Control system stability root locus

Control System
Stability: Root Locus
Dr. Nilesh Bhaskarrao Bahadure
nbahadure@gmail.com
https://www.sites.google.com/site/nileshbbahadure/home
June 29, 2021
Dr. Nilesh Bhaskarrao Bahadure (PhD) Control System June 29, 2021 1 / 81

Overview I
1 Root Locus
What is Root Locus?
Concept of Root Locus
Angle and Magnitude Condition of Root Locus
2 Rules for Constructing Root Locus
Symmetricity of root locus
The starting and termination of root locus
Root Locus Points
Angle of Asymptotes
Centroid
Example:01
Example:02
Breakaway Point
Example:03
Intersection
Example:04

Overview II
Angle of Departure and Angle of Arrival
General Steps to Draw Root Locus
3 Problems on Root Locus
Example:05
Example:06
Example:07
4 Thank You

Root locus

Root Locus in Control System
What is Root Locus?

What is Root Locus?
A graphical method used for analyzing the location and movement of
poles in the s-plane with the variation in the gain factor of the
system is known as Root Locus. This technique is used to check the
stability of the closed-loop control system.

What is Root Locus?
A graphical method used for analyzing the location and movement of
poles in the s-plane with the variation in the gain factor of the
system is known as Root Locus. This technique is used to check the
stability of the closed-loop control system.
So, in simple terms, a graph plotted for roots of a characteristic equation
by varying the system parameter (generally gain) from 0 to infinity is
called Root Locus.

Suppose we have a closed-loop system as represented below:
Figure : Closed loop system

For the above given closed-loop system, gain K is a variable parameter
which is a part of the forward path of the system.

So, the transfer function for the above system will be given as:
T(s) =
KG(s)
1 + KG(s)H(s)

So, the transfer function for the above system will be given as:
T(s) =
KG(s)
1 + KG(s)H(s)
We are already aware of the fact the characteristic equation of the
closed-loop system is represented by equating the denominator of the
transfer function to 0.

Concept of Root Locus...

So, with the system gain K, the characteristic equation will be given as:
1 + KG(s)H(s) = 0

1 + KG(s)H(s) = 0
This shows that the roots of the characteristic equation given above now
depends on the variable ’K’.

1 + KG(s)H(s) = 0
So, this leads to taking us to the conclusion that if gain K is varied from
−∞ to +∞ then every individual value of k within this range will provide
a different set of locations of the poles in the s-plane.

1 + KG(s)H(s) = 0
Hence, root locus is defined as the locus of the poles of the closed-loop
control system achieved for the various values of K ranging between −∞
to +∞. However, it is generally assumed to be between 0 to ∞.

1 + KG(s)H(s) = 0
Hence, root locus is defined as the locus of the poles of the closed-loop
control system achieved for the various values of K ranging between −∞
to +∞. However, it is generally assumed to be between 0 to ∞.
Thus, the technique helps in determining the stability of the system and so
is utilized as a stability criterion in control theory.

The characteristic equation of a general closed-loop negative feedback
system is given as:
1 + G(s)H(s) = 0
So,
G(s)H(s) = −1

system is given as:
1 + G(s)H(s) = 0
So,
G(s)H(s) = −1
As s-plane is complex in nature, constituting both real and imaginary
value. Thus, writing the above equation as:
G(s)H(s) = −1 + j0

system is given as:
1 + G(s)H(s) = 0
So,
G(s)H(s) = −1
As s-plane is complex in nature, constituting both real and imaginary
value. Thus, writing the above equation as:
G(s)H(s) = −1 + j0
This means G(s)H(s) is also complex. Thus, the above-given equation
must be satisfied for each individual value of s in order to be present on
the root locus.

Further, the two conditions of root locus are:
1 Angle condition
2 Magnitude condition
Angle condition
As we know,
G(s)H(s) = −1 + j0
On equating angles, we will get,
∠G(s)H(s) = ±(2r + 1)180o
: r = 0, 1, 2 − −

1 Angle condition
Angle condition
As we know,
G(s)H(s) = −1 + j0
∠G(s)H(s) = ±(2r + 1)180o
: r = 0, 1, 2 − −
It is to be noted here that −1 + j0 = 1∠ ± 180o. However, −1 + j0 will be
present on the negative real axis of the s-plane.

