Chapter 8, Problem 8.21P

To determine

(a)

The response $x\left(t\right)$ for the model equation $3\ddot{x}+21\dot{x}+30x=f\left(t\right)$, where $f\left(t\right)$ is a unit step function.

Answer to Problem 8.21P

The response $x\left(t\right)$ is $x(t)=\frac{1}{30}-\frac{1}{18}{e}^{-2t}+\frac{1}{45}{e}^{-5t}$.

Explanation of Solution

Given:

The given model equation is as:

$3\ddot{x}+21\dot{x}+30x=f\left(t\right)$

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:

$3\ddot{x}+21\dot{x}+30x=f\left(t\right)$

By taking the Laplace of this equation that is,

$\begin{array}{l}3\ddot{x}+21\dot{x}+30x=f\left(t\right)\\ \Rightarrow 3\ddot{x}+21\dot{x}+30x=u_s\left(t\right)\because f\left(t\right)=u_s\left(t\right)\\ \Rightarrow 3\left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+21\left(sX(s)-x(0)\right)+30X(s)=\frac{1}{s}\end{array}$

$\begin{array}{l}\Rightarrow 3\left(s^{2}X(s)-0-0\right)+21\left(sX(s)-0\right)+30X(s)=\frac{1}{s}\because x(0)=0,\dot{x}\left(0\right)=0\\ \Rightarrow X(s).(3s^2+21s+30)=\frac{1}{s}\\ \Rightarrow X(s)=\frac{1}{3s(s^2+7s+10)}=\frac{1}{3s\left(s+2\right)\left(s+5\right)}\end{array}$

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

$\begin{array}{l}X(s)=\frac{1}{3s\left(s+2\right)\left(s+5\right)}=\frac{A}{s}+\frac{B}{\left(s+2\right)}+\frac{C}{\left(s+5\right)}\\ \Rightarrow \frac{1}{3s\left(s+2\right)\left(s+5\right)}=\frac{A\left(s+5\right)\left(s+2\right)+Bs\left(s+5\right)+Cs\left(s+2\right)}{s\left(s+2\right)\left(s+5\right)}\\ \Rightarrow \frac{1}{3}=s^2\left(A+B+C\right)+s\left(7A+5B+2C\right)+10A\\ \Rightarrow A+B+C=0,7A+5B+2C=0,A=\frac{1}{30}\\ \Rightarrow A=\frac{1}{30},B=-\frac{1}{18},C=\frac{1}{45}\end{array}$

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

$\begin{array}{l}X(s)=\frac{1}{30s}-\frac{1}{18\left(s+2\right)}+\frac{1}{45\left(s+5\right)}\\ \Rightarrow x(t)=\frac{1}{30}-\frac{1}{18}{e}^{-2t}+\frac{1}{45}{e}^{-5t}\end{array}$.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{1}{30}-\frac{1}{18}{e}^{-2t}+\frac{1}{45}{e}^{-5t}$.

To determine

(b)

The response $x\left(t\right)$ for the model equation $5\ddot{x}+20\dot{x}+20x=f\left(t\right)$, where $f\left(t\right)$ is a unit step function.

Answer to Problem 8.21P

The response $x\left(t\right)$ is $x(t)=\frac{1}{20}-\frac{1}{20}{e}^{-2t}-\frac{1}{10}t{e}^{-2t}$.

Explanation of Solution

Given:

The given model equation is as:

$5\ddot{x}+20\dot{x}+20x=f\left(t\right)$

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:

$5\ddot{x}+20\dot{x}+20x=f\left(t\right)$

By taking the Laplace of this equation that is,

$\begin{array}{l}5\ddot{x}+20\dot{x}+20x=f\left(t\right)\\ \Rightarrow 5\ddot{x}+20\dot{x}+20x=u_s\left(t\right)\because f\left(t\right)=u_s\left(t\right)\\ \Rightarrow 5\left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+20\left(sX(s)-x(0)\right)+20X(s)=\frac{1}{s}\\ \Rightarrow 5\left(s^{2}X(s)-0-0\right)+20\left(sX(s)-0\right)+20X(s)=\frac{1}{s}\because x(0)=0,\dot{x}\left(0\right)=0\\ \Rightarrow X(s).(5s^2+20s+20)=\frac{1}{s}\end{array}$

