresponding gains 7.36, 12.79 and 39.64.
Note that the third pole cannot be complex
as the CLTF is third order, ie. the third pole
must be on the real axis.
The poles are at s = -9.25, -8.6 and 1.8 respectively.
Root Locus
10
8
6
4
System:System:
sys
sys
Gain: 7.4
Gain: 12.8
Pole: -9.25
Pole: -8.61
Damping:
Damping:
1
1
Overshoot
Overshoot
(%): 0 (%): 0
Frequency
Frequency
(rad/sec):
(rad/sec):
9.25
8.61
Imag Axis
2
0
System: sys
Gain: 39.2
Pole: -1.8
Damping: 1
Overshoot (%): 0
Frequency (rad/sec): 1.8
-2
-4
-6
-8
-10
-10
-9
-8
-7
-6
-5
-4
-3
-2
-1
0
Real Axis
23
Using Ts 
4

; Tp 
n
d
and the real and imaginary parts of the dominant pole.
The velocity error constant is
sK ( s  1.5)
K (1.5)

s ( s  1)( s  10) (1)(10)
Subst. various values of K gives the K v values in the Table below.
K v  lim sG ( s )  lim
s 0
s 0
Second- and third-order
responses
a. Case 2; b. Case 3
Cases 1 and 2 have the third
pole far away from the
complex pair. However,
there is no approx. pole zero
cancellation.
Case 3, the third pole is
closer to the zero, so a 2nd
order approx can be
considered valid.
The plots are relatively
close.
A step input is used to show the second order dynamics
and validity of the second order approximation.
We will now re-evaluate for the third pole analytically
knowing the gain at the corresponding dominant pair.
As an example, Case 3 will be used.
24
In case 3, K  39.64. Using the magnitude criterion
assuming the third pole p exists somewhere between the
pole at  10 and the zero at  1.5. The point in question is p.
L1  10  p
L2  p  1
L3  p
L4  p  1.5
(10  p )( p  1) p
 39.64
( p  1.5)
Solving for p gives
K
p  1.795 which is the third pole.
The same can be done for the finding
the other third poles. Note the RL does NOT
exist on portions of the real axis where the sum
of poles and zeros to the left is even.
The closed loop TF is using K  39.64
C (s)
39.64 s  59.64

R( S ) s 3  11s 2  49.64 s  59.64
1
(discuss Matlab)
s2
The figure shows that there is a steady state error e()
Since R( s ) 
of (4.5  4.33)  0.17.
NOTE THAT THIS IS =
1
Kv
For Case 3,  K  39.64  K v  5.9
e( ) 
1
 0.17 which matches the Matlab plot.
5.9
Step Response
6
5
System: sys
Time (sec): 4.5
Amplitude: 4.33
Amplitude
4
3
2
1
0
0
1
2
3
4
5
6
Time (sec)
25
Example
Given a unity feedback system that has a forward TF
K
G (s) 
, do the following
( s  2)( s  4)( s  6)
(a) Obtain a RL using Matlab
(b) Using a 2nd order approx. determine the value of
K to give a 10% overshoot for a unit step input
(c) Determine settling and peak times
(d) The natural frequency
(e) The SSE for the value of K .
(f) Determine the validity of the 2nd order approx.
Repeat part (b) using the magnitude criterion.
This is to be done at the point  2.028  j 2.768
From the poles
L1  (2.028  2) 2  (2.768) 2  2.76814
L2  (2.028  4) 2  (2.768) 2  3.399
L3  (2.028  6) 2  (2.768) 2  4.841
There are no zeros
K
L1 L2 L3
 45.55
1
26
(d) n  2.0282  2.7682  3.432
(e) The system is Type 0, the position error constant is
K
45.55
lim G ( s )  K p 

 0.949
s 0
2*4*6
48
1
Therefore, e() 
 0.51
1 Kp
(f)
In this case, K  45.55. Using the magnitude criterion
assuming the third pole p exists somewhere to the left of the
pole at  6. The point in question is p.
L1  p  6
L2  p  4
L3  p  2
( p  6)( p  4)( p  2)
 45.55
1
Solving for p gives
K
p  7.943 which is the third pole.
Since this third close loop pole is NOT 5 times or more
the magnitude of the real part of the closed loop dominant
pole, the second approx. NOT valid.
Root Locus
10
8
6
4
System: sys
Gain: 44.6
Pole: -7.92
Damping: 1
Overshoot (%): 0
Frequency (rad/sec): 7.92
Imag Axis
2
0
-2
-4
-6
-8
-10
-16
-14
-12
-10
-8
-6
-4
-2
0
2
Real Axis
27