System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
bartleby

Concept explainers

Question
Book Icon
Chapter 6, Problem 6.40P
To determine

The effect of the damping constant c on the armature-controlled motor’s characteristic roots and on its speed response to a step voltage input. Also, to find the time taken by the motor to attain constant speed and check whether the oscillations will occur in the motor response or not.

Expert Solution & Answer
Check Mark

Answer to Problem 6.40P

The obtained characteristic roots for the various values of damping constant are as follows:

For c=0Nms/rad, s=25(10616)3,s=25(16+106)3

For c=0.01Nms/rad, s=11756+i253116,s=11756i253116

For c=0.1Nms/rad, s=25(333191)3,s=25(91+3331)3

Thus, we found that for the values of damping constant, c=0,0.1Nms/rad the roots lie on left side of the imaginary axis in s-plane. And, these roots are real and distinct in nature. While, for damping constant, c=0.01Nms/rad the roots are also on the left of the imaginary axis in s-plane, but these are complex in nature.

The obtained response to step voltage input for the various values of damping constant are as follows:

For c=0Nms/rad, ω(t)=20e133.33tcosh(85.797t)31.08e133.33tsinh(85.797t)+20

For c=0.01Nms/rad, ω(t)=4.762e195.833tcos(73.479t)12.691e195.833tsin(73.479t)+10021

For c=0.1Nms/rad, ω(t)=0.606e758.33tcosh(480.957t)0.956e758.33tsinh(480.957t)+2033

Here, for the values of damping constant, c=0,0.1Nms/rad the speed responses are stable and free from oscillations. While, for the damping constant, c=0.01Nms/rad, the response is oscillatory and finally settles for a steady-state in future that is stable.

The time taken for the oscillations to decay for various values of damping constants is as follows:

For c=0Nms/rad, there won’t be any oscillations as the response includes hyperbolic functions. However, the response reaches the steady state at t=4τ=4133.33=0.03seconds.

For c=0.01Nms/rad, the time taken by the oscillations to decay is t=4τ=4195.833=0.02seconds.

For c=0.1Nms/rad, there won’t be any oscillations as the response includes hyperbolic functions. However, the response reaches the steady state at t=4τ=4758.33=0.005seconds.

Explanation of Solution

Given information:

The given parameters for the armature-controlled motor are as follows:

KT=Kb=0.05Nm/A,La=3×103H,I=8×105kgm2,Ra=0.8Ω

Also, the various values of damping constants are as follows:

c=0Nms/rad,0.01Nms/rad,0.1Nms/rad.

Concept Used:

The motor speed response with armature voltage as input is as follows:

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKT.

Calculation:

For c=0Nms/rad

Since,

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the values of all the known parameters such that

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKTΩ(s)Va(s)=0.053×103×8×105×s2+(0.8×8×105+0)s+0+0.05×0.05

Ω(s)Va(s)=0.0524×108s2+6.4×105s+25×104Ω(s)Va(s)=14.8×106s2+1.28×103s+5×102

For the unit step input voltage response, we have

Ω(s)Va(s)=14.8×106s2+1.28×103s+5×102Ω(s)=1s(4.8×106s2+1.28×103s+5×102)Ω(s)=106s(4.8s2+1280s+50000)

On taking partial fraction expansion for the above expression, we have

Ω(s)=1000000s(4.8s2+1280s+50000)=As+Bs+C4.8s2+1280s+500001000000s(4.8s2+1280s+50000)=A(4.8s2+1280s+50000)+s(Bs+C)s(4.8s2+1280s+50000)

On comparing the numerator of both sides

1000000=A(4.8s2+1280s+50000)+s(Bs+C) A=20,B=96,C=25600

Therefore,

Ω(s)=96s256004.8s2+1280s+50000+20sΩ(s)=20(s+133.33)((s+133.33)27361.11)2666.671((s+133.33)27361.11)+20s

On taking the inverse Laplace transform of the above obtained expression, we get

ω(t)=20e133.33tcosh(85.797t)31.08e133.33tsinh(85.797t)+20

That shows the response doesn’t contain any oscillations. However, the steady-state for the response could be obtained at t=4τ=4133.33=0.03seconds.

The characteristic polynomial for the system is as follows:

LaIs2+(RaI+cLa)s+cRa+KbKT=0

On keeping the values, we have

LaIs2+(RaI+cLa)s+cRa+KbKT=03×103×8×105s2+(0.8×8×105+0)s+0+0.05×0.05=024×108s2+6.4×105s+25×104=024s2+6400s+250000=03s2+800s+31250=0

Thus, the roots are as follows:

s=25(10616)3,s=25(16+106)3

For c=0.01Nms/rad

Since,

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the values of all the known parameters such that

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKTΩ(s)Va(s)=0.053×103×8×105s2+(0.8×8×105+0.01×3×103)s+0.01×0.8+0.05×0.05Ω(s)Va(s)=0.0524×108s2+(6.4×105+3×105)s+105×104Ω(s)Va(s)=0.0524×108s2+9.4×105s+105×104

Ω(s)Va(s)=14.8×106s2+1.88×103s+21×102

For the unit step input voltage response, we have

Ω(s)Va(s)=14.8×106s2+1.88×103s+21×102Ω(s)=1s(4.8×106s2+1.88×103s+21×102)Ω(s)=106s(4.8s2+1880s+210000)

On taking partial fraction expansion for the above expression, we have

Ω(s)=106s(4.8s2+1880s+210000)=As+Bs+C4.8s2+1880s+210000106s(4.8s2+1880s+210000)=A(4.8s2+1880s+210000)+s(Bs+C)s(4.8s2+1880s+210000)

On comparing the numerator of both sides

1000000=A(4.8s2+1880s+210000)+s(Bs+C) A=10021,B=22.857,C=8952.381

Therefore,

Ω(s)=22.857s8952.3814.8s2+1880s+210000+10021s

Ω(s)=4.762(s+195.833)((s+195.833)2+5399.31)932.5461((s+195.833)2+5399.31)+10021s

On taking the inverse Laplace transform of the above obtained expression, we get

ω(t)=4.762e195.833tcos(73.479t)12.691e195.833tsin(73.479t)+10021

That shows the response does contain oscillations. Thus, the time taken by the oscillations to decay is t=4τ=4195.833=0.02seconds.

