System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 2, Problem 2.38P
To determine

(a)

The response x(t) for the model equation 7x˙+5x=4g˙(t).Also, compare the values of x(0)andx(0+).

Expert Solution
Check Mark

Answer to Problem 2.38P

The response x(t) for the model equations is as:

x(t)=257e57t.

The response x(t) changes from t=0tot=0+ such that

x(0)=3andx(0+)=257.

Explanation of Solution

Given:

The given model equation is as:

7x˙+5x=4g˙(t)

Where, g(t)=us(t) and the initial conditions are

x(0)=3andg(0)=0.

Concept Used:

First, evaluate the response x(t) from the given model equation using the Laplace transform. Then, find the right-hand limit of x(t) at t=0.

Calculation:

The model equation to be solved is as:

7x˙+5x=4g˙(t)

Where, g(t)=us(t)

Therefore, g˙(t)=δ(t)

And 7x˙+5x=4δ(t)

On taking Laplace transform, that is

7x˙+5x=4δ(t)7(sX(s)x(0))+5X(s)=47(sX(s)3)+5X(s)=4X(s)=25(7s+5)

On taking the inverse Laplace of this,

x(t)=257e57t

Therefore,

x(0+)=limt0+257e57t=257

Thus the response x(t) changes from 3 to 257 as time varies from t=0tot=0+.

Conclusion:

The obtained response x(t) is as:

x(t)=257e57t

And x(t) changes to 257 from 3 as time varies from t=0tot=0+.

To determine

(b)

The response x(t) for the model equation 7x˙+5x=4g˙(t)+6g(t). Compare the values of x(0)andx(0+).

Expert Solution
Check Mark

Answer to Problem 2.38P

The response x(t) for the model equation is as:

x(t)=65+8335e57t.

The response x(t) changes from t=0tot=0+ such that

x(0)=3andx(0+)=257.

Explanation of Solution

Given:

The given model equation is as:

7x˙+5x=4g˙(t)+6g(t)

Where, g(t)=us(t) and the initial conditions are

x(0)=3andg(0)=0.

Concept Used:

First, evaluate the response x(t) from the given model equation using the Laplace transform. Then, find the right-hand limit of x(t) at t=0.

Calculation:

The model equation to be solved is as:

7x˙+5x=4g˙(t)+6g(t)

Where, g(t)=us(t)

Therefore, g˙(t)=δ(t)

And 7x˙+5x=4δ(t)+6us(t)

On taking Laplace transform, that is

7x˙+5x=4δ(t)+6us(t)7(sX(s)x(0))+5X(s)=4+6s7(sX(s)3)+5X(s)=4+6sX(s)(7s+5)=25+6sX(s)=83351(s+57)+651s

On taking the inverse Laplace of this,

x(t)=65+8335e57t

Therefore,

x(0+)=limt0+(65+8335e57t)=(65+8335)=257

Thus the response x(t) changes from 3 to 257 as time varies from t=0tot=0+.

Conclusion:

The obtained response x(t) is as:

x(t)=65+8335e57t

And x(t) changes to 257 from 3 as time varies from t=0tot=0+.

To determine

(c)

The response x(t) for the model equation 3x¨+30x˙+63x=4g˙(t). Compare the values of x(0)andx(0+).Also, do the same for x˙(0)andx˙(0+).

Expert Solution
Check Mark

Answer to Problem 2.38P

The response x(t) for the model equation is as:

x(t)=5512e3t3112e7t.

The response x(t) doesn’t change from t=0tot=0+ such that

x(0)=x(0+)=2

While the response x˙(t) changes from t=0tot=0+ such that

x˙(0)=3andx˙(0+)=133.

Explanation of Solution

Given:

The given model equations are as:

3x¨+30x˙+63x=4g˙(t)

Where, g(t)=us(t) and the initial conditions are

x(0)=2,x˙(0)=3andg(0)=0.

Concept Used:

First, evaluate the response x(t) from the given model equation using the Laplace transform. Then, find the right-hand limit of x(t) and x˙(t) at t=0.

Calculation:

The model equation to be solved is as:

3x¨+30x˙+63x=4g˙(t)

On taking Laplace transform, that is

3x¨+30x˙+63x=4δ(t)3(s2X(s)sx(0)x˙(0))+30(sX(s)x(0))+63X(s)=43(s2X(s)2s3)+30(sX(s)2)+63X(s)=4X(s)(3s2+30s+63)=6s+73X(s)=(6s+73)(3s2+30s+63)=55121(s+3)31121(s+7)

On taking the inverse Laplace of this,

x(t)=5512e3t3112e7t

On taking the derivative,

x˙(t)=554e3t+21712e7t

Therefore,

x(0+)=limt0+(5512e3t3112e7t)=(55123112)=2412=2

Similarly,

x˙(0+)=limt0+(554e3t+21712e7t)=(554+21712)=5212=133

Thus the response x(t) doesn’t change as time varies from t=0tot=0+.

Whereas, the response x˙(t) changes from 3 to 133 as time varies from t=0tot=0+.

Conclusion:

The obtained response x(t) is as:

x(t)=5512e3t3112e7t

And x(t) remains unchanged as time varies from t=0tot=0+.

