Chapter 2, Problem 2.23P

To determine

(a)

The steady state, transient, free and forced responses for the following nonhomogenous equation $\ddot{x}+8\dot{x}+15x=30$.

Answer to Problem 2.23P

Steady state response is 2.

Transient state response is$22e^{-3t}-14e^{-5t}$

Free response is $27e^{-3t}-17e^{-5t}$.

Forced response is $2-5e^{-3t}+3e^{-5t}$.

Explanation of Solution

Given:

The given equation is as:

$\ddot{x}+8\dot{x}+15x=30$

With initial conditions as follows:

$x(0)=10$ and $\dot{x}(0)=4$.

Concept Used:

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

$\ddot{x}+8\dot{x}+15x=30$

Taking Laplace of the above equation that is,

$\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+8\left(sX(s)-x(0)\right)+15X(s)=\frac{30}{s}$

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

Therefore,

$(s)=\frac{10s^2+84s+30}{s\left(s^2+8s+15\right)}=\frac{10s^2+84s+30}{s(s+3)(s+5)}$

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

$X(s)=2\cdot \frac{1}{s}+22\cdot \frac{1}{(s+3)}-14\cdot \frac{1}{(s+5)}$

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:

$\ddot{x}+8\dot{x}+15x=0$

Therefore,

$X(s)=\frac{10s+84}{(s+3)(s+5)}$

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

$X(s)=\frac{30}{s(s+3)(s+5)}$

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

Calculation:

The equation to be solved is as:

$\ddot{x}+8\dot{x}+15x=30$

By taking the Laplace of this equation that is,

$\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+8\left(sX(s)-x(0)\right)+15X(s)=\frac{30}{s}$

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

$\begin{array}{l}\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+8\left(sX(s)-x(0)\right)+15X(s)=\frac{30}{s}\\ \Rightarrow X(s)\left(s^2+8s+15\right)=\frac{30}{s}+10s+84\\ \Rightarrow X(s)=\frac{10s^2+84s+30}{s\left(s^2+8s+15\right)}=\frac{10s^2+84s+30}{s(s+3)(s+5)}\end{array}$

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

$X(s)=\frac{10s^2+84s+30}{s(s+3)(s+5)}=2\cdot \frac{1}{s}+22\cdot \frac{1}{(s+3)}-14\cdot \frac{1}{(s+5)}$

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

$x(t)=2+22e^{-3t}-14e^{-5t}$

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 is$22e^{-3t}-14e^{-5t}$

At zero initial conditions, using the partial fraction expansion $X(s)=\frac{30}{s(s+3)(s+5)}=2\cdot \frac{1}{s}-5\cdot \frac{1}{(s+3)}+3\cdot \frac{1}{(s+5)}$

Therefore,

$x(t)=2-5e^{-3t}+3e^{-5t}$

This is the forced response.

At zero input condition, using the partial fraction expansion.

$X(s)=\frac{10s+84}{(s+3)(s+5)}=27\cdot \frac{1}{(s+3)}-17\cdot \frac{1}{(s+5)}$

Therefore,

$x(t)=27e^{-3t}-17e^{-5t}$

This is the free response.

Conclusion:

Steady state response is 2.

Transient state response is$22e^{-3t}-14e^{-5t}$

Free response is $27e^{-3t}-17e^{-5t}$.

Forced response is $2-5e^{-3t}+3e^{-5t}$.

To determine

(b)

The steady state, transient, free and forced responses for the following nonhomogenous equation $\ddot{x}+10\dot{x}+25x=75$.

Answer to Problem 2.23P

Steady state response is $3$.

Transient state response is $39t{e}^{-5t}+7e^{-5t}$

Free response is $10e^{-5t}+54t{e}^{-5t}$.

Forced response is $3-15t{e}^{-5t}-3e^{-5t}$.

Explanation of Solution

Given:

The given equation is as:

$\ddot{x}+10\dot{x}+25x=75$

With initial conditions as follows:

$x(0)=10$ and $\dot{x}(0)=4$.

Concept Used:

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

$\ddot{x}+10\dot{x}+25x=75$

Taking Laplace of the above equation that is,

$\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+10\left(sX(s)-x(0)\right)+25X(s)=\frac{75}{s}$

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

Therefore,

$X(s)=\frac{10s^2+104s+75}{s\left(s^2+10s+25\right)}=\frac{10s^2+104s+75}{s{(s+5)}^2}$

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

$X(s)=\frac{10s^2+104s+75}{s{(s+5)}^2}=3\cdot \frac{1}{s}+39\cdot \frac{1}{{(s+5)}^2}+7\cdot \frac{1}{(s+5)}$

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:

$\ddot{x}+10\dot{x}+25x=0$

Therefore,

$X(s)=\frac{10s+104}{{(s+5)}^2}$

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

$X(s)=\frac{75}{s{(s+5)}^2}$

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

Calculation:

