Unit 3
Stability Analysis
For closed Loop system
Fig 1 Closed Loop system
Where:
G(s) – forward amplification
H(s) _ Reverse amplification
Above system is with negative feedback.
Positive feedback is used only in oscillators and other use is not known as discussing only negative feedback.
[ R(s) – c(s) H (S) ] G(s) = c(s)
R (s) G(s) = [1+G(s) H(s)] c(s)
C(s)/R(S) = G(S)/1+G(s) H(s)
S0, the transfer function of closed Loop system is
C(s)/ R(S) = G(s)/ 1+G(S) H(S) --------(2)
1+ G(S) H(s) = 0 [ characteristic Equation]
Key takeaway
The characteristic equation for a closed loop system is given as
1+ G(S) H(s) = 0
The transfer function of closed Loop system is
C(s)/ R(S) = G(s)/ 1+G(S) H(S)
If numerator of above equation (2) is equalized to zero we get zeros of closed Loop Transfer function [G(S) = 0]
If dominator of equation (2) is equaled to zero we get polls of the closed loop transfer function [ 1+G(s) H(S) = 0]
Fig 2 Location of Poles in s-plane and the response
Concept of Stability:
A stable system always gives bounded output for bounded input and the system is known as BIBO stable
A linear time invariant (LTI) system is stable if,
The system is BIBO stable
In absence of the input the output tends towards zero
For system;
For R(s) = 1
C(S) = 
R (t) = 
(t) C (t) = t
Fig 3 Input and output for 
So, system unstable.
= 
C(t) = e-10t C(t)
Fig 4 BIBO stable figure
Effect of location of Poles on stability
A) Real and negative G(s)= 
Fig 5 Output for G(s)= 
Stable
B) Poles on right half of S- plane
G(s) = 
Unstable:- Unbounded Output
Fig 6 Input and Output for G(s) = 
(C)Poles at origin
G(s)=
Marginally stable
Fig 7 Input and output for G(s)=
4) Complex poles on left half of S-plane
S1, S2= -α ± iw
= 
C(t) = k e- αt Cos wt
Fig 8 Pole location and output for C(t) = k e- αt Cos wt
At t - ∞ C(t) – 0 so, system stable
5) Complex poles on right half of S-plane
S1,S2 = α ± iw.
C(t) = k eαt COS wt.
C(t) increases exponentially with t ∞
So, unstable.
6) Poles on iw axis:
S1,S2 = ± iw.
C(t) = k COS wt.
Fig 9 Output response of C(t) = k COS wt.
The system is marginally stable than sustained oscillations.
Key takeaway
Consider a system with characteristic equation
aoSm + a1 Sm-1 ……………..am=0.
1) All the coefficients of above equation should have same sign.
2) There should be no missing term.
The above two condition are necessary but not sufficient condition for stability
C) Real and negative G(s)= 
Fig 10 Output for G(s)= 
D) Poles on right half of S- plane
G(s) = 
Unstable:- Unbounded Output
Fig 11 Input and Output for G(s) = 
(C)Poles at origin
G(s)=
Marginally stable
Fig 12 Input and output for G(s)=
4) Complex poles on left half of S-plane
S1,S2= -α ± iw
= 
C(t) = k e- αt Cos wt
Fig 13 Input and output of C(t) = k e- αt Cos wt
At t - ∞ C(t) – 0 so, system stable
5) Complex poles on right half of S-plane
S1,S2 = α ± iw.
C(t) = k eαt COSwt
C(t) increases exponentially with t ∞
So, unstable.
6) Poles on iw axis:
S1,S2 = ± iw.
C(t) = k COS wt.
Fig 13 Output of C(t) = k COS wt
The system is marginally stable than sustained so illations.
Key takeaway
Consider a system with characteristic equation
aoSm + a1 Sm-1 ……………..am=0.
3) All the coefficients of above equation should have same sign.
4) There should be no missing term.
The above two condition are necessary but not sufficient condition for stability
It states that the system is stable if and only if all the elements is the first column have the same algebraic sign. If all elements are not of the same sign then the number of sign changes of elements in first column equals the number of roots of the characteristic equation in the right half of S-plane.
