Chapter 15, Problem 36P

(a)

To determine

Find the inverse Laplace transform for the given function $X\left(s\right)=\frac{3}{s^2\left(s+2\right)\left(s+3\right)}$.

Answer to Problem 36P

The inverse Laplace transform $x\left(t\right)$ for the given function is $\left(-\frac{5}{12}+\frac{1}{2}t+\frac{3}{4}{e}^{-2t}-\frac{1}{3}{e}^{-3t}\right)u\left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$X\left(s\right)=\frac{3}{s^2\left(s+2\right)\left(s+3\right)}$        (1)

Formula used:

Write the general expression for the inverse Laplace transform.

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

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

${ℒ}^{-1}\left[\frac{1}{s+a}\right]=e^{-at}u\left(t\right)$        (3)

${ℒ}^{-1}\left[\frac{n!}{{\left(s+a\right)}^{n+1}}\right]=t^n{e}^{-at}u\left(t\right)$        (4)

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

Here,

$\omega$ is the angular frequency, and

$s$ is a complex variable.

Calculation:

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

$X\left(s\right)=3\left[\frac{1}{s^2\left(s+2\right)\left(s+3\right)}\right]=3\left[\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s+2}+\frac{D}{s+3}\right]$        (6)

Here,

A, B, C, and D are the constants.

Find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{1}{s^2\left(s+2\right)\left(s+3\right)}=\frac{A}{s}+\frac{B}{s^2}+\frac{C}{s+2}+\frac{D}{s+3}\\ \frac{1}{s^2\left(s+2\right)\left(s+3\right)}=\frac{As\left(s+2\right)\left(s+3\right)+B\left(s+2\right)\left(s+3\right)+C{s}^2\left(s+3\right)+D{s}^2\left(s+2\right)}{s^2\left(s+2\right)\left(s+3\right)}\\ 1=As\left(s^2+2s+3s+6\right)+B\left(s^2+2s+3s+6\right)+C{s}^3+3C{s}^2+D{s}^3+2D{s}^2\\ 1=A{s}^3+2A{s}^2+3A{s}^2+6As+B{s}^2+2Bs+3Bs+6B+C{s}^3+3C{s}^2+D{s}^3+2D{s}^2\end{array}$

Reduce the equation as follows,

$1=A{s}^3+5A{s}^2+6As+B{s}^2+5Bs+6B+C{s}^3+3C{s}^2+D{s}^3+2D{s}^2$

$1=\left(A+C+D\right)s^3+\left(5A+B+3C+2D\right)s^2+\left(6A+5B\right)s+6B$        (7)

Equating the coefficients of $s^3$ in equation (7).

$A+C+D=0$        (8)

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

$0=5A+B+3C+2D$        (9)

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

$0=6A+5B$        (10)

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

$1=6B$

$B=\frac{1}{6}$        (11)

Substitute equation (11) in equation (10) to find the constant A.

$\begin{array}{l}0=6A+5\left(\frac{1}{6}\right)\\ 6A=-\frac{5}{6}\end{array}$

$A=-\frac{5}{36}$        (12)

Substitute equation (12) in equation (8).

$\begin{array}{l}-\frac{5}{36}+C+D=0\\ C+D=\frac{5}{36}\end{array}$

$C=\frac{5}{36}-D$        (13)

Substitute equation (11), (12), and (13) in equation (9).

$\begin{array}{l}0=5\left(-\frac{5}{36}\right)+\left(\frac{1}{6}\right)+3\left(\frac{5}{36}-D\right)+2D\\ 0=-\frac{25}{36}+\frac{1}{6}+\frac{5}{12}-3D+2D\\ D=-\frac{25}{36}+\frac{1}{6}+\frac{5}{12}\end{array}$

$D=-\frac{1}{9}$        (14)

Substitute equation (14) in equation (13).

$C=\frac{5}{36}+\frac{1}{9}$

$C=\frac{1}{4}$        (15)

Substitute equation (11), (12), (14) and (15) in equation (6) to find $X\left(s\right)$.

