Unit-4
Stability of Linear Control Systems
Q1. Sketch the root locus for given open-loop transfer function G(S) = .
Sol:- 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 the 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 the real axis at the 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 the open-loop transfer function.
G(S) = K(S + 6)/S(S + 4)
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 center 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
Q7). 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.
Q8. For using feedback open-loop transfer function G(s) =
find a range of k for stability
Sol:- The 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 < .
Q9. 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
Q10) Check if all roots of 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
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