The characteristic polynomial for the system is as follows:

LaIs2+(RaI+cLa)s+cRa+KbKT=0

On keeping the values, we have

LaIs2+(RaI+cLa)s+cRa+KbKT=0t=4τ=4195.833=0.02seconds3×103×8×105s2+(0.8×8×105+0.01×3×103)s+0.01×0.8+0.05×0.05=024×108s2+9.4×105s+105×104=024s2+9400s+1050000=03s2+1175s+131250=0

Thus, the roots are as follows:

s=11756+i253116,s=11756i253116

For c=0.1Nms/rad

Since,

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the values of all the known parameters such that

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKT

Ω(s)Va(s)=0.053×103×8×105s2+(0.8×8×105+0.1×3×103)s+0.1×0.8+0.05×0.05Ω(s)Va(s)=0.0524×108s2+(6.4×105+30×105)s+825×104Ω(s)Va(s)=0.0524×108s2+36.4×105s+825×104

Ω(s)Va(s)=14.8×106s2+7.28×103s+165×102

For the unit step input voltage response, we have

Ω(s)Va(s)=14.8×106s2+7.28×103s+165×102Ω(s)=1s(4.8×106s2+7.28×103s+165×102)Ω(s)=106s(4.8s2+7280s+1650000)

On taking partial fraction expansion for the above expression, we have

Ω(s)=1000000s(4.8s2+7280s+1650000)=As+Bs+C4.8s2+7280s+16500001000000s(4.8s2+7280s+1650000)=A(4.8s2+7280s+1650000)+s(Bs+C)s(4.8s2+7280s+1650000)

On comparing the numerator of both sides

1000000=A(4.8s2+7280s+1650000)+s(Bs+C) A=2033,B=-2.909andC=-4412.121

Therefore,

Ω(s)=2.909s4412.1214.8s2+7280s+1650000+2033s

Ω(s)=0.606(s+758.33)((s+758.33)2231319.444)459.611((s+758.33)2231319.444)+2033s

On taking the inverse Laplace transform of the above obtained expression, we get

ω(t)=0.606e758.33tcosh(480.957t)0.956e758.33tsinh(480.957t)+2033

That shows the response doesn’t contain any oscillations. However, the steady-state for the response could be obtained at t=4τ=4758.33=0.005seconds.

The characteristic polynomial for the system is as follows:

LaIs2+(RaI+cLa)s+cRa+KbKT=0

On keeping the values, we have

LaIs2+(RaI+cLa)s+cRa+KbKT=03×103×8×105s2+(0.8×8×105+0.1×3×103)s+0.1×0.8+0.05×0.05=024×108s2+36.4×105s+825×104=024s2+36400s+8250000=03s2+4550s+1031250=0

Thus, the roots are as follows:

s=25(333191)3,s=25(91+3331)3.

Conclusion:

The obtained characteristic roots for the various values of damping constant are as follows:

For c=0Nms/rad, s=25(10616)3,s=25(16+106)3

For c=0.01Nms/rad, s=11756+i253116,s=11756i253116

For c=0.1Nms/rad, s=25(333191)3,s=25(91+3331)3

Thus, we found that for the values of damping constant, c=0or0.1Nms/rad the roots lie on the left side of the imaginary axis in s-plane. And, these roots are real and distinct in nature. While, for damping constant, c=0.01Nms/rad the roots are also on the left of the imaginary axis in s-plane, but these are complex in nature.

The obtained response to step voltage input for the various values of damping constant are as follows:

For c=0Nms/rad, ω(t)=20e133.33tcosh(85.797t)31.08e133.33tsinh(85.797t)+20

For c=0.01Nms/rad,

ω(t)=4.762e195.833tcos(73.479t)12.691e195.833tsin(73.479t)+10021

For c=0.1Nms/rad,

ω(t)=0.606e758.33tcosh(480.957t)0.956e758.33tsinh(480.957t)+2033

Here, for the values of damping constant, c=0or0.1Nms/rad the speed responses are stable and free from oscillations. While, for the damping constant, c=0.01Nms/rad the response is oscillatory and finally settles for a steady-state in future that is stable.

The time taken for the oscillations to decay for various values of damping constants is as follows:

For c=0Nms/rad, there won’t be any oscillations as the response includes hyperbolic functions. However, the response reaches the steady state at t=4τ=4133.33=0.03seconds.

For c=0.01Nms/rad, the time taken by the oscillations to decay is t=4τ=4195.833=0.02seconds.

For c=0.1Nms/rad, there won’t be any oscillations as the response includes hyperbolic functions. However, the response reaches the steady state at t=4τ=4758.33=0.005seconds.

Thus, we see that for values of damping constants c=0,0.1(Nms/rad), the characteristic roots are real and lie on the left-hand side of the s-plane. While for the damping constant c=0.01Nms/rad, the characteristic roots are complex in nature.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 6 Solutions

System Dynamics

Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY