Chapter 15, Problem 39P

Determine f(t) if:

1. (a) $F(s)=\frac{2s^3+4s^2+1}{(s^2+2s+17)(s^2+4s+20)}$
2. (b) $F(s)=\frac{s^3+4}{(s^2+9)(s^2+6s+3)}$

To determine

Find the inverse Laplace transform for the function $F\left(s\right)=\frac{2s^3+4s^2+1}{\left(s^2+2s+17\right)\left(s^2+4s+20\right)}$.

Answer to Problem 39P

The inverse Laplace transform for the given function is

$f\left(t\right)=\left[-1.6e^{-t}\cos \left(4t\right)-4.05e^{-t}\sin \left(4t\right)+3.6e^{-2t}\cos \left(4t\right)+3.45e^{-2t}\sin \left(4t\right)\right]u\left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$F\left(s\right)=\frac{2s^3+4s^2+1}{\left(s^2+2s+17\right)\left(s^2+4s+20\right)}$        (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.

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

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

Here,

$\omega$ is the angular frequency,

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

$s$ is a complex variable.

Calculation:

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

$F\left(s\right)=\frac{2s^3+4s^2+1}{\left(s^2+2s+17\right)\left(s^2+4s+20\right)}=\frac{As+B}{s^2+2s+17}+\frac{Cs+D}{s^2+4s+20}$        (5)

Here,

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

Find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{2s^3+4s^2+1}{\left(s^2+2s+17\right)\left(s^2+4s+20\right)}=\frac{As+B}{s^2+2s+17}+\frac{Cs+D}{s^2+4s+20}\\ \frac{2s^3+4s^2+1}{\left(s^2+2s+17\right)\left(s^2+4s+20\right)}=\frac{\left(As+B\right)\left(s^2+4s+20\right)+\left(Cs+D\right)\left(s^2+2s+17\right)}{\left(s^2+2s+17\right)\left(s^2+4s+20\right)}\\ 2s^3+4s^2+1=\left(As+B\right)\left(s^2+4s+20\right)+\left(Cs+D\right)\left(s^2+2s+17\right)\\ 2s^3+4s^2+1=\left(\begin{array}{l}A{s}^3+4A{s}^2+20As+B{s}^2+4Bs+20B+C{s}^3+2C{s}^2+17Cs\\ +D{s}^2+2Ds+17D\end{array}\right)\end{array}$

Reduce the equation as follows,

$2s^3+4s^2+1=\left[\begin{array}{l}\left(A+C\right)s^3+\left(4A+B+2C+D\right)s^2+\left(20A+4B+17C+2D\right)s\\ +\left(20B+17D\right)\end{array}\right]$        (6)

Equating the coefficients of $s^3$ in equation (6).

$A+C=2$

$C=2-A$        (7)

Equating the coefficients of $s^2$ in equation (6).

$\begin{array}{l}4A+B+2C+D=4\\ 4A=4-B-2C-D\\ A=\frac{4-B-2C-D}{4}\end{array}$

$A=1-0.25B-0.5C-0.25D$        (8)

Equating the coefficients of $s$ in equation (6).

$20A+4B+17C+2D=0$        (9)

Equating the coefficients of constant term in equation (6).

$\begin{array}{l}20B+17D=1\\ 20B=1-17D\\ B=\frac{1-17D}{20}\end{array}$

$B=0.05-0.85D$        (10)

Substitute equation (7) and (10) in equation (9).

$\begin{array}{l}20A+4\left(0.05-0.85D\right)+17\left(2-A\right)+2D=0\\ 20A+0.2-3.4D+34-17A+2D=0\\ 3A-1.4D=-34-0.2\\ 1.4D=3A+34.2\end{array}$

Rearrange the equation as follows,

$D=\frac{3A+34.2}{1.4}$

$D=2.1429A+24.4286$        (11)

Substitute equation (11) in equation (10).

$\begin{array}{c}B=0.05-0.85\left(2.1429A+24.4286\right)\\ =0.05-1.8215A-20.7643\end{array}$

$B=-20.7143-1.8215A$        (12)

Substitute equation (7), (11), and (12) in equation (8) to find the constant A.

$\begin{array}{l}A=1-0.25\left(-20.7143-1.8215A\right)-0.5\left(2-A\right)-0.25\left(2.1429A+24.4286\right)\\ A=1+5.1786+0.4554A-1+0.5A-0.5357A-6.1072\\ A-0.4554A-0.5A+0.5357A=1+5.1786-1-6.1072\\ 0.5803A=-0.9286\end{array}$

Reduce the equation as follows,

$A=-\frac{0.9286}{0.5803}$

$A=-1.6$        (13)

Substitute equation (13) in equation (12) to find the constant B.

