Stability: Methods of determining stability, Routh Hurwitz Criterion, Root Locus, Frequency Domain Analysis: Resonant Peak, Resonant frequency and Bandwidth of the second order system, Effect of adding a zero and a pole to the forward path, Nyquist Stability Criterion, Relative Stability: Gain Margin and Phase Margin, Bode Plot.

1. Control Systems
1
Department of Electronics & Communication Engineering,
Madan Mohan Malaviya University of Technology, Gorakhpur
Subject Code: BEC-26 Third Year ECE
Unit-IV
Shadab A. Siddique
Assistant Professor

2. UNIT IV
✓ Analysis of stable, unstable, critically stable and conditionally stable
✓ Relative Stability
✓ Root locations in S-plane for stable and unstable system
✓ Routh’s Stability Criterion : Different cases and conditions
✓ Statement Method
✓ Numerical Problems
✓ Root Locus
UNIT-IV Stability: Methods of determining stability, Routh Hurwitz Criterion, Root Locus,
Frequency Domain Analysis: Resonant Peak, Resonant frequency and Bandwidth of the
second order system, Effect of adding a zero and a pole to the forward path, Nyquist
Stability Criterion, Relative Stability: Gain Margin and Phase Margin, Bode Plot.
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Shadab. A. Siddique

3. A linear time invariant system is stable if following conditions are satisfied:
➢ A bounded input is given to the system, the response of the system is bounded and
controllable.
➢ In the absence of the inputs, the output should tend to zero as time increases.
Stable System
Bounded i/p bounded o/p for
stable system
Location of roots for
stable system
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Shadab. A. Siddique

4. ➢ A linear time invarient system comes under the class of unstable system if the system is
excited by a bounded input, response is unbounded.
➢ This means once any input is given system output goes on increasing & designer does
not have any control on it
Unstable System
Bounded i/p Unbounded o/p
for unstable system
Location of roots for
unstable system
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Shadab. A. Siddique

5. ➢ When the input is given to a linear time invarient system, for critically stable systems
the output does not go on increasing infinitely nor does it go to zero as time increases.
➢ The output usually oscillates in a finite range or remains steady at some value.
➢ Such systems are not stable as their response does not decay to zero. Neither they are
defined as unstable because their output does not go on increasing infinitely.
Critically Stable System
Bounded i/p & o/p response for
critically stable system
Location of roots
for critically stable system
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Shadab. A. Siddique

6. ➢ A system may be absolutely stable i.e. it may have passed the Routh stability test.
➢ As a result their response decays to zero under zero input conditions.
➢ The ratio at which these decay to zero is important to check the concept of “Relative
stability”
➢ When the poles are located far away from j axis in LHP of s-plane, the response decays
to zero much faster, as compared to the poles close to j-axis.
➢ The more the poles are located far away from j-axis the more is the system relatively
stable.
Relative Stability
Response
comparison
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Shadab. A. Siddique

7. Relative Stability
Location of poles comparison

−
j
−j




System B
Complex conjugate
poles
System A
Complex conjugate
poles
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Shadab. A. Siddique

8. Relative Stability

−
j
−j
Relative Stability improves as one moves
away from jw axis
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Shadab. A. Siddique

9. Routh’s Stability Criterion
m m−1
C(s)
=
b0s + b1s
R(s) a0 sn
+ a1sn−1
+ ...............+ bm
+ ...............+ an
In this criterion, the coefficients of denominator are arranged in an
Array called “Routh’s Array”;
a0sn
+ a1sn−1
+ ............... + an = 0
For the transfer function;
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Shadab. A. Siddique

10. The coefficients of sn
and sn−1
row are directly written from the given
equation.
a0 sn
+ a1 sn − 1
+ ............... + an = 0
s n
sn−1
sn−2
sn−3
a0 a2
a1 a3
a4
a5
a6
a7
b1
The Routh’s array as below; For next row i.e. sn−2
;
a1.a2 − a3a0
b1 =
a1
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Shadab. A. Siddique

11. The coefficients of sn
and sn−1
row are directly written from the given
equation.
a0 sn
+ a1 sn − 1
+ ............... + an = 0
s n
sn−1
sn−2
sn−3
a0 a2 a4 a6
a1 a3 a5 a7
b1 b2
The Routh’s array as below; For next row i.e. sn−2
;
b1 =
a1 .a2 − a 3 a 0
a 1
a1.a4 − a0.a5
a1
b2 =
s0
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Shadab. A. Siddique

