System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
Question
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Chapter 8, Problem 8.22P
To determine

(a)

The response x(t) for the model equation 3x¨+21x˙+30x=4t.

Expert Solution
Check Mark

Answer to Problem 8.22P

The response x(t) is x(t)=215t775+19e2t4225e5t.

Explanation of Solution

Given:

The given model equation is as:

3x¨+21x˙+30x=4t

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=4t

By taking the Laplace of this equation that is,

3x¨+21x˙+30x=4t3(s2X(s)sx(0)x˙(0))+21(sX(s)x(0))+30X(s)=4s23(s2X(s)00)+21(sX(s)0)+30X(s)=4s2 x(0)=0,x˙(0)=0

X(s).(3s2+21s+30)=4s2X(s)=43s2(s2+7s+10)=43s2(s+2)(s+5)

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

X(s)=43s2(s+2)(s+5)=As2+Bs+C(s+2)+D(s+5)43s2(s+2)(s+5)=A(s+5)(s+2)+Bs(s+5)(s+2)+Cs2(s+5)+Ds2(s+2)s2(s+2)(s+5)43=s3(B+C+D)+s2(A+7B+5C+2D)+s(7A+10B)+10A B+C+D=0,A+7B+5C+2D=0,7A+10B=0,A=430 A=215,B=775,C=19,D=4225

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

X(s)=215s2775s+19(s+2)4225(s+5)x(t)=215t775+19e2t4225e5t.

Conclusion:

The total response x(t) for the model equation x(t)=215t775+19e2t4225e5t.

To determine

(b)

The response x(t) for the model equation 5x¨+20x˙+20x=7t.

Expert Solution
Check Mark

Answer to Problem 8.22P

The response x(t) is x(t)=720t720+720e2t+720te2t.

Explanation of Solution

Given:

The given model equation is as:

5x¨+20x˙+20x=7t

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=7t

By taking the Laplace of this equation that is,

5x¨+20x˙+20x=7t5(s2X(s)sx(0)x˙(0))+20(sX(s)x(0))+20X(s)=7s25(s2X(s)00)+20(sX(s)0)+20X(s)=7s2 x(0)=0,x˙(0)=0X(s).(5s2+20s+20)=7s2X(s)=75s2(s2+4s+4)=75s2(s+2)2

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

X(s)=75s2(s+2)2=As2+Bs+C(s+2)+D(s+2)275s2(s+2)2=A(s+2)2+Bs(s+2)2+Cs2(s+2)+Ds2s2(s+2)275=s3(B+C)+s2(A+4B+2C+D)+s(4A+4B)+4A B+C=0,A+4B+2C+D=0,4A+4B=0,A=720 A=720,B=720,C=720,D=720

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

X(s)=720s2720s+720(s+2)+720(s+2)2x(t)=720t720+720e2t+720te2t.

Conclusion:

The total response x(t) for the model equation x(t)=720t720+720e2t+720te2t.

To determine

(c)

The response x(t) for the model equation 2x¨+8x˙+58x=5t.

Expert Solution
Check Mark

Answer to Problem 8.22P

The response x(t) is x(t)=558t10841+10841e2tcos5t211682e2tsin5t.

Explanation of Solution

Given:

The given model equation is as:

2x¨+8x˙+58x=5t

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=5t

By taking the Laplace of this equation that is,

2x¨+8x˙+58x=5t2(s2X(s)sx(0)x˙(0))+8(sX(s)x(0))+58X(s)=5s22(s2X(s)00)+8(sX(s)0)+58X(s)=5s2 x(0)=0,x˙(0)=0X(s).(2s2+8s+58)=5s2X(s)=52s2(s2+4s+29)

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

X(s)=52s2(s2+4s+29)=As2+Bs+Cs+D(s2+4s+29)52s2(s2+4s+29)=A(s2+4s+29)+Bs(s2+4s+29)+(Cs+D)s2s2(s2+4s+29)52=s3(B+C)+s2(A+4B+D)+s(4A+29B)+29A

B+C=0,A+4B+D=0,4A+29B=0,A=558 A=558,B=10841,C=10841,D=651682

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

X(s)=558s210841s+11682(20s65)(s2+4s+29)X(s)=558s210841s+10841(s+2)((s+2)2+(5)2)2116825((s+2)2+(5)2)x(t)=558t10841+10841e2tcos5t211682e2tsin5t.

Conclusion:

The total response x(t) for the model equation x(t)=558t10841+10841e2tcos5t211682e2tsin5t.

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