System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 6, Problem 6.41P
To determine

(a)

The step response of ia(t) and ω(t) for the armature voltage va=10V.

Expert Solution
Check Mark

Answer to Problem 6.41P

The responses are as follows:

ia(t)=0.125e201t2cos(40399t2)+2487.6875240399e201t2sin(40399t2)+25202

ω(t)=5000101e201t2cos(40399t2)100500010140399e201t2sin(40399t2)+5000101.

Explanation of Solution

Given:

The given parameters for an armature-controlled motor are as:

Kb=KT=0.2Nm/A, c=5×104Nms/rad, Ra=0.8Ω, La=4×103H, I=5×104kgm2 and va=10V.

Concept Used:

For an armature-controlled motor, the output responses armature current ia(t) and motor speed ω(t) is related to input response armature voltage va as follows:

Ia(s)Va(s)=Is+cLaIs2+(RaI+cLa)s+cRa+KbKT

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

Calculation:

Since,

Ia(s)Va(s)=Is+cLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the values of all the parameters, we get

Ia(s)Va(s)=Is+cLaIs2+(RaI+cLa)s+cRa+KbKTIa(s)Va(s)=5×104s+5×1044×103×5×104s2+(0.8×5×104+5×104×4×103)s+5×104×0.8+0.2×0.2Ia(s)Va(s)=5×104(s+1)2×106s2+(4×104+2×106)s+4×104+0.04Ia(s)Va(s)=(s+1)0.004s2+0.804s+80.8

Now, since input armature voltage is a unit step response of strength 10V.

Therefore,

Ia(s)=(s+1)(0.004s2+0.804s+80.8)Va(s)Ia(s)=(s+1)(0.004s2+0.804s+80.8)10s

On using the partial fraction expansion in order to simplify above expression, we have

Ia(s)=(s+1)s(0.004s2+0.804s+80.8)10s=As+Bs+C(0.004s2+0.804s+80.8)10(s+1)s(0.004s2+0.804s+80.8)=A(0.004s2+0.804s+80.8)+s(Bs+C)s(0.004s2+0.804s+80.8)

On comparing the numerator on both sides, we get

10s+10=(0.004A+B)s2+(0.804A+C)s+80.8A A=25202,B=0.0005,C=9.9005

Thus,

Ia(s)=As+Bs+C(0.004s2+0.804s+80.8)Ia(s)=252021s+9.90050.0005s(0.004s2+0.804s+80.8)Ia(s)=252021s0.125s+2012(s+2012)2+403994+2487.68751(s+2012)2+403994

On taking the inverse Laplace of above obtained expression, the response ia(t) is,

ia(t)=0.125e201t2cos(40399t2)+2487.6875240399e201t2sin(40399t2)+25202

Similarly, for the response ω(t),

Since,

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the parameters in this transfer function,

Ω(s)Va(s)=KTLaIs2+(RaI+cLa)s+cRa+KbKTΩ(s)Va(s)=0.24×103×5×104s2+(0.8×5×104+5×104×4×103)s+5×104×0.8+0.2×0.2Ω(s)Va(s)=0.22×106s2+(4×104+2×106)s+4×104+0.04

Ω(s)Va(s)=1105s2+(2×103+105)s+2×103+0.2Ω(s)Va(s)=105s2+201s+20200

Now, since input armature voltage is a unit step response of strength 10V.

Therefore,

Ω(s)Va(s)=105s2+201s+20200Ω(s)=105(s2+201s+20200)Va(s)Ω(s)=106(s2+201s+20200)1s

On using the partial fraction expansion in order to simplify the above expression, we have

Ω(s)=106(s2+201s+20200)1s=As+Bs+C(s2+201s+20200)106s(s2+201s+20200)=A(s2+201s+20200)+(Bs+C)ss(s2+201s+20200)

Compare the denominator of both sides

Thus,

106=A(s2+201s+20200)+(Bs+C)s A=5000101,B=5000101andC=1005000101

Ω(s)=106(s2+201s+20200)1s=5000s1005000101(s2+201s+20200)+5000101s

Ω(s)=5000101s+2012(s+2012)2+4039945025001011(s+2012)2+403994+5000101s

On taking the inverse Laplace of the above obtained expression, the response ω(t) is,

ω(t)=5000101e201t2cos(40399t2)100500010140399e201t2sin(40399t2)+5000101.

Conclusion:

The responses ia(t) and ω(t) for the armature-controlled motor is,

ia(t)=0.125e201t2cos(40399t2)+2487.6875240399e201t2sin(40399t2)+25202

ω(t)=5000101e201t2cos(40399t2)100500010140399e201t2sin(40399t2)+5000101.

To determine

(b)

The step response of ia(t) and ω(t) for the load torque TL=0.2Nm.

