M.

Renuga / AP – ECE / AAMEC UNIT -4 EC3351 – Control Systems

SYLLABUS: UNIT IV -CONCEPTS OF STABILITY ANALYSIS 9

Concept of stability-Bounded - Input Bounded - Output stability-Routh stability criterion-Relative
stability-Root locus concept-Guidelines for sketching root locus-Nyquist stability criterion.
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1. Stability : When the system is excited by a bounded input, the output is bounded.
1.1 Routh-Hurwitz Criterion:

For a system to be stable, it is necessary and sufficient condition that each term of first column of
Routh array of its characteristic equation be positive if ao .

If this condition is not met, the system is unstable and number of sign changes of the terms of the first
column of the Routh array corresponds to the number of roots of the characteristics equation in the right
half of the s-plane.

1.2 Problem:

1. Using Routh criterion , determine the stability of the system represented by the characteristics
equation , 16s + 5 = 0. Comment on the location of the roots of characteristics
equation.
Routh array: (Order =4, 4 roots)
: 1 18 5 ... Row 1

: 8 16 ... Row 2
(Row -2 by 8)
: 1 2

: 16 5 ... Row 3

: 1.7 ... Row 4

: 5 ... Row 5

Result: In the first column, all the elements are positive and there is no sign change. the system is stable.
All the four roots are lying on the left half of s-plane
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2. Construct Routh array and determine the stability of the system whose characteristic equation
is . Also determine the number of roots lying
on right half of s-plane, left half of s-plane and on imaginary axis. (May -13, 16 marks)
(Order =6, 6 roots)
: 1 8 20 16

: 2 12 16 ( by 2)

: 1 6 8
: 1 6 8
: 0 0
: 1 3
: 3 8
: 0.33

: 8
Result : Marginally stable

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3. The characteristic equation of the system is ,

( Order = 5, 5 roots)
Routh array
: 1 2 3

: 1 2 5

: -2

: 5

: 5

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1. 4. Determine the range of K for stability of unity feedback system whose open loop transfer
function is G(s) = (Dec -12, 8 marks)

The closed loop transfer function is, = =

The characteristics equation is s (s +1)(s+2) +K = 0; s ( +3s +2) +K = 0

+ +2s +K = 0

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5. Determine the range of K for stability of unity feedback system whose open loop transfer
function is q(s) =

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2. ROOT LOCUS: The path taken by a root of characteristics equation, when open loop gain K is
varied from 0 to infinite .

2.1 Construction of Root Locus: (note: ordinary graph sheet)

Step 1: Locate the poles and zeros o f G( s )H ( s ) on the s- plane.
The poles are marked by ” ; Zeros are marked by ”
Step 2: Determine the root loci on the real axis.
In constructing the root loci on the real axis, choose a test point on it. If the total number of real
poles and real zeros to the right of this test point is odd, then this point lies on a root locus. If it
is even the test point does not lie on the root locus.

Step 3: Determine the asymptotes of 'root loci and centroid.

Angles of asymptotes = ; q = 0, 1, ……. n– m
where , n = number of finite poles of G(s) H (s)
m = number of finite zeros of G(s) H(s)
Centroid =
Step 4: Find the breakaway and break-in points.
Step 5: Determine the angle of departure (angle of arrival) of the root locus from a complex pole (at a
complex zero).
Angle of departure:

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Angle of arrival:

Step 6: Find the points where the root loci may cross the imaginary axis.
Step 7: Taking a series of test points in the broad neighborhood of the origin of the s plane, sketch the
root loci.

Step 8: Determine closed-loop poles. The value of gain K t any point on the root locus can be
determined from magnitude condition.
K =

Ex. 1: A unity feedback control system has an open loop transfer function, G(s) = .
Sketch the root locus. (May-9, 16 marks)
zero
Solution:
Step1: To locate poles and Zeros: pole
The poles of open loop transfer function are the roots of the equation, s (s2+4s+13) = 0.
a=1, b= 4, c=13 in quadratic Eqn.
√ √
= = –2 j3

Let us denote the poles as P1, P2, and P3.

