Chapter 2, Problem 2.8P

To determine

(a)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.8P

Inverse Laplace transform of the function $F\left(s\right)=\frac{6}{s^2+9}$ is $f\left(t\right)=2\sin 3t$.

Explanation of Solution

Given:

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

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(\sin at\right)=L\left(\frac{e^{iat}-e^{-iat}}{2i}\right)=\frac{a}{s^2+a^2}\\ L\left(\cos at\right)=L\left(\frac{e^{iat}+e^{-iat}}{2}\right)=\frac{s}{s^2+a^2}\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

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

Calculation:

Here, we have

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

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$\begin{array}{l}f\left(t\right)=L^{-1}\left(F\left(s\right)\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{6}{s^2+9}\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{2\left(3\right)}{s^2+3^2}\right)\\ \Rightarrow f\left(t\right)=2\sin 3t\left[\because L\left(\sin at\right)=\frac{a}{s^2+a^2}\right]\end{array}$.

To determine

(b)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.8P

Inverse Laplace transform of the function $F\left(s\right)=\frac{5}{s^2+4}+\frac{4s}{s^2+4}$ is $f\left(t\right)=\frac{5}{2}\sin \left(2t\right)+4\cos \left(2t\right)$.

Explanation of Solution

Given:

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

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(\sin at\right)=L\left(\frac{e^{iat}-e^{-iat}}{2i}\right)=\frac{a}{s^2+a^2}\\ L\left(\cos at\right)=L\left(\frac{e^{iat}+e^{-iat}}{2}\right)=\frac{s}{s^2+a^2}\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

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

Calculation:

Here, we have

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

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$\begin{array}{l}f\left(t\right)=L^{-1}\left(\frac{5}{s^2+4}+\frac{4s}{s^2+4}\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{5}{s^2+2^2}\right)+L^{-1}\left(\frac{4s}{s^2+2^2}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{2}{L}^{-1}\left(\frac{2}{s^2+2^2}\right)+4L^{-1}\left(\frac{s}{s^2+2^2}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{2}\sin \left(2t\right)+4\cos \left(2t\right)\left[\because L\left(\sin at\right)=\frac{a}{s^2+a^2}\&L\left(\cos at\right)=\frac{s}{s^2+a^2}\right]\end{array}$.

To determine

(c)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.8P

Inverse Laplace transform of the function $F\left(s\right)=\frac{6}{s^2+4s+13}$ is $f\left(t\right)=2e^{-2t}\sin \left(3t\right)$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{6}{s^2+4s+13}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(\sin at\right)=L\left(\frac{e^{iat}-e^{-iat}}{2i}\right)=\frac{a}{s^2+a^2}\\ L\left(\cos at\right)=L\left(\frac{e^{iat}+e^{-iat}}{2}\right)=\frac{s}{s^2+a^2}\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

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

Calculation:

Here, we have

$F\left(s\right)=\frac{6}{s^2+4s+13}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$\begin{array}{l}f\left(t\right)=L^{-1}\left(\frac{6}{s^2+4s+13}\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{6}{s^2+2\left(2\right)s+4+9}\right)\\ \Rightarrow f\left(t\right)=2L^{-1}\left(\frac{3}{{\left(s+2\right)}^2+3^2}\right)\\ \Rightarrow f\left(t\right)=2e^{-2t}\sin \left(3t\right)\left[\because L\left(\sin at\right)=\frac{a}{s^2+a^2}\&L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right)\right]\end{array}$.

To determine

(d)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.8P

Inverse Laplace transform of the function $F\left(s\right)=\frac{5}{s\left(s+3\right)}$ is $f\left(t\right)=\frac{5}{3}\left(1-e^{-3t}\right)$.

Explanation of Solution

Given:

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

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

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

Calculation:

Here, we have

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

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$\begin{array}{l}f\left(t\right)=L^{-1}\left(\frac{5}{s\left(s+3\right)}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{3}{L}^{-1}\left(\frac{3}{s\left(s+3\right)}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{3}{L}^{-1}\left(\frac{s+3-s}{s\left(s+3\right)}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{3}{L}^{-1}\left(\frac{1}{s}-\frac{1}{\left(s+3\right)}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{3}{L}^{-1}\left(\frac{1}{s}\right)-\frac{5}{3}{L}^{-1}\left(\frac{1}{s+3}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{3}\left(1\right)-\frac{5}{3}{e}^{-3t}\left[\because L\left(1\right)=\frac{1}{s}\&L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right)\right]\\ \Rightarrow f\left(t\right)=\frac{5}{3}\left(1-e^{-3t}\right)\end{array}$.