$B=-20.7143-1.8215\left(-1.6\right)$

$B=-17.8$        (14)

Substitute equation (13) in equation (11) to find the constant D.

$\begin{array}{c}D=2.1429\left(-1.6\right)+24.4286\\ =20.9999\end{array}$

$D\cong 21$        (15)

Substitute equation (13) in equation (7).

$C=2-\left(-1.6\right)$

$C=3.6$        (16)

Substitute equation (13), (14), (15), and (16) in equation (5) to find $F\left(s\right)$.

$\begin{array}{c}F\left(s\right)=\frac{-1.6s-17.8}{s^2+2s+17}+\frac{3.6s+21}{s^2+4s+20}\\ =\frac{-1.6s-1.6-16.2}{s^2+1s+1s+1+16}+\frac{3.6s+7.2+13.8}{s^2+2s+2s+4+16}\\ =\frac{-1.6\left(s+1\right)-16.2}{s\left(s+1\right)+1\left(s+1\right)+4^2}+\frac{3.6\left(s+2\right)+13.8}{s\left(s+2\right)+2\left(s+2\right)+4^2}\\ =\frac{-1.6\left(s+1\right)}{\left(s+1\right)\left(s+1\right)+4^2}-\frac{\left(4.05\times 4\right)}{\left(s+1\right)\left(s+1\right)+4^2}+\frac{3.6\left(s+2\right)}{\left(s+2\right)\left(s+2\right)+4^2}+\frac{\left(3.45\times 4\right)}{\left(s+2\right)\left(s+2\right)+4^2}\end{array}$

Reduce the equation as follows,

$F\left(s\right)=-\frac{1.6\left(s+1\right)}{{\left(s+1\right)}^2+4^2}-\frac{\left(4.05\times 4\right)}{{\left(s+1\right)}^2+4^2}+\frac{3.6\left(s+2\right)}{{\left(s+2\right)}^2+4^2}+\frac{\left(3.45\times 4\right)}{{\left(s+2\right)}^2+4^2}$        (17)

Apply inverse Laplace transform given in equation (2) to equation (17). Therefore,

$f\left(t\right)={ℒ}^{-1}\left[-\frac{1.6\left(s+1\right)}{{\left(s+1\right)}^2+4^2}-\frac{\left(4.05\times 4\right)}{{\left(s+1\right)}^2+4^2}+\frac{3.6\left(s+2\right)}{{\left(s+2\right)}^2+4^2}+\frac{\left(3.45\times 4\right)}{{\left(s+2\right)}^2+4^2}\right]$

$\begin{array}{l}f\left(t\right)=-1.6{ℒ}^{-1}\left[\frac{s+1}{{\left(s+1\right)}^2+4^2}\right]-4.05{ℒ}^{-1}\left[\frac{4}{{\left(s+1\right)}^2+4^2}\right]+3.6{ℒ}^{-1}\left[\frac{s+2}{{\left(s+2\right)}^2+4^2}\right]\\ +3.45{ℒ}^{-1}\left[\frac{4}{{\left(s+2\right)}^2+4^2}\right]\end{array}$        (18)

Apply inverse Laplace transform function of equation (3) and (4) in equation (18).

$\begin{array}{c}f\left(t\right)=\left[\begin{array}{c}-1.6e^{-t}\cos \left(4t\right)u\left(t\right)-4.05e^{-t}\sin \left(4t\right)u\left(t\right)+3.6e^{-2t}\cos \left(4t\right)u\left(t\right)\\ +3.45e^{-2t}\sin \left(4t\right)u\left(t\right)\end{array}\right]\\ =\left[-1.6e^{-t}\cos \left(4t\right)-4.05e^{-t}\sin \left(4t\right)+3.6e^{-2t}\cos \left(4t\right)+3.45e^{-2t}\sin \left(4t\right)\right]u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform for the given function is

$f\left(t\right)=\left[-1.6e^{-t}\cos \left(4t\right)-4.05e^{-t}\sin \left(4t\right)+3.6e^{-2t}\cos \left(4t\right)+3.45e^{-2t}\sin \left(4t\right)\right]u\left(t\right)$.

To determine

Find the inverse Laplace transform for the given function $F\left(s\right)=\frac{s^2+4}{\left(s^2+9\right)\left(s^2+6s+3\right)}$.

Answer to Problem 39P

The inverse Laplace transform for the given function is

$f\left(t\right)=\left[0.08333\cos \left(3t\right)+0.02778\sin \left(3t\right)+0.0944e^{-0.551t}-0.1778e^{-5.449t}\right]u\left(t\right)$.