Consider the following characteristic equation:
a0 Sn + a1 Sn-1 ………….an = 0 where a0,a1,,,,,,,,,,,,,,,,,,,,an have same sign and are non-zero.
Step1 Arrange coefficients in rows
Row1 ao a2 a4
Row2 a1 a3 a5
Step2 Find third row from above two rows
Row1 a0 a2 a4
Row2 a1 a3 a5
Row3 a1 a3 a5
a1 = 
= 
a3 = 
= 
Continue the same procedure to find new rows.
Q1) for the given polynomials below determine the stability of the system
S4+2S3+3S2+4S+5=0
1) Arranging Coefficient in Rows.
For row S2 first term
S2 = 
= 1
For row S2 Second term
S2 = 
= 5
For row S1:
S1 = 
= -6
For row S0
S0 = 
= 5
As there are two sign change in the first column, so, there are two roots or right half of S-plane making system unstable.
Q2. Using Routh criterion determine the stability of the system with characteristic equation S4+8S3+18S2+16S+S = 0
Sol:- Arrange in rows.
For row S2 first element
S1 = 
= 16
Second terms = 
= 5
For S1
First element = 
= 13.5
For S0
First element = 
= 5
As there is no sign change for first column so all roots are is left half of S-plane and hence system is stable.
Special Cases of Routh Hurwitz Criterions
- When first element of any row is zero.
In this case the zero is replaced by a very small positive number E and rest of the array is evaluated.
Eg.(1) Consider the following equation
S3+S+2 = 0
Replacing 0 by E
Now when E 
0, values in column 1 becomes
Two sign changes hence two roots on right side of S-plans
II) When any one row is having all its terms zero.
When array one row of Routh Hurwitz table is zero, it shown that the X is attests one pair of roots which lies radially opposite to each other in this case the array can be completed by auxiliary polynomial. It is the polynomial row first above row zero.
Consider following example
S3 + 5S2 + 6S + 30
For forming auxiliary equation, selecting row first above row hang all terms zero.
A(s) = 5S2 e 30
= 10s e0.
Again, forming Routh array
No sign change in column one the roots of Auxiliary equation A(s)=5s2+ 30-0
5s2+30 = 0
S2 α 6= 0
S = ± j 
Both lie on imaginary axis so system is marginally stable.
Q3. Determine the stability of the system represent by following characteristic equations using Routh criterion
1) S4 + 3s3 + 8s2 + 4s +3 = 0
2) S4 + 9s3 + 4S2 – 36s -32 = 0
1) S4+3s3+8s2+4s+3=0
S4 | 1 | 8 | 3 |
S3 | 3 | 4 | 0 |
S2 | 6.66 | 3 |
|
S1 | 2.650 | 0 |
|
S0 | 3 |
|
|
No sign change in first column to no rows on right half of S-plane system stable.
S4+9S3+4S2-36S-32 = 0
Special case II of Routh Hurwitz criterion forming auxiliary equation
A1 (s) = 8S2 – 32 = 0
= 16S – 0 =0
S4 | 1 | 4 | -32 |
S3 | 9 | -36 | 0 |
S2 | 8 | -32 |
|
| 16 | 0 |
|
S0 | -32 |
|
|
One sign change so, one root lies on right half S-plane hence system is unstable.
Q4. For using feedback open loop transfer function G(s) = 
Find range of k for stability
Soln: Findlay characteristics equation .
CE = 1+G (s) H(s) = 0
H(s) =1 using feedback
CE = 1+ G(s)
1+ 
= 0
S(S+1)(S+3)(S+4)+k = 0
(S2+5)(S2+7Sα12)αK = 0
S4α7S3α1252+S3α7S2α125αK = 0
S4+8S3α19S2+125+k = 0
By Routh Hurwitz Criterion
S4 | 1 | 19 | K |
S3 | 8 | 12 | 0 |
S2 | 17.5 | K |
|
S1 |  | 0 |
|
S0 | k |
|
|
For system to be stable the range of K is 0< K < 
Q5. The characteristic equation for certain feedback control system is given. Find range of K for system to be stable.
Soln
S4+4S3α12S2+36SαK = 0
For stability K>0
> 0
K < 27
Range of K will be 0 < K < 27
Relative Stability:
Routh stability criterion deals about absolute stability of any closed loop system. For relative stability we need to shift the S-plane and the apply the Routh criterion.
