Chapter 15, Problem 35P

Obtain f(t) for the following transforms:

1. (a) $F\left(s\right)=\frac{\left(s+3\right)e^{-6s}}{\left(s+1\right)\left(s+2\right)}$
2. (b) $F\left(s\right)=\frac{4-e^{-2s}}{s^2+5s+4}$
3. (c) $F\left(s\right)=\frac{s{e}^{-s}}{\left(s+3\right)\left(s^2+4\right)}$

To determine

Find the inverse Laplace transform $f\left(t\right)$ for the given function $F\left(s\right)=\frac{\left(s+3\right)e^{-6s}}{\left(s+1\right)\left(s+2\right)}$.

Answer to Problem 35P

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

Explanation of Solution

Given data:

The Laplace transform function is,

$F\left(s\right)=\frac{\left(s+3\right)e^{-6s}}{\left(s+1\right)\left(s+2\right)}$ (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)

Write the general expression for the time shift property.

${\mathcal{L}}^{-1}\left[e^{-as}F\left(s\right)\right]=f\left(t-a\right)u\left(t-a\right)$ (4)

Calculation:

Consider the function from equation (1),

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

Substitute $G\left(s\right)$ for $\frac{s+3}{\left(s+1\right)\left(s+2\right)}$ in equation (1).

$F\left(s\right)=G\left(s\right)e^{-6s}$ (5)

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

$G\left(s\right)=\frac{s+3}{\left(s+1\right)\left(s+2\right)}=\frac{A}{s+1}+\frac{B}{s+2}$ (6)

Here,

A and B are the constants.

Find the constants by using residue method.

Constant A:

$A={\left(s+1\right)G\left(s\right)|}_{s+1=0}$ (7)

Substitute equation (6) in equation (7) to find the constant A.

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

Constant B:

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

Substitute equation (6) in equation (8) to find the constant B.

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

Substitute $2$ for A and $-1$ for B in equation (6) to find $G\left(s\right)$.

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

$G\left(s\right)=\frac{2}{s+1}-\frac{1}{s+2}$ (9)

Substitute equation (9) in equation (5).

$F\left(s\right)=\left(\frac{2}{s+1}-\frac{1}{s+2}\right)e^{-6s}$

$F\left(s\right)=2\frac{e^{-6s}}{s+1}-\frac{e^{-6s}}{s+2}$ (10)

Apply inverse Laplace transform of equation (2) in equation (10).

$f\left(t\right)={\mathcal{L}}^{-1}\left[2\frac{e^{-6s}}{s+1}-\frac{e^{-6s}}{s+2}\right]$

$f\left(t\right)=2{\mathcal{L}}^{-1}\left[\frac{e^{-6s}}{s+1}\right]-{\mathcal{L}}^{-1}\left[\frac{e^{-6s}}{s+2}\right]$ (11)

Apply inverse Laplace transform function of equation (3) and time shift property of equation (4) in equation (11).

$\begin{array}{c}f\left(t\right)=2e^{-\left(t-6\right)}u\left(t-6\right)-e^{-2\left(t-6\right)}u\left(t-6\right)\\ =\left(2e^{-\left(t-6\right)}-e^{-2\left(t-6\right)}\right)u\left(t-6\right)\end{array}$

Conclusion:

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

To determine

Find the inverse Laplace transform $f\left(t\right)$ for the given function $F\left(s\right)=\frac{4-e^{-2s}}{s^2+5s+4}$.

Answer to Problem 35P

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

Explanation of Solution

Given data:

The Laplace transform function is,

$F\left(s\right)=\frac{4-e^{-2s}}{s^2+5s+4}$ (12)

Calculation:

Consider the function from equation (12),

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

Substitute $G\left(s\right)$ for $\frac{1}{s^2+5s+4}$ in equation (12).

$F\left(s\right)=\left(4-e^{-2s}\right)G\left(s\right)$ (13)

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

$G\left(s\right)=\frac{1}{\left(s+1\right)\left(s+4\right)}=\frac{A}{s+1}+\frac{B}{s+4}$ (14)

Here,

A and B are the constants.

Find the constants by using residue method.

Constant A:

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

Substitute equation (14) in equation (15) to find the constant A.

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

Constant B:

$B={\left(s+4\right)G\left(s\right)|}_{s+4=0}$ (16)

Substitute equation (14) in equation (16) to find the constant B.

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

Substitute $\frac{1}{3}$ for A and $-\frac{1}{3}$ for B in equation (14) to find $G\left(s\right)$.

$G\left(s\right)=\frac{\left(\frac{1}{3}\right)}{s+1}+\frac{\left(-\frac{1}{3}\right)}{s+4}$ (17)

Substitute equation (17) in equation (13).

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

$F\left(s\right)=\frac{4}{3}\left(\frac{1}{s+1}-\frac{1}{s+4}\right)-\frac{1}{3}\left(\frac{e^{-2s}}{s+1}-\frac{e^{-2s}}{s+4}\right)$ (18)

Apply inverse Laplace transform of equation (2) to equation (18).

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

$f\left(t\right)=\frac{4}{3}{\mathcal{L}}^{-1}\left[\frac{1}{s+1}-\frac{1}{s+4}\right]-\frac{1}{3}{\mathcal{L}}^{-1}\left[\frac{e^{-2s}}{s+1}-\frac{e^{-2s}}{s+4}\right]$ (19)

Apply the time shift property of equation (4) and inverse Laplace transform function of equation (3) in equation (19).

