CONTROL SYSTEM

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

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CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

UNIT III:- Stability: Routh-Hurwitz stability analysis Characteristics equation of closed loop system root
loci, construction of loci, Effect of adding, poles and Zeros on the loci, Stability by root loci.
3.1Stability:
A system is stable if its output is bounded for any input, in other words A system is stable if its response
to a bounded disturbing signal approaches to Zero as time tends to infinity.
A system is unstable if its response to a bounded disturbing signal results in an output of infinite
amplitude of an oscillatory signal.
If the output response to a bounded input signal results in constant amplitude oscillating then such a
system is called marginally stable.
In stability analysis the characteristic equation is play very important role .from the roots of characteristic
equation some of the conclusion are as follow.

1. Stable System: If all the roots of the characteristic equation lie on the left half of the 'S' plane
then the system is said to be a stable system.

2. Marginally Stable System: If all the roots of the system lie on the imaginary axis of the 'S' plane
then the system is said to be marginally stable.

3. Unstable System: If all the roots of the system lie on the right half of the 'S' plane then the system
is said to be an unstable system.

3.2 Statement of Routh-Hurwitz Criterion
Routh Hurwitz criterion states that any system can be stable if and only if all the roots of the first
column have the same sign and if it does not has the same sign or there is a sign change then the
number of sign changes in the first column is equal to the number of roots of the characteristic
equation in the right half of the s-plane i.e. equals to the number of roots with positive real parts.
3.2.1Necessary but not sufficient conditions for Stability
We have to follow some conditions to make any system stable, or we can say that there are some
necessary conditions to make the system stable.
Consider a system with characteristic equation:

a0sm+a1sm-1+………………………+am=0

1. All the coefficients of the equation should have the same sign.
2. There should be no missing term.
If all the coefficients have the same sign and there are no missing terms, we have no guarantee that the
system will be stable. For this, we use Routh Hurwitz Criterion to check the stability of the system. If
the above-given conditions are not satisfied, then the system is said to be unstable. This criterion is given
by A. Hurwitz and E.J. Routh.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

3.2.2 Advantages of Routh- Hurwitz Criterion
1. We can find the stability of the system without solving the equation.
2. We can easily determine the relative stability of the system.
3. By this method, we can determine the range of K for stability.
4. By this method, we can also determine the point of intersection for root locus with an
imaginary axis.

3.2.3.Limitations of Routh- Hurwitz Criterion
1. This criterion is applicable only for a linear system.
2. It does not provide the exact location of poles on the right and left half of the S plane.
3. In case of the characteristic equation, it is valid only for real coefficients.

3.3 The Routh- Hurwitz Criterion
Consider the following characteristic Polynomial

a0sn+a1sn-1+………………………+an=0

When the coefficients a0, a1, ......................an are all of the same sign, and none is zero.
Step 1: Arrange all the coefficients of the above equation in two rows:

Step 2: From these two rows we will form the third row:

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Step 3: Now, we shall form fourth row by using second and third row:

Step 4: We shall continue this procedure of forming a new rows:

Example 1

Check the stability of the system whose characteristic equation is given by
s4 + 2s3+6s2+4s+1 = 0

Solution

Obtain the arrow of coefficients as follows

s4 1 61
s3 2 4

s2 [(2x6)-(1x4)]/2=4 [(2x1)-(1x0)]/2=1

s1 [(4x4)-(2x1)]/4=3.5 -

s0 4

Since all the coefficients in the first column are of the same sign, i.e., positive, the given equation has no
roots with positive real parts; therefore, the system is said to be stable.
Special Cases of Routh Array
We may come across two types of situations, while forming the Routh table. It is difficult to complete the
Routh table from these two situations.
The two special cases are −

 The first element of any row of the Routh array is zero.
 All the elements of any row of the Routh array are zero.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Let us now discuss how to overcome the difficulty in these two cases, one by one.

Case :1
 First Element of any row of the Routh array is zero

If any row of the Routh array contains only the first element as zero and at least one of the remaining
elements have non-zero value, then replace the first element with a small positive integer, ϵ. And then

continue the process of completing the Routh table. Now, find the number of sign changes in the first
column of the Routh table by substituting ϵ tends to zero.