12. s n
sn−1
sn−2
sn−3
a0 a2
a1 a3
a6
a7
b1
a4
a5
b2 b3
The Routh’s array as below;
a0 sn
+ a1 sn − 1
+ ............... + an = 0
For next row i.e. sn−2
;
b1 =
a1 .a2 − a 3 a 0
a 1
a1 .a4 − a .0 a5
b2 =
a1
a1.a6 − a0 . a7
b3 =
a1
s0
The coefficients of sn
and sn−1
row are directly written from the given
equation.
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Shadab. A. Siddique

13. only previous two rows are used i.e. and
Now the same technique is used, for the next row i.e. sn−3row, but
s n
sn−1
sn−2
sn−3
a0 a2
a1 a3
a6
a7
a4
a5
b2 b3
b1
c1
The Routh’s array as below;
For next row i.e. sn−2
a1 .a2 − a 3 a 0
b1 =
a1
a1 .a4 − a0 . a5
b2 =
a1
a1.a6 − a0 . a7
b3 =
a1
sn−1
sn−2
For next row i.e. sn−3
b1.a3−b2a1
c1 =
b1
s0
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Shadab. A. Siddique

14. only previous two rows are used i.e. and
Now the same technique is used, for the next row i.e. sn−3
row, but
s n
sn−1
sn−2
sn−3
s0
a0 a2
a1 a3
a6
a7
b1
c1
a4
a5
b2 b3
c2
The Routh’s array as below;
For next row i.e. sn−2
a1 .a2 − a 3 a 0
b1 =
a1
a1 .a4 − a0 . a5
b2 =
a1
a1.a6 − a0 . a7
a1
b3 =
sn−1
sn−2
For next row i.e. sn−3
b1 .a3 − b 2 a 1
b1
c1 =
c2 =
b1.a5−b3a1
b1 14
Shadab. A. Siddique

15. ➢ Each column will reduce by
one as we move down the
array.
➢ This process is obtained till
last row is obtained.
s n
sn−1
sn−2
sn−3
s0
a0 a2
a1 a3
a6
a7
b1
c1
a4
a5
b2 b3
c2
an
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Shadab. A. Siddique

16. Routh’s Criterion
➢ The necessary & sufficient
conditions for a system to be
stable is all terms in the first
column at Routh’s Array should
have same sign
➢ There should not be any sign
change in first column.
s n
sn−1
sn−2
sn−3
s0
a0 a2
a1 a3
a6
a7
b1
c1
a4
a5
b2 b3
c2
an
Must have same sign
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Shadab. A. Siddique

17. ➢ When there are sign changes in
the first column of Routh’s array
then the system is unstable.
➢ There are roots in RHP.
➢ The number of sign changes
equal the number of roots in
RHP.
s n
sn−1
sn−2
sn−3
s0
a6
a7
a4
a5
b2 b3
c2
an
+ a0 a2
+a1 a3
- b1
+c1
Routh’s Criterion
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Shadab. A. Siddique

18. Example 1
s3
a0
+ 6s2
+ 12s + 8 = 0
a1 a2 a3
s 3
s2
s1
s0
1
6
b1
an
12
8
b2
a1
a1.a2−a3.a0
b1= s 3
s2
s1
s0
1
6
b1
an
12
8
b2
6
b1=
(612)−(81)
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Shadab. A. Siddique

19. s 3
s2
s1
s0
1
6
12
8
b2
10.67
an
0
0
a1
a1.a4 −a0.a5
b2 =
Example 1 cont……
s 3
s2
s1
s0
1
6
12
8
b2
10.67
an
0
0
6
10−60
b2 =
19
Shadab. A. Siddique

20. s 3
s2
s1
s0
an
1
6
12
8
10.67
0
0
10.67
10.678−60
b2 =
0
Example 1 cont……
s 3
s2
s1
s0
1
6
12
8
10.67
0
0
0
8
s3
+ 6s2
+ 12s + 8 = 0
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Shadab. A. Siddique

21. s 3
s2
s1
s0
1 12 0
6 8 0
10.67 0
8
Elements in first column are of same sign
As all elements in first column
are of same sign. Hence system
is stable
Example 1 cont……
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Shadab. A. Siddique

