Chapter 15, Problem 30P

(a)

To determine

Find the inverse Laplace transform for the function $F_1\left(s\right)=\frac{6s^2+8s+3}{s\left(s^2+2s+5\right)}$.

Answer to Problem 30P

The inverse Laplace transform for the given function is $f_1\left(t\right)=\left[\frac{3}{5}+\frac{27}{5}{e}^{-t}\cos 2t+\frac{7}{10}{e}^{-t}\sin 2t\right]u\left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$F_1\left(s\right)=\frac{6s^2+8s+3}{s\left(s^2+2s+5\right)}$ (1)

Formula used:

Write the general expression for the inverse Laplace transform.

$f\left(t\right)={\mathcal{L}}^{-1}\left[F\left(s\right)\right]$ (2)

Write the general expression to find the inverse Laplace transform function.

${\mathcal{L}}^{-1}\left[\frac{s+a}{{\left(s+a\right)}^2+{\omega }^2}\right]=e^{-at}\cos \omega tu\left(t\right)$ (3)

${\mathcal{L}}^{-1}\left[\frac{\omega }{{\left(s+a\right)}^2+{\omega }^2}\right]=e^{-at}\sin \omega tu\left(t\right)$ (4)

${\mathcal{L}}^{-1}\left[\frac{1}{s}\right]=u\left(t\right)$ (5)

Here,

$\omega$ is the angular frequency,

$s+a$ is the frequency shift or frequency translation, and

$s$ is a complex variable.

Calculation:

Expand $F_1\left(s\right)$ using partial fraction.

$F_1\left(s\right)=\frac{6s^2+8s+3}{s\left(s^2+2s+5\right)}=\frac{A}{s}+\frac{Bs+C}{s^2+2s+5}$ (6)

Here,

A, B, and C are the constants.

Now, to find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{6s^2+8s+3}{s\left(s^2+2s+5\right)}=\frac{A}{s}+\frac{Bs+C}{s^2+2s+5}\\ \frac{6s^2+8s+3}{s\left(s^2+2s+5\right)}=\frac{A\left(s^2+2s+5\right)+\left(Bs+C\right)s}{s\left(s^2+2s+5\right)}\\ 6s^2+8s+3=A\left(s^2+2s+5\right)+\left(Bs+C\right)s\\ 6s^2+8s+3=A{s}^2+2As+5A+B{s}^2+Cs\end{array}$

Reduce the equation as follows,

$6s^2+8s+3=\left(A+B\right)s^2+\left(2A+C\right)s+5A$ (7)

Equating the coefficients of $s^2$ in equation (7) on the both sides.

$6=A+B$ (8)

Equating the coefficients of $s$ in equation (7) on the both sides.

$8=2A+C$ (9)

Equating the coefficients of constant term in equation (7).

$3=5A$

$A=\frac{3}{5}$ (10)

Substitute equation (10) in equation (8) to find the constant B.

$\begin{array}{l}6=\frac{3}{5}+B\\ B=6-\frac{3}{5}\\ B=\frac{30-3}{5}\\ B=\frac{27}{5}\end{array}$

Substitute equation (10) in equation (9) to find the constant C.

$\begin{array}{l}8=2\left(\frac{3}{5}\right)+C\\ C=8-\frac{6}{5}\\ C=\frac{40-6}{5}\\ C=\frac{34}{5}\end{array}$

Substitute $\frac{3}{5}$ for A, $\frac{27}{5}$ for B and $\frac{34}{5}$ for C in equation (6) to find $F_1\left(s\right)$.

$\begin{array}{c}{F}_1\left(s\right)=\frac{\left(\frac{3}{5}\right)}{s}+\frac{\left(\frac{27}{5}\right)s+\left(\frac{34}{5}\right)}{s^2+2s+5}\\ =\left(\frac{3}{5}\right)\frac{1}{s}+\left(\frac{1}{5}\right)\frac{27s+34}{s^2+2s+1+4}\\ =\left(\frac{3}{5}\right)\frac{1}{s}+\left(\frac{1}{5}\right)\frac{27s+27+7}{{\left(s+1\right)}^2+2^2}\left\{\because a^2+2ab+b^2={\left(a+b\right)}^2\right\}\\ =\left(\frac{3}{5}\right)\frac{1}{s}+\left(\frac{1}{5}\right)\frac{27s+27}{{\left(s+1\right)}^2+2^2}+\left(\frac{1}{5}\right)\frac{7}{{\left(s+1\right)}^2+2^2}\end{array}$

Reduce the equation as follows,

$F_1\left(s\right)=\left(\frac{3}{5}\right)\frac{1}{s}+\left(\frac{27}{5}\right)\frac{s+1}{{\left(s+1\right)}^2+2^2}+\left(\frac{1}{5}\right)\frac{7\left(\frac{2}{2}\right)}{{\left(s+1\right)}^2+2^2}$

