System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 11, Problem 11.13P
To determine

(a)

The root locus of the given characteristic equation s(s+5)+K=0.

Expert Solution
Check Mark

Explanation of Solution

Given:

1+G(s)H(s)=s(s+5)+KforK0.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as 1+G(s)H(s)=0

1+G(s)H(s)=s(s+5)+KforK0

On solving we get,

s2+5s+K=0

Where,

G(s)=Ks2+5sH(s)=1

Therefore, Poles are

s2+5ss(s+5)s=0,5

P=2

And zero is Z=0

Total number of branches is

N=PZN=20N=2

Centroid is calculated as:

σ=(Realpartofopenlooppoles)(Realpartofopenloopzeros)(PZ)

Where poles are s=0,5 and zero is Z=0

σ=(05)(0)(20)σ=2.5

Angle of asymptotes is calculated as:

Angle of asymptotes

θk=(2k+1)180PZ

Where P = no. of poles

Z = No. of zeros

P=2Z=0k=PZ1

k=0,1

On replacing the values

θ0=(2×0+1)18020=90°θ1=(2×1+1)18020=270°

The PZ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

1+G(s)H(s)=01+G(s)H(s)=s(s+5)+K=0s2+5s+K=0K=s25sdKds=2s5

To calculate breakaway point, replace dKds=0

So, 2s5=0s=52s=2.5

Breakaway point is s=2.5

Calculating the value of K to determine the roots of the equation

s2+5s+K=0s21Ks150s05K1×05=K0

For Stability K>0

Auxiliary equation is defined as:

s2+K=0

Replace the value of K the roots of the equation are calculated

s2+0=0s=0

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

System Dynamics, Chapter 11, Problem 11.13P , additional homework tip  1

Fig.1

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.1.

To determine

(b)

The root locus of the given characteristic equation s2+3s+3+K(s+3)=0.

Expert Solution
Check Mark

Explanation of Solution

Given:

1+G(s)H(s)=s2+3s+3+K(s+3)forK0.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as 1+G(s)H(s)=0

1+G(s)H(s)=s2+3s+3+K(s+3)forK0

Where,

G(s)=K(s+3)s2+3s+3H(s)=1

Therefore, Poles are

s2+3s+3=0s=1.5±0.87i

P=2

And zero is s+3=0s=3Z=1

Total number of branches is

N=PZN=21N=1

Centroid is calculated as:

σ=(Realpartofopenlooppoles)(Realpartofopenloopzeros)(PZ)

Where poles are s=1.5±1.66i and zero is s=3

σ=(0+(1.51.5)(3))(21)σ=0

Angle of asymptotes is calculated as:

Angle of asymptotes

θk=(2k+1)180PZ

Where P = no. of poles

Z = No. of zeros

P=2Z=1k=PZ1

k=0

On replacing the values

θ0=(2×0+1)18021=180°

The PZ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

1+G(s)H(s)=01+G(s)H(s)=s2+3s+3+K(s+3)=0K(s+3)=s23s3K=s23s3(s+3)dKds=s2+6s+6(s+3)2

To calculate breakaway point, replace dKds=0

So,

s2+6s+6s1=1.27s2=4.73

Calculating the value of K to determine the roots of the equation

s2+(3+K)s+(3+3K)=0s21(3+3K)s1(3+K)0s0(3+3K)0

For Stability K>0

K=3,1

Auxiliary equation is defined as:

s2+3+3K=0

Replace the value of K the roots of the equation are calculated

For K=1s=0

For K=3s=±2.45 neglected

This is the point on the imaginary axis

Angle of departure is calculated where there are either poles or zero is imaginary.

ΦD=180+(ΦzΦp)

Φp = Sum of angles from poles

Φz = Sum of angles from zeros

Φ=ΦzΦpΦp=90Φz=0Φ=90ΦD=180+ΦΦD=90

Root locus plot of the characteristic equation is

System Dynamics, Chapter 11, Problem 11.13P , additional homework tip  2

Fig.2

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.2.

To determine

(c)

The root locus of the given characteristic equation s(s2+3s+3)+K=0.

