Chapter 15, Problem 31P

Find f(t) for each F(s):

1. (a) $\frac{10s}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}$
2. (b) $\frac{2s^2+4s+1}{\left(s+1\right){\left(s+2\right)}^3}$
3. (c) $\frac{s+1}{\left(s+2\right)\left(s^2+2s+5\right)}$

To determine

Find the inverse Laplace transform $f\left(t\right)$ for the given function $F\left(s\right)=\frac{10s}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}$.

Answer to Problem 31P

The inverse Laplace transform for $f\left(t\right)$ the given function is $\left(-5e^{-t}+20e^{-2t}-15e^{-3t}\right)u\left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$F\left(s\right)=\frac{10s}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}$ (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.

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

Here,

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

$s$ is a complex variable.

Calculation:

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

$F\left(s\right)=\frac{10s}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}=\frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{s+3}$ (4)

Here,

A, B, and C are the constants.

Now, to find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{10s}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}=\frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{s+3}\\ \frac{10s}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}=\frac{A\left(s+2\right)\left(s+3\right)+B\left(s+1\right)\left(s+3\right)+C\left(s+1\right)\left(s+2\right)}{\left(s+1\right)\left(s+2\right)\left(s+3\right)}\\ 10s=A\left(s+2\right)\left(s+3\right)+B\left(s+1\right)\left(s+3\right)+C\left(s+1\right)\left(s+2\right)\\ 10s=A\left(s^2+5s+6\right)+B\left(s^2+4s+3\right)+C\left(s^2+3s+2\right)\end{array}$

Reduce the equation as follows,

$10s=A{s}^2+5As+6A+B{s}^2+4Bs+3B+C{s}^2+3Cs+2C$

$10s=\left(A+B+C\right)s^2+\left(5A+4B+3C\right)s+\left(6A+3B+2C\right)$ (5)

Equating the coefficients of $s^2$ in equation (5).

$0=A+B+C$

$C=-A-B$ (6)

Equating the coefficients of $s$ in equation (5).

$10=5A+4B+3C$ (7)

Equating the coefficients of constant term in equation (5).

$0=6A+3B+2C$ (8)

Substitute equation (6) in equation (8).

$\begin{array}{l}0=6A+3B+2\left(-A-B\right)\\ 0=6A+3B-2A-2B\\ 0=4A+B\end{array}$

$B=-4A$ (9)

Substitute equation (9) in equation (6).

$\begin{array}{c}C=-A-\left(-4A\right)\\ =-A+4A\end{array}$

$C=3A$ (10)

Substitute equation (9) and (10) in equation (7) to find the constant A.

$\begin{array}{l}10=5A+4\left(-4A\right)+3\left(3A\right)\\ 10=5A-16A+9A\\ 10=-2A\end{array}$

Rearrange the equation as follows,

$A=-\frac{10}{2}$

$A=-5$ (11)

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

$\begin{array}{c}C=3\left(-5\right)\\ =-15\end{array}$

Substitute equation (11) in equation (9) to find the constant B.

$\begin{array}{c}B=-4\left(-5\right)\\ =20\end{array}$

Substitute $-5$ for A, $20$ for B, and $-15$ for C in equation (4) to find $F\left(s\right)$.

$F\left(s\right)=\frac{-5}{s+1}+\frac{20}{s+2}+\frac{-15}{s+3}$ (12)

Apply inverse Laplace transform of equation (2) to equation (12).

$f\left(t\right)={\mathcal{L}}^{-1}\left[\frac{-5}{s+1}+\frac{20}{s+2}+\frac{-15}{s+3}\right]$

$f\left(t\right)=-5{\mathcal{L}}^{-1}\left[\frac{1}{s+1}\right]+20{\mathcal{L}}^{-1}\left[\frac{1}{s+2}\right]-15{\mathcal{L}}^{-1}\left[\frac{1}{s+3}\right]$ (13)

Apply inverse Laplace transform function of equation (3) in equation (13).

$\begin{array}{c}f\left(t\right)=-5e^{-t}u\left(t\right)+20e^{-2t}u\left(t\right)-15e^{-3t}u\left(t\right)\\ =\left(-5e^{-t}+20e^{-2t}-15e^{-3t}\right)u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform $f\left(t\right)$ for the given function is $\left(-5e^{-t}+20e^{-2t}-15e^{-3t}\right)u\left(t\right)$.

To determine

Find the inverse Laplace transform $f\left(t\right)$ for the given function $F\left(s\right)=\frac{2s^2+4s+1}{\left(s+1\right){\left(s+2\right)}^3}$.

Answer to Problem 31P

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

Explanation of Solution

Given data:

The Laplace transform function is,

$F\left(s\right)=\frac{2s^2+4s+1}{\left(s+1\right){\left(s+2\right)}^3}$ (14)

Formula used:

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

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

Calculation:

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

$F\left(s\right)=\frac{2s^2+4s+1}{\left(s+1\right){\left(s+2\right)}^3}=\frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{{\left(s+2\right)}^2}+\frac{D}{{\left(s+2\right)}^3}$ (16)

Here,

A, B, and C are the constants.