$F_1\left(s\right)=\left(\frac{3}{5}\right)\frac{1}{s}+\left(\frac{27}{5}\right)\frac{s+1}{{\left(s+1\right)}^2+2^2}+\left(\frac{7}{10}\right)\frac{2}{{\left(s+1\right)}^2+2^2}$ (11)

Apply inverse Laplace transform given in equation (2) to equation (11). Therefore,

$\begin{array}{c}{f}_1\left(t\right)={\mathcal{L}}^{-1}\left[F_1\left(s\right)\right]\\ ={\mathcal{L}}^{-1}\left[\left(\frac{3}{5}\right)\frac{1}{s}+\left(\frac{27}{5}\right)\frac{s+1}{{\left(s+1\right)}^2+2^2}+\left(\frac{7}{10}\right)\frac{2}{{\left(s+1\right)}^2+2^2}\right]\\ ={\mathcal{L}}^{-1}\left[\left(\frac{3}{5}\right)\frac{1}{s}\right]+{\mathcal{L}}^{-1}\left[\left(\frac{27}{5}\right)\frac{s+1}{{\left(s+1\right)}^2+2^2}\right]+{\mathcal{L}}^{-1}\left[\left(\frac{7}{10}\right)\frac{2}{{\left(s+1\right)}^2+2^2}\right]\end{array}$

$f_1\left(t\right)=\frac{3}{5}{\mathcal{L}}^{-1}\left[\frac{1}{s}\right]+\frac{27}{5}{\mathcal{L}}^{-1}\left[\frac{s+1}{{\left(s+1\right)}^2+2^2}\right]+\frac{7}{10}{\mathcal{L}}^{-1}\left[\frac{2}{{\left(s+1\right)}^2+2^2}\right]$ (12)

Apply inverse Laplace transform function given in equation (3), (4) and (5) to equation (12).

$\begin{array}{c}{f}_1\left(t\right)=\frac{3}{5}u\left(t\right)+\frac{27}{5}{e}^{-t}\cos 2tu\left(t\right)+\frac{7}{10}{e}^{-t}\sin 2tu\left(t\right)\\ =\left[\frac{3}{5}+\frac{27}{5}{e}^{-t}\cos 2t+\frac{7}{10}{e}^{-t}\sin 2t\right]u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform for the given function is $f_1\left(t\right)=\left[\frac{3}{5}+\frac{27}{5}{e}^{-t}\cos 2t+\frac{7}{10}{e}^{-t}\sin 2t\right]u\left(t\right)$.

(b)

To determine

Find the inverse Laplace transform for the given function $F_2\left(s\right)=\frac{s^2+5s+6}{{\left(s+1\right)}^2\left(s+4\right)}$.

Answer to Problem 30P

The inverse Laplace transform for the given function is $f_2\left(t\right)=\left[\frac{7}{9}{e}^{-t}+\frac{2}{3}t{e}^{-t}+\frac{2}{9}{e}^{-4t}\right]u\left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$F_2\left(s\right)=\frac{s^2+5s+6}{{\left(s+1\right)}^2\left(s+4\right)}$ (13)

Formula used:

Write the general expressions to find the inverse Laplace transform function.

${\mathcal{L}}^{-1}\left[\frac{1}{{\left(s+a\right)}^2}\right]=t{e}^{-at}u\left(t\right)$ (14)

${\mathcal{L}}^{-1}\left[\frac{1}{s+a}\right]=e^{-at}u\left(t\right)$ (15)

Calculation:

Expand $F_2\left(s\right)$ using partial fraction.

$F_2\left(s\right)=\frac{s^2+5s+6}{{\left(s+1\right)}^2\left(s+4\right)}=\frac{D}{{\left(s+1\right)}^2}+\frac{E}{s+1}+\frac{F}{s+4}$ (16)

Here,

D, E, and F are the constants.

Now, to find the constants by using residue method.

Constant D:

$D={{\left(s+1\right)}^2{F}_2\left(s\right)|}_{s+1=0}$ (17)

Substitute equation (13) in equation (17) to find the constant D.

$\begin{array}{c}D={{\left(s+1\right)}^2\times \frac{s^2+5s+6}{{\left(s+1\right)}^2\left(s+4\right)}|}_{s=-1}\\ ={\frac{s^2+5s+6}{s+4}|}_{s=-1}\\ =\frac{{\left(-1\right)}^2+5\left(-1\right)+6}{-1+4}\\ =\frac{2}{3}\end{array}$

Constant E:

$E={\frac{d}{ds}\left[{\left(s+1\right)}^2{F}_2\left(s\right)\right]|}_{s+1=0}$ (18)

Substitute equation (13) in equation (18) to find the constant E.