Fig 14 Location of Pole for relative stability
The above fig 10 shows the characteristic equation is modified by shifting the origin of S-plane to S1= -
.
S = Z-S1
After substituting new valve of S =(Z-S1) applying Routh stability criterion, the number of sign changes in first column is the number of roots on right half of S-plane
Q6. Check if all roots of equation
S3+6S2+25S+38 = 0, have real poll more negative than -1.
Soln:-
No sign change in first column, hence all roots are in left half of S-plane.
Replacing S = Z-1. In above equation
(Z-1)3+6(Z-1)2+25(Z-1)+38 = 0
Z3+ Z23+16Z+18=0
No sign change in first column roots lie on left half of Z-plane hence all roots of original equation in S-domain lie to left half 0f S = -1
Key takeaway
Special Cases of Routh Hurwitz Criterions
- When first element of any row is zero.
In this case the zero is replaced by a very small positive number E and rest of the array is evaluated.
II) When any one row is having all its terms zero.
When array one row of Routh Hurwitz table is zero, it shown that the X is attests one pair of roots which lies radially opposite to each other in this case the array can be completed by auxiliary polynomial. It is the polynomial row first above row zero.
The root locus is a graphical produce for determining the stability of a control system which is determined by the location of the poles. The poles are nothing but the roots of the characteristic equation.
Features of Root Locus
- It is symmetrical to the real axis.
- Number of branches of root locus is equal to number of max (poles or zero).
- Starting point (k=0) End point(k—>∞)
(open loop poles) (open loop zeros)
4. A point on the real axis lies on the root locus if number of open loop poles or zero to the right side of that point is odd in number.
5. Value of K anywhere on the root locus is given as
K = 
6. I)If poles > zeros then (Þ-z) branches will terminate at ∞ (where k=∞)
II)If Z > P, then (Z-P) branches will start from ∞ (K = 0)
7. When P>Z, (P-Z) branches will terminate at ∞ (open loop zeros). But by which path. So the path is shown by asymptotes and this asymptotes is given by
Asymptote = 
q = o,1,2……(P-Z-1)
8. There asymptotes intersect the real axis at a single point and this point is known as centroid.
Centroid = 
9. Break away and break in point when root locus lie between two poles its called break in point.
Centroid and Breakaway points are not same
Differentials the characteristic equation and equate to zero 
10. Angle of arrival and angle of departure this print is used when the roots are complex.
Angle of departure - for complex poles
Angle of arrival – for complex zero.
11. Intersection of root locus with imaginary axis can be calculated by Routh Hurwitz. By calculating valve of k at intersection point (we can comment about system stability) so by knowing values of k at intersection point (imaginary axis) the valve of s at that point can also be calculated.
Absolutely stable: If the root low (all the branches) lies within the left side of S-plane.
Conditionally Stable: If some part of the root locus lies on the left half and same
Part on the right of S-plane then it is conditionally stable.
Unstable: If the root locus lies completely on right side of S-plane then it is unstable.
The values of S which satisfy both the angle and magnitude conditions are the roots of the characteristic equation.
Angle condition:
LG(S)H(S) = +-1800(2 KH) (K = 1,2,3,--)
If angle is odd multiple of 1800 it satisfies above condition.
Magnitude condition:
| G(S)H(S) = 1 | at any point on root locus. The magnitude condition can be applied only if angle condition is satisfied.
Dominant pole is a pole which is more near to origin than other poles in the system. The poles near to the imaginary axis are called the dominant poles. Or, get the closed-loop TF from Open loop TF. Determine the poles of the denominators. The poles which have very small real parts or near to the imaginary axis have small damping ratio. These poles are the dominant poles of the system. The dominant pole approximation is a method for approximating a high order system with a system of lower order if the location of the real part of some of the system poles are sufficiently close to the origin compared to the other poles. Dominant pole is significantly required in stability analysis, because it is that location which gives an idea where the root locus is progressing- towards right or towards left. It is also called near poles.