Example 2
Let us find the stability of the control system having characteristic equation s4+2s3+s2+2s+1=0
Step 1 − Verify the necessary condition for the Routh-Hurwitz stability.All the coefficients of the
characteristic polynomial, s4+2s3+s2+2s+1=0 are positive. So, the control system satisfied the necessary

condition. 1
Step 2 − Form the Routh array for the given characteristic polynomial.

s4 1 1
s3 2 2
s2 [(2x1)-(1x2)]/2=0 [(2x1)-(1x0)]/2=1
s1
s0

Special case (i) − Only the first element of row s2 is zero. So, replace it by ϵ

and continue the process of completing the Routh table.
s4 1 1 1
s3 2 2
s2 ϵ 1
s1 (2 ϵ-2)/ ϵ =2-2/ ϵ -
s0 1

Step 3 − Verify the sufficient condition for the Routh-Hurwitz stability. As ϵ tends to zero, the Routh

table becomes like this. 1 1 1
s4 2 2
s3 ϵ 1
s2 -∞ -
s1 1
s0

There are two sign changes in the first column of Routh table. Hence, the control system is unstable.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Case :2
 All the Elements of any row of the Routh array are zero

In this case, follow these two steps −

 Write the auxilary equation, A(s) of the row, which is just above the row of zeros.

 Differentiate the auxiliary equation, A(s) with respect to s. Fill the row of zeros with these

coefficients.

Example 3

Let us find the stability of the control system having characteristic equation,
s5+3s4+s3+3s2+s+3=0
Step 1 − Verify the necessary condition for the Routh-Hurwitz stability.All the coefficients of the given

characteristic polynomial are positive. So, the control system satisfied the necessary condition.
Step 2 − Form the Routh array for the given characteristic polynomial.

s5 1 1 1
s4 3 3 3
s3 [(3x1)-(1x3)]/3=0 [(3x1)-(1x3)]/3=0 [(3x1)-(1x3)]/3=0
s2
s1
s0
Special case (ii) − All the elements of row s3 are zero. So, write the auxiliary equation, A(s) of the row s4
A(s)=3s4+3s2+3

Differentiate the above equation with respect to s.

dA(s)/ds=12s3+6s

Place these coefficients in row s3

s5 1 1 1
3
s4 3 3 -

s3 12 6

s2 [(12x3)-(3x6)]/12=1.5 [(12x3)-(3x0)]/12=3

s1 [(1.5x6)-(12x3)]/1.5=(-18) -

s0 3

Step 3 − Verify the sufficient condition for the Routh-Hurwitz stability.

There are two sign changes in the first column of Routh table. Hence, the control system is unstable.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

In the Routh-Hurwitz stability criterion, we can know whether the closed loop poles are in on left half of
the ‘s’ plane or on the right half of the ‘s’ plane or on an imaginary axis. So, we can’t find the nature of
the control system. To overcome this limitation, there is a technique known as the root locus.
Practice Problem:
1. s3+20s2+9s+100=0
2. s4+2s3+3s2+4s+5=0
3. s5+s4+2s3+2s2+11s+10=0
4. s6+2s5+8s4+12s3+20s2+16s+16=0
5. The open loop transfer function of unity feedback is given as

G(s)=K/[(s+2)(s+4)(s2+6s+25)]By applying the Routh Hurwitz Criterion discuss the stability
of Closed loop system as a function of K. Determine the value of K which will cause
sustained oscillation in the closed loop system what are corresponding oscillation frequency.

3.4Root Locus

The root locus is a graphical representation in s-domain and it is symmetrical about the real axis.
Because the open loop poles and zeros exist in the s-domain having the values either as real or as
complex conjugate pairs..
3.4.1 Rules for Construction of Root Locus
Follow these rules for constructing a root locus.
Rule 1 − Locate the open loop poles and zeros in the‘s’ plane.
Rule 2 − Starting Point-The root locus Starts from the open loop Poles.
Rule 3 − Terminating/Ending Point-The root locus branches terminate either open loop zeroes or infinity
Rule 4 − Find the number of root locus branches.