The equation to be solved is as:

$\ddot{x}+10\dot{x}+25x=75$

By taking the Laplace of this equation that is,

$\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+10\left(sX(s)-x(0)\right)+25X(s)=\frac{75}{s}$

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

$\begin{array}{l}\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+10\left(sX(s)-x(0)\right)+25X(s)=\frac{75}{s}\\ \Rightarrow X(s)\left(s^2+10s+25\right)=\frac{75}{s}+10s+104\\ \Rightarrow X(s)=\frac{10s^2+104s+75}{s\left(s^2+10s+25\right)}=\frac{10s^2+104s+75}{s{(s+5)}^2}\end{array}$

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

$X(s)=\frac{10s^2+104s+75}{s{(s+5)}^2}=3\cdot \frac{1}{s}+39\cdot \frac{1}{{(s+5)}^2}+7\cdot \frac{1}{(s+5)}$

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

$x(t)=3+39t{e}^{-5t}+7e^{-5t}$

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

Steady state response is $3$.

$39t{e}^{-5t}+7e^{-5t}$ Transient state response is

At zero initial conditions, using the partial fraction expansion

$X(s)=\frac{75}{s{(s+5)}^2}=\frac{3}{s}-15\cdot \frac{1}{{(s+5)}^2}-3\cdot \frac{1}{(s+5)}$

Therefore,

$x(t)=3-15t{e}^{-5t}-3e^{-5t}$

This is the forced response.

At zero input condition, using the partial fraction expansion.

$X(s)=\frac{10s+104}{{(s+5)}^2}=\frac{10}{(s+5)}+\frac{54}{{(s+5)}^2}$

Therefore,

$x(t)=10e^{-5t}+54t{e}^{-5t}$

This is the free response.

Conclusion:

Steady state response is $3$.

Transient state response is $39t{e}^{-5t}+7e^{-5t}$

Free response is $10e^{-5t}+54t{e}^{-5t}$.

Forced response is $3-15t{e}^{-5t}-3e^{-5t}$.

To determine

(c)

The steady state, transient, free and forced responses for the following nonhomogenous equation $\ddot{x}+25x=100$.

Answer to Problem 2.23P

Steady state response is $4$.

Transient state response is$6\cos 5t+\frac{4}{5}\sin 5t$

Free response is $10\cos 5t+\frac{4}{5}\sin 5t$.

Forced response is $4-4\cos 5t$.

Explanation of Solution

Given:

The given equation is as:

$\ddot{x}+25x=100$.

With initial conditions as follows:

$x(0)=10$ and $\dot{x}(0)=4$.

Concept Used:

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

$\ddot{x}+25x=100$

Taking Laplace of the above equation that is,

$\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+25X(s)=\frac{100}{s}$

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

Therefore,

$X(s)=\frac{10s^2+4s+100}{s(s^2+5^2)}=\frac{10s^2+4s+100}{s(s^2+5^2)}$

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

$X(s)=\frac{4}{s}+6\cdot \frac{s}{(s^2+5^2)}+\frac{4}{5}\cdot \frac{5}{(s^2+5^2)}$

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:

$\ddot{x}+25x=0$

Therefore,

$X(s)=\frac{10s+4}{(s^2+5^2)}$

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

$X(s)=\frac{100}{s(s^2+25)}$

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

Calculation:

The equation to be solved is as:

$\ddot{x}+25x=100$

By taking the Laplace of this equation that is,

$\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+25X(s)=\frac{100}{s}$

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

$\begin{array}{l}\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+25X(s)=\frac{100}{s}\\ \Rightarrow X(s)(s^2+25)=\frac{100}{s}+10s+4\\ \Rightarrow X(s)=\frac{10s^2+4s+100}{s(s^2+5^2)}=\frac{10s^2+4s+100}{s(s^2+5^2)}\end{array}$

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

$X(s)=\frac{10s^2+4s+100}{s(s^2+5^2)}=\frac{4}{s}+6\cdot \frac{s}{(s^2+5^2)}+\frac{4}{5}\cdot \frac{5}{(s^2+5^2)}$

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

$x(t)=4+6\cos 5t+\frac{4}{5}\sin 5t$

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 $6\cos 5t+\frac{4}{5}\sin 5t$.

At zero initial conditions, using the partial fraction expansion.

$X(s)=\frac{100}{s(s^2+25)}=\frac{4}{s}-4\cdot \frac{s}{(s^2+5^2)}$

Therefore,

$x(t)=4-4\cos 5t$

This is the forced response.

At zero input condition, using the partial fraction expansion.

$X(s)=\frac{10s+4}{(s^2+5^2)}=10\cdot \frac{s}{(s^2+5^2)}+\frac{4}{5}\cdot \frac{5}{(s^2+5^2)}$

Therefore,

$x(t)=10\cos 5t+\frac{4}{5}\sin 5t$

This is the free response.