In fact, there is a region of boundary for considering the significant and insignificant region. If we considering a pole at σ1, then its insignificant pole, say σ2, must be 5 or 10 times far away from of that σ1
Fig 15 Location of poles and dominant pole region
Key takeaway
There is a region of boundary for considering the significant and insignificant region. If we considering a pole at σ1, then its insignificant pole, say σ2, must be 5 or 10 times far away from of that σ1
Effect of Addition of Poles:
1) The root locus is shifted towards imaginary axis.
2) The system becomes oscillatory.
3) The stability of system decreases.
4) The settling time increases.
5) The range of k reduces.
Effect of addition of zeros
1) The root locus shifts away from imaginary axis.
2) Stability of system increases.
3) The settling time decreases.
4) The gain margin increases.
5) The system becomes less oscillatory.
The below section will explain the above points clearly.
Q1. Sketch the root locus for given open loop transfer function G(S) = 
.
Soln:- 1) G(s) = 
Number of Zeros = 0
Number of polls S = (0, -1+j, -1-j) = (3).
1) Number of Branches = max (P, Z) = max (3, 0) = 3.
2) As there are no zeros in the system so, all branches terminate at infinity.
3) As P>Z, branches terminate at infinity through the path shown by asymptotes
Asymptote = 
× 180° q = 0, 1, 2………..(p-z-1)
P=3, Z=0.
q= 0, 1, 2.
For q=0
Asymptote = 1/3 × 180° = 60°
For q=1
Asymptote = 
× 180°
= 180°
For q=2
Asymptote = 
× 180° = 300°
Asymptotes = 60°,180°,300°.
4) Asymptote intersects real axis at centroid
Centroid =
= 
Centroid = -0.66
5) As poles are complex so angle of departure
øD = (2q+1)×180°+ø
ø = ∠Z –∠P.
Calculating ø for S=0
Join all the other poles with S=0
ø = ∠Z –∠P.
= 0-(315°+45°)
= -360°
ØD = (2q + 1)180 + ø.
= 180° - 360°
ØD = -180° (for q=0)
= 180° (for q=1)
=540° (for q=2)
Calculation ØD for pole at (-1+j)
ø = ∠Z –∠P.
= 0 –(135°+90°)
= -225°
ØD = (2q+1) 180°+ø.
= 180-225°
= -45°
ØD = -45° (for q = 0)
= 315° (for q = 1)
= 675° (for q =2)
6) The crossing point on imaginary axis can be calculated by Routh Hurwitz the characteristic equation is.
1+G(s) H (s) = 0
1+
S (S2+2s+2)+k = 0
S3+2s2+2s+K = 0
S3 | 1 | 1 | 0 |
S2 | 2 | K | 0 |
S1 |  | 0 |
|
S0 | K |
|
|
|
|
|
|
For stability 
> 0. And K > 0.
0<K<2.
So, when K=2 root locus crosses imaginary axis
S3 + 2S2 + 2S + 2 =0
For k
Sn-1 = 0 n : no. Of intersection
S2-1 = 0 at imaginary axis
S1 = 0
= 0
K<4
For Sn = 0 for valve of S at that K
S2 = 0
2S2 + K = 0
2S2 + 2 = 0
2(S2 +1) = 0
32 = -2
S = ± j 
The root locus plot is shown in figure 16.
Fig 16 Root Locus for G(S) = 
Q2. Sketch the root locus plot for the following open loop transfer function
G(s) = 
- Number of zero = 0, number of poles = 3
- As P>Z, branches will terminate at infinity
- There are no zeros so all branches will terminate at infinity.
- The path for the branches is shown by asymptote
Asymptote = 
×180°. q=0,1,………p-z-1
P=3, Z=0
q= 0,1,2.
For q = 0
Asymptote = 
× 180° = 60.
For q=1
Asymptote = 
× 180° = 180°
For q=2
Asymptote = 
× 180° = 300°
5. Asymptote intersect real axis at centroid
Centroid = 
= 
= -1
6. As root locus lies between poles S= 0, and S= -1
So, calculating breakaway point.
= 0
The characteristic equation is
1+ G(s) H (s) = 0.
1+ 
= 0
K = -(S3+3S2+2s)
= 3S2+6s+2 = 0
3s2+6s+2 = 0
S = -0.423, -1.577.
