Chapter 14, Problem 31E

(a)

To determine

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

Answer to Problem 31E

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

Explanation of Solution

Given data:

Consider the Laplace transform function is,

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

Calculation:

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

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

Here,

A, B, and C are the constants.

Find the constants by using algebraic method.

Consider the partial fraction,

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

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

Put $s=0$ in equation (4),

$2=A\left(0+2\right)\left(0+1\right)+B{\left(0+2\right)}^2\left(0+1\right)+C{\left(0+2\right)}^3$

$2=2A+4B+8C$        (5)

Put $s=-1$ in equation (4),

$\begin{array}{l}-1+2=A\left(-1+2\right)\left(-1+1\right)+B{\left(-1+2\right)}^2\left(-1+1\right)+C{\left(-1+2\right)}^3\\ 1=0+0+C{\left(1\right)}^3\\ C=1\end{array}$

Put $s=1$ in equation (4),

$\begin{array}{l}1+2=A\left(1+2\right)\left(1+1\right)+B{\left(1+2\right)}^2\left(1+1\right)+C{\left(1+2\right)}^3\\ 3=A\left(3\right)\left(2\right)+B\left(9\right)\left(2\right)+C\left(27\right)\\ 3=6A+18B+27C\end{array}$

$1=2A+6B+9C$        (6)

Subtract equation (5) and (6),

$1=-2B-C$

Substitute 1 for C in the above equation.

$\begin{array}{l}1=-2B-1\\ -2B=2\\ B=-1\end{array}$

Substitute $-1$ for B and 1 for C in equation (5) as follows,

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

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

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

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

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

$f\left(t\right)={ℒ}^{-1}\left[-\frac{1}{{\left(s+2\right)}^2}-\frac{1}{\left(s+2\right)}+\frac{1}{s+1}\right]$        (8)

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

${ℒ}^{-1}\left[\frac{1}{s+a}\right]=e^{-at}u\left(t\right)$        (9)

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

${ℒ}^{-1}\left[\frac{1}{{\left(s+a\right)}^2}\right]=t{e}^{-at}u\left(t\right)$        (10)

Apply inverse Laplace transform function of equation (9) and (10), in equation (8).

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

Conclusion:

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

(b)

To determine

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

Answer to Problem 31E

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

Explanation of Solution

Given data:

Consider the Laplace transform function is,

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

Calculation:

The equation (11) can be rewritten as follows,

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

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

$F\left(s\right)=\frac{s}{{\left(s+2\right)}^3}$        (12)

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

$F\left(s\right)=\frac{s}{{\left(s+2\right)}^3}=\frac{A}{\left(s+2\right)}+\frac{B}{{\left(s+2\right)}^2}+\frac{C}{{\left(s+2\right)}^3}$        (13)

Here,

A, B, C are the constants.

Find the constants by using algebraic method.

Consider the partial fraction,

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

$s=A{\left(s+2\right)}^2+B\left(s+2\right)+C$        (14)

Put $s=-2$ in equation (14).

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

Expanding equation (14) as follows,

$s=A{s}^2+2As+4A+Bs+2B+C$        (15)

Substitute $-2$ for C in equation (15).

$s=A{s}^2+2As+4A+Bs+2B-2$

Equating the coefficient of $s^2$ in equation (15).

$A=0$

Equating the coefficient of constant term in equation (15).

$\begin{array}{l}0=4A+2B-2\\ 0=2A+B-1\\ 2A+B=1\end{array}$

Susbtitute 0 for A in the above equation.

$\begin{array}{l}2\left(0\right)+B=1\\ B=1\end{array}$

Substitute 0 for A, 1 for B, and $-2$ for C in equation (13) to find $F\left(s\right)$.

$F\left(s\right)=0+\frac{1}{{\left(s+2\right)}^2}+\frac{-2}{{\left(s+2\right)}^3}$

$F\left(s\right)=\frac{1}{{\left(s+2\right)}^2}+\frac{-2}{{\left(s+2\right)}^3}$        (16)

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

${ℒ}^{-1}\left[\frac{1}{{\left(s+a\right)}^n}\right]=\frac{t^{n-1}}{\left(n-1\right)!}{e}^{-at}u\left(t\right)$        (17)

Apply inverse Laplace transform function of equation (10) and (17), in equation (16).

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

Conclusion:

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

(c)

To determine

Find the inverse Laplace transform for the given function $\frac{1}{s^2+8s+7}$.

Answer to Problem 31E

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

Explanation of Solution

Given data:

Consider the Laplace transform function is,

$F\left(s\right)=\frac{1}{s^2+8s+7}$        (18)

Calculation:

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

$F\left(s\right)=\frac{1}{s^2+8s+7}=\frac{A}{s+7}+\frac{B}{s+1}$        (19)

Here,

A, B, C, and D are the constants.ind the constants by using algebraic method.

Consider the partial fraction,

$\frac{1}{s^2+8s+7}=\frac{A\left(s+1\right)+B\left(s+7\right)}{\left(s+1\right)\left(s+7\right)}$

$1=A\left(s+1\right)+B\left(s+7\right)$        (20)

Substitute $s=-1$ in equation (20).

$\begin{array}{l}1=A\left(-1+1\right)+B\left(-1+7\right)\\ 1=0+6B\\ B=\frac{1}{6}\end{array}$

Substitute $s=-7$ in equation (20).

$\begin{array}{l}1=A\left(-7+1\right)+B\left(-7+7\right)\\ 1=-6A\\ A=\frac{-1}{6}\end{array}$

Substitute $\frac{-1}{6}$ for A and $\frac{1}{6}$ for B in equation (19) to find $F\left(s\right)$.

$F\left(s\right)=-\frac{1}{6\left(s+7\right)}+\frac{1}{6\left(s+1\right)}$        (21)

Apply inverse Laplace transform of equation (2) in equation (21).

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

$f\left(t\right)={ℒ}^{-1}\left[-\frac{1}{6\left(s+7\right)}+\frac{1}{6\left(s+1\right)}\right]$        (22)

Apply inverse Laplace transform function of equation (9) in equation (22).

$f\left(t\right)=-\frac{1}{6}{e}^{-7t}u\left(t\right)+\frac{1}{6}{e}^{-t}u\left(t\right)$

Conclusion:

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

(d)

To determine

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

Answer to Problem 31E

The given functions are verified with MATLAB.

Explanation of Solution

Calculation:

Consider the function given in Part (a).

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

The MATLAB code for the given function:

syms s t 

ilaplace(1/(s+2)/(s+2)/(s+1))

MATLAB output:

$\text{ans}=e^{-t}-e^{-2t}-t{e}^{-2t}$

Consider the function given in Part (b).

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

The MATLAB code for the given function:

syms s t

ilaplace(s/(s+2)/(s+2)/(s+2))

MATLAB output:

$\text{ans}=t{e}^{-2t}-t^2{e}^{-2t}$

Consider the function given in Part (c).

$F\left(s\right)=\frac{1}{s^2+8s+7}$

The MATLAB code for the given function:

syms s t

ilaplace(1/(s*s+8*s+7))

MATLAB output:

$\text{ans}=\frac{e^{-t}}{6}-\frac{e^{-7t}}{6}$

Conclusion:

Thus, the given functions are verified with MATLAB.