Chapter 8, Problem 8.22P

To determine

(a)

The response $x\left(t\right)$ for the model equation $3\ddot{x}+21\dot{x}+30x=4t$.

Answer to Problem 8.22P

The response $x\left(t\right)$ is $x(t)=\frac{2}{15}t-\frac{7}{75}+\frac{1}{9}{e}^{-2t}-\frac{4}{225}{e}^{-5t}$.

Explanation of Solution

Given:

The given model equation is as:

$3\ddot{x}+21\dot{x}+30x=4t$

With initial conditions as follows:

$x(0)=0$ and $\dot{x}(0)=0$

Concept Used:

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

Calculation:

Equation to be solved is as:

$3\ddot{x}+21\dot{x}+30x=4t$

By taking the Laplace of this equation that is,

$\begin{array}{l}3\ddot{x}+21\dot{x}+30x=4t\\ \Rightarrow 3\left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+21\left(sX(s)-x(0)\right)+30X(s)=\frac{4}{s^2}\\ \Rightarrow 3\left(s^{2}X(s)-0-0\right)+21\left(sX(s)-0\right)+30X(s)=\frac{4}{s^2}\because x(0)=0,\dot{x}\left(0\right)=0\end{array}$

$\begin{array}{l}\Rightarrow X(s).(3s^2+21s+30)=\frac{4}{s^2}\\ \Rightarrow X(s)=\frac{4}{3s^2(s^2+7s+10)}=\frac{4}{3s^2\left(s+2\right)\left(s+5\right)}\end{array}$

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

$\begin{array}{l}X(s)=\frac{4}{3s^2\left(s+2\right)\left(s+5\right)}=\frac{A}{s^2}+\frac{B}{s}+\frac{C}{\left(s+2\right)}+\frac{D}{\left(s+5\right)}\\ \Rightarrow \frac{4}{3s^2\left(s+2\right)\left(s+5\right)}=\frac{A\left(s+5\right)\left(s+2\right)+Bs\left(s+5\right)\left(s+2\right)+C{s}^2\left(s+5\right)+D{s}^2\left(s+2\right)}{s^2\left(s+2\right)\left(s+5\right)}\\ \Rightarrow \frac{4}{3}=s^3\left(B+C+D\right)+s^2\left(A+7B+5C+2D\right)+s\left(7A+10B\right)+10A\\ \Rightarrow B+C+D=0,A+7B+5C+2D=0,7A+10B=0,A=\frac{4}{30}\\ \Rightarrow A=\frac{2}{15},B=-\frac{7}{75},C=\frac{1}{9},D=-\frac{4}{225}\end{array}$

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

$\begin{array}{l}X(s)=\frac{2}{15s^2}-\frac{7}{75s}+\frac{1}{9\left(s+2\right)}-\frac{4}{225\left(s+5\right)}\\ \Rightarrow x(t)=\frac{2}{15}t-\frac{7}{75}+\frac{1}{9}{e}^{-2t}-\frac{4}{225}{e}^{-5t}\end{array}$.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{2}{15}t-\frac{7}{75}+\frac{1}{9}{e}^{-2t}-\frac{4}{225}{e}^{-5t}$.

To determine

(b)

The response $x\left(t\right)$ for the model equation $5\ddot{x}+20\dot{x}+20x=7t$.

Answer to Problem 8.22P

The response $x\left(t\right)$ is $x(t)=\frac{7}{20}t-\frac{7}{20}+\frac{7}{20}{e}^{-2t}+\frac{7}{20}t{e}^{-2t}$.

Explanation of Solution

Given:

The given model equation is as:

$5\ddot{x}+20\dot{x}+20x=7t$

With initial conditions as follows:

$x(0)=0$ and $\dot{x}(0)=0$

Concept Used:

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

Calculation:

Equation to be solved is as:

$5\ddot{x}+20\dot{x}+20x=7t$

By taking the Laplace of this equation that is,

$\begin{array}{l}5\ddot{x}+20\dot{x}+20x=7t\\ \Rightarrow 5\left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+20\left(sX(s)-x(0)\right)+20X(s)=\frac{7}{s^2}\\ \Rightarrow 5\left(s^{2}X(s)-0-0\right)+20\left(sX(s)-0\right)+20X(s)=\frac{7}{s^2}\because x(0)=0,\dot{x}\left(0\right)=0\\ \Rightarrow X(s).(5s^2+20s+20)=\frac{7}{s^2}\\ \Rightarrow X(s)=\frac{7}{5s^2(s^2+4s+4)}=\frac{7}{5s^2{\left(s+2\right)}^2}\end{array}$