$\Rightarrow X(s)=\frac{1}{5s(s^2+4s+4)}=\frac{1}{5s{\left(s+2\right)}^2}$

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

$\begin{array}{l}X(s)=\frac{1}{5s{\left(s+2\right)}^2}=\frac{A}{s}+\frac{B}{\left(s+2\right)}+\frac{C}{{\left(s+2\right)}^2}\\ \Rightarrow \frac{1}{5s{\left(s+2\right)}^2}=\frac{A{\left(s+2\right)}^2+Bs\left(s+2\right)+Cs}{s{\left(s+2\right)}^2}\\ \Rightarrow \frac{1}{5}=s^2\left(A+B\right)+s\left(4A+2B+C\right)+4A\\ \Rightarrow A+B=0,4A+2B+C=0,A=\frac{1}{20}\\ \Rightarrow A=\frac{1}{20},B=-\frac{1}{20},C=-\frac{1}{10}\end{array}$

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

$\begin{array}{l}X(s)=\frac{1}{20s}-\frac{1}{20\left(s+2\right)}-\frac{1}{10{\left(s+2\right)}^2}\\ \Rightarrow x(t)=\frac{1}{20}-\frac{1}{20}{e}^{-2t}-\frac{1}{10}t{e}^{-2t}\end{array}$.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{1}{20}-\frac{1}{20}{e}^{-2t}-\frac{1}{10}t{e}^{-2t}$.

To determine

(c)

The response $x\left(t\right)$ for the model equation $2\ddot{x}+8\dot{x}+58x=f\left(t\right)$, where $f\left(t\right)$ is a unit step function.

Answer to Problem 8.21P

The response $x\left(t\right)$ is $x(t)=\frac{1}{58}-\frac{1}{58}{e}^{-2t}\cos 5t-\frac{1}{145}{e}^{-2t}\sin 5t$.

Explanation of Solution

Given:

The given model equation is as:

$2\ddot{x}+8\dot{x}+58x=f\left(t\right)$

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:

$2\ddot{x}+8\dot{x}+58x=f\left(t\right)$

By taking the Laplace of this equation that is,

$\begin{array}{l}2\ddot{x}+8\dot{x}+58x=f\left(t\right)\\ \Rightarrow 2\ddot{x}+8\dot{x}+58x=u_s\left(t\right)\because f\left(t\right)=u_s\left(t\right)\\ \Rightarrow 2\left(s^{2}X(s)-sx(0)-\dot{x}\left(0\right)\right)+8\left(sX(s)-x(0)\right)+58X(s)=\frac{1}{s}\\ \Rightarrow 2\left(s^{2}X(s)-0-0\right)+8\left(sX(s)-0\right)+58X(s)=\frac{1}{s}\because x(0)=0,\dot{x}\left(0\right)=0\\ \Rightarrow X(s).(2s^2+8s+58)=\frac{1}{s}\\ \Rightarrow X(s)=\frac{1}{2s(s^2+4s+29)}\end{array}$

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

$\begin{array}{l}X(s)=\frac{1}{2s(s^2+4s+29)}=\frac{A}{s}+\frac{Bs+C}{(s^2+4s+29)}\\ \Rightarrow \frac{1}{2s(s^2+4s+29)}=\frac{A(s^2+4s+29)+s\left(Bs+C\right)}{s(s^2+4s+29)}\\ \Rightarrow \frac{1}{2}=s^2\left(A+B\right)+s\left(4A+C\right)+29A\\ \Rightarrow A+B=0,4A+C=0,A=\frac{1}{58}\\ \Rightarrow A=\frac{1}{58},B=-\frac{1}{58},C=-\frac{2}{29}\end{array}$

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

$\begin{array}{l}X(s)=\frac{1}{58s}-\frac{1}{58}\cdot \frac{\left(s+2\right)}{\left({\left(s+2\right)}^2+{\left(5\right)}^2\right)}-\frac{1}{145}\cdot \frac{5}{\left({\left(s+2\right)}^2+{\left(5\right)}^2\right)}\\ \Rightarrow x(t)=\frac{1}{58}-\frac{1}{58}{e}^{-2t}\cos 5t-\frac{1}{145}{e}^{-2t}\sin 5t\end{array}$.

Conclusion:

The total response $x\left(t\right)$ for the model equation $x(t)=\frac{1}{58}-\frac{1}{58}{e}^{-2t}\cos 5t-\frac{1}{145}{e}^{-2t}\sin 5t$.