1 Angle condition
Angle condition
As we know,
G(s)H(s) = −1 + j0
∠G(s)H(s) = ±(2r + 1)180o
: r = 0, 1, 2 − −
It is to be noted here that −1 + j0 = 1∠ ± 180o. However, −1 + j0 will be
present on the negative real axis of the s-plane.
Thus, can be traced as magnitude value 1 and ∠ ± 180o,

Angle and Magnitude Condition of Root Locus...

±540o − − ± (2r + 1)180o, as shown below.
Figure :

Figure :
Hence, for the angle condition, ∠G(s)H(s) for any of the roots of the
general characteristic equation will be ±(2r + 1)180o i.e., odd multiples of
180o.

Figure :
Hence, for the angle condition, ∠G(s)H(s) for any of the roots of the
general characteristic equation will be ±(2r + 1)180o i.e., odd multiples of
180o.
This signifies to be present on the root locus, the point must necessarily
satisfy the angle condition. That means the calculated angle of G(s)H(s)

Magnitude Condition

Magnitude Condition
Further for the magnitude condition, the magnitude of both RHS and LHS
must be equated for the equation G(s)H(s) = −1
Hence,
|G(s)H(s)| = | − 1 + j0| = 1

Magnitude Condition
Hence,
|G(s)H(s)| = | − 1 + j0| = 1
It is to be noted that with the unknown value of K, we will not be able to
determine |G(s)H(s)| at any point in s-plane. However, if we get an idea
regarding the existence of a point in s-plane on the root locus then it must
also satisfy the magnitude condition.

Magnitude Condition
Hence,
|G(s)H(s)| = | − 1 + j0| = 1
So, the value of K for a point whose presence on the root locus is obtained
by the angle condition can be determined by the magnitude condition.

Magnitude Condition
Hence,
|G(s)H(s)| = | − 1 + j0| = 1
So, the value of K for a point whose presence on the root locus is obtained
by the angle condition can be determined by the magnitude condition.
And the obtained value of K signifies the gain of the system for which a
point on root locus acts as the root of the characteristic equation.

The roots of the characteristic equation of the system can be either real,
complex or combination of both.

So, suppose we have n number of roots for K between 0 to infinity, the
root locus on the s-plane must exhibit symmetricity about the real axis of
the s-plane.

So, suppose we have n number of roots for K between 0 to infinity, the
root locus on the s-plane must exhibit symmetricity about the real axis of
the s-plane.
The root locus is always symmetrical about real axis

This rule states, the root locus must begin from open-loop poles and must
specifically terminate on either open-loop zero or infinity.

Now, under this specific rule, we have to consider two conditions,
depending upon the number of open-loop poles i.e., P and number of
open-loop zeros i.e., Z.

Let P = Number of Poles and Z = Number of Zeros
Case - I: P > Z

Case - I: P > Z
If the number of open-loop poles is greater than the number of open-loop
zeros i.e., P > Z.

Case - I: P > Z
If the number of open-loop poles is greater than the number of open-loop
zeros i.e., P > Z.
Then the number of branches, N will be equivalent to the number of
poles, P.
If P > Z, No. of Branches = No. of Poles, i.e. N = P
so, P − Z that much branches will approach to infinity.

Case - II: P < Z

Case - II: P < Z
If the number of open-loop poles is less than the number of open-loop
zeros i.e., P < Z.

Case - II: P < Z
zeros i.e., P < Z.
Then the number of branches, N will be equivalent to the number of
zeros, Z.
If P < Z, No. of Branches = No. of Zeros, i.e. N = Z
so, Z − P that much branches starts from the infinity.

Case - III: P = Z

Case - III: P = Z
zeros i.e., P = Z.

Case - III: P = Z
zeros i.e., P = Z.
Then the number of branches, N will be equivalent to the number of zeros
or poles, P or Z.
If P = Z, No. of Branches = No. of Zeros = No. of Zeros, i.e.
N = Z = P

Case - III: P = Z
zeros i.e., P = Z.
Then the number of branches, N will be equivalent to the number of zeros
or poles, P or Z.
If P = Z, No. of Branches = No. of Zeros = No. of Zeros, i.e.
N = Z = P
It is to be noted here that in both the conditions the
direction of branches will be from poles towards the zeros.