So, breakaway point is at S=-0.423
Because root locus is between S= 0 and S= -1
7. The intersection of root locus with imaginary axis is given by Routh criterion.
Characteristics equation is
S3+3S3+2s+K = 0
| 1 | 2 |
S2 | 3 | K |
S1 |  | 0 |
S0 | K |
|
For k
Sn-1= 0 n: no. Of intersection with imaginary axis
n=2
S1 = 0
= 0
K < 6 Valve of S at the above valve of K
Sn = 0
S2 = 0
3S2 + K =0
3S2 +6 = 0
S2 + 2 = 0
S = ± 
j
Fig 17 Root Locus for G(s) = 
The root locus plot is shown in fig. 17.
Q3. Plot the root locus for the given open loop transfer function
G(s) = 
- Number of zeros = 0 number of poles = 4
P = (S=0,-1,-1+j,-1-j) = 4
2. As P>Z all the branches will terminated at infinity.
3. As no zeros so all branches terminate at infinity.
4. The path for branches is shown by asymptote.
Asymptote = 
q = 0,1,…..(Þ-z-1)
q=0,1,2,3. (P-Z = 4-0)
For q=0
Asymptote = 
×180° =45°
For q=1
Asymptote = 
×180° =135°
For q=2
Asymptote = 
×180° =225°
For q=3
Asymptote = 
×180° =315°
5. Asymptote intersects real axis at unmarried
Centroid = 
Centroid = 
= 
= -0.75
6) As poles are complex so angle of departure is
ØD = (2q+1) ×180 + ø ø = ∠Z –∠P.
A) Calculating Ø for S=0
ø = ∠Z –∠P.
= 0 –[315° + 45°]
Ø = -360°
For q = 0
ØD = (2q+1) 180° + Ø
= 180 - 360°
ØD = -180°
b) Calculating Ø for S=-1+j
ø = ∠Z –∠P.
= 0-[135° + 90° + 90°]
Ø = -315°
For q=0
ØD= (2q+1) 180° +Ø
= 180° -315°
ØD = -135°
ØD for S=1+j will be ØD = 45°
7) As the root locus lie between S=0 and S=-1
So, the breakaway point is calculated
1+ G(s)H(s) = 0
1+ 
= 0
(S2+S)(S2 +2S+2) + K =0
K = -[S4+S3+2s3+2s2+2s2+2s]
= 4S3+9S2+8S+2=0
S = -0.39, -0.93, -0.93.
The breakaway point is at S = -0.39 as root locus exists between S= 0 and S=-1
8) Intersection of root locus with imaginary axis is given by Routh Hurwitz
I + G(s) H(s) = 0
K+S4+3S3+4S2+2S=0
S4 | 1 | 4 | K |
S3 | 3 | 2 |
|
S2 | 3.33 | K |
|
S1 |  |
|
|
S0 | K |
|
|
For system to be stable
>0
6.66>3K
0<K<2.22.
For K = 2.22
3.3352+K =0
3.3352 + 2.22 = 0
S2 = -0.66
S = ± j 0.816.
The root locus plot is shown in figure 18.
Fig 18 Root Locus for G(s) = 
Q4. Plot the root locus for open loop system
G(s) = 
1) Number of zero = 0 number of poles = 4 located at S=0, -2, -1+j, -1-j.
2) As no zeros are present so all branches terminated at infinity.
3) As P>Z, the path for branches is shown by asymptote
Asymptote = 
q = 0,1,2……p-z-1
For q = 0
Asymptote = 45°
q=1
Asymptote = 135°
q=2
Asymptote = 225°
q=3
Asymptote = 315°
4) Asymptote intersects real axis at centroid.
Centroid = 
= 
Centroid = -1.
5) As poles are complex so angle of departure is
ØD=(2q+1)180° + Ø
ø = ∠Z –∠P
= 0-[135°+45°+90°]
= 180°- 270°
ØD = -90°
6) As root locus lies between two poles so calculating point. The characteristic equation is
1+ G(s)H(s) = 0
1+
= 0.
K = -[S4+2S3+2S2+2S3+4S2+4S]
K = -[S4+4S3+6S2+4S]
= 0
= 4s3+12s2+12s+4=0
S = -1
So, breakaway point is at S = -1
7) Intersection of root locus with imaginary axis is given by Routh Hurwitz.