To determine

(e)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.8P

Inverse Laplace transform of the function $F\left(s\right)=\frac{10}{\left(s+3\right)\left(s+7\right)}$ is $f\left(t\right)=\frac{5}{2}\left(e^{-3t}-e^{-7t}\right)$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{10}{\left(s+3\right)\left(s+7\right)}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

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

Calculation:

Here, we have

$F\left(s\right)=\frac{10}{\left(s+3\right)\left(s+7\right)}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$f\left(t\right)=L^{-1}\left(\frac{10}{\left(s+3\right)\left(s+7\right)}\right)\left(1\right)$

Now, we find the partial fraction of $\frac{10}{\left(s+3\right)\left(s+7\right)}$ such that

$\frac{10}{\left(s+3\right)\left(s+7\right)}=\frac{A}{\left(s+3\right)}+\frac{B}{\left(s+7\right)}\left(2\right)$

Here, A and B will be

$\begin{array}{l}A=\underset{s\to -3}{\lim }\left(s+3\right)\frac{10}{\left(s+3\right)\left(s+7\right)}=\frac{10}{4}\left(3\right)\\ B=\underset{s\to -7}{\lim }\left(s+7\right)\frac{10}{\left(s+3\right)\left(s+7\right)}=-\frac{10}{4}\left(4\right)\end{array}$

Using (3) and (4) in (1), we get

$\frac{10}{\left(s+3\right)\left(s+7\right)}=\frac{10}{4\left(s+3\right)}-\frac{10}{4\left(s+7\right)}\left(5\right)$

Using (5) in (1), we get

$\begin{array}{l}f\left(t\right)=L^{-1}\left(\frac{10}{4\left(s+3\right)}-\frac{10}{4\left(s+7\right)}\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{10}{4\left(s+3\right)}\right)-L^{-1}\left(\frac{10}{4\left(s+7\right)}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{2}{L}^{-1}\left(\frac{1}{s+3}\right)-\frac{5}{2}{L}^{-1}\left(\frac{1}{s+7}\right)\\ \Rightarrow f\left(t\right)=\frac{5}{2}{e}^{-3t}-\frac{5}{2}{e}^{-7t}\left[\because L\left(1\right)=\frac{1}{s}\&L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right)\right]\\ \Rightarrow f\left(t\right)=\frac{5}{2}\left(e^{-3t}-e^{-7t}\right)\end{array}$.

To determine

(f)

The inverse Laplace transform $f\left(t\right)$ of the given function.

Answer to Problem 2.8P

Inverse Laplace transform of the function $F\left(s\right)=\frac{2s+8}{\left(s+3\right)\left(s+7\right)}$ is $f\left(t\right)=\frac{1}{2}{e}^{-3t}+\frac{3}{2}{e}^{-7t}$.

Explanation of Solution

Given:

$F\left(s\right)=\frac{2s+8}{\left(s+3\right)\left(s+7\right)}$.

Concept Used:

Laplace transform of function $f\left(t\right)$ is defined as

$L\left(f\left(t\right)\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{e}^{-st}f\left(t\right)dt}$

Where $f\left(t\right)$ is piecewise continuo function defined for all $t\ge 0$.

$\begin{array}{l}L\left(t^n\right)=\frac{\overline{)n+1}}{s^{n+1}}=\frac{\overline{)n}}{s^{n+1}}\\ L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right);L\left(f\left(t\right)\right)=F\left(s\right)\\ L\left(t^{n}f\left(t\right)\right)={\left(-1\right)}^n\frac{d^n}{d{s}^n}\left(F\left(s\right)\right);L\left(f\left(t\right)\right)=F\left(s\right)\end{array}$

Inverse Laplace transform of given function $F\left(s\right)$ is given by

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

Calculation:

Here, we have

$F\left(s\right)=\frac{2s+8}{\left(s+3\right)\left(s+7\right)}$

Thus, the inverse Laplace transform $f\left(t\right)$ of the given function is given as

$f\left(t\right)=L^{-1}\left(\frac{2s+8}{\left(s+3\right)\left(s+7\right)}\right)\left(1\right)$

Now, we find the partial fraction of $\frac{2s+8}{\left(s+3\right)\left(s+7\right)}$ such that

$\frac{2s+8}{\left(s+3\right)\left(s+7\right)}=\frac{A}{\left(s+3\right)}+\frac{B}{\left(s+7\right)}\left(2\right)$

Here, A and B will be

$\begin{array}{l}A=\underset{s\to -3}{\lim }\left(s+3\right)\frac{2s+8}{\left(s+3\right)\left(s+7\right)}=\frac{2\left(-3\right)+8}{4}=\frac{1}{2}\left(3\right)\\ B=\underset{s\to -7}{\lim }\left(s+7\right)\frac{2s+8}{\left(s+3\right)\left(s+7\right)}=\frac{2\left(-7\right)+8}{-4}=\frac{3}{2}\left(4\right)\end{array}$

Using (3) and (4) in (1), we get

$\frac{2s+8}{\left(s+3\right)\left(s+7\right)}=\frac{1}{2\left(s+3\right)}+\frac{3}{2\left(s+7\right)}\left(5\right)$

Using (5) in (1), we get

$\begin{array}{l}f\left(t\right)=L^{-1}\left(\frac{1}{2\left(s+3\right)}+\frac{3}{2\left(s+7\right)}\right)\\ \Rightarrow f\left(t\right)=L^{-1}\left(\frac{1}{2\left(s+3\right)}\right)+L^{-1}\left(\frac{3}{2\left(s+7\right)}\right)\\ \Rightarrow f\left(t\right)=\frac{1}{2}{L}^{-1}\left(\frac{1}{s+3}\right)+\frac{3}{2}{L}^{-1}\left(\frac{1}{s+7}\right)\\ \Rightarrow f\left(t\right)=\frac{1}{2}{e}^{-3t}+\frac{3}{2}{e}^{-7t}\left[\because L\left(1\right)=\frac{1}{s}\&L\left(e^{-at}f\left(t\right)\right)=F\left(s+a\right)\right]\end{array}$.