Chapter 14, Problem 30E

(a)

To determine

Find the inverse Laplace transform for the given function $\frac{1}{s^2+9s+20}$.

Answer to Problem 30E

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

Explanation of Solution

Given data:

Consider the Laplace transform function is,

$F\left(s\right)=\frac{1}{s^2+9s+20}$        (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)

Calculation:

The equation (1) can be rewritten as follows,

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

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

$F\left(s\right)=\frac{1}{\left(s+5\right)\left(s+4\right)}=\frac{A}{\left(s+5\right)}+\frac{B}{\left(s+4\right)}$        (4)

Here,

A and B are the constants.

Find the constants by using algebraic method.

Consider the partial fraction,

$\begin{array}{l}\frac{1}{\left(s+5\right)\left(s+4\right)}=\frac{A}{\left(s+5\right)}+\frac{B}{\left(s+4\right)}\\ \frac{1}{\left(s+5\right)\left(s+4\right)}=\frac{A\left(s+4\right)+B\left(s+5\right)}{\left(s+5\right)\left(s+4\right)}\\ 1=A\left(s+4\right)+B\left(s+5\right)\end{array}$

$1=As+4A+Bs+5B$        (5)

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

$A+B=0$

$A=-B$        (6)

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

$1=4A+5B$        (7)

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

$\begin{array}{l}1=4\left(-B\right)+5B\\ B=1\end{array}$

Substitute 1 for B in equation (6) to find the constant A.

$A=-1$

Substitute $-1$ for A, and 1 for B in equation (4) to find $F\left(s\right)$.

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

Apply inverse Laplace transform of equation (2) in equation (8).

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

$f\left(t\right)={ℒ}^{-1}\left[\frac{-1}{\left(s+5\right)}+\frac{1}{\left(s+4\right)}\right]$        (9)

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

${ℒ}^{-1}\left[\frac{1}{s+a}\right]=e^{-at}u\left(t\right)$        (10)

Apply inverse Laplace transform function of equation (10) in equation (9).

$f\left(t\right)=-e^{-5t}u\left(t\right)+e^{-4t}u\left(t\right)$

Conclusion:

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

(b)

To determine

Find the inverse Laplace transform for the given function $\frac{4}{s^3+18s^2+17s}$.

Answer to Problem 30E

The inverse Laplace transform for the given function is $\frac{4}{17}u\left(t\right)+\frac{1}{68}{e}^{-17t}u\left(t\right)-\frac{1}{4}{e}^{-t}u\left(t\right)$.

Explanation of Solution

Given data:

Consider the Laplace transform function is,

$F\left(s\right)=\frac{4}{s^3+18s^2+17s}$        (11)

Calculation:

The equation (11) can be rewritten as follows,

$F\left(s\right)=\frac{4}{s\left(s^2+18s+17\right)}$

$F\left(s\right)=\frac{4}{s\left(s+17\right)\left(s+1\right)}$        (12)

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

$F\left(s\right)=\frac{4}{s\left(s+17\right)\left(s+1\right)}=\frac{A}{s}+\frac{B}{s+17}+\frac{C}{s+1}$        (13)

Here,

A, B, C are the constants.

Find the constants by using algebraic method.

Consider the partial fraction,

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

$4=\left(A+B+C\right)s^2+\left(A+17A+B+17C\right)s+17A$        (14)

Equating the coefficients of constant term in equation (14).

$\begin{array}{l}17A=4\\ A=\frac{1}{17}\end{array}$

Equating the coefficients of $s^2$ in equation (14).

$A+B+C=0$

$B=-C-A$        (15)

Equating the coefficients of $s$ in equation (14).

$18A+B+17C=0$        (16)

Substitute equation (15) in equation (16),

$\begin{array}{l}18A-C-A+17C=0\\ 17A+16C=0\end{array}$

Substitute $\frac{4}{17}$ for A in the above equation to find constant C.

$\begin{array}{l}17\left(\frac{4}{17}\right)+16C=0\\ 4+16C=0\\ C=\frac{-4}{16}\\ C=\frac{-1}{4}\end{array}$

Substitute $\frac{-1}{4}$ for C and $\frac{4}{17}$ for A in equation 15 to find the constant B.

