Chapter 15, Problem 32P

(a)

To determine

Find the inverse Laplace transform for the given function $\frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+2\right)\left(s+4\right)}$.

Answer to Problem 32P

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

Explanation of Solution

Given data:

The Laplace transform function is ,

$\frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+2\right)\left(s+4\right)}$ (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.

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

${\mathcal{L}}^{-1}\left[\frac{1}{s}\right]=u\left(t\right)$ (4)

Here,

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

$s$ is a complex variable.

Calculation:

Consider the given function,

$F\left(s\right)=\frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+2\right)\left(s+4\right)}$ (5)

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

$F\left(s\right)=\frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+2\right)\left(s+4\right)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+4}$ (6)

Here,

A, B, and C are the constants.

Now, to find the constants by using residue method.

Constant A:

$A={sF\left(s\right)|}_{s=0}$ (7)

Substitute equation (5) in equation (7) to find the constant A.

$\begin{array}{c}A={s\times \frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+2\right)\left(s+4\right)}|}_{s=0}\\ ={\frac{8\left(s+1\right)\left(s+3\right)}{\left(s+2\right)\left(s+4\right)}|}_{s=0}\\ =\frac{8\left(0+1\right)\left(0+3\right)}{\left(0+2\right)\left(0+4\right)}\\ =3\end{array}$

Constant B:

$B={\left(s+2\right)F\left(s\right)|}_{s+2=0}$ (8)

Substitute equation (5) in equation (8) to find the constant B.

$\begin{array}{c}B={\left(s+2\right)\times \frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+2\right)\left(s+4\right)}|}_{s=-2}\\ ={\frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+4\right)}|}_{s=-2}\\ =\frac{8\left(-2+1\right)\left(-2+3\right)}{\left(-2\right)\left(-2+4\right)}\\ =2\end{array}$

Constant C:

$C={\left(s+4\right)F\left(s\right)|}_{s+4=0}$ (9)

Substitute equation (5) in equation (9) to find the constant C.

$\begin{array}{c}C={\left(s+4\right)\times \frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+2\right)\left(s+4\right)}|}_{s=-4}\\ ={\frac{8\left(s+1\right)\left(s+3\right)}{s\left(s+2\right)}|}_{s=-4}\\ =\frac{8\left(-4+1\right)\left(-4+3\right)}{\left(-4\right)\left(-4+2\right)}\\ =3\end{array}$

Substitute $3$ for A, $2$ for B and $3$ for C in equation (6) to find $F\left(s\right)$.

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

Substitute $\frac{3}{s}+\frac{2}{s+2}+\frac{3}{s+4}$ for $F\left(s\right)$ in equation (2) to find $f\left(t\right)$.

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

$f\left(t\right)=3{\mathcal{L}}^{-1}\left[\frac{1}{s}\right]+2{\mathcal{L}}^{-1}\left[\frac{1}{s+2}\right]+3{\mathcal{L}}^{-1}\left[\frac{1}{s+4}\right]$ (10)

Apply inverse Laplace transform function given in equation (3) and (4) to equation (8).

$f\left(t\right)=3u\left(t\right)+2e^{-2t}+3e^{-4t}$

Conclusion:

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

(b)

To determine

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

Answer to Problem 32P

The inverse Laplace transform for the given function is $7e^{-t}-6\left(1+2t\right)e^{-2t}$.

Explanation of Solution

Given data:

The Laplace transform function is,

$\frac{s^2-2s+4}{\left(s+1\right){\left(s+2\right)}^2}$

Formula used:

Write the general expression for the inverse Laplace transform.

$g\left(t\right)={\mathcal{L}}^{-1}\left[G\left(s\right)\right]$ (11)

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

${\mathcal{L}}^{-1}\left[\frac{1}{{\left(s+a\right)}^2}\right]=t{e}^{-at}$ (12)

Calculation:

Consider the given function,

$G\left(s\right)=\frac{s^2-2s+4}{\left(s+1\right){\left(s+2\right)}^2}$ (13)

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

$G\left(s\right)=\frac{s^2-2s+4}{\left(s+1\right){\left(s+2\right)}^2}=\frac{D}{s+1}+\frac{E}{{\left(s+2\right)}^2}+\frac{F}{s+2}$ (14)

Here,

D, E, and F are the constants.