$X\left(s\right)=3\left[\frac{\left(-\frac{5}{36}\right)}{s}+\frac{\left(\frac{1}{6}\right)}{s^2}+\frac{\left(\frac{1}{4}\right)}{s+2}+\frac{\left(-\frac{1}{9}\right)}{s+3}\right]$

$X\left(s\right)=-\frac{5}{12}\left(\frac{1}{s}\right)+\frac{1}{2}\left(\frac{1}{s^2}\right)+\frac{3}{4}\left(\frac{1}{s+2}\right)-\frac{1}{3}\left(\frac{1}{s+3}\right)$        (16)

Apply inverse Laplace transform of equation (2) in equation (16).

$\begin{array}{c}x\left(t\right)={ℒ}^{-1}\left[X\left(s\right)\right]\\ ={ℒ}^{-1}\left[-\frac{5}{12}\left(\frac{1}{s}\right)+\frac{1}{2}\left(\frac{1}{s^2}\right)+\frac{3}{4}\left(\frac{1}{s+2}\right)-\frac{1}{3}\left(\frac{1}{s+3}\right)\right]\end{array}$

$x\left(t\right)=-\frac{5}{12}{ℒ}^{-1}\left[\frac{1}{s}\right]+\frac{1}{2}{ℒ}^{-1}\left[\frac{1}{s^2}\right]+\frac{3}{4}{ℒ}^{-1}\left[\frac{1}{s+2}\right]-\frac{1}{3}{ℒ}^{-1}\left[\frac{1}{s+3}\right]$        (17)

Apply inverse Laplace transform function of equation (3), (4), (5) in equation (17).

$\begin{array}{c}x\left(t\right)=-\frac{5}{12}u\left(t\right)+\frac{1}{2}tu\left(t\right)+\frac{3}{4}{e}^{-2t}u\left(t\right)-\frac{1}{3}{e}^{-3t}u\left(t\right)\\ =\left(-\frac{5}{12}+\frac{1}{2}t+\frac{3}{4}{e}^{-2t}-\frac{1}{3}{e}^{-3t}\right)u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform $x\left(t\right)$ for the given function is $\left(-\frac{5}{12}+\frac{1}{2}t+\frac{3}{4}{e}^{-2t}-\frac{1}{3}{e}^{-3t}\right)u\left(t\right)$.

(b)

To determine

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

Answer to Problem 36P

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

Explanation of Solution

Given data:

The Laplace transform function is,

$Y\left(s\right)=\frac{2}{s{\left(s+1\right)}^2}$        (18)

Calculation:

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

$Y\left(s\right)=2\frac{1}{s{\left(s+1\right)}^2}=2\left[\frac{A}{s}+\frac{B}{s+1}+\frac{C}{{\left(s+1\right)}^2}\right]$        (19)

Here,

A, B, and C are the constants.

Find the constants by using algebraic method.

Consider the partial fraction,

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

Reduce the equation as follows,

$1=A{s}^2+2As+A+B{s}^2+Bs+Cs$

$1=\left(A+B\right)s^2+\left(2A+B+C\right)s+A$        (20)

Equating the coefficients of $s^2$ in equation (20).

$0=A+B$        (21)

Equating the coefficients of $s$ in equation (20).

$0=2A+B+C$        (22)

Equating the coefficients of constant term in equation (20) to find the constant A.

$A=1$        (23)

Substitute equation (23) in equation (21) to find the constant B.

$0=1+B$

$B=-1$        (24)

Substitute equation (23) and (24) in equation (22).

$\begin{array}{l}0=2\left(1\right)+\left(-1\right)+C\\ 0=1+C\end{array}$

$C=-1$        (25)

Substitute equation (23), (24), and (25) in equation (19) to find $Y\left(s\right)$.

$Y\left(s\right)=2\left[\frac{1}{s}-\frac{1}{s+1}-\frac{1}{{\left(s+1\right)}^2}\right]$        (26)

Apply inverse Laplace transform of equation (2) in equation (26).

$y\left(t\right)={ℒ}^{-1}\left[Y\left(s\right)\right]$

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

$y\left(t\right)=2\left[{ℒ}^{-1}\left[\frac{1}{s}\right]-{ℒ}^{-1}\left[\frac{1}{s+1}\right]-{ℒ}^{-1}\left[\frac{1}{{\left(s+1\right)}^2}\right]\right]$        (27)

Apply inverse Laplace transform function of equation (3), (4), (5) in equation (27).