Root Locus Points
Root Locus Points

Root Locus Points
Root Locus Points
A point on the real axis lies on the root locus, if some of number of poles
and zeros on real axis to the RH side of this point is odd
For example,
G(s)H(s) =
K(s + 1)(s + 4)
s(s + 2)(s + 6)
Figure :

Root Locus Points
Root Locus Points
A point on the real axis lies on the root locus, if some of number of poles
and zeros on real axis to the RH side of this point is odd
For example,
G(s)H(s) =
K(s + 1)(s + 4)
s(s + 2)(s + 6)
Figure :
Complex conjugate roots should not be considered while

Angle of Asymptotes
Angle of Asymptotes

Angle of Asymptotes
Angle of Asymptotes
The branches that approach infinity do so in a fashion of straight line
which is called asymptotes of root locus. These are symmetrical about the
real axis.

Angle of Asymptotes
Angle of Asymptotes
The branches that approach infinity do so in a fashion of straight line
which is called asymptotes of root locus. These are symmetrical about the
real axis.
So, the angles at which the asymptotes of the root locus are formed are
given as:
θ =
(2q + 1)180o
P − Z
where, q=0,1,2 ... P-Z-1

Centroid:Location of asymptotes in the s-plane
Centroid

Centroid
For asymptotes, it is said that there is a common point at which all the
asymptotes undergo intersection on the real axis. This point is referred as
centroid and is represented by σ.

Centroid
So, the coordinates where the centroid is present on the real axis is
determined by:
σ =
P
Real part of Poles −
P
Real part of Zeros
P − Z

Centroid
So, the coordinates where the centroid is present on the real axis is
determined by:
σ =
P
P
Real part of Zeros
P − Z
It is to be noted for centroid that it may or may not be located on the
root locus but is always present on either the negative or positive real axis
as it is always of real nature.

Example:01
Example:
G(s)H(s) =
K
(s + 1)(s + 2 + 2j)(s + 2 − 2j)
Calculate angle of asymptotes and the centroid

Example:01
Solution:
Z = No. of Zeros = 0
P = No. of Poles = 3
s = −1, −2 − 2j, −2 + 2j
Asymptotes, θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2...P − Z − 1 = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
= 60o
θ2 =
(2 ∗ 1 + 1)180o
3
= 180o
θ3 =
(2 ∗ 2 + 1)180o
3
= 300o

Example:01...
Solution:
Centroids,
σ =
P
P
Real part of Zeros
P − Z
σ =
−1 − 2 − 2
3 − 0
=
−5
3
= −1.67

Example:01...
Figure : Plot of Angle of Asymptotes and Centroid

Example:02
Example:
G(s)H(s) =
K(s + 10)
s(s + 1)(s + 2)
Calculate angle of asymptotes and the centroid

Example:02
Solution:
Z = No. of Zeros = 1
P = No. of Poles = 3
s = −10
s = 0, −1, −2
Asymptotes, θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2...P − Z − 1 = 0, 1
P − Z = 3 − 1 = 2
θ1 =
(2 ∗ 0 + 1)180o
2
= 90o
θ2 =
(2 ∗ 1 + 1)180o
2
= 270o

Example:02...
Solution:
Centroids,
σ =
P
P
Real part of Zeros
P − Z
σ =
(0 − 1 − 2) − (−10)
3 − 1
=
7
2
= 3.5

Example:02...
Figure : Plot of Angle of Asymptotes and Centroid

Breakaway Point
Breakaway Point

Breakaway Point
Breakaway Point
Breakaway point is always present on the root locus and indicates a point
where various roots of the characteristic equation occur keeping K at a
specific value.

Breakaway Point
Breakaway Point
Breakaway point is always present on the root locus and indicates a point
where various roots of the characteristic equation occur keeping K at a
specific value.
To understand this, consider the plot shown below:
Figure :

Breakaway Point

Breakaway Point
Here the section between pole 0 and -2 is present on the root locus thus
according to rule at least one breakaway point is present here.