Now, to find the constants by using residue method.

Constant D:

$D={\left(s+1\right)G\left(s\right)|}_{s+1=0}$ (15)

Substitute equation (13) in equation (15) to find the constant D.

$\begin{array}{c}D={\left(s+1\right)\times \frac{s^2-2s+4}{\left(s+1\right){\left(s+2\right)}^2}|}_{s=-1}\\ ={\frac{s^2-2s+4}{{\left(s+2\right)}^2}|}_{s=-1}\\ =\frac{{\left(-1\right)}^2-2\left(-1\right)+4}{{\left(-1+2\right)}^2}\\ =7\end{array}$

Constant E:

$E={{\left(s+2\right)}^{2}G\left(s\right)|}_{s=-2}$ (16)

Substitute equation (13) in equation (16) to find the constant E.

$\begin{array}{c}E={{\left(s+2\right)}^2\times \frac{s^2-2s+4}{\left(s+1\right){\left(s+2\right)}^2}|}_{s=-2}\\ ={\frac{s^2-2s+4}{\left(s+1\right)}|}_{s=-2}\\ =\frac{{\left(-2\right)}^2-2\left(-2\right)+4}{\left(-2+1\right)}\\ =-12\end{array}$

Constant F:

$F={\frac{d}{ds}\left[{\left(s+2\right)}^{2}G\left(s\right)\right]|}_{s=-2}$ (17)

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

$\begin{array}{c}F={\frac{d}{ds}\left[{\left(s+2\right)}^2\times \frac{s^2-2s+4}{\left(s+1\right){\left(s+2\right)}^2}\right]|}_{s=-2}\\ ={\frac{d}{ds}\left[\frac{s^2-2s+4}{s+1}\right]|}_{s=-2}\\ ={\frac{\left(s+1\right)\left(2s-2\right)-\left(s^2-2s+4\right)\left(1\right)}{{\left(s+1\right)}^2}|}_{s=-2}\\ =\frac{\left(-2+1\right)\left(2\left(-2\right)-2\right)-\left({\left(-2\right)}^2-2\left(-2\right)+4\right)\left(1\right)}{{\left(-2+1\right)}^2}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}F=\frac{\left(-1\right)\left(-6\right)-\left(12\right)\left(1\right)}{{\left(-1\right)}^2}\\ =-6\end{array}$

Substitute $7$ for D, $-12$ for E, and $-6$ for F in equation (14) to find $G\left(s\right)$.

$G\left(s\right)=\frac{7}{s+1}+\frac{-12}{{\left(s+2\right)}^2}+\frac{-6}{s+2}$

Substitute $\frac{7}{s+1}+\frac{-12}{{\left(s+2\right)}^2}+\frac{-6}{s+2}$ for $G\left(s\right)$ in equation (11) to find $g\left(t\right)$.

$\begin{array}{c}g\left(t\right)={\mathcal{L}}^{-1}\left[\frac{7}{s+1}+\frac{-12}{{\left(s+2\right)}^2}+\frac{-6}{s+2}\right]\\ ={\mathcal{L}}^{-1}\left[\frac{7}{s+1}\right]+{\mathcal{L}}^{-1}\left[\frac{-12}{{\left(s+2\right)}^2}\right]+{\mathcal{L}}^{-1}\left[\frac{-6}{s+2}\right]\end{array}$

$g\left(t\right)=7{\mathcal{L}}^{-1}\left[\frac{1}{s+1}\right]-12{\mathcal{L}}^{-1}\left[\frac{1}{{\left(s+2\right)}^2}\right]-6{\mathcal{L}}^{-1}\left[\frac{1}{s+2}\right]$ (18)

Apply inverse Laplace transform function given in equation (3) and (12) to equation (18).

$\begin{array}{c}g\left(t\right)=7e^{-t}-12t{e}^{-2t}-6e^{-2t}\\ =7e^{-t}-6\left(2t+1\right)e^{-2t}\\ =7e^{-t}-6\left(1+2t\right)e^{-2t}\end{array}$

Conclusion:

Thus, the inverse Laplace transform for the given function is $7e^{-t}-6\left(1+2t\right)e^{-2t}$.

(c)

To determine

Find the inverse Laplace transform for the given function $\frac{s^2+1}{\left(s+3\right)\left(s^2+4s+5\right)}$.

Answer to Problem 32P

The inverse Laplace transform for the given function is $5e^{-3t}-4e^{-2t}\cos \left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$\frac{s^2+1}{\left(s+3\right)\left(s^2+4s+5\right)}$

Formula used:

Write the general expression for the inverse Laplace transform.

$h\left(t\right)={\mathcal{L}}^{-1}\left[H\left(s\right)\right]$ (19)

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

${\mathcal{L}}^{-1}\left[\frac{s+a}{{\left(s+a\right)}^2+{\omega }^2}\right]=e^{-at}\cos \omega t$ (20)

Calculation:

Consider the given function,

$H\left(s\right)=\frac{s^2+1}{\left(s+3\right)\left(s^2+4s+5\right)}$ (21)

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

$H\left(s\right)=\frac{s^2+1}{\left(s+3\right)\left(s^2+4s+5\right)}=\frac{A}{s+3}+\frac{Bs+C}{s^2+4s+5}$ (22)

Here,

A, B, and C are the constants.

Now, to find the constants by using algebraic method.

Consider the partial fraction,

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

Reduce the equation as follows,

$s^2+1=\left(A+B\right)s^2+\left(4A+3B+C\right)s+\left(5A+3C\right)$ (23)

Equating the coefficients of $s^2$ in equation (23).

$A+B=1$

$B=1-A$ (24)

Equating the coefficients of $s$ in equation (23).

$4A+3B+C=0$ (25)

Equating the coefficients of constant term in equation (23).

$5A+3C=1$ (26)

Substitute equation (24) in equation (25).

$\begin{array}{l}4A+3\left(1-A\right)+C=0\\ 4A+3-3A+C=0\\ A+C=-3\end{array}$

$C=-3-A$ (27)

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

$\begin{array}{l}5A+3\left(-3-A\right)=1\\ 5A-9-3A=1\\ 2A=1+9\\ A=5\end{array}$

Substitute 5 for A in equation (24) to find the constant B.

$\begin{array}{c}B=1-5\\ =-4\end{array}$

Substitute 5 for A in equation (27) to find the constant C.

$\begin{array}{c}C=-3-5\\ =-8\end{array}$

Substitute $5$ for A, $-4$ for B and $-8$ for C in equation (22) to find $H\left(s\right)$.

$\begin{array}{c}H\left(s\right)=\frac{5}{s+3}+\frac{-4s-8}{s^2+4s+5}\\ =\frac{5}{s+3}+\frac{-4s-8}{s^2+4s+4+1}\\ =\frac{5}{s+3}+\frac{-4\left(s+2\right)}{{\left(s+2\right)}^2+1}\left\{\because {\left(a+b\right)}^2=a^2+2ab+b^2\right\}\end{array}$

$H\left(s\right)=\frac{5}{s+3}-\frac{4\left(s+2\right)}{{\left(s+2\right)}^2+1}$ (28)

Substitute $\frac{5}{s+3}-\frac{4\left(s+2\right)}{{\left(s+2\right)}^2+1}$ for $H\left(s\right)$ in equation (19) to find $h\left(t\right)$.

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

$h\left(t\right)=5{\mathcal{L}}^{-1}\left[\frac{1}{s+3}\right]-4{\mathcal{L}}^{-1}\left[\frac{\left(s+2\right)}{{\left(s+2\right)}^2+1}\right]$ (29)

Apply inverse Laplace transform function given in equation (3) and (20) to equation (29).

$h\left(t\right)=5e^{-3t}-4e^{-2t}\cos \left(t\right)$

Conclusion:

Thus, the inverse Laplace transform for the given function is $5e^{-3t}-4e^{-2t}\cos \left(t\right)$.