System Dynamics
System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 2, Problem 2.23P
To determine

(a)

The steady state, transient, free and forced responses for the following nonhomogenous equation x¨+8x˙+15x=30.

Expert Solution
Check Mark

Answer to Problem 2.23P

Steady state response is 2.

Transient state response is22e3t14e5t

Free response is 27e3t17e5t.

Forced response is 25e3t+3e5t.

Explanation of Solution

Given:

The given equation is as:

x¨+8x˙+15x=30

With initial conditions as follows:

x(0)=10 and x˙(0)=4.

Concept Used:

The given nonhomogenous equation is first transformed into the time domain using the Laplace transform as shown:

x¨+8x˙+15x=30

Taking Laplace of the above equation that is,

(s2X(s)sx(0)x˙(0))+8(sX(s)x(0))+15X(s)=30s

Now, the expression for X(s) is computed by simplifying the above equation.

Therefore,

(s)=10s2+84s+30s(s2+8s+15)=10s2+84s+30s(s+3)(s+5)

Now, in order to simplify the above equation, the partial fraction expansion could be used so that the expression becomes as:

X(s)=21s+221(s+3)141(s+5)

Then, the inverse Laplace of X(s) is evaluated for determining the steady state and transient state responses.

And for the free response the input would be taken zero or say the right-hand side part of the given equation is kept zero as shown:

x¨+8x˙+15x=0

Therefore,

X(s)=10s+84(s+3)(s+5)

For the forced response the initial conditions are kept zero such that:

X(s)=30s(s+3)(s+5)

Then the inverse of these two functions is calculated for finding the respective response.

Calculation:

The equation to be solved is as:

x¨+8x˙+15x=30

By taking the Laplace of this equation that is,

(s2X(s)sx(0)x˙(0))+8(sX(s)x(0))+15X(s)=30s

Therefore, X(s) is found as shown below:

(s2X(s)sx(0)x˙(0))+8(sX(s)x(0))+15X(s)=30sX(s)(s2+8s+15)=30s+10s+84X(s)=10s2+84s+30s(s2+8s+15)=10s2+84s+30s(s+3)(s+5)

Using the partial fraction expansion, the transfer function could be simplified as shown below:

X(s)=10s2+84s+30s(s+3)(s+5)=21s+221(s+3)141(s+5)

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

x(t)=2+22e3t14e5t

Thus the above response consists of two parts, that is, the steady-state response and the transient state response.

Steady state response is 2.

Transient state response is22e3t14e5t

At zero initial conditions, using the partial fraction expansion X(s)=30s(s+3)(s+5)=21s51(s+3)+31(s+5)

Therefore,

x(t)=25e3t+3e5t

This is the forced response.

At zero input condition, using the partial fraction expansion.

X(s)=10s+84(s+3)(s+5)=271(s+3)171(s+5)

Therefore,

x(t)=27e3t17e5t

This is the free response.

Conclusion:

Steady state response is 2.

Transient state response is22e3t14e5t

Free response is 27e3t17e5t.

Forced response is 25e3t+3e5t.

To determine

(b)

The steady state, transient, free and forced responses for the following nonhomogenous equation x¨+10x˙+25x=75.

Expert Solution
Check Mark

Answer to Problem 2.23P

Steady state response is 3.

Transient state response is 39te5t+7e5t

Free response is 10e5t+54te5t.

Forced response is 315te5t3e5t.

Explanation of Solution

Given:

The given equation is as:

x¨+10x˙+25x=75

With initial conditions as follows:

x(0)=10 and x˙(0)=4.

Concept Used:

The given nonhomogenous equation is first transformed into the time domain using the Laplace transform as shown:

x¨+10x˙+25x=75

Taking Laplace of the above equation that is,

(s2X(s)sx(0)x˙(0))+10(sX(s)x(0))+25X(s)=75s

Now, the expression for X(s) is computed by simplifying the above equation.

