System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
Question
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Chapter 8, Problem 8.21P
To determine

(a)

The response x(t) for the model equation 3x¨+21x˙+30x=f(t), where f(t) is a unit step function.

Expert Solution
Check Mark

Answer to Problem 8.21P

The response x(t) is x(t)=130118e2t+145e5t.

Explanation of Solution

Given:

The given model equation is as:

3x¨+21x˙+30x=f(t)

With initial conditions as follows:

x(0)=0 and x˙(0)=0

Concept Used:

Laplace transform is used for obtaining the response of the model equation.

Calculation:

Equation to be solved is as:

3x¨+21x˙+30x=f(t)

By taking the Laplace of this equation that is,

3x¨+21x˙+30x=f(t)3x¨+21x˙+30x=us(t) f(t)=us(t)3(s2X(s)sx(0)x˙(0))+21(sX(s)x(0))+30X(s)=1s

3(s2X(s)00)+21(sX(s)0)+30X(s)=1s x(0)=0,x˙(0)=0X(s).(3s2+21s+30)=1sX(s)=13s(s2+7s+10)=13s(s+2)(s+5)

On using partial fraction expansion, this expression could be expressed as follows:

X(s)=13s(s+2)(s+5)=As+B(s+2)+C(s+5)13s(s+2)(s+5)=A(s+5)(s+2)+Bs(s+5)+Cs(s+2)s(s+2)(s+5)13=s2(A+B+C)+s(7A+5B+2C)+10A A+B+C=0,7A+5B+2C=0,A=130 A=130,B=118,C=145

Now, on taking the inverse Laplace of the above transfer function, we get

X(s)=130s118(s+2)+145(s+5)x(t)=130118e2t+145e5t.

Conclusion:

The total response x(t) for the model equation x(t)=130118e2t+145e5t.

To determine

(b)

The response x(t) for the model equation 5x¨+20x˙+20x=f(t), where f(t) is a unit step function.

Expert Solution
Check Mark

Answer to Problem 8.21P

The response x(t) is x(t)=120120e2t110te2t.

Explanation of Solution

Given:

The given model equation is as:

5x¨+20x˙+20x=f(t)

With initial conditions as follows:

x(0)=0 and x˙(0)=0

Concept Used:

Laplace transform is used for obtaining the response of the model equation.

Calculation:

Equation to be solved is as:

5x¨+20x˙+20x=f(t)

By taking the Laplace of this equation that is,

5x¨+20x˙+20x=f(t)5x¨+20x˙+20x=us(t) f(t)=us(t)5(s2X(s)sx(0)x˙(0))+20(sX(s)x(0))+20X(s)=1s5(s2X(s)00)+20(sX(s)0)+20X(s)=1s x(0)=0,x˙(0)=0X(s).(5s2+20s+20)=1s

X(s)=15s(s2+4s+4)=15s(s+2)2

On using partial fraction expansion, this expression could be expressed as follows:

X(s)=15s(s+2)2=As+B(s+2)+C(s+2)215s(s+2)2=A(s+2)2+Bs(s+2)+Css(s+2)215=s2(A+B)+s(4A+2B+C)+4A A+B=0,4A+2B+C=0,A=120 A=120,B=120,C=110

Now, on taking the inverse Laplace of the above transfer function, we get

X(s)=120s120(s+2)110(s+2)2x(t)=120120e2t110te2t.

Conclusion:

The total response x(t) for the model equation x(t)=120120e2t110te2t.

To determine

(c)

The response x(t) for the model equation 2x¨+8x˙+58x=f(t), where f(t) is a unit step function.

Expert Solution
Check Mark

Answer to Problem 8.21P

The response x(t) is x(t)=158158e2tcos5t1145e2tsin5t.

Explanation of Solution

Given:

The given model equation is as:

2x¨+8x˙+58x=f(t)

With initial conditions as follows:

x(0)=0 and x˙(0)=0

Concept Used:

Laplace transform is used for obtaining the response of the model equation.

Calculation:

Equation to be solved is as:

2x¨+8x˙+58x=f(t)

By taking the Laplace of this equation that is,

2x¨+8x˙+58x=f(t)2x¨+8x˙+58x=us(t) f(t)=us(t)2(s2X(s)sx(0)x˙(0))+8(sX(s)x(0))+58X(s)=1s2(s2X(s)00)+8(sX(s)0)+58X(s)=1s x(0)=0,x˙(0)=0X(s).(2s2+8s+58)=1sX(s)=12s(s2+4s+29)

On using partial fraction expansion, this expression could be expressed as follows:

X(s)=12s(s2+4s+29)=As+Bs+C(s2+4s+29)12s(s2+4s+29)=A(s2+4s+29)+s(Bs+C)s(s2+4s+29)12=s2(A+B)+s(4A+C)+29A A+B=0,4A+C=0,A=158 A=158,B=158,C=229

Now, on taking the inverse Laplace of the above transfer function, we get

X(s)=158s158(s+2)((s+2)2+(5)2)11455((s+2)2+(5)2)x(t)=158158e2tcos5t1145e2tsin5t.

Conclusion:

The total response x(t) for the model equation x(t)=158158e2tcos5t1145e2tsin5t.

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