$\begin{array}{c}B=-\left(\frac{-1}{4}\right)-\left(\frac{4}{17}\right)\\ =\frac{1}{4}-\frac{4}{17}\\ =\frac{17-16}{68}\\ =\frac{1}{68}\end{array}$

Substitute $\frac{-1}{4}$ for C, $\frac{1}{68}$ for B, and $\frac{4}{17}$ for A in equation (13) to find $F\left(s\right)$.

$F\left(s\right)=\frac{4}{17s}+\frac{1}{68\left(s+17\right)}-\frac{1}{4\left(s+1\right)}$        (17)

Apply inverse Laplace transform of equation (2) in equation (8).

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

$f\left(t\right)={ℒ}^{-1}\left[\frac{4}{17s}+\frac{1}{68\left(s+17\right)}-\frac{1}{4\left(s+1\right)}\right]$        (18)

Apply inverse Laplace transform function of equation (10) in equation (18).

$f\left(t\right)=\frac{4}{17}u\left(t\right)+\frac{1}{68}{e}^{-17t}u\left(t\right)-\frac{1}{4}{e}^{-t}u\left(t\right)$

Conclusion:

Thus, the inverse Laplace transform for the given function is $\frac{4}{17}u\left(t\right)+\frac{1}{68}{e}^{-17t}u\left(t\right)-\frac{1}{4}{e}^{-t}u\left(t\right)$.

(c)

To determine

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

Answer to Problem 30E

The inverse Laplace transform for the given function is $\frac{3}{40}u\left(t\right)-\frac{1}{4}{e}^{-t}u\left(t\right)-\frac{1}{8}{e}^{-4t}u\left(t\right)+\frac{1}{20}{e}^{-5t}u\left(t\right)+\frac{1}{4}{e}^{-2t}u\left(t\right)$.

Explanation of Solution

Given data:

Consider the Laplace transform function is,

$F\left(s\right)=\frac{3}{s\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}$        (19)

Calculation:

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

$F\left(s\right)=\frac{3}{s\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}=\frac{A}{s}+\frac{B}{s+1}+\frac{C}{s+4}+\frac{D}{s+5}+\frac{E}{s+2}$        (20)

Here,

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

Find the constants by using algebraic method.

Consider the partial fraction,

$\frac{3s\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}{s\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}=\left[\begin{array}{l}\frac{As\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}{s}\\ +\frac{Bs\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}{s+1}\\ +\frac{Cs\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}{s+4}\\ +\frac{Ds\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}{s+5}\\ +\frac{Es\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)}{s+2}\end{array}\right]$$\begin{array}{l}3=A\left(s+1\right)\left(s+4\right)\left(s+5\right)\left(s+2\right)+Bs\left(s+4\right)\left(s+5\right)\left(s+2\right)\\ +Cs\left(s+1\right)\left(s+5\right)\left(s+2\right)+Ds\left(s+1\right)\left(s+4\right)\left(s+2\right)\\ +Es\left(s+1\right)\left(s+4\right)\left(s+5\right)\end{array}$        (21)

Substitute $s=0$ in equation (21).

$\begin{array}{l}3=A\left(1\right)\left(4\right)\left(5\right)\left(2\right)+B\left(0\right)+C\left(0\right)+D\left(0\right)+E\left(0\right)\\ 3=40A\\ A=\frac{3}{40}\end{array}$

Substitute $s=-1$ in equation (21).

$\begin{array}{l}3=A\left(0\right)+B\left(-1\right)\left(-1+4\right)\left(-1+5\right)\left(-1+2\right)+C\left(0\right)+D\left(0\right)+E\left(0\right)\\ 3=B(-1)(3)(4)(1)\\ 3=-12B\\ B=\frac{-1}{4}\end{array}$

Substitute $s=-4$ in equation (21).