Determination of Breakaway Point
1 Consider the characteristic equation
1 + G(s)H(s) = 0

1 + G(s)H(s) = 0
2 From this equation separate the terms of K and s i.e. K = f (s)

1 + G(s)H(s) = 0
3 Differentiate above equation wrt s and equate to 0 dK
ds = 0

1 + G(s)H(s) = 0
3 Differentiate above equation wrt s and equate to 0 dK
ds = 0
4 Roots of equation dK
ds = 0 gives breakaway point

Validation of Breakaway Point

To decide validity of breakaway point, put breakaway point values in
equation of K to get values of K

To decide validity of breakaway point, put breakaway point values in
equation of K to get values of K
If the value of K is positive then that root is valid, if value of K is negative
then that root will be invalid

Example:03
Example:
Determine coordinate of valid break away points
G(s)H(s) =
K
s(s + 1)(s + 4)

Example:03
Solution:
1 + G(s)H(s) = 1 +
K
s(s + 1)(s + 4)
= 0
1 + G(s)H(s) = s(s + 1)(s + 4) + K = 0
s[s2
+ 5s + 4] + K = 0
s3
+ 5s2
+ 4s + K = 0
K = −s3
− 5s2
− 4s (1)
dK
ds
= −3s2
− 10s − 4 = 0
3s2
+ 10s + 4 = 0
s = −0.46, −2.86

Example:03...
Solution:
Put value of s in Equation 1
K = 0.87, −6.06
for s = −0.46, the value of K is positive, and hence it is a valid breakaway
point on the root locus.

Intersection of Root Locus with the Imaginary Axis
1 Consider the characteristic equation 1 + G(s)H(s) = 0

2 Construct Routh array in terms of K

3 Determine K marginal i.e. value of K which creates one of rows of
Routh array as row of zero

4 Construct auxiliary equation i.e. K(s) = 0

4 Construct auxiliary equation i.e. K(s) = 0
5 Roots of auxiliary equation, they are nothing but intersection point of
root locus with imaginary axis.

Example:04
Example:
Determine intersection point on imaginary axis
G(s)H(s) =
K
s(s + 1)(s + 4)

Example:04
Solution:
1 + G(s)H(s) = 1 +
K
s(s + 1)(s + 4)
= 0
1 + G(s)H(s) = s(s + 1)(s + 4) + K = 0
s[s2
+ 5s + 4] + K = 0
s3
+ 5s2
+ 4s + K = 0
Routh array,
s3 1 4 0
s2 5 K 0
s1 20−K
5 0
s0 K

Example:04...
Solution:
By considering K marginal, K = 20, (20−K
5 = 0) then
s1 = 0
A(s) = 5s2
+ K = 5s2
+ 20
s2
= −
20
5
= −4
s = ±2j

Angle of Departure at Complex Poles

Angle of Departure at Complex Poles
As branch always leave from an open loop pole then we have to calculate
angle which departs from complex conjugate pole. This angle at which it
departs from complex pole is called angle of departure denoted by φd ,
calculated by,
φd = 180o
− φ
where, φ =
P
φP −
P
φZ Where,
P
φP = contribution by angle made by remaining open loop poles and
P
φZ = contribution by angle made by remaining open loop zeros

General steps to solve the problem on Root locus
1 Get general information about number of open loop poles, zeros, and number of
branches etc.

branches etc.
2 Now, the poles-zero plot must be drawn. Once the s-plane plot is formed then the
sections of real axis where the root locus exists is determined. Also, through
general predictions, the minimum number of breakaway points is predicted
simultaneously.

branches etc.
simultaneously.
3 Calculate angle of asymptotes

branches etc.
simultaneously.
4 Determine centroid. Now a draw a sketch till the centroid to get the rough idea
about the construction of locus.

branches etc.
simultaneously.
5 Further, the breakaway point is to be determined using the method for its
determination. And in case, it appears to be complex conjugates then its validity
must be checked using angle condition.

branches etc.
simultaneously.
6 Calculate intersection points of root locus with imaginary axis

branches etc.
simultaneously.
7 Calculate angle of departure/arrival if applicable

branches etc.
simultaneously.
8 Combine step 1 to 7 and draw final sketch of root locus

branches etc.
simultaneously.
8 Combine step 1 to 7 and draw final sketch of root locus
9 Predict stability, and performance of given system by using root locus

Videos on LaTeX, CorelDraw and Many More

Example:05
Example:
The transfer function of the closed system is given as:
G(s)H(s) =
K
s(s + 5)(s + 10)
Construct the root locus for this system and predict the stability of the
same.