While x˙(t) attains a value of 133 from 3 as time increases from t=0tot=0+.

To determine

(d)

The response x(t) for the model equation 3x¨+30x˙+63x=4g˙(t)+6g(t). Compare the values of x(0)andx(0+).Also, do the same for x˙(0)andx˙(0+).

Expert Solution
Check Mark

Answer to Problem 2.38P

The response x(t) for the model equation is as:

x(t)=22121184e7t+5312e3t.

The response x˙(t) decreases from t=0tot=0+ such that

x(0)=4andx(0+)=2

While the response x˙(t) also decreases from t=0tot=0+ such that

x˙(0)=7andx˙(0+)=133.

Explanation of Solution

Given:

The given model equation is as:

3x¨+30x˙+63x=4g˙(t)+6g(t)

Where, g(t)=us(t) and the initial conditions are

x(0)=4,x˙(0)=7andg(0)=0.

Concept Used:

First, evaluate the response x(t) from the given model equation using the Laplace transform. Then, find the right-hand limit of x(t) and x˙(t) at t=0.

Calculation:

The model equation to be solved is as:

3x¨+30x˙+63x=4g˙(t)+6g(t)

On taking Laplace transform, that is

3x¨+30x˙+63x=4δ(t)+6us(t)3(s2X(s)sx(0)x˙(0))+30(sX(s)x(0))+63X(s)=4+6s3(s2X(s)2s3)+30(sX(s)2)+63X(s)=4+6sX(s)(3s2+30s+63)=6s2+73s+6sX(s)=(6s2+73s+6)s(3s2+30s+63)=(6s2+73s+6)3s(s+3)(s+7)X(s)=2211s211841(s+7)+53121(s+3)

On taking the inverse Laplace of this,

x(t)=22121184e7t+5312e3t

On taking the derivative,

x˙(t)=21112e7t534e3t

Therefore,

x(0+)=limt0+(22121184e7t+5312e3t)=(22121184+5312)=16824=2

Similarly,

x˙(0+)=limt0+(21112e7t534e3t)=(21112534)=5212=133

Thus the response x(t) decreases from 4 to2 as time varies from t=0tot=0+.

In addition, the response x˙(t) also decreases from 7 to 133 as time varies from t=0tot=0+.

Conclusion:

The obtained response x(t) is as:

x(t)=22121184e7t+5312e3t

And x(t) decreases from 4 to2 as time varies from t=0tot=0+.

While x˙(t) attains a value of 133 from 7 as time increases from t=0tot=0+.

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Chapter 2 Solutions

System Dynamics

Ch. 2 - Prob. 2.11PCh. 2 - Obtain the inverse Laplace transform xt for each...Ch. 2 - Solve the following problems: 5x=7tx0=3...Ch. 2 - Solve the following: 5x+7x=0x0=4 5x+7x=15x0=0...Ch. 2 - Solve the following problems: x+10x+21x=0x0=4x0=3...Ch. 2 - Solve the following problems: x+7x+10x=20x0=5x0=3...Ch. 2 - Solve the following problems: 3x+30x+63x=5x0=x0=0...Ch. 2 - Solve the following problems where x0=x0=0 ....Ch. 2 - Invert the following transforms: 6ss+5 4s+3s+8...Ch. 2 - Invert the following transforms: 3s+2s2s+10...Ch. 2 - Prob. 2.21PCh. 2 - Compare the LCD method with equation (2.4.4) for...Ch. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - (a) Prove that the second-order system whose...Ch. 2 - For each of the following models, compute the time...Ch. 2 - Prob. 2.27PCh. 2 - Prob. 2.28PCh. 2 - Prob. 2.29PCh. 2 - If applicable, compute , , n , and d for the...Ch. 2 - Prob. 2.31PCh. 2 - For each of the following equations, determine the...Ch. 2 - Prob. 2.33PCh. 2 - Obtain the transfer functions Xs/Fs and Ys/Fs for...Ch. 2 - a. Obtain the transfer functions Xs/Fs and Ys/Fs...Ch. 2 - Prob. 2.36PCh. 2 - Solve the following problems for xt . Compare the...Ch. 2 - Prob. 2.38PCh. 2 - Prob. 2.39PCh. 2 - Prob. 2.40PCh. 2 - Determine the general form of the solution of the...Ch. 2 - a. Use the Laplace transform to obtain the form of...Ch. 2 - Prob. 2.43PCh. 2 - Prob. 2.44PCh. 2 - Obtain the inverse transform in the form xt=Asint+...Ch. 2 - Use the Laplace transform to solve the following...Ch. 2 - Express the oscillatory part of the solution of...Ch. 2 - Prob. 2.48PCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52PCh. 2 - Prob. 2.53PCh. 2 - 2.54 The Taylor series expansion for tan t...Ch. 2 - 2.55 Derive the initial value theorem: Ch. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Use MATLAB to obtain the inverse transform of the...Ch. 2 - Use MATLAB to obtain the inverse transform of the...Ch. 2 - Use MATLAB to solve for and plot the unit-step...Ch. 2 - Use MATLAB to solve for and plot the unit-impulse...Ch. 2 - Use MATLAB to solve for and plot the impulse...Ch. 2 - Use MATLAB to solve for and plot the response of...
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