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