Now, to find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{2s^2+4s+1}{\left(s+1\right){\left(s+2\right)}^3}=\frac{A}{s+1}+\frac{B}{s+2}+\frac{C}{{\left(s+2\right)}^2}+\frac{D}{{\left(s+2\right)}^3}\\ \frac{2s^2+4s+1}{\left(s+1\right){\left(s+2\right)}^3}=\frac{A{\left(s+2\right)}^3+B\left(s+1\right){\left(s+2\right)}^2+C\left(s+1\right)\left(s+2\right)+D\left(s+1\right)}{\left(s+1\right){\left(s+2\right)}^3}\\ 2s^2+4s+1=A{\left(s+2\right)}^3+B\left(s+1\right){\left(s+2\right)}^2+C\left(s+1\right)\left(s+2\right)+D\left(s+1\right)\\ 2s^2+4s+1=A\left(s+2\right)\left(s^2+4s+4\right)+B\left(s+1\right)\left(s^2+4s+4\right)+C\left(s^2+3s+2\right)+Ds+D\end{array}$

Reduce the equation as follows,

$\begin{array}{l}2s^2+4s+1=A\left(s+2\right)\left(s^2+4s+4\right)+B\left(s+1\right)\left(s^2+4s+4\right)+C\left(s^2+3s+2\right)+Ds+D\\ 2s^2+4s+1=\left[\begin{array}{l}A\left(s^3+4s^2+4s+2s^2+8s+8\right)+B\left(s^3+4s^2+4s+s^2+4s+4\right)\\ +C{s}^2+3Cs+2C+Ds+D\end{array}\right]\\ 2s^2+4s+1=\left[\begin{array}{l}A{s}^3+4A{s}^2+4As+2A{s}^2+8As+8A+B{s}^3+4B{s}^2\\ +4Bs+B{s}^2+4Bs+4B+C{s}^2+3Cs+2C+Ds+D\end{array}\right]\\ 2s^2+4s+1=\left[\begin{array}{l}\left(A+B\right)s^3+\left(4A+2A+4B+B+C\right)s^2\\ +\left(4A+8A+4B+4B+3C+D\right)s+\left(8A+4B+2C+D\right)\end{array}\right]\end{array}$

Reduce the equation as follows,

$2s^2+4s+1=\left[\begin{array}{l}\left(A+B\right)s^3+\left(6A+5B+C\right)s^2+\left(12A+8B+3C+D\right)s\\ +\left(8A+4B+2C+D\right)\end{array}\right]$ (17)

Equating the coefficients of $s^3$ in equation (17).

$0=A+B$

$B=-A$ (18)

Equating the coefficients of $s^2$ in equation (17).

$2=6A+5B+C$ (19)

Substitute equation (18) in equation (19).

$\begin{array}{l}2=6A+5\left(-A\right)+C\\ 2=6A-5A+C\\ 2=A+C\end{array}$

$C=2-A$ (20)

Equating the coefficients of $s$ in equation (17).

$4=12A+8B+3C+D$ (21)

Equating the coefficients of constant term in equation (17).

$1=8A+4B+2C+D$ (22)

Substitute equation (18) and (20) in equation (21).

$\begin{array}{l}4=12A+8\left(-A\right)+3\left(2-A\right)+D\\ 4=12A-8A+6-3A+D\\ 4-6=A+D\end{array}$

$D=-2-A$ (23)

Substitute equation (18), (20) and (23) in equation (22) to find the constant A.

$\begin{array}{l}1=8A+4\left(-A\right)+2\left(2-A\right)+\left(-2-A\right)\\ 1=8A-4A+4-2A-2-A\\ A=1-4+2\end{array}$

$A=-1$ (24)

Substitute equation (24) in equation (18) to find the constant B.

$B=-\left(-1\right)$

$B=1$ (25)

Substitute equation (24) in equation (23) to find the constant D.

$\begin{array}{c}D=-2-\left(-1\right)\\ =-1\end{array}$

Substitute equation (24) in equation (20) to find the constant C.

$\begin{array}{c}C=2-\left(-1\right)\\ =3\end{array}$

Substitute $-1$ for A, $1$ for B, $3$ for C, and $-1$ for D in equation (16) to find $F\left(s\right)$.

$F\left(s\right)=\frac{-1}{s+1}+\frac{1}{s+2}+\frac{3}{{\left(s+2\right)}^2}+\frac{-1}{{\left(s+2\right)}^3}$ (26)

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

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

$f\left(t\right)=-{\mathcal{L}}^{-1}\left[\frac{1}{s+1}\right]+{\mathcal{L}}^{-1}\left[\frac{1}{s+2}\right]+3{\mathcal{L}}^{-1}\left[\frac{1}{{\left(s+2\right)}^2}\right]-\frac{1}{2}{\mathcal{L}}^{-1}\left[\frac{2}{{\left(s+2\right)}^3}\right]$ (27)

Apply inverse Laplace transform function of equation (3) and (15) in equation (27).