Example:05
Solution:
Step: I
Firstly, writing the characteristic equation of the above system,
s(s + 5)(s + 10) = 0
So, from the above equation, we get,
s = 0, −5, −10

Example:05
Solution:
Step: I
s(s + 5)(s + 10) = 0
s = 0, −5, −10
Thus, P = 3, Z = 0 and since P > Z therefore, the number of branches
will be equal to the number of poles.
So, N = P = 3
Thus, under this condition, the branches will start from the locations of 0,
-5 and -10 in the s-plane and will approach infinity.

Step: II - Pole Zero Plot
Initially, with general prediction, we can say that the point -5 on the real
axis has an odd sum of total number of poles and zeros to the right of it.
Thus, in between 0 and -5 there will be one breakaway point
Figure :

Step: III - Calculate angle of asymptotes

Now, let us calculate the angle of asymptotes with the formula given
below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o
θ2 =
(2 ∗ 1 + 1)180o
3
=
3 ∗ 180
3
= 180o

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o
θ2 =
(2 ∗ 1 + 1)180o
3
=
3 ∗ 180
3
= 180o
θ3 =
(2 ∗ 2 + 1)180o
3
=
5 ∗ 180
3
= 300o

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o
θ2 =
(2 ∗ 1 + 1)180o
3
=
3 ∗ 180
3
= 180o
θ3 =
(2 ∗ 2 + 1)180o
3
=
5 ∗ 180
3
= 300o
So, these three are the angle possessed by asymptotes approaching infinity.

Step: IV - Determine Centroid

Now, let us check where the centroid lies on the real axis by using the
formula given below:
σ =
P
Sum of Real part of Poles −
P
Sum of Real part of Zeros
P − Z
σ =
−0 − 5 − 10
3
=
−15
3
= −5

σ =
P
P
P − Z
σ =
−0 − 5 − 10
3
=
−15
3
= −5
The figure below represents a rough sketch of the plot that is obtained by
the above analysis

Step: V - Breakaway Points

Earlier we have predicted that one breakaway point will be present in the
section between points 0 and -5. So, now using the method to determine
the breakaway point we will check the validity of the breakaway point.
1 + G(s)H(s) = 0
1 +
K
s(s + 5)(s + 10)
= 0
s(s + 5)(s + 10) + K = 0
s[s2
+ 15s + 50] + K = 0
s3
+ 15s2
+ 50s + K = 0
K = −s3
− 15s2
− 50s (2)

Step: V - Breakaway Points...

In this method, roots obtained on differentiating K with respect to s and
equating it to 0, will be the breakaway point.

Therefore,
dK
ds
= 0
d(−s3 − 15s2 − 50s)
ds
= 0
−3s2
− 30s − 50 = 0
3s2
+ 30s + 50 = 0

Therefore,
dK
ds
= 0
d(−s3 − 15s2 − 50s)
ds
= 0
−3s2
− 30s − 50 = 0
3s2
+ 30s + 50 = 0
Thus, on solving, roots obtained will be -2.113 and -7.88

Put s=-2.113 in Equation 3
K = −(−2.113)3
− 15(−2.113)2
− 50(−2.113)
K = +9.43 − 66.97 + 105.65
K = 48.11

K = −(−2.113)3
− 15(−2.113)2
− 50(−2.113)
K = +9.43 − 66.97 + 105.65
K = 48.11
K = −(−7.88)3
− 15(−7.88)2
− 50(−7.88)
K = +489 − 931 + 394
K = −48

K = −(−2.113)3
− 15(−2.113)2
− 50(−2.113)
K = +9.43 − 66.97 + 105.65
K = 48.11
K = −(−7.88)3
− 15(−7.88)2
− 50(−7.88)
K = +489 − 931 + 394
K = −48
K is positive at s = −2.113, hence, s = −2.113 is valid.