Explanation of Solution

Given data:

The Laplace transform function is,

$F\left(s\right)=\frac{s^2+4}{\left(s^2+9\right)\left(s^2+6s+3\right)}$        (19)

Formula used:

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

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

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

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

Calculation:

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

$F\left(s\right)=\frac{s^2+4}{\left(s^2+9\right)\left(s^2+6s+3\right)}=\frac{As+B}{s^2+9}+\frac{Cs+D}{s^2+6s+3}$        (23)

Here,

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

Find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{s^2+4}{\left(s^2+9\right)\left(s^2+6s+3\right)}=\frac{As+B}{s^2+9}+\frac{Cs+D}{s^2+6s+3}\\ \frac{s^2+4}{\left(s^2+9\right)\left(s^2+6s+3\right)}=\frac{\left(As+B\right)\left(s^2+6s+3\right)+\left(Cs+D\right)\left(s^2+9\right)}{\left(s^2+9\right)\left(s^2+6s+3\right)}\\ s^2+4=\left(As+B\right)\left(s^2+6s+3\right)+\left(Cs+D\right)\left(s^2+9\right)\\ s^2+4=A{s}^3+6A{s}^2+3As+B{s}^2+6Bs+3B+C{s}^3+9Cs+D{s}^2+9D\end{array}$

Reduce the equation as follows,

$s^2+4=\left(A+C\right)s^3+\left(6A+B+D\right)s^2+\left(3A+6B+9C\right)s+\left(3B+9D\right)$        (24)

Equating the coefficients of $s^3$ in equation (24).

$A+C=0$

$A=-C$        (25)

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

$6A+B+D=1$        (26)

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

$3A+6B+9C=0$        (27)

Substitute equation (25) in equation (27).

$\begin{array}{l}3\left(-C\right)+6B+9C=0\\ 6B+6C=0\\ B+C=0\end{array}$

$B=-C$        (28)

Substitute equation (28) in equation (25).

$A=B$        (29)

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

$3B+9D=4$        (30)

Rearrange the equation (30) as follows,

$\begin{array}{l}3B=4-9D\\ B=\frac{4-9D}{3}\end{array}$

$B=\frac{4}{3}-3D$        (31)

Substitute equation (29) in equation (30).

$\begin{array}{l}3A+9D=4\\ 3A=4-9D\\ A=\frac{4-9D}{3}\end{array}$

$A=\frac{4}{3}-3D$        (32)

Substitute equation (31) and (32) in equation (26).

$\begin{array}{l}6\left(\frac{4}{3}-3D\right)+\left(\frac{4}{3}-3D\right)+D=1\\ 8-18D+\frac{4}{3}-3D+D=1\\ -18D-3D+D=1-8-\frac{4}{3}\\ -20D=-\frac{25}{3}\end{array}$

Reduce the equation as follows,

$\begin{array}{c}D=\frac{25}{3\times 20}\\ =\frac{25}{60}\end{array}$

$D=\frac{5}{12}$        (33)

Substitute equation (33) in equation (32).

$\begin{array}{c}A=\frac{4}{3}-3\left(\frac{5}{12}\right)\\ =\frac{4}{3}-\frac{5}{4}\end{array}$

$A=\frac{1}{12}$        (34)

Substitute equation (34) in equation (29).

$B=\frac{1}{12}$        (35)

Substitute equation (34) in equation (25).

$\frac{1}{12}=-C$

$C=-\frac{1}{12}$        (36)

Substitute equation (33), (34), (35), and (36) in equation (23) to find $F\left(s\right)$.

$F\left(s\right)=\frac{\left(\frac{1}{12}\right)s+\left(\frac{1}{12}\right)}{s^2+9}+\frac{\left(-\frac{1}{12}\right)s+\left(\frac{5}{12}\right)}{s^2+6s+3}$

$12F\left(s\right)=\frac{s+1}{s^2+9}+\frac{-s+5}{s^2+6s+3}$        (37)

Expand $\frac{-s+5}{s^2+6s+3}$ using partial fraction in equation (37)

The roots of $s^2+6s+3$ is calculated as follows,

$\begin{array}{c}s=\frac{-6\pm \sqrt{6^2-4\left(1\right)\left(3\right)}}{2\left(1\right)}\left\{\because x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\right\}\\ =\frac{-6\pm \sqrt{24}}{2}\\ =\frac{-6+\sqrt{24}}{2},\frac{-6-\sqrt{24}}{2}\\ =-0.551,-5.449\end{array}$

Therefore, the roots of $s^2+6s+3$ is $\left(s+0.551\right)\left(s+5.449\right)$.

Consider the partial fraction,

$\frac{-s+5}{s^2+6s+3}=\frac{-s+5}{\left(s+0.551\right)\left(s+5.449\right)}=\frac{E}{s+0.551}+\frac{F}{s+5.449}$        (38)

Here,

E and F are the constants.