Expert Solution
Check Mark

Explanation of Solution

Given:

1+G(s)H(s)=s(s2+3s+3)+KforK0.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as 1+G(s)H(s)=0

1+G(s)H(s)=s(s2+3s+3)+KforK0

Where,

G(s)=Ks(s2+3s+3)H(s)=1

Therefore, Poles are

s(s2+3s+3)=0s=0,1.5±0.87i

P=3

And zero is Z=0

Total number of branches is

N=PZN=30N=3

Centroid is calculated as:

σ=(Realpartofopenlooppoles)(Realpartofopenloopzeros)(PZ)

Where poles are s=0,1.5±1.66i and zero is Z=0

σ=(01.51.5)(0)(30)σ=1

Angle of asymptotes is calculated as:

Angle of asymptotes

θk=(2k+1)180PZ

Where P = no. of poles

Z = No. of zeros

P=3Z=0k=PZ1

k=0,1,2

On replacing the values

θ0=(2×0+1)18030=60°θ1=(2×1+1)18030=180°θ2=(2×2+1)18030=300°

The PZ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

1+G(s)H(s)=01+G(s)H(s)=s(s2+3s+3)+K=0s(s2+3s+3)+K=0K=s(s2+3s+3)K=s33s23sdKds=3s26s3

To calculate breakaway point, replace dKds=0

So,

3s26s3=03s2+6s+3=0s2+2s+1=0s=1,1

Breakaway point is s=1

Calculating the value of K to determine the roots of the equation:

s3+3s2+3s+K=0s313s23Ks13×3K30s0K0

For Stability K>0

9K3=0K=9

Auxiliary equation is defined as:

3s2+K=0

Replace the value of K the roots of the equation are calculated

3s2+K=0K=9s2=93=3s=±1.732i

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

System Dynamics, Chapter 11, Problem 11.13P , additional homework tip  3

Fig.3

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.3.

To determine

(d)

The root locus of the given characteristic equation s(s+5)(s+7)+K=0.

Expert Solution
Check Mark

Explanation of Solution

Given:

1+G(s)H(s)=s(s+5)(s+7)+KforK0.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as 1+G(s)H(s)=0

1+G(s)H(s)=s(s+5)(s+7)+KforK0

On solving we get,

s(s+5)(s+7)+K=0(s2+7s)(s+5)+K=0s3+13s2+35s+K=0

Therefore, Poles are

s(s+5)(s+7)=0s=0,5,7

P=3

And zero is Z=0

Total number of branches is

N=PZN=30N=3

Centroid is calculated as:

σ=(Realpartofopenlooppoles)(Realpartofopenloopzeros)(PZ)

Where poles are s=0,5,7 and zero is Z=0

σ=(057)(0)(30)σ=4

Angle of asymptotes is calculated as:

Angle of asymptotes

θk=(2k+1)180PZ

Where P = no. of poles

Z = No. of zeros

P=3Z=0k=PZ1

k=0,1,2

On replacing the values

θ0=(2×0+1)18030=60°θ1=(2×1+1)18030=180°θ2=(2×2+1)18030=300°

The PZ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

1+G(s)H(s)=01+G(s)H(s)=s(s+5)(s+7)+K=01+G(s)H(s)=s(s+5)(s+7)+K=0s(s+5)(s+7)+K=0(s2+7s)(s+5)+K=0s3+12s2+35s+K=0K=s312s235sdKds=3s224s35

To calculate breakaway point, replace dKds=0

So,

3s224s35=03s2+24s+35=0s1=1.92s2=6.1

Breakaway point is s=1.92

Calculating the value of K to determine the roots of the equation

s3+12s2+35s+K=0s3135s212Ks112×35K120s0K0

For Stability K>0

12×35K12=0K=420

Auxiliary equation is defined as:

12s2+K=0

Replace the value of K the roots of the equation are calculated

12s2+420=0s=±5.92i

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

System Dynamics, Chapter 11, Problem 11.13P , additional homework tip  4

Fig.4

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig4.

To determine

(e)

The root locus of the given characteristic equation s(s+3)+K(s+4)=0.