$\begin{array}{c}y\left(t\right)=2\left(u\left(t\right)-e^{-t}u\left(t\right)-t{e}^{-t}u\left(t\right)\right)\\ =2u\left(t\right)-2e^{-t}u\left(t\right)-2t{e}^{-t}u\left(t\right)\\ =\left(2-2e^{-t}-2t{e}^{-t}\right)u\left(t\right)\end{array}$

Conclusion:

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

(c)

To determine

Find the inverse Laplace transform for the given function $Z\left(s\right)=\frac{5}{s\left(s+1\right)\left(s^2+6s+10\right)}$.

Answer to Problem 36P

The inverse Laplace transform $z\left(t\right)$ for the given function is $0.5\left[1-2e^{-t}+e^{-3t}\cos \left(t\right)+e^{-3t}\sin \left(t\right)\right]u\left(t\right)$.

Explanation of Solution

Given data:

Consider the Laplace transform function is,

$Z\left(s\right)=\frac{5}{s\left(s+1\right)\left(s^2+6s+10\right)}$        (28)

Formula used:

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

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

${ℒ}^{-1}\left[\cos \omega tu\left(t\right)\right]=\frac{s}{s^2+{\omega }^2}$        (30)

Calculation:

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

$Z\left(s\right)=5\frac{1}{s\left(s+1\right)\left(s^2+6s+10\right)}=5\left[\frac{A}{s}+\frac{B}{s+1}+\frac{Cs+D}{s^2+6s+10}\right]$        (31)

Here,

A, B, C, and D are the constants.

Find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{1}{s\left(s+1\right)\left(s^2+6s+10\right)}=\frac{A}{s}+\frac{B}{s+1}+\frac{Cs+D}{s^2+6s+10}\\ \frac{1}{s\left(s+1\right)\left(s^2+6s+10\right)}=\frac{A\left(s+1\right)\left(s^2+6s+10\right)+Bs\left(s^2+6s+10\right)+\left(Cs+D\right)s\left(s+1\right)}{s\left(s+1\right)\left(s^2+6s+10\right)}\\ 1=A\left(s+1\right)\left(s^2+6s+10\right)+B\left(s^3+6s^2+10s\right)+\left(Cs+D\right)\left(s^2+s\right)\\ 1=A{s}^3+6A{s}^2+10As+A{s}^2+6As+10A+B{s}^3+6B{s}^2+10Bs+C{s}^3+C{s}^2+D{s}^2+Ds\end{array}$

Reduce the equation as follows,

$1=\left(A+B+C\right)s^3+\left(7A+6B+C+D\right)s^2+\left(16A+10B+D\right)s+10A$        (32)

Equating the coefficients of constant term in equation (32) to find the constant A.

$1=10A$

$A=\frac{1}{10}$        (33)

Equating the coefficients of $s^3$ in equation (32).

$0=A+B+C$        (34)

Equating the coefficients of $s^2$ in equation (32).

$7A+6B+C+D=0$        (35)

Equating the coefficients of $s$ in equation (32).

$16A+10B+D=0$        (36)

Substitute equation (33) in equation (36).

$16\left(\frac{1}{10}\right)+10B+D=0$

$D=-\frac{16}{10}-10B$        (37)

Substitute equation (33) in equation (34).

$0=\frac{1}{10}+B+C$

$B=-\frac{1}{10}-C$        (38)

Substitute equation (33), (37) and (38) in equation (35).

$\begin{array}{l}7\left(\frac{1}{10}\right)+6\left(-\frac{1}{10}-C\right)+C+\left(-\frac{16}{10}-10B\right)=0\\ \frac{7}{10}-\frac{6}{10}-6C+C-\frac{16}{10}-10B=0\\ \frac{1}{10}-5C-\frac{16}{10}-10B=0\end{array}$

$5C+10B=-\frac{15}{10}$        (39)

Substitute equation (38) in equation (39).