Therefore,

X(s)=10s2+104s+75s(s2+10s+25)=10s2+104s+75s(s+5)2

Now, in order to simplify the above equation, the partial fraction expansion could be used so that the expression becomes as:

X(s)=10s2+104s+75s(s+5)2=31s+391(s+5)2+71(s+5)

Then, the inverse Laplace of X(s) is evaluated for determining the steady state and transient state responses.

And for the free response the input would be taken zero or say the right-hand side part of the given equation is kept zero as shown:

x¨+10x˙+25x=0

Therefore,

X(s)=10s+104(s+5)2

For the forced response the initial conditions are kept zero such that:

X(s)=75s(s+5)2

Then the inverse of these two functions is calculated for finding the respective response.

Calculation:

The equation to be solved is as:

x¨+10x˙+25x=75

By taking the Laplace of this equation that is,

(s2X(s)sx(0)x˙(0))+10(sX(s)x(0))+25X(s)=75s

Therefore, X(s) is found as shown below:

(s2X(s)sx(0)x˙(0))+10(sX(s)x(0))+25X(s)=75sX(s)(s2+10s+25)=75s+10s+104X(s)=10s2+104s+75s(s2+10s+25)=10s2+104s+75s(s+5)2

Using the partial fraction expansion, the transfer function could be simplified as shown below:

X(s)=10s2+104s+75s(s+5)2=31s+391(s+5)2+71(s+5)

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

x(t)=3+39te5t+7e5t

Thus the above response consists of two parts, that is, the steady-state response and the transient state response.

Steady state response is 3.

39te5t+7e5t Transient state response is

At zero initial conditions, using the partial fraction expansion

X(s)=75s(s+5)2=3s151(s+5)231(s+5)

Therefore,

x(t)=315te5t3e5t

This is the forced response.

At zero input condition, using the partial fraction expansion.

X(s)=10s+104(s+5)2=10(s+5)+54(s+5)2

Therefore,

x(t)=10e5t+54te5t

This is the free response.

Conclusion:

Steady state response is 3.

Transient state response is 39te5t+7e5t

Free response is 10e5t+54te5t.

Forced response is 315te5t3e5t.

To determine

(c)

The steady state, transient, free and forced responses for the following nonhomogenous equation x¨+25x=100.

Expert Solution
Check Mark

Answer to Problem 2.23P

Steady state response is 4.

Transient state response is6cos5t+45sin5t

Free response is 10cos5t+45sin5t.

Forced response is 44cos5t.

Explanation of Solution

Given:

The given equation is as:

x¨+25x=100.

With initial conditions as follows:

x(0)=10 and x˙(0)=4.

Concept Used:

The given nonhomogenous equation is first transformed into the time domain using the Laplace transform as shown:

x¨+25x=100

Taking Laplace of the above equation that is,

(s2X(s)sx(0)x˙(0))+25X(s)=100s

Now, the expression for X(s) is computed by simplifying the above equation.

Therefore,

X(s)=10s2+4s+100s(s2+52)=10s2+4s+100s(s2+52)

Now, in order to simplify the above equation, the partial fraction expansion could be used so that the expression becomes as:

X(s)=4s+6s(s2+52)+455(s2+52)

Then, the inverse Laplace of X(s) is evaluated for determining the steady state and transient state responses.

And for the free response the input would be taken zero or say the right-hand side part of the given equation is kept zero as shown:

x¨+25x=0

Therefore,

X(s)=10s+4(s2+52)

For the forced response the initial conditions are kept zero such that:

X(s)=100s(s2+25)

Then the inverse of these two functions is calculated for finding the respective response.

Calculation:

The equation to be solved is as:

x¨+25x=100

By taking the Laplace of this equation that is,

(s2X(s)sx(0)x˙(0))+25X(s)=100s

Therefore, X(s) is found as shown below:

(s2X(s)sx(0)x˙(0))+25X(s)=100sX(s)(s2+25)=100s+10s+4X(s)=10s2+4s+100s(s2+52)=10s2+4s+100s(s2+52)

Using the partial fraction expansion, the transfer function could be simplified as shown below:

X(s)=10s2+4s+100s(s2+52)=4s+6s(s2+52)+455(s2+52)

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

x(t)=4+6cos5t+45sin5t

Thus the above response consists of two parts, that is, the steady-state response and the transient state response.