22. Example 2
Comment on stability.
s5
s4
s3
s2
s1
s0 an
1 4 2
2 6 5
b1 b2
c1
d1
c2
s5
+2s4
+4s3
+6s2
+2s+5=0
b1=
(24)−(61)
2
b1 = 1
b2=
(22)−(51)
2
b2 = −0.5
22
Shadab. A. Siddique

23. Example 2 cont…..
s5
s4
s3
s2
s1
s0
1
2
5
4
6
−0.5
1
2
c1 c2
d1
an
c1=
(16)−(−0.52)
1
c1 = 7
c2 =
(15)−(02)
1
c2 = 5
d1 =
(7−0.5)−(51)
7
d 1 = −1.21
(5−1.21)−(70)
−1.21
an=
an = 5
23
Shadab. A. Siddique

24. s5
s4
s3
s2
s1
s0
1
2
5
4
6
−0.5
1
2
5
5
7
−1.21
There are two sign changes +7
to -1.21 and -1.21 to +5. Hence
two roots are in RHP S-plane
and system is unstable.
Example 2 cont…..
24
Shadab. A. Siddique

25. Example 3
Comment on stability.
s5
s4
s3
s2
s1
s0 an
1 2 3
1 2 5
b1 b2
c1
d1
c2
s5
+s4
+2s3
+2s2
+3s+5=0
b1=
(12)−(21)
1
b1 = 0
b2=
(13)−(51)
1
b2 = −2
25
Shadab. A. Siddique

26. Example 3 cont…..
Comment on stability. s5
+s4
+2s3
+2s2
+3s+5=0
s5
s4
s3
s2
s1
s0
1 2 3
1 2 5
0 −2
c1 c2
d1
an
Case I – “When the first of any row is zero and the rest
of the row is non-zero.”
26
Shadab. A. Siddique

27. Routh’s Criterion Special Cases
Case I – “When the first of any row is zero and the rest of
the row is non-zero.” Here the next row cannot be formed
as division by 0 will take place.
Method to Overcome: A method to overcome above problem is to
replace s by and complete the Routh’s test for z.
z
1
27
Shadab. A. Siddique

28. Example 3 cont…..
5 4 3 2
1 1 1 1 1
z z z z z
( ) + ( ) + 2 ( ) + 2 ( ) + 3 ( ) + 5 = 0
Replace s by (1/z)
Take L.C.M
z 5
1 + z + 2 z 2
+ 2 z 3
+ 3 z 4
+ 5 z 5
= 0
1+ z + 2z2
+ 2z3
+ 3z4
+ 5z5
=0
5z5
+ 3z4
+ 2z3
+ 2z2
+ z +1 =0
Use above characteristics equation and complete Routh’s Test
28
Shadab. A. Siddique

29. Example 3 cont…..
z4
z3
z1
z0
4
3
−
2
3
−
5 2 1
1
2
3
1
z5
1
z2
2
2
1
There are two sign changes
in first column. Hence two
roots are in RHP S-plane
and system is unstable
29
Shadab. A. Siddique

30. Example 4
Comment on stability. s4
+6s3
+11s2
+6s+10=0
s4
s3
s2
1
s
s0
b1 b2
10
0
11
6
1
6
c1
an
b1=
(611)−(16)
6
b1 = 10
b2=
(610)−(10)
6
b1 = 10
c1=
(106)−(106)
10
c1 = 0
30
Shadab. A. Siddique

31. Example 4 cont….
s4
s3
s2
s1
s0
10 10
10
0
11
6
1
6
0
an
c1=
(106)−(106)
10
c1 = 0
Case II – “When all elements in any one row is zero.”
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Shadab. A. Siddique

32. Routh’s Criterion Special Cases
Case II – “When all elements in any one row is zero.”
Method to Overcome:
✓ Here form an auxillary equation with the help of the coefficients
of the coefficients of the row just above the row of zeros.
✓ Take the derivative of this equation and replace it’s coefficients in
the present row of zeros.
✓ Then proceed for Routh’s test.
32
Shadab. A. Siddique