$\begin{array}{l}5C+10\left(-\frac{1}{10}-C\right)=-\frac{15}{10}\\ 5C-1-10C=-\frac{15}{10}\\ -5C=-\frac{15}{10}+1\\ -5C=-\frac{5}{10}\end{array}$

Reduce the equation as follows,

$C=\frac{1}{10}$        (40)

Substitute equation (40) in equation (38) to find the constant B.

$\begin{array}{c}B=-\frac{1}{10}-\frac{1}{10}\\ =-0.2\end{array}$

$B=-\frac{1}{5}$        (41)

Substitute equation (41) in equation (37) to find the constant D.

$\begin{array}{c}D=-\frac{16}{10}-10\left(-\frac{1}{5}\right)\\ =-\frac{16}{10}+\frac{10}{5}\\ =0.4\end{array}$

$D=\frac{4}{10}$        (42)

Substitute equation (33),(40), (41), and (42) in equation (31) to find $Z\left(s\right)$.

$\begin{array}{c}Z\left(s\right)=5\left[\frac{\left(\frac{1}{10}\right)}{s}-\frac{\left(\frac{1}{5}\right)}{s+1}+\frac{\left(\frac{1}{10}\right)s+\left(\frac{4}{10}\right)}{s^2+6s+9+1}\right]\\ =\frac{1}{2}\left(\frac{1}{s}\right)-\frac{1}{s+1}+\frac{\left(\frac{5}{10}\right)s+\left(\frac{20}{10}\right)}{{\left(s+3\right)}^2+1}\\ =\frac{1}{2}\left(\frac{1}{s}\right)-\frac{1}{s+1}+\frac{\left(\frac{1}{2}\right)s+2}{{\left(s+3\right)}^2+1}\\ =\frac{1}{2}\left(\frac{1}{s}\right)-\frac{1}{s+1}+\frac{1}{2}\frac{s+4}{{\left(s+3\right)}^2+1}\end{array}$

Reduce the equation as follows,

$Z\left(s\right)=\frac{1}{2}\left(\frac{1}{s}\right)-\frac{1}{s+1}+\frac{1}{2}\frac{s+3+1}{{\left(s+3\right)}^2+1}$

$Z\left(s\right)=\frac{1}{2}\left(\frac{1}{s}\right)-\frac{1}{s+1}+\frac{1}{2}\frac{s+3}{{\left(s+3\right)}^2+1}+\frac{1}{2}\frac{1}{{\left(s+3\right)}^2+1}$        (43)

Apply inverse Laplace transform of equation (2) in equation (43).

$\begin{array}{c}z\left(t\right)={ℒ}^{-1}\left[Z\left(s\right)\right]\\ ={ℒ}^{-1}\left[\frac{1}{2}\left(\frac{1}{s}\right)-\frac{1}{s+1}+\frac{1}{2}\frac{s+3}{{\left(s+3\right)}^2+1}+\frac{1}{2}\frac{1}{{\left(s+3\right)}^2+1}\right]\end{array}$

$z\left(t\right)=\frac{1}{2}{ℒ}^{-1}\left[\frac{1}{s}\right]-{ℒ}^{-1}\left[\frac{1}{s+1}\right]+\frac{1}{2}{ℒ}^{-1}\left[\frac{s+3}{{\left(s+3\right)}^2+1}\right]+\frac{1}{2}{ℒ}^{-1}\left[\frac{1}{{\left(s+3\right)}^2+1}\right]$        (44)

Apply inverse Laplace transform function of equation (3), (19) in equation (44).

$\begin{array}{c}z\left(t\right)=\frac{1}{2}u\left(t\right)-e^{-t}u\left(t\right)+\frac{1}{2}{e}^{-3t}\cos \left(t\right)u\left(t\right)+\frac{1}{2}{e}^{-3t}\sin \left(t\right)u\left(t\right)\\ =0.5\left[1-2e^{-t}+e^{-3t}\cos \left(t\right)+e^{-3t}\sin \left(t\right)\right]u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform $z\left(t\right)$ for the given function is $0.5\left[1-2e^{-t}+e^{-3t}\cos \left(t\right)+e^{-3t}\sin \left(t\right)\right]u\left(t\right)$.