K = −(−2.113)3
− 15(−2.113)2
− 50(−2.113)
K = +9.43 − 66.97 + 105.65
K = 48.11
K = −(−7.88)3
− 15(−7.88)2
− 50(−7.88)
K = +489 − 931 + 394
K = −48
K is positive at s = −2.113, hence, s = −2.113 is valid.
Now, we have to check the at what point the root locus intersects with
the imaginary axis. Thus, for this routh array is used.

Step: VI - Calculate intersection points of root locus with imaginary
axis

axis
Here a proper method is used where the characteristic equation is used and routh array
in terms of K is formed.
s3
+ 15s2
+ 50s + K = 0
Thus, the Rouths Array:
s3
1 50
s2
15 K
s1 750−K
15
0
s0
K

axis
s3
+ 15s2
+ 50s + K = 0
s3
1 50
s2
15 K
s1 750−K
15
0
s0
K
Now, to find Kmar , which is the value of K from one of the rows of routh’s array as row
of zeros, except the row s0
.

axis
s3
+ 15s2
+ 50s + K = 0
s3
1 50
s2
15 K
s1 750−K
15
0
s0
K
.
Considering, row s1
,
750 − K
15
= 0
K = 750

axis
s3
+ 15s2
+ 50s + K = 0
s3
1 50
s2
15 K
s1 750−K
15
0
s0
K
.
Considering, row s1
,
750 − K
15
= 0
K = 750
Thus,
Km = 750

axis...

axis...
Further, with the help of coefficients of the row which is present above the
row of zero, an auxiliary equation A(s) = 0 is constructed. In this case,

axis...
A(s) = 15s2
+ K = 15s2
+ 750
s2
= −50
s = ±7.07j

axis...
A(s) = 15s2
+ K = 15s2
+ 750
s2
= −50
s = ±7.07j
Thus, these are the intersection points of the root locus with the
imaginary axis.

axis...
A(s) = 15s2
+ K = 15s2
+ 750
s2
= −50
s = ±7.07j
imaginary axis.
Also, as the poles are not complex thus angles of departure not needed.
Hence, at the breakaway point, the root locus breaks at ±90o.

axis...

axis...
So, the complete root locus is given below:
Figure :

Step: VII - Stability Prediction

From the above sketch (Figure 9), the stability of the system can be
analyzed that for K between 0 to 750 the system is completely stable as
the complete root locus lies in the left half of s-plane. While at K = 750,
the system is marginally stable.

From the above sketch (Figure 9), the stability of the system can be
analyzed that for K between 0 to 750 the system is completely stable as
the complete root locus lies in the left half of s-plane. While at K = 750,
the system is marginally stable.
While, for K between 750 to ∞, the system is unstable as the dominant
roots proceed towards the right half of s-plane.

Example:06
Example:
G(s)H(s) =
K
s(s2 + 2s + 2)
same.

Example:06
Solution:
Step: I
s(s2
+ 2s + 2) = 0
s = 0, −1 ± j
s = 0, −1 + j, −1 − j

Example:06
Solution:
Step: I
s(s2
+ 2s + 2) = 0
s = 0, −1 ± j
s = 0, −1 + j, −1 − j
So, N = P = 3

The pole-zero plot is given below:
Figure :

Figure :
Here, it is clear that branch originating from s = 0 approaches −∞. And
general predictions clear that there is no breakaway point here.

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o
θ2 =
(2 ∗ 1 + 1)180o
3
=
3 ∗ 180
3
= 180o

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o
θ2 =
(2 ∗ 1 + 1)180o
3
=
3 ∗ 180
3
= 180o
θ3 =
(2 ∗ 2 + 1)180o
3
=
5 ∗ 180
3
= 300o

σ =
P
P
P − Z
σ =
−0 − 1 − 1
3
=
−2
3
= −0.67

σ =
P
P
P − Z
σ =
−0 − 1 − 1
3
=
−2
3
= −0.67
the above analysis

Step: IV - Determine Centroid...

Step: IV - Determine Centroid...
With angle θ2, one branch from s = 0 approaches to infinity and with θ1
and θ3, branches starting from −1 + j and −1 − j respectively, approach
infinity.