Here, P1 = 0, P2 = –2 + j3, and P3 = –2 – j3.
The poles are marked by X(cross) in Graph sheet. No zeros in numerator
Step2: To find the root locus on real axis
There is only one pole on real axis at the origin. Hence if we choose any test point on the
negative real axis then to the right of that point the total number of real poles and zeros is one, which is
an odd number.
Hence the entire negative real axis will be part of root locus. The root locus on real axis is shown
as bold line in Fig.
(Note: For the given transfer function one root locus branch will start at the pole in the origin and meet the zero
at infinity through the negative real axis.)
Step3: To find angles of asymptotes and centroid
Since there are 3 poles, the number of root locus branches are three. There is no finite zero.
Hence all the three root locus branches ends at infinity. The number of asymptotes required at
three.
Angles of asymptotes = ; q = 0, 1, ……. n– m
Here, n = 3, and m = 0. q = 0, 1, 2, 3.
When q = 0, Angles = =

When q = 1, Angles = = =

When q = 2, Angles = = =

When q = 3, Angles = = =
(Note: It is enough if you calculate the required number of angles.
Here it is given by first three values of angles. The remaining
values will be repetitions of the previous values.)

Centroid = = = = –1.33

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The centroid is marked on real axis from the centroid, the angles of asymptotes are marked using a
protractor. The asymptotes are drawn as dotted lines as shown in Fig.
Step4: To find the break–away and break–in points

The closed loop transfer function,

The characteristic equation is, =0

K= =
On differentiating the equation of K with respect to s, we get:

Put

a=3, b= 8, c=13 in quadratic Eqn.

√ √
= = –1.33 j1.6
Check for K : When, s = –1.33 + j1.6, the value of K is given by,
K =
positive and real.
Also it can be shown that when s = –1.33 –j1.6 the value of K is not equal to real and positive.
Since, the values of K for, s = –1.33 j1.6, are not real and positive, these points are not an actual
breakaway or break–in points. The root locus has neither break–away nor break–in point.
Step5: To find the angle of departure
Let us consider the complex pole p2 as shown in Fig. Draw vectors from all other poles to the
pole p2 as shown in Fig. Let the angles of these vectors be 1 and 2.
Here, 1 = – tan–1(3/2) = ; 2 =
Angle of departure from the complex pole p2 =
= + ) =

The angle of departure at complex pole p3 is negative of the

angle of departure at complex pole A.
Angle of departure at pole p3 = +
Mark the angles of departure at complex poles using
protractor.
Step6: To find the crossing point on imaginary axis
The characteristics equation is given by, s3 + 4s2+ 13s+K =0
3 2
Putting s = (jω): (jω) + 4(jω) + 13(jω)+K =0
3 2
–jω –4ω +13jω + K =0
Imaginary part = 0 Real part =0
3 2
–jω +13jω = 0 –4ω + K = 0
ω = j3.6 K = 52
The crossing point of root locus is j3.6. The value of K at this point is K = 52 (This is the limiting
value of K for the stability of the system).
Result
Since No break away and break in points, draw the locus starts at complex poles (P2= ,
P3 = ), and crosses imaginary axis at j3.6.

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Ex. 2: Sketch the root locus of the system whose open loop transfer function, G(s) = .
Find the value of K so that the damping ratio of the closed loop system is 0.5.
Solution: May-12, 16 marks
Step1: To locate poles and Zeros
The poles of open loop transfer function are the roots of the equation, s = 0.
Here, P1 = 0, P 2 = –2 and P 3 = – 4.
The poles are marked by X(cross) as shown in Fig. No zeros in numerator
Step2: To find the root locus on real axis
There are three poles on the real axis. at the origin.
Choose a test point on real axis between s = 0 and s = –2. To the right of this point the total
number of real poles and zeros is one, which is an odd number. Hence the real axis between s=0
and s=–2 will be a part of root locus.
Choose a test point on real axis between s = –2 and s = –4. To the right of this point the total
number of real poles and zeros is two, which is an even number. Hence the real axis between s=–2
and s=–4 will not be a part of root locus.
Choose a test point on real axis to the left of s = – 4. To the right of this point the total number of
real poles and zeros is three, which is an odd number. Hence the real axis from s= – 4 to will
be a part of root locus.
The root locus on real axis is shown as bold line in Fig.
Step3: To find angles of asymptotes and centroid
Since there are 3 poles, the number of root locus branches are three. There is no finite zero.
Hence all the three root locus branches ends at infinity. The number of asymptotes required at three.
Angles of asymptotes = ; q = 0, 1, ……. n– m
Here, n = 3, and m = 0. q = 0, 1, 2, 3.
When q = 0, Angles = =