Expert Solution
Check Mark

Explanation of Solution

Given:

1+G(s)H(s)=s(s+3)+K(s+4)forK0.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as 1+G(s)H(s)=0

1+G(s)H(s)=s(s+3)+K(s+4)forK0

Where,

G(s)=K(s+4)s(s+3)H(s)=1

Therefore, Poles are

s(s+3)s=0,3

P=2

And zero is s+4=0s=4Z=1

Total number of branches is

N=PZN=21N=1

Centroid is calculated as:

σ=(Realpartofopenlooppoles)(Realpartofopenloopzeros)(PZ)

Where poles are s=0,3 and zero is s=4

σ=(03)(4)(21)σ=1

Angle of asymptotes is calculated as:

Angle of asymptotes

θk=(2k+1)180PZ

Where P = no. of poles

Z = No. of zeros

P=2Z=1k=PZ1

k=0

On replacing the values

θ0=(2×0+1)18021=180°

The PZ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

1+G(s)H(s)=01+G(s)H(s)=s(s+3)+K(s+4)=0s2+3s+K(s+4)=0K=s23s(s+4)dKds=s28s12(s+4)2

To calculate breakaway point, replace dKds=0

So,

s28s12=0s2+8s+12=0s=2s=6

Breakaway point is s=2

Calculating the value of K to determine the roots of the equation

s2+(3+K)s+4K=0s214Ks1(3+K)0s0(3+K)4K1×0(3+K)=4K0

For Stability K>0

(3+K)=0K=3 & 4K=0K=0

Auxiliary equation is defined as:

s2+4K=0

Replace the value of K=3 the roots of the equation are calculated

s24×3=0s=12s=±3.46

This is the point on the imaginary axis

Root locus plot of the characteristic equation is

System Dynamics, Chapter 11, Problem 11.13P , additional homework tip  5

Fig.5

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig.5.

To determine

(f)

The root locus of the given characteristic equation s(s+6)+K(s4)=0.

Expert Solution
Check Mark

Explanation of Solution

Given:

1+G(s)H(s)=s(s+6)+K(s4)=0.

Concept Used:

Root Locus technique.

Calculation:

The characteristic equation is defined as 1+G(s)H(s)=0

1+G(s)H(s)=s(s+6)+K(s4)=0

Where,

G(s)=K(s4)s(s+6)H(s)=1

Therefore, Poles are

s(s+6)=0s=0,6

P=2

And zero is

s4=0s=4Z=1

Total number of branches is

N=PZN=21N=1

Centroid is calculated as:

σ=(Realpartofopenlooppoles)(Realpartofopenloopzeros)(PZ)

Where poles are s=0,6 and zero is s=4

σ=(06)(4)(21)σ=10

Angle of asymptotes is calculated as:

Angle of asymptotes

θk=(2k+1)180PZ

Where P = no. of poles

Z = No. of zeros

P=2Z=1k=PZ1

k=0

On replacing the values

θ0=(2×0+1)18021=180°

The PZ branches will terminate at infinity along certain lines known as asymptotes of root locus.

Breakaway point is calculated as:

1+G(s)H(s)=01+G(s)H(s)=s(s+6)+K(s4)=0K(s4)=s26sK=s26s(s4)dKds=s2+8s+24(s4)2

To calculate breakaway point, replace dKds=0

So,

s2+8s+24(s4)2=0s2+8s+24=0s28s24=0s=10.32,2.33

Breakaway point is s=2.33

Calculating the value of K to determine the roots of the equation

s(s+6)+K(s4)=0s2+(6+K)s4K=0s214Ks1(6+K)0s04K0

For Stability K>0

6+K=0K=64K=0K=0

Auxiliary equation is defined as:

s24K=0

Replace the value of K the roots of the equation are calculated

For

K=6s=±4.89i

For

K=0s=0 neglected

Root locus plot of the characteristic equation is

System Dynamics, Chapter 11, Problem 11.13P , additional homework tip  6

Fig.6

Conclusion:

Root has been plotted for the given characteristic equation is shown in Fig6.

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