$\begin{array}{c}E={\frac{d}{ds}\left[\frac{s^2+5s+6}{s+4}\right]|}_{s=-1}\\ ={\frac{\left(s+4\right)\left(2s+5+0\right)-\left(s^2+5s+6\right)\left(1+0\right)}{{\left(s+4\right)}^2}|}_{s=-1}\\ =\frac{\left(-1+4\right)\left(2\left(-1\right)+5\right)-\left({\left(-1\right)}^2+5\left(-1\right)+6\right)\left(1\right)}{{\left(-1+4\right)}^2}\\ =\frac{7}{9}\end{array}$

Constant F:

$F={\left(s+4\right)F_2\left(s\right)|}_{s+4=0}$ (19)

Substitute equation (13) in equation (19) to find the constant F.

$\begin{array}{c}F={\left(s+4\right)\times \frac{s^2+5s+6}{{\left(s+1\right)}^2\left(s+4\right)}|}_{s=-4}\\ ={\frac{s^2+5s+6}{{\left(s+1\right)}^2}|}_{s=-4}\\ =\frac{{\left(-4\right)}^2+5\left(-4\right)+6}{{\left(-4+1\right)}^2}\\ =\frac{2}{9}\end{array}$

Substitute $\frac{2}{3}$ for D, $\frac{7}{9}$ for E, and $\frac{2}{9}$ for F in equation (16) to find $F_2\left(s\right)$.

$F_2\left(s\right)=\frac{\left(\frac{2}{3}\right)}{{\left(s+1\right)}^2}+\frac{\left(\frac{7}{9}\right)}{s+1}+\frac{\left(\frac{2}{9}\right)}{s+4}$ (20)

Apply inverse Laplace transform given in equation (2) to equation (20). Therefore,

$\begin{array}{c}{f}_2\left(t\right)={\mathcal{L}}^{-1}\left[F_2\left(s\right)\right]\\ ={\mathcal{L}}^{-1}\left[\frac{\left(\frac{2}{3}\right)}{{\left(s+1\right)}^2}+\frac{\left(\frac{7}{9}\right)}{s+1}+\frac{\left(\frac{2}{9}\right)}{s+4}\right]\\ ={\mathcal{L}}^{-1}\left[\frac{\left(\frac{2}{3}\right)}{{\left(s+1\right)}^2}\right]+{\mathcal{L}}^{-1}\left[\frac{\left(\frac{7}{9}\right)}{s+1}\right]+{\mathcal{L}}^{-1}\left[\frac{\left(\frac{2}{9}\right)}{s+4}\right]\end{array}$

$f_2\left(t\right)=\frac{2}{3}{\mathcal{L}}^{-1}\left[\frac{1}{{\left(s+1\right)}^2}\right]+\frac{7}{9}{\mathcal{L}}^{-1}\left[\frac{1}{s+1}\right]+\frac{2}{9}{\mathcal{L}}^{-1}\left[\frac{1}{s+4}\right]$ (21)

Apply inverse Laplace transform function given in equation (14) and (15) to equation (21).

$\begin{array}{c}{f}_2\left(t\right)=\frac{2}{3}t{e}^{-t}u\left(t\right)+\frac{7}{9}{e}^{-t}u\left(t\right)+\frac{2}{9}{e}^{-4t}u\left(t\right)\\ =\frac{7}{9}{e}^{-t}u\left(t\right)+\frac{2}{3}t{e}^{-t}u\left(t\right)+\frac{2}{9}{e}^{-4t}u\left(t\right)\\ =\left[\frac{7}{9}{e}^{-t}+\frac{2}{3}t{e}^{-t}+\frac{2}{9}{e}^{-4t}\right]u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform for the given function is $f_2\left(t\right)=\left[\frac{7}{9}{e}^{-t}+\frac{2}{3}t{e}^{-t}+\frac{2}{9}{e}^{-4t}\right]u\left(t\right)$.

(c)

To determine

Find the inverse Laplace transform for the function $F_3\left(s\right)=\frac{10}{\left(s+1\right)\left(s^2+4s+8\right)}$.

Answer to Problem 30P

The inverse Laplace transform for the given function is $f_3\left(t\right)=\left(2e^{-t}-2e^{-2t}\cos \left(2t\right)-2e^{-2t}\sin \left(2t\right)\right)u\left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$F_3\left(s\right)=\frac{10}{\left(s+1\right)\left(s^2+4s+8\right)}$ (22)

Calculation:

Expand $F_3\left(s\right)$ using partial fraction.