When q = 1, Angles = =
(Note: It is enough if you calculate the required number of angles. Here it is given by first three values of angles.
The remaining values will be repetitions of the previous values).
Centroid = = =
The centroid is marked on real axis from the centroid, the angles of asymptotes are marked using a
protractor. The asymptotes are drawn as dotted lines as shown in Fig.
Step4: To find the break–away and break–in points

The closed loop transfer function,

The characteristic equation is, =0 s( +K=0

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=0 K=
On differentiating the equation of K with respect to s, we get:

Putting,

a=1, b= 12, c=8 in quadratic Eqn.

√

√
s= = –0.845 or –3.154
Check for K : When, s = –0.845, the value of K is given by,
K = = 3.08
Since K, is positive and real for, s = -0.845, this point is a actual breakaway point.
When, s = –3.154, the value of K is given by,
K = = – 3.08
Since K, is negative for s = -3.154, this is not a actual breakaway point.
The break-away point is marked on the negative real axis as shown in Fig.
Step5: To find the angle of departure
Since there are no complex pole or zero, we need not find angle of departure or arrival.
Step6: To find the crossing point on imaginary axis
The characteristic equation is, =0
3 2
Putting s = (jω): (jω) + 6(jω) + 8(jω)+K = 0
–jω3 –6ω2+8jω + K =0
Equating the imaginary part to zero Equating real part to zero
–jω3 +8jω = 0 –6ω2+ K = 0
3
–jω = – 8jω K = 6ω2 = 6 x 8= 48
ω2 = 8 ω = 2.8

The crossing point of root locus is j2.8. K = 48 .

Result
Draw the locus using break away point( s = - 0.845) , and crosses imaginary axis at j2.8.

To find the value of K corresponding to = 0.5

Given that = 0.5
-1
Let = cos = cos-1 0.5 = 60o

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Draw a line OP, such that the angle between line OP and negative real axis is 60o ( ) as
shown in Fig. 4.23.2. The meeting point of the root locus gives the denominator pole. sd.
Let Ksd be the value of K corresponding to the point s = sd
Ksd =

= 1.3 x 1.75 x 3.5 = 7.96 8

Note: The length of vectors are measured to scale.

ζ = 0.5
P

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Example 3: The open loop transfer function of a unity feedback control system, G(s) = .
Sketch the root locus. zero
Solution:
Step1: To locate poles and Zeros pole
The poles of open loop transfer function are the roots of the equation, s (s2+4s+11) = 0.
a=1, b= 4, c=11 in quadratic Eqn.
√ √
= = –2 j2.64
Here poles, p1 = 0, p2 = –2 + j2.64 and p3 = –2 – j2.64.
; s = - 9 (numerator)
Here Zero ,z = - 9

The poles are marked by X(cross) and zero by “o”(circle) as shown in Fig.
Step2: To find the root locus on real axis
One pole and one zero lie on real axis.
Choose a test point to the left of s = 0, then to the right of this point, the total number of real
poles and zeros is one, which is an odd number. Hence the portion of real axis from s=0 and s=–9
will be a part of root locus.
If we choose a test point to the left of s = – 9, then to the right of this point, the total number of
real poles and zeros is two, which is an even number. Hence the real axis from s= – 9 to will
not be a part of root locus.
The root locus on real axis is shown as bold line in Fig.

Step3: To find angles of asymptotes and centroid

Since there are 3 poles, the number of root locus branches are three. One root locus branch starts
at the pole at origin and travel along negative real axis to meet the zero at s = -9. The other two root
locus branches meet the zeros at infinity. The number of asymptotes required at two.

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Angles of asymptotes = ; q = 0, 1, …….

n– m
Here, n = 3, and m = 1. q = 0, 1, 2, 3.
When q = 0, Angles = =

When q = 1, Angles = = =

When q = 2, Angles = = =
Note: It is enough if you calculate the required number of
angles. Here it is given by first three values of angles. The
remaining values will be repetitions of the previous values.

