Chapter 14, Problem 33E

(a)

To determine

The value of $L^{-1}\left\{G\left(s\right)\right\}$.

Answer to Problem 33E

The value of $L^{-1}\left\{G\left(s\right)\right\}$ is $\underset{\_}{24t{e}^{-4t}u\left(t\right)+6e^{-4t}u\left(t\right)-6e^{-2t}u\left(t\right)}$.

Explanation of Solution

Given data:

The function is given as,

$G\left(s\right)=\frac{3s}{{\left(\frac{s}{2}+2\right)}^2\left(s+2\right)}$

Calculation:

The function is simplified as,

$\begin{array}{c}G\left(s\right)=\frac{3s}{{\left(\frac{s}{2}+2\right)}^2\left(s+2\right)}\\ =\frac{12s}{{\left(s+4\right)}^2\left(s+2\right)}\end{array}$ (1)

Apply the partial fraction on the above expression.

$\frac{12s}{{\left(s+4\right)}^2\left(s+2\right)}=\frac{A}{{\left(s+4\right)}^2}+\frac{B}{\left(s+4\right)}+\frac{C}{\left(s+2\right)}$ (2)

The equation is written as,

$12s=A\left(s+2\right)+B\left(s+4\right)\left(s+2\right)+C{\left(s+4\right)}^2$ (3)

Substitute $-2$ for $s$ in equation (3).

$\begin{array}{l}-24=A\left(0\right)+B\left(0\right)+C{\left(-2+4\right)}^2\\ -24=4C\\ C=-6\end{array}$

Substitute $-4$ for $s$ in equation (3).

$\begin{array}{l}-48=A\left(-4+2\right)+B\left(0\right)+C\left(0\right)\\ -48=-2A\\ A=24\end{array}$

The simplification of equation (3) is written as,

$12s=\left(B+C\right)s^2+\left(A+6B+8C\right)s+\left(2A+8B+16C\right)$ (4)

Equate the coefficient of $s^2$ of equation (4).

$B+C=0$

Substitute $-6$ for $C$ in the above expression.

$\begin{array}{l}B-6=0\\ B=6\end{array}$

Substitute $24$ for $A$, $6$ for $B$ and $-6$ for $C$ in equation (2).

$\begin{array}{c}\frac{12s}{{\left(s+4\right)}^2\left(s+2\right)}=\frac{24}{{\left(s+4\right)}^2}+\frac{6}{\left(s+4\right)}+\frac{-6}{\left(s+2\right)}\\ =\frac{24}{{\left(s+4\right)}^2}+\frac{6}{\left(s+4\right)}-\frac{6}{\left(s+2\right)}\end{array}$

Substitute $\frac{24}{{\left(s+4\right)}^2}+\frac{6}{\left(s+4\right)}-\frac{6}{\left(s+2\right)}$ for $\frac{12s}{{\left(s+4\right)}^2\left(s+2\right)}$ in equation (1).

$G\left(s\right)=\frac{24}{{\left(s+4\right)}^2}+\frac{6}{\left(s+4\right)}-\frac{6}{\left(s+2\right)}$

The Laplace transform of $e^{-at}u\left(t\right)$ is written as,

$L\left[e^{-at}u\left(t\right)\right]=\frac{1}{s+a}$

The Laplace transform of $t{e}^{-at}u\left(t\right)$ is written as,

$L\left[t{e}^{-at}u\left(t\right)\right]=\frac{1}{{\left(s+a\right)}^2}$

The properties for Laplace transform are written as,

$L\left\{f_1\left(t\right)+f_2\left(t\right)\right\}=L\left\{f_1\left(t\right)\right\}+L\left\{f_2\left(t\right)\right\}$

$L^{-1}\left\{kF\left(s\right)\right\}=k{L}^{-1}\left\{F\left(s\right)\right\}$

The inverse Laplace of the given function is written as,

$g\left(t\right)=L^{-1}\left\{G\left(s\right)\right\}$ (5)

Substitute $\frac{24}{{\left(s+4\right)}^2}+\frac{6}{\left(s+4\right)}-\frac{6}{\left(s+2\right)}$ for $G\left(s\right)$ in equation (5).