We know that the root locus branches start at the open loop poles and end at open loop zeros. So,
the number of root locus branches N is equal to the number of finite open loop poles P or the
number of finite open loop zeros Z, whichever is greater.
Mathematically, we can write the number of root locus branches N as
N=P if P≥Z
N=Z if P<Z
Rule 5 – Existence of root locus on real axis .
If odd number of the open loop poles and zeros exist to the left side of a point on the real axis,
then that point is on the root locus branch.
Rule 6 – Angle of the angle of asymptotes θ is

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

θ= [(2q+1)1800 ]/ P−Z
Where,

q= 0,1,2,....,(P−Z)−1
Asymptotes give the direction of these root locus branches. The intersection point of asymptotes
on the real axis is known as centroid.
We can calculate the centroid x by using this formula,
x = [∑P−∑ Z ]/P−Z
∑P=Sum of Poles
∑ Z =Sum of Zeroes
P=No of Poles
Z=No of Zeroes
Rule 7 − Find the intersection points of root locus branches with an imaginary axis.
We can calculate the point at which the root locus branch intersects the imaginary axis and the
value of K at that point by using the Routh array method and special case (ii).
 If all elements of any row of the Routh array are zero, then the root locus branch intersects the
imaginary axis and vice-versa.
 Identify the row in such a way that if we make the first element as zero, then the elements of the
entire row are zero. Find the value of K for this combination.
 Substitute this K value in the auxiliary equation. You will get the intersection point of the root
locus branch with an imaginary axis.
Rule 8 − Find Break-away and Break-in points.
 If there exists a real axis root locus branch between two open loop poles, then there will be a
break-away point in between these two open loop poles.
 If there exists a real axis root locus branch between two open loop zeros, then there will be a
break-in point in between these two open loop zeros.
Follow these steps to find break-away and break-in points.
 Write K in terms of s from the characteristic equation 1+G(s)H(s)=0
 Differentiate K with respect to s and make it equal to zero. Substitute these values of s in the
above equation.
 The values of s for which the K value is positive are the break points.
Rule 9 − Find the angle of departure and the angle of arrival.
The Angle of departure and the angle of arrival can be calculated at complex conjugate open loop
poles and complex conjugate open loop zeros respectively.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

The formula for the angle of departure ϕd is ϕd=1800−ϕ

The formula for the angle of arrival ϕa is ϕa=1800+ϕ

Where, ϕ=∑ϕP−∑ϕZ

∑ϕP=is the sum of all the angles subtended by remaining poles

∑ϕZ = is the sum of all the angles subtended by remaining poles

Example 4

Sketch the Root locus of the system having Open Loop transfer function is given by

G(s)H(s)=K/s(s+1)(s+3)
Step 1 − Locate the open loop poles and zeros in the‘s’ plane. The given open loop transfer function

has three poles at s=0,s=−1 and s=−3. It doesn’t have any zero.
Step 2 − Starting Point-The root locus Starts from the open loop Poles. s=0,s=−1 and s=−3
Step 3 − Ending Point-The root locus branches terminate at infinity. because It doesn’t have any zero
Step 4 − Find the number of root locus branches. , the number of root locus branches is equal to the

number of poles of the open loop transfer function.
N=3 if P≥Z

Step 5 – Existence of root locus on real axis .If odd number of the open loop poles and zeros exist to

the left side of a point on the real axis, then that point is on the root locus branch.

Fig 3.1 root locus of K/s(s+1)(s+3)

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=−3.
If odd number of the open loop poles and zeros exist to the left side of a point on the real axis, then
that point is on the root locus branch.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Step 6 – Angle of the angle of asymptotes θ is
θ= [(2q+1)1800 ]/ P−Z
Where,
q= 0,1,2,....,(P−Z)−1
q= 0,1,2,....,(3−0)−1=2
q=0,1,2
q=1 θ1= [(2.0+1)1800 ]/ (3−0) =600
q=2 θ2= [(2.1+1)1800 ]/ (3−0) =1800
q=3 θ3= [(2.2+1)1800 ]/ (3−0) =3000

The angle of asymptotes are θ=600,1800 and 3000.
Asymptotes give the direction of these root locus branches. The intersection point of asymptotes on
the real axis is known as centroid.