$F_3\left(s\right)=\frac{10}{\left(s+1\right)\left(s^2+4s+8\right)}=\frac{A}{s+1}+\frac{Bs+C}{s^2+4s+8}$ (23)

Here,

A, B, and C are the constants.

Now, to find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{10}{\left(s+1\right)\left(s^2+4s+8\right)}=\frac{A}{s+1}+\frac{Bs+C}{s^2+4s+8}\\ \frac{10}{\left(s+1\right)\left(s^2+4s+8\right)}=\frac{A\left(s^2+4s+8\right)+\left(Bs+C\right)\left(s+1\right)}{\left(s+1\right)\left(s^2+4s+8\right)}\\ 10=A\left(s^2+4s+8\right)+\left(Bs+C\right)\left(s+1\right)\\ 10=A{s}^2+4As+8A+B{s}^2+Bs+Cs+C\end{array}$

Reduce the equation as follows,

$10=\left(A+B\right)s^2+\left(4A+B+C\right)s+\left(8A+C\right)$ (24)

Equating the coefficients of $s^2$ in equation (24) on both sides.

$0=A+B$

$B=-A$ (25)

Equating the coefficients of $s$ in equation (24) on the both sides.

$0=4A+B+C$ (26)

Equating the coefficients of constant term in equation (24) on both sides.

$10=8A+C$

$C=10-8A$ (27)

Substitute equation (25) and (27) in equation (26) to find the constant A.

$\begin{array}{l}0=4A+\left(-A\right)+\left(10-8A\right)\\ 0=4A-A+10-8A\\ 0=-5A+10\\ 5A=10\end{array}$

Rearrange the equation as follows,

$\begin{array}{c}A=\frac{10}{5}\\ =2\end{array}$

Substitute $2$ for A in equation (25) to find the constant B.

$B=-2$

Substitute $2$ for A in equation (27) to find the constant C.

$\begin{array}{c}C=10-8\left(2\right)\\ =10-16\\ =-6\end{array}$

Substitute $2$ for A, $-2$ for B and $-6$ for C in equation (23) to find $F_3\left(s\right)$.

$\begin{array}{c}{F}_3\left(s\right)=\frac{2}{s+1}+\frac{\left(-2\right)s+\left(-6\right)}{s^2+4s+8}\\ =\frac{2}{s+1}+\frac{-2s-6}{s^2+4s+4+4}\\ =\frac{2}{s+1}+\frac{-2s-4-2}{{\left(s+2\right)}^2+2^2}\left\{\because a^2+2ab+b^2={\left(a+b\right)}^2\right\}\\ =\frac{2}{s+1}+\frac{-2\left(s+2\right)-2}{{\left(s+2\right)}^2+2^2}\end{array}$

Reduce the equation as follows,

$F_3\left(s\right)=\frac{2}{s+1}-\frac{2\left(s+2\right)}{{\left(s+2\right)}^2+2^2}-\frac{2}{{\left(s+2\right)}^2+2^2}$ (28)

Apply inverse Laplace transform given in equation (2) to equation (28). Therefore,

$\begin{array}{c}{f}_3\left(t\right)={\mathcal{L}}^{-1}\left[F_3\left(s\right)\right]\\ ={\mathcal{L}}^{-1}\left[\frac{2}{s+1}-\frac{2\left(s+2\right)}{{\left(s+2\right)}^2+2^2}-\frac{2}{{\left(s+2\right)}^2+2^2}\right]\\ ={\mathcal{L}}^{-1}\left[\frac{2}{s+1}\right]-{\mathcal{L}}^{-1}\left[\frac{2\left(s+2\right)}{{\left(s+2\right)}^2+2^2}\right]-{\mathcal{L}}^{-1}\left[\frac{2}{{\left(s+2\right)}^2+2^2}\right]\end{array}$

$f_3\left(t\right)=2{\mathcal{L}}^{-1}\left[\frac{1}{s+1}\right]-2{\mathcal{L}}^{-1}\left[\frac{s+2}{{\left(s+2\right)}^2+2^2}\right]-{\mathcal{L}}^{-1}\left[\frac{2}{{\left(s+2\right)}^2+2^2}\right]$ (29)

Apply inverse Laplace transform function given in equation (3), (4) and (15) to equation (29).

$\begin{array}{c}{f}_3\left(t\right)=2e^{-t}u\left(t\right)-2e^{-2t}\cos \left(2t\right)u\left(t\right)-2e^{-2t}\sin \left(2t\right)u\left(t\right)\\ =\left(2e^{-t}-2e^{-2t}\cos \left(2t\right)-2e^{-2t}\sin \left(2t\right)\right)u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform for the given function is $f_3\left(t\right)=\left(2e^{-t}-2e^{-2t}\cos \left(2t\right)-2e^{-2t}\sin \left(2t\right)\right)u\left(t\right)$.