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

$\begin{array}{l}X(s)=\frac{7}{5s^2{\left(s+2\right)}^2}=\frac{A}{s^2}+\frac{B}{s}+\frac{C}{\left(s+2\right)}+\frac{D}{{\left(s+2\right)}^2}\\ \Rightarrow \frac{7}{5s^2{\left(s+2\right)}^2}=\frac{A{\left(s+2\right)}^2+Bs{\left(s+2\right)}^2+C{s}^2\left(s+2\right)+D{s}^2}{s^2{\left(s+2\right)}^2}\\ \Rightarrow \frac{7}{5}=s^3\left(B+C\right)+s^2\left(A+4B+2C+D\right)+s\left(4A+4B\right)+4A\\ \Rightarrow B+C=0,A+4B+2C+D=0,4A+4B=0,A=\frac{7}{20}\\ \Rightarrow A=\frac{7}{20},B=-\frac{7}{20},C=\frac{7}{20},D=\frac{7}{20}\end{array}$

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

$\begin{array}{l}X(s)=\frac{7}{20s^2}-\frac{7}{20s}+\frac{7}{20\left(s+2\right)}+\frac{7}{20{\left(s+2\right)}^2}\\ \Rightarrow x(t)=\frac{7}{20}t-\frac{7}{20}+\frac{7}{20}{e}^{-2t}+\frac{7}{20}t{e}^{-2t}\end{array}$.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{7}{20}t-\frac{7}{20}+\frac{7}{20}{e}^{-2t}+\frac{7}{20}t{e}^{-2t}$.

To determine

(c)

The response $x\left(t\right)$ for the model equation $2\ddot{x}+8\dot{x}+58x=5t$.

Answer to Problem 8.22P

The response $x\left(t\right)$ is $x(t)=\frac{5}{58}t-\frac{10}{841}+\frac{10}{841}{e}^{-2t}\cos 5t-\frac{21}{1682}{e}^{-2t}\sin 5t$.

Explanation of Solution

Given:

The given model equation is as:

$2\ddot{x}+8\dot{x}+58x=5t$

With initial conditions as follows:

$x(0)=0$ and $\dot{x}(0)=0$

Concept Used:

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

Calculation:

Equation to be solved is as:

$2\ddot{x}+8\dot{x}+58x=5t$

By taking the Laplace of this equation that is,

$\begin{array}{l}2\ddot{x}+8\dot{x}+58x=5t\\ \Rightarrow 2\left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+8\left(sX(s)-x(0)\right)+58X(s)=\frac{5}{s^2}\\ \Rightarrow 2\left(s^{2}X(s)-0-0\right)+8\left(sX(s)-0\right)+58X(s)=\frac{5}{s^2}\because x(0)=0,\dot{x}\left(0\right)=0\\ \Rightarrow X(s).(2s^2+8s+58)=\frac{5}{s^2}\\ \Rightarrow X(s)=\frac{5}{2s^2(s^2+4s+29)}\end{array}$

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

$\begin{array}{l}X(s)=\frac{5}{2s^2(s^2+4s+29)}=\frac{A}{s^2}+\frac{B}{s}+\frac{Cs+D}{(s^2+4s+29)}\\ \Rightarrow \frac{5}{2s^2(s^2+4s+29)}=\frac{A(s^2+4s+29)+Bs(s^2+4s+29)+\left(Cs+D\right)s^2}{s^2(s^2+4s+29)}\\ \Rightarrow \frac{5}{2}=s^3\left(B+C\right)+s^2\left(A+4B+D\right)+s\left(4A+29B\right)+29A\end{array}$

$\begin{array}{l}\Rightarrow B+C=0,A+4B+D=0,4A+29B=0,A=\frac{5}{58}\\ \Rightarrow A=\frac{5}{58},B=-\frac{10}{841},C=\frac{10}{841},D=-\frac{65}{1682}\end{array}$

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

$\begin{array}{l}X(s)=\frac{5}{58s^2}-\frac{10}{841s}+\frac{1}{1682}\cdot \frac{\left(20s-65\right)}{(s^2+4s+29)}\\ \Rightarrow X(s)=\frac{5}{58s^2}-\frac{10}{841s}+\frac{10}{841}\cdot \frac{\left(s+2\right)}{\left({\left(s+2\right)}^2+{\left(5\right)}^2\right)}-\frac{21}{1682}\cdot \frac{5}{\left({\left(s+2\right)}^2+{\left(5\right)}^2\right)}\\ \Rightarrow x(t)=\frac{5}{58}t-\frac{10}{841}+\frac{10}{841}{e}^{-2t}\cos 5t-\frac{21}{1682}{e}^{-2t}\sin 5t\end{array}$.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{5}{58}t-\frac{10}{841}+\frac{10}{841}{e}^{-2t}\cos 5t-\frac{21}{1682}{e}^{-2t}\sin 5t$.