Expert Solution
Check Mark

Answer to Problem 6.41P

The responses are follows:

ia(t)=100101e201t2cos(40399t2)2010010140399e201t2sin(40399t2)+100101

ω(t)=3.9604e201t2cos(40399t2)+3.960440399e201t2sin(40399t2)+400101.

Explanation of Solution

Given:

The given parameters for an armature-controlled motor are as:

Kb=KT=0.2Nm/A, c=5×104Nms/rad, Ra=0.8Ω, La=4×103H, I=5×104kgm2 and va=10V.

Concept Used:

For an armature-controlled motor, the output responses armature current ia(t) and motor speed ω(t) is related to input load torque TL as follows:

Ia(s)TL(s)=KbLaIs2+(RaI+cLa)s+cRa+KbKT

Ω(s)TL(s)=Las+RaLaIs2+(RaI+cLa)s+cRa+KbKT.

Calculation:

Since,

Ia(s)TL(s)=KbLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the values of all the parameters, we get

Ia(s)TL(s)=KbLaIs2+(RaI+cLa)s+cRa+KbKTIa(s)TL(s)=0.24×103×5×104s2+(0.8×5×104+5×104×4×103)s+5×104×0.8+0.2×0.2

Ia(s)TL(s)=0.22×106s2+(4×104+2×106)s+4×104+0.04Ia(s)TL(s)=1105s2+(2×103+105)s+2×103+0.2Ia(s)TL(s)=105s2+201s+20200

Now, since input load torque is a unit step response of strength 0.2Nm.

Therefore,

Ia(s)TL(s)=105s2+201s+20200Ia(s)=105s2+201s+20200TL(s)Ia(s)=2×104s(s2+201s+20200)

On using the partial fraction expansion in order to simplify above expression, we have

Ia(s)=2×104s(s2+201s+20200)Ia(s)=2×104s(s2+201s+20200)=As+Bs+C(s2+201s+20200)2×104s(s2+201s+20200)=A(s2+201s+20200)+(Bs+C)ss(s2+201s+20200)

On comparing the numerator on both sides, we get

2×104=A(s2+201s+20200)+(Bs+C)s A=100101,B=100101,C=20100101

Thus,

Ia(s)=106(s2+201s+20200)1s=100s20100101(s2+201s+20200)+100101sIa(s)=100101s+2012(s+2012)2+403994100501011(s+2012)2+403994+100101s

On taking the inverse Laplace of the above obtained expression, the response ia(t) is as shown

ia(t)=100101e201t2cos(40399t2)2010010140399e201t2sin(40399t2)+100101

Similarly, for the response ω(t),

Since,

Ω(s)TL(s)=Las+RaLaIs2+(RaI+cLa)s+cRa+KbKT

Keeping the parameters in this transfer function,

Ω(s)TL(s)=Las+RaLaIs2+(RaI+cLa)s+cRa+KbKTΩ(s)TL(s)=4×103s+0.84×103×5×104s2+(0.8×5×104+5×104×4×103)s+5×104×0.8+0.2×0.2Ω(s)TL(s)=4×103s+0.82×106s2+(4×104+2×106)s+4×104+0.04Ω(s)TL(s)=(s+200)0.0005s2+0.1005s+10.1

Now, since input load torque is a unit step response of strength 0.2Nm.

Therefore,

Ω(s)TL(s)=(s+200)0.0005s2+0.1005s+10.1Ω(s)=(s+200)0.0005s2+0.1005s+10.1TL(s)Ω(s)=0.2(s+200)s(0.0005s2+0.1005s+10.1)

On using the partial fraction expansion in order to simplify above expression, we have

Ω(s)=0.2(s+200)s(0.0005s2+0.1005s+10.1)=As+Bs+C(0.0005s2+0.1005s+10.1)0.2(s+200)s(0.0005s2+0.1005s+10.1)=A(0.0005s2+0.1005s+10.1)+s(Bs+C)s(0.0005s2+0.1005s+10.1)

Compare the denominator of both sides

Thus,

0.2s+40=(0.004A+B)s2+(0.804A+C)s+80.8A A=400101,B=0.0019,C=0.1980

Ω(s)=As+Bs+C(0.0005s2+0.1005s+10.1)Ω(s)=0.0019802s0.198020.0005s2+0.1005s+10.1+400101sΩ(s)=3.9604s+2012(s+2012)2+403994+1.98021(s+2012)2+403994+400101s

On taking the inverse Laplace of the above obtained expression, the response ω(t) is as shown

ω(t)=3.9604e201t2cos(40399t2)+3.960440399e201t2sin(40399t2)+400101.

Conclusion:

The responses ia(t) and ω(t) for the armature-controlled motor is as follows:

ia(t)=100101e201t2cos(40399t2)2010010140399e201t2sin(40399t2)+100101

ω(t)=3.9604e201t2cos(40399t2)+3.960440399e201t2sin(40399t2)+400101.

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Chapter 6 Solutions

System Dynamics

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