$\begin{array}{l}3=A\left(0\right)+B\left(0\right)+C\left(-4\right)\left(-4+1\right)\left(-4+5\right)\left(-4+2\right)+D\left(0\right)+E\left(0\right)\\ 3=C\left(-4\right)\left(-3\right)\left(1\right)\left(-2\right)\\ 3=-24\\ C=\frac{-1}{8}\end{array}$

Substitute $s=-5$ in equation (21).

$\begin{array}{l}3=A\left(0\right)+B\left(0\right)+C\left(0\right)+D\left(-5\right)\left(-5+1\right)\left(-5+4\right)\left(-5+2\right)+E\left(0\right)\\ 3=D\left(-5\right)\left(-4\right)\left(-1\right)\left(-3\right)\\ 3=D60\\ D=\frac{1}{20}\end{array}$

Substitute $s=-2$ in equation (21).

$\begin{array}{l}3=A\left(0\right)+B\left(0\right)+C\left(0\right)+D\left(0\right)+E\left(-2\right)\left(-2+1\right)\left(-2+4\right)\left(-2+5\right)\\ 3=E\left(-2\right)\left(-1\right)\left(2\right)\left(3\right)\\ 3=12E\\ E=\frac{1}{4}\end{array}$

Substitute $\frac{3}{40}$ for A, $\frac{-1}{4}$ for B, $\frac{-1}{8}$ for C, $\frac{1}{20}$ for D, and $\frac{1}{4}$ for E in equation (20) to find $F\left(s\right)$.

$F\left(s\right)=\frac{3}{40s}-\frac{1}{4\left(s+1\right)}-\frac{1}{8\left(s+4\right)}+\frac{1}{20\left(s+5\right)}+\frac{1}{4\left(s+2\right)}$        (22)

Apply inverse Laplace transform of equation (2) in equation (8).

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

$f\left(t\right)={ℒ}^{-1}\left[\frac{3}{40s}-\frac{1}{4\left(s+1\right)}-\frac{1}{8\left(s+4\right)}+\frac{1}{20\left(s+5\right)}+\frac{1}{4\left(s+2\right)}\right]$        (23)

Apply inverse Laplace transform function of equation (10) in equation (22).

$f\left(t\right)=\frac{3}{40}u\left(t\right)-\frac{1}{4}{e}^{-t}u\left(t\right)-\frac{1}{8}{e}^{-4t}u\left(t\right)+\frac{1}{20}{e}^{-5t}u\left(t\right)+\frac{1}{4}{e}^{-2t}u\left(t\right)$

Conclusion:

Thus, the inverse Laplace transform for the given function is $\frac{3}{40}u\left(t\right)-\frac{1}{4}{e}^{-t}u\left(t\right)-\frac{1}{8}{e}^{-4t}u\left(t\right)+\frac{1}{20}{e}^{-5t}u\left(t\right)+\frac{1}{4}{e}^{-2t}u\left(t\right)$.

(d)

To determine

Verify the functions given in Part (a), Part (b), and Part (c) with MATLAB.

Answer to Problem 30E

The given functions are verified with MATLAB.

Explanation of Solution

Calculation:

Consider the function given in Part (a).

$F\left(s\right)=\frac{1}{s^2+9s+20}$

The MATLAB code for the given function:

syms s t

ilaplace (1/(s*s + 9*s + 20))

MATLAB output:

$\text{ans}=e^{-4t}-e^{-5t}$

Consider the function given in Part (b).

$F\left(s\right)=\frac{4}{s^3+18s^2+17s}$

The MATLAB code for the given function:

syms s t

ilaplace((4/(s*s*s + 18*s*s + 17*s)))

MATLAB output:

$\text{ans}=\frac{e^{-17t}}{68}-{\frac{e}{4}}^{-t}+\frac{4}{17}$

Consider the function given in Part (c).

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

The MATLAB code for the given function:

syms s t

ilaplace(3/s/(s+1)/(s+4)/(s+5)/(s+2))

MATLAB output:

$\text{ans}=\frac{e^{-2t}}{4}-\frac{e^{-t}}{4}-\frac{e^{-4t}}{8}+\frac{e^{-5t}}{20}+\frac{3}{40}$

Conclusion:

Thus, the given functions are verified with MATLAB.