– –
Centroid = = = = 2.5
The centroid is marked on real axis from the centroid, the angles of asymptotes are marked using a
protractor. The asymptotes are drawn as dotted lines as shown in Fig.
Step4: To find the break–away and break–in points
From the location of poles and zero and from the knowledge of typical sketches of root locus, it can
be concluded that there is no possibility of breakaway or break-in points.
Step5: To find the angle of departure
Let us consider the complex pole p2 as shown in Fig. Draw vectors from all other poles and
zero to the pole p2 as shown in Fig.4.24.2. Let the angles of these vectors be 1 , 2and 3.
Here, 1 = – tan–1(2.64/2) = ; 2 = ; 3 = tan–1(2.64/7) =
Angle of departure from the complex pole p2 = +
= + )+
p2 =
Angle of departure at pole p3 =+

Mark the angles of departure at complex poles using protractor.

Step6: To find the crossing point on imaginary axis

The closed loop transfer function,

The characteristic equation is the denominator polynomial of C(s)/R(s).

Therefore the characteristic equation is ,
=0 =0
3 2
Putting s = (jω): (jω) + 4(jω) + 11(jω)+K(jω) +9K = 0
–jω3 –4ω2+j11ω + jKω + 9K =0
On equating the imaginary part to zero On equating the imaginary part to zero
–jω3 +j11ω+j Kω = 0 –4ω2+ 9K = 0
3
–jω = –j11ω - jKω 9K = 4 ω2
ω2 = 11+K Put, ω2 = 11 + K.
2
Put K = 8.8, ω = 11+8.8 = 19.8 9K = 4 (11+K) = 44 + 4K
ω= √ j4.4 9K - 4K = 44K
5K = 44 K = 44 / 5 = 8.8
The crossing point of root locus is 4.4. The value of K at this point is K = 8.8 (This is the
limiting value of K for the stability of the system).

Result

Since No break away and break in points, draw the locus starts at complex poles (P2
,P3= ), and crosses imaginary axis at j4.4.

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3. Nyquist stability Criterion:

Nyquist stability criterion for a general case when G(s)H(s) has poles and /or zeros on the j axis.]:
In the system shown in Fig., if the open-loop transfer function G(s)H(s) has k poles in the right-half s
plane, then for stability the G(s)H(s) locus, as a representative point s traces on the modified Nyquist
path in the clockwise direction, must encircle the -1 + j0 point k times in the counter-clockwise
direction.

3.1 Stability analysis:

In examining the stability of linear control systems using the Nyquist stability criterion,we see that
three possibilities can occur:
1. There is no encirclement of the -1+ j0 point. This implies that the system is stable if there are no
poles of G( s )H( s)in the right-half s plane; otherwise, the system is unstable.

2. There are one or more counterclockwise encirclements of the -1 + j0 point. In this case the
system is stable if the number of counterclockwise encirclements is the same as the number of poles
of G(s) H (s) in the right-half s plane; otherwise, the system is unstable.

3. There are one or more clockwise encirclements of the -1 + j0 point. In this case the system is
unstable. In the following examples, we assume that the values of the gain K and the time constants
are all positive.
Note

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1.Draw the Nyquist plot for the system whose open loop transfer function is, G(s)H(s) =
Determine the range of K for which closed loop system is stable. (may-12, 16 marks)
Solution: (note : change into 1+s term)
Given: G(s) H(s) = = =
( ) ( )

A single pole presented at origin. It has four sections C1, C2, C3 and
C4. The mapping of each section is performed separately and the overall
Nyquist plots is obtained by combining the individual sections.