1 + G(s)H(s) = 0
1 +
K
s(s2 + 2s + 2)
= 0
s3
+ 2s2
+ 2s + K = 0
K = −s3
− 2s2
− 2s (3)

Therefore,
dK
ds
= 0
d(−s3 − 2s2 − 2s)
ds
= 0
−3s2
− 4s − 2 = 0
3s2
+ 4s + 2 = 0

Therefore,
dK
ds
= 0
d(−s3 − 2s2 − 2s)
ds
= 0
−3s2
− 4s − 2 = 0
3s2
+ 4s + 2 = 0
Thus, on solving, roots obtained will be s = −0.67 ± j0.47

Therefore,
dK
ds
= 0
d(−s3 − 2s2 − 2s)
ds
= 0
−3s2
− 4s − 2 = 0
3s2
+ 4s + 2 = 0
Thus, on solving, roots obtained will be s = −0.67 ± j0.47
Now, as here we are having complex conjugates, thus, checking the
validity of these points as breakaway point by using angle condition.

Figure :

Figure :
As it is not an odd multiple of 180o thus, this point is not present on the
root locus, hence there is no breakaway point here.

axis

axis
s3
+ 2s2
+ 2s + K = 0
s3
1 2
s2
2 K
s1 4−K
2
0
s0
K

axis
s3
+ 2s2
+ 2s + K = 0
s3
1 2
s2
2 K
s1 4−K
2
0
s0
K
.

axis
s3
+ 2s2
+ 2s + K = 0
s3
1 2
s2
2 K
s1 4−K
2
0
s0
K
.
Considering, row s1
,
4 − K
2
= 0
K = 4

axis
s3
+ 2s2
+ 2s + K = 0
s3
1 2
s2
2 K
s1 4−K
2
0
s0
K
.
Considering, row s1
,
4 − K
2
= 0
K = 4
Thus,
Km = 4

axis...

axis...
A(s) = 2s2
+ K = 2s2
+ 4
s2
= −2
s = ±1.414j

axis...
A(s) = 2s2
+ K = 2s2
+ 4
s2
= −2
s = ±1.414j
imaginary axis.

Step: VII - Calculate Angle of Departure

At complex pole, −1 + j
φd = tan−1
(
1
−1
)
φd = −45

φd = tan−1
(
1
−1
)
φd = −45
At complex pole, −1 − j
φd = tan−1
(
−1
−1
)
φd = +45

φd = tan−1
(
1
−1
)
φd = −45
At complex pole, −1 − j
φd = tan−1
(
−1
−1
)
φd = +45
So, with the above-determined values and parameters, the complete root
locus sketch obtained is given below:

Step: VIII - Plot of Complete Root Locus

Step: VIII - Plot of Complete Root Locus
Figure :

Step: IX - Stability Prediction

Now talking about stability, for K between 0 to 4, the roots are present on
the left half of s-plane, representing a completely stable system.

K = +4 makes the system marginally stable due to the presence of
dominant roots on the imaginary axis. While for K > 4, the system
becomes unstable as dominant roots lie in the right half of s-plane.

Thus, in this way by plotting the root locus, the stability of the system can
be determined.

Example:07
Example:
G(s)H(s) =
K
s(s + 1)(s + 5)
same.

Example:07
Solution:
Step: I
s(s + 1)(s + 5) = 0
s = 0, −1, −5

Example:07
Solution:
Step: I
s(s + 1)(s + 5) = 0
s = 0, −1, −5
So, N = P = 3

Figure :

Figure :
The three poles are located are shown in the above figure. The line
segment between s=-1 and s=0 is one branch of root locus on real axis.
And the other branch of the root locus on the real axis is the line segment
to the left of s=-5

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o
θ2 =
(2 ∗ 1 + 1)180o
3
=
3 ∗ 180
3
= 180o

below:
θ =
(2q + 1)180o
P − Z
where, q = 0, 1, 2, ...P − Z − 1
so, q = 0, 1, 2
θ1 =
(2 ∗ 0 + 1)180o
3
=
180
3
= 60o
θ2 =
(2 ∗ 1 + 1)180o
3
=
3 ∗ 180
3
= 180o
θ3 =
(2 ∗ 2 + 1)180o
3
=
5 ∗ 180
3
= 300o