Centroid x by using this formula,
x = [∑P−∑ Z ]/P−Z

∑P=Sum of Poles=-3-1-0=-4
∑ Z =Sum of Zeroes=0
P=No of Poles=3
Z=No of Zeros=0
x = [-4−0 ]/(3−0)=-1.33
The centroid and three asymptotes are shown in the following figure.

Fig 3.2 root locus of K/s(s+1)(s+3)

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Step 7 − Find the intersection points of root locus branches with an imaginary axis.
Since two asymptotes have the angles of 600 and 3000, two root locus branches intersect the imaginary

axis. By using the Routh array method and special case (ii),we can calculate the root locus branches

intersects the imaginary axis.

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

1+K/s(s+1)(s+3)=0

s(s+1)(s+3)+K=0
s3+4s2+3s+K=0

Form the Routh array for the given characteristic polynomial.
s3 1 3
s2 4 K
s1 (12-K)/4 -
s0 K

The value of K at which the root locus plot crosses the imaginary axis is determined by equating the first

term in s1 row to zero, therefore

(12-K)/4=0

K=12

At K=12 the root locus plot intersects the imaginary axis and the value of s at the intersection point is
determined by solving auxiliary equation formed from the s2 terms in routh array , therefore

4s2+K=0 4s2+12=0

s=-1.732j,1.732j
Step 8 − Find Break-away

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

1+K/s(s+1)(s+3)=0

s(s+1)(s+3)+K=0
s3+4s2+3s+K=0
K=-(s3+4s2+3s)
dK/ds=-(3s2+8s+3)=0

s=-0.45,-2.21
There will be one break-away point on the real axis root locus branch between the poles s=−1 and s=0.
By following the procedure given for the calculation of break-away point, we will get it as s=−0.45.

The root locus diagram for the given control system is shown in the following figure.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig 3.3 root locus of K/s(s+1)(s+3)

In this way, you can draw the root locus diagram of any control system and observe the movement of
poles of the closed loop transfer function.
Step 9 − Find the angle of departure and the angle of arrival.

Not required in this problem because we don’t have any complex pole or zero.
Example 5
Sketch the Root locus of the system having Open Loop transfer function is given by

G(s)H(s)=K/s(s+4)(s2+4s+13)
Step 1 − Locate the open loop poles and zeros in the‘s’ plane. The given open loop transfer function
has four poles at s=0,s=−4 , s= -2+3j and s= -2-3j. It doesn’t have any zero.
Step 2 − Starting Point-The root locus Starts from the open loop Poles. s=0,s=−4 , s= -2+3j &s= -2-3j
Step 3 − Terminating/Ending Point-The root locus branches terminate at infinity. because It doesn’t
have any zero
Step 4 − Find the number of root locus branches. , the number of root locus branches is equal to the
number of poles of the open loop transfer function.

N=4 if P≥Z
Step 5 – Existence of root locus on real axis .
If odd number of the open loop poles and zeros exist to the left side of a point on the real axis, then that
point is on the root locus branch.
Step 6 – Angle of the angle of asymptotes θ is

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

θ= [(2q+1)1800 ]/ P−Z

Where,
q= 0,1,2,....,(4−Z)−1
q= 0,1,2,....,(4−0)−1=3

q=0,1,2,3
q=0 θ1= [(2.0+1)1800 ]/ (4−0) =450
q=1 θ2= [(2.1+1)1800 ]/ (4−0) =1350
q=2 θ3= [(2.2+1)1800 ]/ (4−0) =2250
q=3 θ4= [(2.2+1)1800 ]/ (4−0) =3150
The angle of asymptotes are θ=450,1350,225 and 3150.