33. Example 4 cont….
s4
s3
s2
s1
s0
10 10
10
0
11
6
1
6
0
an
Here s1 row breaks down.
Hence write auxiliary equation
for s2 .
A(s) =10s2
+10
(Note each term of next column
differs by degree of 2)
Take derivative of auxiliary equation
Use these for s row coefficients
d
A(s)= 20s
ds
33
Shadab. A. Siddique

34. Example 4 cont….
s4
3
s
s2
s1
s0
10 10
10
0
11
6
1
6
20
an
an =
(2010)−(100)
20
an = 10
s4
s3
s2
s1
s0
10 10
10
0
11
6
1
6
20
10
As no sign change in first
column; system is stable
34
Shadab. A. Siddique

35. Example 5
Comment on stability.
s6
s5
s4
s3
s2
s1
s0
s6
+3s5
+5s4
+9s3
+8s2
+6s+4=0
(35)−(91)
3
b1=
b1 = 2
b2=
(38)−(61)
3
b2 = 6
4
1 5 8
3 9 6
b1 b2 b3
c1 c2
d1
e1
an
b3 =
(34)−(01)
3
b3 = 4 35
Shadab. A. Siddique

36. Example 5 cont….
s6
s5
s4
s3
s2
s1
s0
1 5 8 4
3 9 6
2 6 4
c1 c2
d1
e1
an
c1 =
(29)−(36)
2
c1 = 0
c2=
(26)−(43)
2
c2 = 0
36
Shadab. A. Siddique

37. Example 5 cont….
s6
s5
s4
s3
s2
s1
s0
5
9
6
1
3
2
0
4
8
6
4
0
d1
e1
an
Here s3 row breaks down.
Hence write auxiliary equation
for s4 .
A(s) = 2s4
+ 6s2
+ 4
(Note each term of next column
differs by degree of 2)
Take derivative of auxiliary equation
d
A(s) = 8s3
+ 1 2 s
ds
Use these for s3 row coefficients.
37
Shadab. A. Siddique

38. Example 5 cont….
s6
s5
s4
s3
s2
s1
s0
1 5 8 4
3 9 6
2 6 4
8 12
d1 d2
e1
an
d1 =
(86)−(122)
8
d1 = 3
d2=
(84)−(02)
8
d 2 = 4
38
Shadab. A. Siddique

39. Example 5 cont….
s6
s5
s4
s3
s2
s1
s0
e1
an
1 5 8 4
3 9 6
2 6 4
8 12
3 4
e1=
(312) −(84)
3
e1 = 4
39
Shadab. A. Siddique

40. Example 5 cont….
s6
s5
s4
s3
s2
s1
s0
4
an
1 5 8 4
3 9 6
2 6 4
8 12
3 4
an=
(44)−(30)
4
an = 4
40
Shadab. A. Siddique

41. Example 5 cont….
s6
s5
s4
s3
s2
s1
s0
1 5 8 4
3 9 6
2 6 4
8 12
3 4
4
4
As no sign change in first
column; system is stable
41
Shadab. A. Siddique

42. Example 6
+
-
R(s) C(s)
100
s2
+10s
Problem: Using routh’s criteria find the stability for given figure.
42
Shadab. A. Siddique

43. Example 6 cont……
100
s2
G(s) =
+10s
H(s) =1

C(s)
=
G(s)
R(s) 1+G(s)H(s)
100

C(s)
=
100
s2
s2
+10s
R(s)
1+
+10s
100
s2
s2
s2
+10s
+10s +100
+10s
=
s2
100
+10s +100
=
Characteristics equation is the denominator of the CLTF
Characteristics equation s2
+10s+100 = 0
43
Shadab. A. Siddique

44. Example 6 cont…….
s2
s1
s0
10
an
1 100
s2
+10s+100 = 0
an =
(10100)−(10)
10
an = 100
s2
s1
s0
10
1 100
As no sign change in first
column; system is stable
100
44
Shadab. A. Siddique

45. Example 7
Determine the range of k for stable system. s4
+ 5s3
+ 5s2
+ 4s + k =0
s4
s3
s2
s1
s0
k
1 5 k
0
4
5
4.2
4.2
16.8−5k
k
45
Shadab. A. Siddique

46. k0
16.8 − 5k
0
4.2
16.8 5k or k<
Example 7 cont……
For stability all elements of first column 1 should be positive
For s0
row
For s1
row
i.e.
and
i.e.
16.8
5
k3.36
Thus combining equations (1) and (2),
------------------------(1)
------------------------(2)
0  k  3.36
This is the range of k stable operation.
46
Shadab. A. Siddique