$\begin{array}{c}f\left(t\right)=-e^{-t}u\left(t\right)+e^{-2t}u\left(t\right)+3t{e}^{-2t}u\left(t\right)-\frac{t^2}{2}{e}^{-2t}u\left(t\right)\\ =\left(-e^{-t}+\left(1+3t-\frac{t^2}{2}\right)e^{-2t}\right)u\left(t\right)\end{array}$

Conclusion:

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

To determine

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

Answer to Problem 31P

The inverse Laplace transform for $f\left(t\right)$ the given function is $\left(-0.2e^{-2t}+0.2e^{-t}\cos \left(2t\right)+0.4e^{-t}\sin \left(2t\right)\right)u\left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

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

Formula used:

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

${\mathcal{L}}^{-1}\left[\frac{s+a}{{\left(s+a\right)}^2+{\omega }^2}\right]=e^{-at}\cos \omega tu\left(t\right)$ (29)

${\mathcal{L}}^{-1}\left[\frac{\omega }{{\left(s+a\right)}^2+{\omega }^2}\right]=e^{-at}\sin \omega tu\left(t\right)$ (30)

Calculation:

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

$F\left(s\right)=\frac{s+1}{\left(s+2\right)\left(s^2+2s+5\right)}=\frac{A}{s+2}+\frac{Bs+C}{s^2+2s+5}$ (31)

Here,

A, B, and C are the constants.

Now, to find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{s+1}{\left(s+2\right)\left(s^2+2s+5\right)}=\frac{A}{s+2}+\frac{Bs+C}{s^2+2s+5}\\ \frac{s+1}{\left(s+2\right)\left(s^2+2s+5\right)}=\frac{A\left(s^2+2s+5\right)+\left(Bs+C\right)\left(s+2\right)}{\left(s+2\right)\left(s^2+2s+5\right)}\\ s+1=A\left(s^2+2s+5\right)+\left(Bs+C\right)\left(s+2\right)\\ s+1=A{s}^2+2As+5A+B{s}^2+2Bs+Cs+2C\end{array}$

Reduce the equation as follows,

$s+1=\left(A+B\right)s^2+\left(2A+2B+C\right)s+\left(2C+5A\right)$ (32)

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

$0=A+B$

$B=-A$ (33)

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

$1=2A+2B+C$ (34)

Substitute equation (33) in equation (34).

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

$C=1$ (35)

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

$1=2C+5A$ (36)

Substitute equation (35) in equation (36).

$\begin{array}{l}1=2\left(1\right)+5A\\ 5A=1-2\\ 5A=-1\end{array}$

$A=\frac{-1}{5}$ (37)

Substitute equation (37) in equation (33).

$\begin{array}{c}B=-\left(\frac{-1}{5}\right)\\ =\frac{1}{5}\end{array}$

Substitute $\frac{-1}{5}$ for A, $\frac{1}{5}$ for B, and $1$ for C in equation (31) to find $F\left(s\right)$.

$\begin{array}{c}F\left(s\right)=\frac{\frac{-1}{5}}{s+2}+\frac{\left(\frac{1}{5}\right)s+1}{s^2+2s+5}\\ =-0.2\frac{1}{s+2}+\frac{\left(\frac{1}{5}\right)\left(s+1\right)}{s^2+2s+5}+\frac{\left(\frac{4}{5}\right)}{s^2+2s+5}\\ =-0.2\frac{1}{s+2}+\frac{1}{5}\frac{s+1}{{\left(s+1\right)}^2+2^2}+\frac{2}{5}\frac{2}{{\left(s+1\right)}^2+2^2}\end{array}$

$F\left(s\right)=-0.2\frac{1}{s+2}+0.2\frac{s+1}{{\left(s+1\right)}^2+2^2}+0.4\frac{2}{{\left(s+1\right)}^2+2^2}$ (38)

Apply inverse Laplace transform of equation (2) to equation (38).

$f\left(t\right)={\mathcal{L}}^{-1}\left[-0.2\frac{1}{s+2}+0.2\frac{s+1}{{\left(s+1\right)}^2+2^2}+0.4\frac{2}{{\left(s+1\right)}^2+2^2}\right]$

$f\left(t\right)=-0.2{\mathcal{L}}^{-1}\left[\frac{1}{s+2}\right]+0.2{\mathcal{L}}^{-1}\left[\frac{s+1}{{\left(s+1\right)}^2+2^2}\right]+0.4{\mathcal{L}}^{-1}\left[\frac{2}{{\left(s+1\right)}^2+2^2}\right]$ (39)

Apply inverse Laplace transform function of equation (3), (29) and (30) in equation (39).

$\begin{array}{c}f\left(t\right)=-0.2e^{-2t}u\left(t\right)+0.2e^{-t}\cos \left(2t\right)u\left(t\right)+0.4e^{-t}\sin \left(2t\right)u\left(t\right)\\ =\left(-0.2e^{-2t}+0.2e^{-t}\cos \left(2t\right)+0.4e^{-t}\sin \left(2t\right)\right)u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform $f\left(t\right)$ for the given function is $\left(-0.2e^{-2t}+0.2e^{-t}\cos \left(2t\right)+0.4e^{-t}\sin \left(2t\right)\right)u\left(t\right)$.