S4+4S3+6S2+4s+K = 0
S4 | 1 | 6 | K |
S3 | 4 | 4 |
|
S2 | 5 | K |
|
S1 |  |
|
|
S0 | K |
|
|
≤ 0
K≤5.
For K=5 valve of S will be.
5S2+K = 0
5S2+5 = 0
S2 +1 = 0
S2 = -1
S = ±j.
The root locus is shown in figure 19.
Fig 19 Root Locus For G(s) = 
Q5. Plot the root locus for open loop transfer function G(s) = 
- Number of zeros = 0. Number of poles = 4 located at S=0, -3, -1+j, -1-j.
- As no. Zero so all branches terminate at infinity.
- The asymptote shows the both to the branches terminating at infinity.
Asymptote = 
q=0,1,….(p-z).
For q = 0
Asymptote = 45
For q = 1
Asymptote = 135
For q = 2
Asymptote = 225
For q = 3
Asymptote = 315
(4). The asymptote intersects real axis at centroid.
Centroid = ∑Real part of poles - ∑Real part of zero / P – Z
= [-3-1-1] – 0 / 4 – 0
Centroid = -1.25
(5). As poles are complex so angle of departure
φD = (29 + 1)180 + φ
ø = ∠Z –∠P.
= 0 – [ 135 + 26.5 + 90 ]
= -251.56
For q = 0
φD = (29 + 1)180 + φ
= 180 – 215.5
φD = - 71.56
(6). Break away point dk / ds = 0 is at S = -2.28.
(7). The intersection of root locus on imaginary axis is given by Routh Hurwitz.
1 + G(S)H(S) = 0
K + S4 + 3S3 + 2S3 + 6S2 + 2S2 + 6S = 0
S4 1 8 K
S3 5 6
S2 34/5 K
S1 40.8 – 5K/6.8
K ≤ 8.16
For K = 8.16 value of S will be
6.8 S2 + K = 0
6.8 S2 + 8.16 = 0
S2 = - 1.2
S = ± j1.09
The plot is shown in figure 20.
Fig 20 Root Locus for G(s) = 
Q.6. Sketch the root locus for open loop transfer function.
G(S) = K(S + 6)/S(S + 4)
- Number of zeros = 1(S = -6)
Number of poles = 2(S = 0, -4)
2. As P > Z one branch will terminate at infinity and the other at S = -6.
3. For Break away and breaking point
1 + G(S)H(S) = 0
1 + K(S + 6)/S(S + 4) = 0
Dk/ds = 0
S2 + 12S + 24 = 0
S = -9.5, -2.5
Breakaway point is at -2.5 and Break in point is at -9.5.
4. Root locus will be in the form of a circle. So finding the centre and radius. Let S = + jw.
G( + jw) = K( + jw + 6)/( + jw)( + jw + 4 ) = +- π
Tan-1 w/ + 6 - tan-1 w/ – tan-1 w / + 4 = - π
Taking tan of both sides.
w/ + w/ + 4 / 1 – w/ w/ + 4 = tan π + w / + 6 / 1 - tan π w/ + 6
w/ + w/ + 4 = w/ + 6[ 1 – w2 / ( + 4) ]
(2 + 4)( + 6) = (2 + 4 – w2)
2 2 + 12 + 4 + 24 = 2 + 4 – w2
22 + 12 + 24 = 2 – w2
2 + 12 – w2 + 24 = 0
Adding 36 on both sides
( + 6)2 + (w + 0)2 = 12
The above equation shows circle with radius 3.46 and center(-6, 0) the plot is shown in figure 21.
Fig 21 Root locus for G(S) = K(S + 6)/S(S + 4)
References:
1. Benjamin C. Kuo, “Automatic control systems”, Prentice Hall of India, 7th Edition.
2. M. Gopal, “Control System – Principles and Design”, Tata McGraw Hill, 4th Edition.
3. N. J. Nagrath and M. Gopal, “Control System Engineering”, New Age International Publishers, 5th Edition.
4. K. Ogata, “Modern Control Engineering”, Prentice Hall India Learning Private Limited; 5th Edition.