Find the constants by using algebraic method.

$\begin{array}{l}\frac{-s+5}{\left(s+0.551\right)\left(s+5.449\right)}=\frac{E}{s+0.551}+\frac{F}{s+5.449}\\ \frac{-s+5}{\left(s+0.551\right)\left(s+5.449\right)}=\frac{E\left(s+5.449\right)+F\left(s+0.551\right)}{\left(s+0.551\right)\left(s+5.449\right)}\\ -s+5=E\left(s+5.449\right)+F\left(s+0.551\right)\\ -s+5=Es+5.449E+Fs+0.551F\end{array}$

Reduce the equation as follows,

$-s+5=\left(E+F\right)s+\left(5.449E+0.551F\right)$        (39)

Equating the coefficients of $s$ in equation (39).

$E+F=-1$

$E=-1-F$        (40)

Equating the coefficients of constant term in equation (39).

$5.449E+0.551F=5$        (41)

Substitute equation (40) in equation (41).

$\begin{array}{l}5.449\left(-1-F\right)+0.551F=5\\ -5.449-5.449F+0.551F=5\\ -5.449F+0.551F=5+5.449\\ -4.898F=10.449\end{array}$

Rearrange the equation as follows,

$F=-\frac{10.449}{4.898}$

$F=-2.133$        (42)

Substitute equation (42) in equation (40) to find the constant E.

$E=-1-\left(-2.133\right)$

$E=1.133$        (43)

Substitute equation (42) and (43) in equation (38).

$\frac{-s+5}{s^2+6s+3}=\frac{1.133}{s+0.551}-\frac{2.133}{s+5.449}$

Substitute $\frac{1.133}{s+0.551}-\frac{2.133}{s+5.449}$ for $\frac{-s+5}{s^2+6s+3}$ in equation (37).

$\begin{array}{l}12F\left(s\right)=\frac{s+1}{s^2+9}+\frac{1.133}{s+0.551}-\frac{2.133}{s+5.449}\\ F\left(s\right)=\frac{1}{12}\left[\frac{s}{s^2+3^2}+\frac{1}{s^2+3^2}+\frac{1.133}{s+0.551}-\frac{2.133}{s+5.449}\right]\end{array}$

$F\left(s\right)=\frac{1}{12}\left[\frac{s}{s^2+3^2}+\frac{\left(\frac{3}{3}\right)}{s^2+3^2}+\frac{1.133}{s+0.551}-\frac{2.133}{s+5.449}\right]$        (44)

Apply inverse Laplace transform given in equation (2) to equation (44). Therefore,

$\begin{array}{c}f\left(t\right)={ℒ}^{-1}\left[\frac{1}{12}\left[\frac{s}{s^2+3^2}+\frac{1}{3}\frac{3}{s^2+3^2}+\frac{1.133}{s+0.551}-\frac{2.133}{s+5.449}\right]\right]\\ f\left(t\right)=\frac{1}{12}{ℒ}^{-1}\left[\frac{s}{s^2+3^2}\right]+\frac{1}{36}{ℒ}^{-1}\left[\frac{3}{s^2+3^2}\right]+\frac{1.133}{12}{ℒ}^{-1}\left[\frac{1}{s+0.551}\right]-\frac{2.133}{12}{ℒ}^{-1}\left[\frac{1}{s+5.449}\right]\end{array}$

$f\left(t\right)=\left\{\begin{array}{l}0.08333{ℒ}^{-1}\left[\frac{s}{s^2+3^2}\right]+0.02778{ℒ}^{-1}\left[\frac{3}{s^2+3^2}\right]+0.0944{ℒ}^{-1}\left[\frac{1}{s+0.551}\right]\\ -0.1778{ℒ}^{-1}\left[\frac{1}{s+5.449}\right]\end{array}\right\}$        (45)

Apply inverse Laplace transform function of equation (20), (21) and (22) in equation (45).

$\begin{array}{c}f\left(t\right)=0.08333\cos \left(3t\right)u\left(t\right)+0.02778\sin \left(3t\right)u\left(t\right)+0.0944e^{-0.551t}u\left(t\right)-0.1778e^{-5.449t}u\left(t\right)\\ =\left[0.08333\cos \left(3t\right)+0.02778\sin \left(3t\right)+0.0944e^{-0.551t}-0.1778e^{-5.449t}\right]u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Laplace transform for the given function is

$f\left(t\right)=\left[0.08333\cos \left(3t\right)+0.02778\sin \left(3t\right)+0.0944e^{-0.551t}-0.1778e^{-5.449t}\right]u\left(t\right)$.