47. Example 8
s3
s2
s1
s0
1 30k
200k
2+k
30k(2+k)−200k
(2 +k)
k
Determine the range of k for stable system. s3
+ s2
(2 + k)+ 30sk + 200k = 0
47

48. Example 8 cont……
30k(k + 2) − 200k
0
2 +k
k  4.67
For stability,
k0
k+20
k−2
i.e.
and
Thus 4.67  k  
This is the range of k stable operation.
i.e.
48
Shadab. A. Siddique

49. Advantages of Routh’s Criterion
➢ It is a simple algebraic method to determine the stability of closed loop without solving
for roots of higher order polynomial of the characteristic equation.
➢ It is not tedious or time consuming.
➢ It progress systematically.
➢ It is frequently used to determine the conditions of absolute & relative stability of a
system.
➢ It can determine range of k for stable operation.
Disadvantages of Routh’s Criterion
➢ It is valid only for real coefficients of characteristics equation. Any coefficient that is a
complex number or contains exponential factors, the test fails.
➢ It is applicable only to the linear systems.
➢ Exact location of poles is not known.
➢ Only idea is obtained about stability. A method to stabilize the system is not suggested.
Application of Routh’s Criterion
➢ The gain is kept in terms of k and Routh’s array is solved to find k for stable
operation.
49
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50. ROOT LOCUS
What is Root Locus ?
✓ Root locus is a graphical representation of the closed loop poles as a system parameter is
varied.
✓ It can be used to describe qualitatively the performance of a system as various parameters are
changed.]
✓ It gives graphic representation of a system’s transient response and also stability.
✓ We can see the range of stability, instability and the conditions that cause a system to break
into oscillation.
INTRODUCTION
50
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51. ANGLE AND MAGNITUDE CONDITION:
51
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52. 52
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53. 53
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54. CONSTRUCTION OF ROOT LOCUS
Rule 1: Symmetrical about real axis
Rule 2: Number of Root locus branches = Number of open loop poles and open loop zeros
Rule 3: Real axis loci
A point on real axis lies on root locus if the sum of poles and zeros to the right hand side of
that point must be odd number.
Root locus point will move on root locus path
To left of an
odd numbers
54
Shadab. A. Siddique

55. Note: Starting and Ending point
The root locus starts at open loop pole with K = 0 and terminates at open loop zeros
with K = ∞
Ending
Starting
55
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56. 56
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57. 57
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58. 58
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59. Rule 6: Break Away Point
✓ It is a point on root locus where two poles meet once the two pole meet, they split i.e they
break away from real axis.
✓ Procedure to calculate Break away point:
1. Write the characteristic equation i.e 1 + G(s)H(s) = 0
2. Arrange characteristic equation in such a way that K = f(s)
3. Differentiate K w.r.t s and equate it to zero, this will give valid and invalid break away
point.
59
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60. 60
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61. Break Away
Point S =-1.45
61
Shadab. A. Siddique

62. Rule 7: Intersection with imaginary axis
This rule gives us point on the imaginary axis which the root locus will cut while
moving to right half of the s-plane.
Intersection of Imaginary axis can be found by using Routh-Hurwitz criterion.
Procedure to calculate Intersection with Imaginary axis:
1. Write the characteristic equation i.e 1 + G(s)H(s) = 0
2. Construct the Routh array
3. Determine K
4. Get the Auxiliary equation
5. Compute the roots of Auxiliary equation. These will be intersecting point with the
imaginary axis.
62
Shadab. A. Siddique

63. 63

64. 64
Shadab. A. Siddique

65. +j1.59
K=9.65
-j1.59
K=9.65
65
Shadab. A. Siddique

66. 66
Shadab. A. Siddique

67. 67
Shadab. A. Siddique

68. 68
Shadab. A. Siddique

69. 69
Shadab. A. Siddique

70. 70
Shadab. A. Siddique

71. 71
Shadab. A. Siddique

72. 72
Shadab. A. Siddique

73. -6 -5 -4 -3 -2 -1 0
-4
-3
-1
-2
0
1
2
RealAxis
ImagAxis
4
3
-2 +
j2.45
-2 -
j2.45
73
Shadab. A. Siddique