Asymptotes give the direction of these root locus branches. The intersection point of asymptotes on

the real axis is known as centroid.

Centroid x by using this formula,
x = [∑P−∑ Z ]/P−Z
∑P=Sum of Poles=-2-3j-2+3j-4-0=-8

∑ Z =Sum of Zeroes=0

P=No of Poles=4

Z=No of Zeros=0
x = [-8−0 ]/(4−0)=-2
Step 7 − Find the intersection points of root locus branches with an imaginary axis.
Since two asymptotes have the angles of 450 and 3150, two root locus branches intersect the imaginary

axis. By using the Routh array method and special case (ii),we can calculte the root locus branches

intersects the imaginary axis.

1+ G(s)H(s)=0
1+K/s(s+4)(s2+4s+13)=0
s(s+4)(s2+4s+13)+K=0
s4+8s3+29s2+52s+K=0

Form the Routh array for the given characteristic polynomial. K
s4 1 29
s3 8 52
s2 22.5 K
s1 52-0.35K

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

s0 K

The value of K at which the root locus plot crosses the imaginary axis is determined by equating the first

term in s1 row to zero, therefore

(52-0.35K)=0

K=148.6

At K=148.6 the root locus plot intersects the imaginary axis and the value of s at the intersection point is

determined by solving auxiliary equation formed from the s2 terms in routh array , therefore

22.5s2+K=0 22.5s2+148.6=0

s=-2.56j, 2.56j
Step 8 − Find Break-away

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

1+K/s(s+4)(s2+4s+13)=0

s(s+4)(s2+4s+13)+K=0

s4+8s3+29s2+52s+K=0

K=-(s4+8s3+9s2+52s+)

dK/ds=-(4s3+24s2+58s+52=0)=0

s=-2,(-2-1.58j), (-2+1.58j)

Step 9 − Find the angle of departure

The formula for the angle of departure ϕd is ϕd=1800−ϕ

Where, ϕ=∑ϕP−∑ϕZ

∑ϕP=is the sum of all the angles subtended by remaining poles

∑ϕZ = is the sum of all the angles subtended by remaining poles

d (-2+3j)=-900

d (-2-3j)= 900

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig 3.4 root locus of K/s(s+4)(s2+4s+13)

Example 6

Sketch the Root locus of the system having Open Loop transfer function is given by

G(s)H(s)=K(s+1)/s(s-1)
Step 1 − Locate the open loop poles and zeros in the‘s’ plane. The given open loop transfer function
has two poles at s=0,s=1 and one zero s=−1.
Step 2 − Starting Point-The root locus Starts from the open loop Poles. s=0 & s=1
Step 3 − Terminating/Ending Point-one root locus branch terminate at zero s=-1 & other at infinity.

Because it have only one zero
Step 4 − Find the number of root locus branches. , the number of root locus branches is equal to the

number of poles of the open loop transfer function.
N=2 if P≥Z

Step 5 – Existence of root locus on real axis .

If odd number of the open loop poles and zeros exist to the left side of a point on the real axis, then that

point is on the root locus branch.
Step 6 – Angle of the angle of asymptotes θ is

θ= [(2q+1)1800 ]/ P−Z

Where,
q= 0,1,2,....,(P−Z)−1
q= 0,1,2,....,(2−1)−1=0

q=0
θ1= [(2.0+1)1800 ]/ (2−1) =1800

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

The angle of asymptotes are θ=1800.

Asymptotes give the direction of these root locus branches. The intersection point of asymptotes on

the real axis is known as centroid.

Centroid x by using this formula,
x = [∑P−∑ Z ]/P−Z
∑P=Sum of Poles=0+1=1
∑ Z =Sum of Zeroes=-1

P=No of Poles=2

Z=No of Zeros=1
x = [1−(-1) ]/(2−1)=2
Step 7 − Find the intersection points of root locus branches with an imaginary axis.