$\begin{array}{c}g\left(t\right)=L^{-1}\left\{\frac{24}{{\left(s+4\right)}^2}+\frac{6}{\left(s+4\right)}-\frac{6}{\left(s+2\right)}\right\}\\ =24L^{-1}\left\{\frac{1}{{\left(s+4\right)}^2}\right\}+6L^{-1}\left\{\frac{1}{\left(s+4\right)}\right\}-6L^{-1}\left\{\frac{1}{\left(s+2\right)}\right\}\\ =24t{e}^{-4t}u\left(t\right)+6e^{-4t}u\left(t\right)-6e^{-2t}u\left(t\right)\end{array}$

Conclusion:

Therefore, the value of $L^{-1}\left\{G\left(s\right)\right\}$ is $\underset{\_}{24t{e}^{-4t}u\left(t\right)+6e^{-4t}u\left(t\right)-6e^{-2t}u\left(t\right)}$.

(b)

To determine

The value of $L^{-1}\left\{G\left(s\right)\right\}$.

Answer to Problem 33E

The value of $L^{-1}\left\{G\left(s\right)\right\}$ is $\underset{\_}{3\delta \left(t\right)+3.75t{e}^{-5t}u\left(t\right)-2.625e^{-5t}u\left(t\right)+2.625e^{-7t}u\left(t\right)}$.

Explanation of Solution

Given data:

The function is given as,

$G\left(s\right)=3-\frac{3s}{\left(2s^2+24s+70\right)\left(s+5\right)}$

Calculation:

The function is simplified as,

$\begin{array}{c}G\left(s\right)=3-\frac{3s}{\left(2s^2+24s+70\right)\left(s+5\right)}\\ =3-\frac{3s}{2\left(s^2+12s+35\right)\left(s+5\right)}\\ =3-\frac{1.5s}{\left(s+5\right)\left(s+7\right)\left(s+5\right)}\\ =3-\frac{1.5s}{\left(s+7\right){\left(s+5\right)}^2}\end{array}$ (6)

Apply the partial fraction on the above expression.

$\frac{1.5s}{\left(s+7\right){\left(s+5\right)}^2}=\frac{A}{{\left(s+5\right)}^2}+\frac{B}{\left(s+5\right)}+\frac{C}{\left(s+7\right)}$ (7)

The equation is written as,

$1.5s=A\left(s+7\right)+B\left(s+7\right)\left(s+5\right)+C{\left(s+5\right)}^2$ (8)

Substitute $-7$ for $s$ in equation (8).

$\begin{array}{l}-10.5=A\left(0\right)+B\left(0\right)+C{\left(-7+5\right)}^2\\ -10.5=4C\\ C=-2.625\end{array}$

Substitute $-5$ for $s$ in equation (8).

$\begin{array}{l}-7.5=A\left(-5+7\right)+B\left(0\right)+C\left(0\right)\\ -7.5=2A\\ A=-3.75\end{array}$

The simplification of equation (8) is written as,

$1.5s=\left(B+C\right)s^2+\left(A+12B+10C\right)s+\left(7A+35B+25C\right)$ (9)

Equate the coefficient of $s^2$ of equation (4).

$B+C=0$

Substitute $-2.625$ for $C$ in the above expression.

$\begin{array}{l}B-2.625=0\\ B=2.625\end{array}$

Substitute $-3.75$ for $A$, $2.625$ for $B$ and $-2.625$ for $C$ in equation (7).

$\begin{array}{c}\frac{1.5s}{\left(s+7\right){\left(s+5\right)}^2}=\frac{-3.75}{{\left(s+5\right)}^2}+\frac{2.625}{\left(s+5\right)}+\frac{-2.625}{\left(s+7\right)}\\ =-\frac{3.75}{{\left(s+5\right)}^2}+\frac{2.625}{\left(s+5\right)}-\frac{2.625}{\left(s+7\right)}\end{array}$

Substitute $-\frac{3.75}{{\left(s+5\right)}^2}+\frac{2.625}{\left(s+5\right)}-\frac{2.625}{\left(s+7\right)}$ for $\frac{1.5s}{\left(s+7\right){\left(s+5\right)}^2}$ in equation (6).