Steady state response is 4.

Transient state response is 6cos5t+45sin5t.

At zero initial conditions, using the partial fraction expansion.

X(s)=100s(s2+25)=4s4s(s2+52)

Therefore,

x(t)=44cos5t

This is the forced response.

At zero input condition, using the partial fraction expansion.

X(s)=10s+4(s2+52)=10s(s2+52)+455(s2+52)

Therefore,

x(t)=10cos5t+45sin5t

This is the free response.

Conclusion:

Steady state response is 4.

Transient state response is6cos5t+45sin5t

Free response is 10cos5t+45sin5t.

Forced response is 44cos5t.

To determine

(d)

The steady state, transient, free and forced responses for the following non homogenous equation x¨+8x˙+65x=130.

Expert Solution
Check Mark

Answer to Problem 2.23P

Steady state response is 2.

Transient state response is 8e4tcos7t+367e4tsin7t

Free response is 10e4tcos7t+447e4tsin7t.

Forced response is 22e4tcos7t87e4tsin7t.

Explanation of Solution

Given:

The given equation is as:

x¨+8x˙+65x=130

With initial conditions as follows:

x(0)=10 and x˙(0)=4.

Concept Used:

The given nonhomogenous equation is first transformed into the time domain using the Laplace transform as shown:

x¨+8x˙+65x=130

Taking Laplace of the above equation that is,

(s2X(s)sx(0)x˙(0))+8(sX(s)x(0))+65X(s)=130s

Now, the expression for X(s) is computed by simplifying the above equation.

Therefore,

X(s)=10s2+84s+130s(s2+8s+65)=10s2+84s+130s((s+4)2+(7)2)

Now, in order to simplify the above equation, the partial fraction expansion could be used so that the expression becomes as:

X(s)=2s+8(s+4)((s+4)2+(7)2)+3677((s+4)2+(7)2)

Then, the inverse Laplace of X(s) is evaluated for determining the steady state and transient state responses.

And for the free response the input would be taken zero or say the right-hand side part of the given equation is kept zero as shown:

x¨+8x˙+65x=0

Therefore,

X(s)=10s+84(s2+8s+65)

For the forced response the initial conditions are kept zero such that:

X(s)=130s(s2+8s+65)

Then the inverse of these two functions is calculated for finding the respective response.

Calculation:

The equation to be solved is as:

x¨+8x˙+65x=130

By taking the Laplace of this equation that is,

(s2X(s)sx(0)x˙(0))+8(sX(s)x(0))+65X(s)=130s

Therefore, X(s) is found as shown below:

(s2X(s)sx(0)x˙(0))+8(sX(s)x(0))+65X(s)=130sX(s)(s2+8s+65)=130s+10s+84X(s)=10s2+84s+130s(s2+8s+65)=10s2+84s+130s((s+4)2+(7)2)

Using the partial fraction expansion, the transfer function could be simplified as shown below:

X(s)=2s+8(s+4)((s+4)2+(7)2)+3677((s+4)2+(7)2)

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

x(t)=2+8e4tcos7t+367e4tsin7t

Thus the above response consists of two parts, that is, the steady-state response and the transient state response.

Steady state response is 2.

Transient state response is8e4tcos7t+367e4tsin7t

At zero initial conditions, using the partial fraction expansion.

X(s)=130s(s2+8s+65)=2s2(s+4)((s+4)2+(7)2)877((s+4)2+(7)2)

Therefore,

x(t)=22e4tcos7t87e4tsin7t

This is the forced response.

At zero input condition, using the partial fraction expansion.

X(s)=10s+84(s2+8s+65)=10(s+4)((s+4)2+(7)2)+4477((s+4)2+(7)2)

Therefore,

x(t)=10e4tcos7t+447e4tsin7t

This is the free response.

Conclusion:

Steady state response is 2.

Transient state response is8e4tcos7t+367e4tsin7t

Free response is 10e4tcos7t+447e4tsin7t.

Forced response is 22e4tcos7t87e4tsin7t.

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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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