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

1+K(s+1)/s(s-1)=0

s(s-1)+K(s+3)=0
s2+(K-1)s+K=0

Form the Routh array for the given characteristic polynomial.

s2 1 K

s1 (K-1) -

s0 K

The value of K at which the root locus plot crosses the imaginary axis is determined by equating the first
term in s1 row to zero, therefore

(K-1)=0

K=1

At K=1 the root locus plot intersects the imaginary axis and the value of s at the intersection point is

determined by solving auxiliary equation formed from the s2 terms in routh array , therefore

s2+K=0 s2+1=0

s=-j,j
Step 8 − Find Break-away point

1+ G(s) H(s) =0 1+K(s+1)/s(s-1) =0

s(s-1)+K(s+1)=0 K=s(s+1)/(s+1)

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

dK/ds=0 , s2+2s-1=0 , s=-0.414,-2.414

Fig 3.5 root locusK(s+1)/s(s-1)

Step 9 − Find the angle of departure and the angle of arrival.
Not required in this problem because we don’t have any complex pole or zero.

Example 7
Sketch the Root locus of the system having Open Loop transfer function is given by
G(s)H(s)=K(s+4)/(s+2)2
Step 1 − Locate the open loop poles and zeros in the‘s’ plane. The given open loop transfer function
has two poles at s=-2,s=-2 and one zero s=−4.
Step 2 − Starting Point-The root locus Starts from the open loop Poles. s=-2, s=−2
Step 3 − Ending Point-one root locus branch terminate at zero s=-4& other at infinity. Because it have
only one zero
Step 4 − Find the number of root locus branches. the number of root locus branches is equal to the
number of poles of the open loop transfer function.
N=2 if P≥Z
Step 5 – Existence of root locus on real axis .If odd number of the open loop poles and zeros exist to
the left side of a point on the real axis, then that point is on the root locus branch.
Step 6 – Angle of the angle of asymptotes θ is
θ= [(2q+1)1800 ]/ P−Z
Where,
q= 0,1,2,....,(P−Z)−1
q= 0,1,2,....,(2−1)−1=0

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

q=0
θ1= [(2.0+1)1800 ]/ (2−1) =1800
The angle of asymptotes are θ=1800.

Asymptotes give the direction of these root locus branches. The intersection point of asymptotes on

the real axis is known as centroid.

Centroid x by using this formula,
x = [∑P−∑ Z ]/P−Z
∑P=Sum of Poles=-2-2=-4

∑ Z =Sum of Zeroes=-4

P=No of Poles=2

Z=No of Zeros=1
x = [-4−(-4) ]/(2−1)=0
Step 7 − Find the intersection points of root locus branches with an imaginary axis.

1+ G(s)H(s)=0
1+K(s+4)/(s+2)2=0
(s+2)2+K(s+4)=0
s2 + 4s+4+Ks+4K=0
s2 +( 4+K)s+4(1+K)=0

Form the Routh array for the given characteristic polynomial.

s2 1 1+K

s1 4+K -

s0 1+K

1+K=0; & 4+K=0

K is negative. root locus branches are not intersecting the imaginary axis but are always located in the

left half of s plane.
Step 8 − Find Break-away

1+ G(s) H(s) =0
1+K(s+4)/(s+2)2 =0
(s+2)2+K(s+4) =0
K=(s+2)2/(s+4)

dK/ds=(s+4) 2(s+2) -(s+2).1 = 0
(s+4)2

(s+2)(s+6)=0

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

s=-2,-6
Step 9 − Find the angle of departure and the angle of arrival.

Not required in this problem because we don’t have any complex pole or zero.

