Unit-4
Stability of Linear Control Systems
The BIBO system means that bounded input yields bounded output. The system which has input bounded for all time will give bounded output for all time.
A continuous-time signal is one that is present for all time.
x(t)=cosωt
The functions that are defined for every value of time is defined as a continuous-time signal. The time is taken between the limits because the value between infinity at some time is not measurable.
Fig 1 continuous-time signal
The pole's location and the asymptotically stable system is shown below.
Fig 2 Location of poles for an Asymptotically stable system
The system is asymptotically stable if all its poles have a negative real part. If the system has no repeated roots than the transfer function can be written as
G(s) = m(s)/(s-p1)(s-p2)….(s-pm)
Where the denominator represents the poles. The mathematical proof is shown below.
The poles can be represented as pk=
k+jwk
kt = 
= 
= -1/
for 
<0
= 
for 
>0
As every pole has 
<0 hence the system is asymptotically stable as required.
Key takeaway
As poles are asymptotically stable so there cannot be any pole on the imaginary axis or the right half of the s-plane.
It states that the system is stable if and only if all the elements in 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 the first column equals the number of roots of the characteristic equation in the right half of the 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 is 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 the first column so all roots are in the left half of the S-plane and hence the system is stable.
Special Cases of Routh Hurwitz Criterions
In this case, the zero is replaced by a very small positive number E, and the rest of the array is evaluated.
Eg.(1) Consider the following equation
S3+S+2 = 0
Now when E 
0, values in column 1 becomes
Two sign changes hence two roots on the right side of S-plans
II) When anyone row is having all its terms zero.
When array one row of Routh Hurwitz table is zero, it is shown that the X is attested one pair of roots which lies radially opposite to each other in this case the array can be completed by the auxiliary polynomial. It is the polynomial row first above row zero.
Consider the 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
| 1 | 6 |
S2 | 5 | 30 |
S1 | 10 |
|
S0 | 30 |
|
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 an imaginary axis so the 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
No sign change in the first column to no rows on the right half of the S-plane system stable.
S4+9S3+4S2-36S-32 = 0
Special case II of Routh Hurwitz criterion forming an auxiliary equation
A1 (s) = 8S2 – 32 = 0
= 16S – 0 =0
One sign changes so, one root lies on the right half S-plane hence system is unstable.
Q4. For using feedback open-loop transfer function G(s) = 
find a 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
For the system to be stable the range of K is 0< K <
.
Q5. The characteristic equation for a certain feedback control system is given. Find a range of K for the 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 with the absolute stability of any closed-loop system. For relative stability, we need to shift the S-plane and apply the Routh criterion.
Fig 2 Location of Pole for relative stability
The above fig 2 shows the characteristic equation is modified by shifting the origin of the S-plane to S1= -
.
S = Z-S1
After substituting new valve of S =(Z-S1) applying the Routh stability criterion, the number of sign changes in the first column is the number of roots on the right half of the S-plane
Q6. Check if all roots of the equation
S3+6S2+25S+38 = 0, have real poll more negative than -1.
Soln:-
No sign change in the first column, hence all roots are in the left half of the S-plane.
Replacing S = Z-1. In the above equation
(Z-1)3+6(Z-1)2+25(Z-1)+38 = 0
Z3+ Z23+16Z+18=0
Z3 | 1 | 16 |
Z2 | 3 | 18 |
Z1 | 10 |
|
Z0 | 18 |
|
No sign change in first column roots lie on the left half of Z-plane hence all roots of the original equation in S-domain lie to left half 0f S = -1
Key takeaway
Special Cases of Routh Hurwitz Criterions
In this case, the zero is replaced by a very small positive number E, and the rest of the array is evaluated.
II) When anyone row is having all its terms zero.
When array one row of Routh Hurwitz table is zero, it is shown that the X is attested one pair of roots which lies radially opposite to each other in this case the array can be completed by the auxiliary polynomial. It is the polynomial row first above row zero.
4.5.1 Introduction
The root locus is 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.
4.5.2 Properties of Root Locus
(open-loop poles) (open loop zeros)
4. A point on the real axis lies on the root locus if many 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 asymptote is given by
Asymptote = 
q = o,1,2……(P-Z-1)
8. These asymptotes intersect the real axis at a single point and this point is known as centroid.
Centroid = 
9. Breakaway and break-in point when the root locus lies between two poles it's called break-in point.
Centroid and Breakaway points are not the same
Differentials the characteristic equation and equate to zero 
10. The angle of arrival and angle of departure this print is used when the roots are complex.
The angle of departure - for complex poles
The angle of arrival – for complex zero.
11. The intersection of root locus with an imaginary axis can be calculated by Routh Hurwitz. By calculating the value of k at intersection point (we can comment about system stability) so by knowing the values of k at intersection point (imaginary axis) the valve of s at that point can also be calculated.
Stable: If the root low (all the branches) lies within the left side of the S-plane.
Conditionally Stable:- If some part of the root locus lies on the left half and the same
The part on the right of the S-plane then is conditionally stable.
Unstable:- If the root locus lies completely on the right side of the 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 the angle is an odd multiple of 1800 it satisfies the above condition.
Magnitude condition:-
| G(S)H(S) = 1 | at any point on the root locus. The magnitude condition can be applied only if the angle condition is satisfied.
4.5.3 Design aspects of Root Locus
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 the real axis at the centroid
Centroid =
= 
Centroid = -0.66
5) As poles are complex so the 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 the 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
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 1.
Q2. Sketch the root locus plot for the following open-loop transfer function
G(s) = 
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 intersects the real axis at the centroid
Centroid = 
= 
= -1
6. As root locus lies between poles S= 0, and S= -1
So, calculating the 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, the breakaway point is at S=-0.423
because root locus is between S= 0 and S= -1
7. The intersection of the root locus with the imaginary axis is given by the Routh criterion.
Characteristics equation is
S3+3S3+2s+K = 0
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
The root locus plot is shown in fig. 2.
Q3. Plot the root locus for the given open-loop transfer function
G(s) = 
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) The intersection of root locus with the imaginary axis is given by Routh Hurwitz
I + G(s) H(s) = 0
K+S4+3S3+4S2+2S=0
For the 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 3.
Figure-3
Q4. Plot the root locus for the 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 are 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 the real axis at the centroid.
Centroid = 
= 
Centroid = -1.
5) As poles are complex so the 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, the breakaway point is at S = -1
7) The intersection of root locus with the imaginary axis is given by Routh Hurwitz.
S4+4S3+6S2+4s+K = 0
≤ 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 4.
Figure-4
Q5. Plot the root locus for the open-loop transfer function G(s) = 
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 the 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 5.
Figure-5
Q.6. Sketch the root locus for open loop transfer function.
G(S) = K(S + 6)/S(S + 4)
Number of poles = 2(S = 0, -4)
2. As P > Zone 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.6.
Figure-6
Reference:
1. I. J. Nagrath& M. Gopal, “Control System Engineering”, New Age International Publishers
2. A. Ambikapathy, Control Systems, Khanna Publishing House, Delhi.
2. Joseph J. Distefano III, Allen R. Stubberud, Ivan J. Williams, “Control Systems” Schaums Outlines Series, 3rdEdition, Tata McGraw Hill, Special Indian Edition 2010.
3. William A. Wolovich, “Automatic Control Systems”, Oxford University Press, 2010.