σ =
P
P
P − Z
σ =
−0 − 1 − 5
3
=
−6
3
= −2

the above analysis
Figure :

1 + G(s)H(s) = 0
1 +
K
s(s + 1)(s + 5)
= 0
s(s + 1)(s + 5) + K = 0
s[s2
+ 6s + 5] + K = 0
s3
+ 6s2
+ 5s + K = 0
K = −s3
− 6s2
− 5s (4)

Therefore,
dK
ds
= 0
d(−s3 − 6s2 − 5s)
ds
= 0
−3s2
− 12s − 5 = 0
3s2
+ 12s + 5 = 0

Therefore,
dK
ds
= 0
d(−s3 − 6s2 − 5s)
ds
= 0
−3s2
− 12s − 5 = 0
3s2
+ 12s + 5 = 0
Thus, on solving, roots obtained will be s = −0.47, −3.52

Put s=- 0.47 in Equation 4
K = −(−0.47)3
− 6(−0.47)2
− 2(−0.47)
K = 0.10 − 1.32 + 0.94
K = −0.28

K = −(−0.47)3
− 6(−0.47)2
− 2(−0.47)
K = 0.10 − 1.32 + 0.94
K = −0.28
K = −(−3.52)3
− 6(−3.52)2
− 2(−3.52)
K = 43.61 − 74.34 + 7.04
K = −23.69

K = −(−0.47)3
− 6(−0.47)2
− 2(−0.47)
K = 0.10 − 1.32 + 0.94
K = −0.28
K = −(−3.52)3
− 6(−3.52)2
− 2(−3.52)
K = 43.61 − 74.34 + 7.04
K = −23.69
K is positive (very close) at s = −0.47, hence, s = −0.47 is valid.

K = −(−0.47)3
− 6(−0.47)2
− 2(−0.47)
K = 0.10 − 1.32 + 0.94
K = −0.28
K = −(−3.52)3
− 6(−3.52)2
− 2(−3.52)
K = 43.61 − 74.34 + 7.04
K = −23.69
K is positive (very close) at s = −0.47, hence, s = −0.47 is valid.
Now, we have to check the at what point the root locus intersects with
the imaginary axis. Thus, for this routh array is used.

Figure :

axis

axis
s3
+ 6s2
+ 5s + K = 0
s3
1 5
s2
6 K
s1 30−K
6
0
s0
K

axis
s3
+ 6s2
+ 5s + K = 0
s3
1 5
s2
6 K
s1 30−K
6
0
s0
K
.

axis
s3
+ 6s2
+ 5s + K = 0
s3
1 5
s2
6 K
s1 30−K
6
0
s0
K
.
Considering, row s1
,
30 − K
6
= 0
K = 30

axis
s3
+ 6s2
+ 5s + K = 0
s3
1 5
s2
6 K
s1 30−K
6
0
s0
K
.
Considering, row s1
,
30 − K
6
= 0
K = 30
Thus,
Km = 30

axis...

axis...
A(s) = 6s2
+ K = 6s2
+ 30
s2
= −5
s = ±j
√
5

axis...
A(s) = 6s2
+ K = 6s2
+ 30
s2
= −5
s = ±j
√
5
imaginary axis.

Step: VII - Plot of Complete Root Locus

Step: VII - Plot of Complete Root Locus
Figure :

Step: VIII - Stability Prediction

Now talking about stability, for K between 0 to 30, the roots are present
on the left half of s-plane, representing a completely stable system.

Thus, in this way by plotting the root locus, the stability of the system can
be determined.

Thank you
Please send your feedback at nbahadure@gmail.com

Thank you
Please send your feedback at nbahadure@gmail.com
For download and more information Click Here

Control system stability root locus

Recommended

Recommended

More Related Content

Similar to Control system stability root locus

Similar to Control system stability root locus (20)

More from Nilesh Bhaskarrao Bahadure

More from Nilesh Bhaskarrao Bahadure (20)

Recently uploaded

Recently uploaded (20)

Control system stability root locus