$\begin{array}{c}G\left(s\right)=3-\left[-\frac{3.75}{{\left(s+5\right)}^2}+\frac{2.625}{\left(s+5\right)}-\frac{2.625}{\left(s+7\right)}\right]\\ =3+\frac{3.75}{{\left(s+5\right)}^2}-\frac{2.625}{\left(s+5\right)}+\frac{2.625}{\left(s+7\right)}\end{array}$

The Laplace transform of $\delta \left(t\right)$ is written as,

$L\left[\delta \left(t\right)\right]=1$

Substitute $3+\frac{3.75}{{\left(s+5\right)}^2}-\frac{2.625}{\left(s+5\right)}+\frac{2.625}{\left(s+7\right)}$ for $G\left(s\right)$ in equation (5).

$\begin{array}{c}g\left(t\right)=L^{-1}\left\{3+\frac{3.75}{{\left(s+5\right)}^2}-\frac{2.625}{\left(s+5\right)}+\frac{2.625}{\left(s+7\right)}\right\}\\ =3L^{-1}\left\{1\right\}+3.75L^{-1}\left\{\frac{1}{{\left(s+5\right)}^2}\right\}-2.625L^{-1}\left\{\frac{1}{\left(s+5\right)}\right\}+2.625L^{-1}\left\{\frac{1}{\left(s+7\right)}\right\}\\ =3\delta \left(t\right)+3.75t{e}^{-5t}u\left(t\right)-2.625e^{-5t}u\left(t\right)+2.625e^{-7t}u\left(t\right)\end{array}$

Conclusion:

Therefore, the value of $L^{-1}\left\{G\left(s\right)\right\}$ is $\underset{\_}{3\delta \left(t\right)+3.75t{e}^{-5t}u\left(t\right)-2.625e^{-5t}u\left(t\right)+2.625e^{-7t}u\left(t\right)}$.

(c)

To determine

The value of $L^{-1}\left\{G\left(s\right)\right\}$.

Answer to Problem 33E

The value of $L^{-1}\left\{G\left(s\right)\right\}$ is $\underset{\_}{2\delta \left(t\right)-e^{-100t}u\left(t\right)+\cos \left(10t\right)u\left(t\right)}$.

Explanation of Solution

Given data:

The function is given as,

$G\left(s\right)=2-\frac{1}{\left(s+100\right)}+\frac{1}{s^2+100}$

Calculation:

The function is simplified as,

$\begin{array}{c}G\left(s\right)=2-\frac{1}{\left(s+100\right)}+\frac{1}{s^2+100}\\ =2-\frac{1}{\left(s+100\right)}+\frac{1}{s^2+{10}^2}\end{array}$

The Laplace transform of $\cos \omega tu\left(t\right)$ is written as,

$L\left[\cos \omega tu\left(t\right)\right]=\frac{1}{s^2+{\omega }^2}$

Substitute $2-\frac{1}{\left(s+100\right)}+\frac{1}{s^2+{10}^2}$ for $G\left(s\right)$ in equation (5).

$\begin{array}{c}g\left(t\right)=L^{-1}\left\{2-\frac{1}{\left(s+100\right)}+\frac{1}{s^2+{10}^2}\right\}\\ =2L^{-1}\left\{1\right\}-L^{-1}\left\{\frac{1}{\left(s+100\right)}\right\}+L^{-1}\left\{\frac{1}{s^2+{10}^2}\right\}\\ =2\delta \left(t\right)-e^{-100t}u\left(t\right)+\cos \left(10t\right)u\left(t\right)\end{array}$

Conclusion:

Therefore, the value of $L^{-1}\left\{G\left(s\right)\right\}$ is $\underset{\_}{2\delta \left(t\right)-e^{-100t}u\left(t\right)+\cos \left(10t\right)u\left(t\right)}$.

(d)

To determine

The value of $L^{-1}\left\{G\left(s\right)\right\}$.

Answer to Problem 33E

The value of $L^{-1}\left\{G\left(s\right)\right\}$ is $\underset{\_}{tu\left(2t\right)}$.

Explanation of Solution

Given data:

The function is given as,

$G\left(s\right)=L\left[tu\left(2t\right)\right]$

Calculation:

Substitute $L\left[tu\left(2t\right)\right]$ for $G\left(s\right)$ in equation (5).

$\begin{array}{c}g\left(t\right)=L^{-1}\left\{L\left[tu\left(2t\right)\right]\right\}\\ =tu\left(2t\right)\end{array}$

Conclusion:

Therefore, the value of $L^{-1}\left\{G\left(s\right)\right\}$ is $\underset{\_}{tu\left(2t\right)}$.