Conclusion:

Steady state response is $4$.

Transient state response is$6\cos 5t+\frac{4}{5}\sin 5t$

Free response is $10\cos 5t+\frac{4}{5}\sin 5t$.

Forced response is $4-4\cos 5t$.

To determine

(d)

The steady state, transient, free and forced responses for the following non homogenous equation $\ddot{x}+8\dot{x}+65x=130$.

Answer to Problem 2.23P

Steady state response is $2$.

Transient state response is $8e^{-4t}\cos 7t+\frac{36}{7}{e}^{-4t}\sin 7t$

Free response is $10e^{-4t}\cos 7t+\frac{44}{7}{e}^{-4t}\sin 7t$.

Forced response is $2-2e^{-4t}\cos 7t-\frac{8}{7}{e}^{-4t}\sin 7t$.

Explanation of Solution

Given:

The given equation is as:

$\ddot{x}+8\dot{x}+65x=130$

With initial conditions as follows:

$x(0)=10$ and $\dot{x}(0)=4$.

Concept Used:

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

$\ddot{x}+8\dot{x}+65x=130$

Taking Laplace of the above equation that is,

$\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+8\left(sX(s)-x(0)\right)+65X(s)=\frac{130}{s}$

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

Therefore,

$X(s)=\frac{10s^2+84s+130}{s\left(s^2+8s+65\right)}=\frac{10s^2+84s+130}{s\left({(s+4)}^2+{(7)}^2\right)}$

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

$X(s)=\frac{2}{s}+8\cdot \frac{(s+4)}{\left({(s+4)}^2+{(7)}^2\right)}+\frac{36}{7}\cdot \frac{7}{\left({(s+4)}^2+{(7)}^2\right)}$

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:

$\ddot{x}+8\dot{x}+65x=0$

Therefore,

$X(s)=\frac{10s+84}{\left(s^2+8s+65\right)}$

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

$X(s)=\frac{130}{s\left(s^2+8s+65\right)}$

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

Calculation:

The equation to be solved is as:

$\ddot{x}+8\dot{x}+65x=130$

By taking the Laplace of this equation that is,

$\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+8\left(sX(s)-x(0)\right)+65X(s)=\frac{130}{s}$

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

$\begin{array}{l}\left(s^{2}X(s)-sx(0)-\dot{x}(0)\right)+8\left(sX(s)-x(0)\right)+65X(s)=\frac{130}{s}\\ \Rightarrow X(s)\left(s^2+8s+65\right)=\frac{130}{s}+10s+84\\ \Rightarrow X(s)=\frac{10s^2+84s+130}{s\left(s^2+8s+65\right)}=\frac{10s^2+84s+130}{s\left({(s+4)}^2+{(7)}^2\right)}\end{array}$

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

$X(s)=\frac{2}{s}+8\cdot \frac{(s+4)}{\left({(s+4)}^2+{(7)}^2\right)}+\frac{36}{7}\cdot \frac{7}{\left({(s+4)}^2+{(7)}^2\right)}$

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

$x(t)=2+8e^{-4t}\cos 7t+\frac{36}{7}{e}^{-4t}\sin 7t$

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 is$8e^{-4t}\cos 7t+\frac{36}{7}{e}^{-4t}\sin 7t$

At zero initial conditions, using the partial fraction expansion.

$X(s)=\frac{130}{s\left(s^2+8s+65\right)}=\frac{2}{s}-2\cdot \frac{(s+4)}{\left({(s+4)}^2+{(7)}^2\right)}-\frac{8}{7}\cdot \frac{7}{\left({(s+4)}^2+{(7)}^2\right)}$

Therefore,

$x(t)=2-2e^{-4t}\cos 7t-\frac{8}{7}{e}^{-4t}\sin 7t$

This is the forced response.

At zero input condition, using the partial fraction expansion.

$X(s)=\frac{10s+84}{\left(s^2+8s+65\right)}=10\cdot \frac{(s+4)}{\left({(s+4)}^2+{(7)}^2\right)}+\frac{44}{7}\cdot \frac{7}{\left({(s+4)}^2+{(7)}^2\right)}$

Therefore,

$x(t)=10e^{-4t}\cos 7t+\frac{44}{7}{e}^{-4t}\sin 7t$

This is the free response.

Conclusion:

Steady state response is $2$.

Transient state response is$8e^{-4t}\cos 7t+\frac{36}{7}{e}^{-4t}\sin 7t$

Free response is $10e^{-4t}\cos 7t+\frac{44}{7}{e}^{-4t}\sin 7t$.

Forced response is $2-2e^{-4t}\cos 7t-\frac{8}{7}{e}^{-4t}\sin 7t$.