Mapping of section C1: varies from 0 to

G(s) H(s) = ….(1)

s=j in Eqn.(1): G(j ) H(j ) = =

= ….(2)
Imaginary term = 0, in Eqn.(2): ( )

= 0; √ ; = 4.472 rad / sec

Taking Real term in Eqn. (2)
= 4.472 in Eqn.(2):
G(j ) H(j ) =
= - 0.00417K

Mapping of section C2: s= ;

G(s) H(s) = = = ….(3)

s= in Eqn.(3): = = = 0 ….(4)

When in Eqn.(4): =0 = - 270o

Note : 270 o + 270 o = 540 o

When in Eqn.(4): =0 = + 270o Curve start from -270o to + 270 o

-270o
90o

180o +ve

0o
-180o -ve

270o
-90o

Mapping of section C3: varies from to 0 (Note: This plot is opposite of C1 plot)

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Mapping of section C4: s= ;

G(s) H(s) = = = ….(5)

s= in Eqn.(5): = = = ….(6)

When in Eqn.(6): = = 90o

Note : 90 o + 90 o = 180 o
When in Eqn.(6): = = -90o Curve start from +90o to - 90 o

Completer Nyquist plot: The entire plot in G(s) H(s)-plane can be obtained by combining the mappings
of individual sections, as shown in Fig.

Stability analysis: When, - 0.00417 K = -1 , the contour passes through (-1+j0)point and corresponding
value of K is the limiting value of K for stability.
- 0.00417 K = -1; K= = 240
When K < 240 : The point -1+j0 is not-encircled. Therefore the closed loop system is stable.
When K > 240 : The point -1+j0 is encircled in clockwise direction twice. Therefore the closed loop
system is unstable.
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2.Construct the nyquist plot for a system whose open loop transfer function is given by G(s)H(s)
= . Find the range of K for stability. ( Dec -10,16 marks)

Solution: (note : change into 1+s term)

Given that, G(s) H(s) =

Three pole presented at origin . It has four sections C1, C2, C3 and C4.
The mapping of each section is performed separately and the overall
Nyquist plots is obtained by combining the individual sections.

Mapping of section C1: varies from 0 to

G(s)H(s) = ….(1)

Let s = in Eqn.(1): ….(2)

= = + = +j ….(3) 𝑗
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Eqn.(3): Taking imaginary term = 0 at

( ) = 0 ( )= 0

Eqn.(3): Taking Real term = 0 at = 1 rad/sec

- = – = – 2K ….(4)

Eqn.(2): Taking magnitude and phase Eqn .

√ √

As ….(5)
As ….(6)
From the Eqns.( 4),(5),(6) the polar plot starts - axis at Infinity ,crosses real axis at -2K and ends at the origin
in the 3rd quadrant .

Mapping of section C2: s= ;

G(s) H(s) = ….(7)

s= in Eqn.(7): = = ….(8)

When , in Eqn.(8) G(s)H(s) = ….(9)

When , in Eqn.(8) G(s)H(s) = ….(10)
Eqns.( 9) and (10) :

Note : 90 o +90 o = 180 o

Curve start from -90o to + 90 o

Mapping of section C3: varies from to 0 (Note: This plot is opposite of C1 plot)

Mapping of section C4: s= ;

G(s)H(s) = ….(10)

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Let, s = in Eqn.(10): = ….(11)

When ,in Eqn (11) G(s)H(s) = ….(12)

When , in Eqn (11) G(s)H(s) = ….(13)

Eqns.( 12) and (13) :

Note : 270 o +270 o = 540 o

Curve start from +270o to - 270 o

Complete Nyquist plot: The entire plot in G(s) H(s)-plane can be obtained by combining the mappings
of individual sections, as shown in Fig.
Stability analysis: When ,-2K =-1, the contour
passes through - 1+j0 point and corresponding
value of K is the limiting value of K for stability.
Limiting value of K =1/2 = 0.5
When K < 0.5 : The point -1+j0 is encircled in
clockwise direction two times. Therefore the
closed loop system is unstable.
When K > 0.5 : The point -1+j0 is encircled in
both clockwise and anticlockwise direction one
time. Hence, net encirclement is zero. Therefore
the closed loop system is unstable.
Result: The system is stable when K > 0.5.
4. Relative Stability:
1. It indicates the closeness of the system to stable region.
2. It is an indication of strength or degree of stability.
4.1 Dominant pole :
1. It is a pair of complex conjugate pole which decides transient response of the system.
2. In higher order systems the dominant poles are very close to origin and all other poles of the
system are widely separated and so they have less effect on transient response of the system.
1. The open loop transfer function of a unity feedback system is given by , G(s ) =
. Derive an expression for gain K in terms of T1, T2 and specified gain margin
Kg.

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