Practice Questions Fig 3.6 root locus K(s+4)/(s+2)2
Plot the root locus of G(s)H(s)=
K(s+2)
(s2+2s+2)

Plot the root locus of G(s)H(s)= K

s(s+4)(s2+4s+20)

Plot the root locus of G(s)H(s)= K(s+1)(s+2)
(s+0.1)(s-1)

Plot the root locus of G(s)H(s)= K(s+4/3)
s2(s+12)

3.5 Effect of adding, poles and Zeros

Addition of poles to the transfer function has the effect of pulling the root locus to the right, making the
system less stable. Addition of zeros to the transfer function has the effect of pulling the root locus to the
left, making the system more stable.This can be proved by Following Examples.
3.5.1 Addition of poles
Consider, G(s)H(s) =K/s(s+2) Corresponding root locus is shown in Fig 3.7

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig3.7 Root locus of K/s(s+2)

Now If pole at s=-4 is added to G(s)H(s) root locus becomes as shown in Fig3.8
G(s)H(s) =K/s(s+2)(s+4)

Fig 3.8 root locus of K/s(s+2)(s+4)

So for any value of K before addition of poles in left half, system is totally stable but after addition of
pole in left half two branches of root locus after some value of K moves in right half of s plane so system
up to this value of K is stable. After this value of K system becomes unstable so stability of system gets
restricted.
If Now one more pole at s=-6 is added to the system, G(s)H(s) =K/s(s+2)(s+4)(s+6)
Breakaway point in section s=0and s=-2 gets shifted towards right as compared to previous case.so
system stability further gets restricted. This is shown in the Fig.3.9

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig 3.9 root locus of K/s(s+2)(s+4)(s+6)

From this we can conclude that due to addition of poles, root locus shifts towards R.H.S. of s plane so
system stability decreases.
3.5.2 Effect of Addition of Open Loop Poles can be summarised as
1.Root locus shifts towards imaginary axis.
2.System stability relatively decreases
3.System becomes more oscillatory in nature.
4. Range of operating values of ‘K’for stability of the system decreases.
3.5.3Addition of Zeros
Let us introduce a zero at s=-4 G(s)H(s) =K/s(s+2), Root locus for G(s)H(s) =K/s(s+2) is shown in Fig
3.7. Root locus for G(s)H(s) =K(s+4)/s(s+2) is shown in Fig 3.10.it can be seen that root locus shift
towards left i.e. towards zero which is added. So as roots move towards left half of s plane relative
stability increases as compared to previous case.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

Fig 3.10 root locus of K(s+4)/s(s+2)

Let us introduce one more zero at s=-6
G(s)H(s) =K(s+4)(s+6)/s(s+2

In this case breakaway point’s shifts towards left half of s plane as shown. so relative stability of
system gets further increased.

As we go on adding zeros to left resultant root loci bend towards left half of s plane. As roots
move towards left half of s plane, relative stability of system improves. Due to addition of zeros towards
left half of s plane system stability increases. Also it increases the range of operating values of K for
system stability.

Fig 3.11 root locus of K(s+4)(s+6)/s(s+2)

3.5.4 Effect of Addition of Open Loop Poles can be summarized as
1. Root locus shifts to left away from imaginary axis.
2. Relatively stability of the system increases.

CONTROL SYSTEM EE/EX-405

By Dr.Amit Shrivastava

Professor
Department of Electrical Engineering

3. System becomes less oscillatory in nature.
4. Range of operating values of ‘K’for stability of the system increases.
3.6 Stability by Root locus
3.6.1 Gain Margin: We define gain margin by which the design value of the gain factor can be
multiplied before the system becomes unstable. Mathematically it is given by the formula

3.6.2 Phase Margin: Phase margin can be calculated from the given formula:

To determine phase margin using above formula for a given value of K .determine gain cross over
frequency  such that G(j)H(j) =1
3.6.3 Magnitude Criteria: At any points on the root locus we can apply magnitude criteria as,

Using this formula we can calculate the value of K at any desired point.
Using Root Locus Plot: The value of K at any s on the root locus is given by

REFERENCES
1. B.S.Manke, Control system Engineering, Khanna Publishers
2. M. Gopal, ‘Control system engineering’, McGraw Hill
3. K. Ogata, ‘Modern Control Engineering’, Pearson
4. D. Roy, Chaudhary,‘Modern Control Systems’, PHI.
5. B.C. Kuo and FaridGolnaraghi, ‘Automatic Control Systems’, Wiley India.
6. www.tutorialspoint.com
7.Electronics guide4u.com
8.U A Bakshi & S.C.Goyal Feedback Control System,Technical Publication.