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

Conclusion:

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

To determine

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

Answer to Problem 35P

The inverse Laplace transform $f\left(t\right)$ for the given function is $\frac{1}{13}\left[-3e^{-3\left(t-1\right)}+3\cos \left(2\left(t-1\right)\right)+2\sin \left(2\left(t-1\right)\right)\right]u\left(t-1\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$F\left(s\right)=\frac{s{e}^{-s}}{\left(s+3\right)\left(s^2+4\right)}$ (20)

Formula used:

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

${\mathcal{L}}^{-1}\left[\frac{s}{s^2+{\omega }^2}\right]=\cos \omega tu\left(t\right)$ (21)

${\mathcal{L}}^{-1}\left[\frac{\omega }{s^2+{\omega }^2}\right]=\sin \omega tu\left(t\right)$ (22)

Calculation:

Consider the function from equation (20),

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

Substitute $G\left(s\right)$ for $\frac{s}{\left(s+3\right)\left(s^2+4\right)}$ in equation (20).

$F\left(s\right)=e^{-s}G\left(s\right)$ (23)

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

$G\left(s\right)=\frac{s}{\left(s+3\right)\left(s^2+4\right)}=\frac{A}{s+3}+\frac{Bs+C}{s^2+4}$ (24)

Here,

A, B, and C are the constants.

Find the constants by using algebraic method.

Consider the partial fraction,

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

Reduce the equation as follows,

$s=\left(A+B\right)s^2+\left(3B+C\right)s+\left(3C+4A\right)$ (25)

Equating the coefficients of $s^2$ in equation (25).

$0=A+B$

$B=-A$ (26)

Equating the coefficients of $s$ in equation (25).

$1=3B+C$ (27)

Equating the coefficients of constant term in equation (25).

$\begin{array}{l}0=3C+4A\\ 4A=-3C\end{array}$

$A=-\frac{3}{4}C$ (28)

Substitute equation (28) in equation (26).

$B=-\left(-\frac{3}{4}C\right)$

$B=\frac{3}{4}C$ (29)

Substitute equation (29) in equation (27).

$\begin{array}{l}1=3\left(\frac{3}{4}C\right)+C\\ 1=\frac{9}{4}C+C\\ C\left(1+\frac{9}{4}\right)=1\\ C\left(\frac{13}{4}\right)=1\end{array}$

Rearrange the equation as follows,

$C=\frac{4}{13}$

Substitute $\frac{4}{13}$ for C in equation (28) to find the constant A.

$\begin{array}{c}A=-\frac{3}{4}\left(\frac{4}{13}\right)\\ =-\frac{3}{13}\end{array}$

Substitute $-\frac{3}{13}$ for A in equation (26) to find the constant B.

$\begin{array}{c}B=-\left(-\frac{3}{13}\right)\\ =\frac{3}{13}\end{array}$

Substitute $-\frac{3}{13}$ for A, $\frac{3}{13}$ for B, and $\frac{4}{13}$ for C in equation (24) to find $G\left(s\right)$.

$G\left(s\right)=\frac{\left(-\frac{3}{13}\right)}{s+3}+\frac{\left(\frac{3}{13}\right)s+\left(\frac{4}{13}\right)}{s^2+4}$

$G\left(s\right)=\frac{1}{13}\left(\frac{-3}{s+3}+\frac{3s+4}{s^2+4}\right)$ (30)

Substitute equation (30) in equation (23).

$\begin{array}{c}F\left(s\right)=e^{-s}\left[\frac{1}{13}\left(\frac{-3}{s+3}+\frac{3s+4}{s^2+4}\right)\right]\\ =e^{-s}\left[\frac{1}{13}\left(-\frac{3}{s+3}+\frac{3s}{s^2+4}+\frac{4}{s^2+4}\right)\right]\\ =e^{-s}\left[\frac{1}{13}\left(-3\frac{1}{s+3}+3\frac{s}{s^2+2^2}+2\frac{2}{s^2+2^2}\right)\right]\end{array}$

$F\left(s\right)=\frac{1}{13}\left[-3e^{-s}\frac{1}{s+3}+3e^{-s}\frac{s}{s^2+2^2}+2e^{-s}\frac{2}{s^2+2^2}\right]$ (31)

Apply inverse Laplace transform of equation (2) to equation (31).

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

$f\left(t\right)=\frac{1}{13}\left[-3{\mathcal{L}}^{-1}\left[e^{-s}\frac{1}{s+3}\right]+3{\mathcal{L}}^{-1}\left[e^{-s}\frac{s}{s^2+2^2}\right]+2{\mathcal{L}}^{-1}\left[e^{-s}\frac{2}{s^2+2^2}\right]\right]$ (32)

Apply inverse Laplace transform function of equation (21), (22) and time shift property of equation (4) in equation (32).

$\begin{array}{c}f\left(t\right)=\frac{1}{13}\left[-3e^{-3\left(t-1\right)}u\left(t-1\right)+3\cos \left(2\left(t-1\right)\right)u\left(t-1\right)+2\sin \left(2\left(t-1\right)\right)u\left(t-1\right)\right]\\ =\frac{1}{13}\left[-3e^{-3\left(t-1\right)}+3\cos \left(2\left(t-1\right)\right)+2\sin \left(2\left(t-1\right)\right)\right]u\left(t-1\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform $f\left(t\right)$ for the given function is $\frac{1}{13}\left[-3e^{-3\left(t-1\right)}+3\cos \left(2\left(t-1\right)\right)+2\sin \left(2\left(t-1\right)\right)\right]u\left(t-1\right)$.