Chapter 3, Problem 3.1E

Find the Laplace transforms of the following functions, using the information in Table 3.1. (However, someof the individual terms in these functions may not have Laplace transforms.)
a. $f\left(t\right)=5+e^{-3t}+t{e}^{-4t}$
b. $f\left(t\right)=\sin \left(4t\right)+t-3+e^{-\left(t-3\right)}+5/t$
c. $f\left(t\right)=t\cos \left(4t\right)+t/5$
d. $f\left(t\right)=\left(t-1\right)\cos \left(4\left(t-1\right)\right)+t^2$

Interpretation Introduction

(a)

Interpretation:

Laplace transform of $f\left(t\right)=5+e^{-3t}+t{e}^{-4t}$ is to be determined.

Concept introduction:

For a function $f\left(t\right)$, the Laplace transform is given by,

$F\left(s\right)=ℒ\left[f\left(t\right)\right]={\displaystyle {\int }_0^{\infty }f\left(f\right)e^{-st}dt}$

Here, $F\left(s\right)$ represents the Laplace transform, $s$ is a variable that is complex and independent, $f\left(t\right)$ is any function of time which is being transformed, and $ℒ$ is the operator which is defined by an integral.

Answer to Problem 3.1E

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

Explanation of Solution

The Laplace transform of the function of type $f\left(t\right)=u\left(t\right)$ which is a unit step function is as follows:

$ℒ\left[u\left(t\right)\right]=\frac{1}{s}$    ...... (1)

Laplace transform of the function of type $f\left(t\right)=e^{-bt}$, where b is a constant, is as follows:

$ℒ\left[e^{-bt}\right]=\frac{1}{s+b}$    ...... (2)

Laplace transform of the function of type $f\left(t\right)=\frac{t^{n-1}{e}^{-bt}}{\left(n-1\right)!},n>0$, where b is a constant, is as follows:

$ℒ\left[\frac{t^{n-1}{e}^{-bt}}{\left(n-1\right)!}\right]=\frac{1}{{\left(s+b\right)}^n}$    ...... (3)

Now, for the given function $f\left(t\right)=5+e^{-3t}+t{e}^{-4t}$, break it into smaller functions as follows:

$\begin{array}{c}f\left(t\right)=f{\left(t\right)}_1+f{\left(t\right)}_2+f{\left(t\right)}_3\\ f{\left(t\right)}_1=5\\ f{\left(t\right)}_2=e^{-3t}\\ f{\left(t\right)}_3=t{e}^{-4t}\end{array}$

For $f{\left(t\right)}_1$, compare it with equation (1) to get the Laplace transform of it as follows:

$\begin{array}{c}ℒ\left[f{\left(t\right)}_1\right]=ℒ\left[5\right]\\ =\frac{5}{s}\end{array}$

For $f{\left(t\right)}_2$, compare it with equation (2) to get the Laplace transform of it. Here, $b=3$.

$\begin{array}{c}ℒ\left[f{\left(t\right)}_2\right]=ℒ\left[e^{-3t}\right]\\ =\frac{1}{s+3}\end{array}$

For $f{\left(t\right)}_3$, compare it with equation (3) to get the Laplace transform of it. Here, $b=4\text{ and }n=2$.

$\begin{array}{c}ℒ\left[f{\left(t\right)}_3\right]=ℒ\left[t{e}^{-4t}\right]\\ =\frac{1}{{\left(s+4\right)}^2}\end{array}$

Combine the Laplace transformsof $f{\left(t\right)}_1,f{\left(t\right)}_2,\text{ and }f{\left(t\right)}_3$ to get the Laplace transform of $f\left(t\right)$ as follows:

$\begin{array}{c}ℒ\left[f\left(t\right)\right]=ℒ\left[f{\left(t\right)}_1\right]+ℒ\left[f{\left(t\right)}_2\right]+ℒ\left[f{\left(t\right)}_3\right]\\ =\frac{5}{s}+\frac{1}{s+3}+\frac{1}{{\left(s+4\right)}^2}\end{array}$

Interpretation Introduction

(b)

Interpretation:

Laplace transform of $f\left(t\right)=\sin \left(4t\right)+t-3+e^{-\left(t-3\right)}+5/t$ is to be determined.

Concept introduction:

For a function $f\left(t\right)$, the Laplace transform is given by,

$F\left(s\right)=ℒ\left[f\left(t\right)\right]={\displaystyle {\int }_0^{\infty }f\left(f\right)e^{-st}dt}$

Here, $F\left(s\right)$ represents the Laplace transform, $s$ is a variable that is complex and independent, $f\left(t\right)$ is any function of time which is being transformed, and $ℒ$ is the operator which is defined by an integral.

Answer to Problem 3.1E

$ℒ\left[f\left(t\right)\right]=\frac{16}{s+16}+\frac{1}{s^2}-\frac{3}{s}+\left(\frac{1}{s+1}\cdot e^3\right)+ℒ\left[5/t\right]$

Explanation of Solution

The Laplace transform of the function of type $f\left(t\right)=u\left(t\right)$ which is a unit step function is:

$ℒ\left[u\left(t\right)\right]=\frac{1}{s}$   ...... (1)

Laplace transform of the function of type $f\left(t\right)=e^{-bt}$, where b is a constant, is:

$ℒ\left[e^{-bt}\right]=\frac{1}{s+b}$   ...... (2)

The Laplace transform of the function of type $f\left(t\right)=\sin \left(\omega t\right)$, where $\omega$ is a constant, is:

$ℒ\left[\sin \left(\omega t\right)\right]=\frac{\omega }{s^2+{\omega }^2}$   ...... (4)

The Laplace transform of the function of type $f\left(t\right)=t$ which is a ramp function is:

$ℒ\left[t\right]=\frac{1}{s^2}$   ...... (5)

Now, for the given function $f\left(t\right)=\sin \left(4t\right)+t-3+e^{-\left(t-3\right)}+5/t$, break it into smaller functions as:

$\begin{array}{c}f\left(t\right)=f{\left(t\right)}_1+f{\left(t\right)}_2+f{\left(t\right)}_3+f{\left(t\right)}_4+f{\left(t\right)}_5\\ f{\left(t\right)}_1=\sin \left(4t\right)\\ f{\left(t\right)}_2=t\\ f{\left(t\right)}_3=-3\\ f{\left(t\right)}_4=e^{-\left(t-3\right)}\\ f{\left(t\right)}_5=5/t\end{array}$

For $f{\left(t\right)}_1$, compare it with equation (4) to get the Laplace transform of it. Here, $\omega =4$

$\begin{array}{c}ℒ\left[f{\left(t\right)}_1\right]=ℒ\left[\sin \left(4t\right)\right]\\ =\frac{4^2}{s^2+4^2}\\ =\frac{16}{s+16}\end{array}$

For $f{\left(t\right)}_2$, compare it with equation (4) to get the Laplace transform of it.

$\begin{array}{c}ℒ\left[f{\left(t\right)}_2\right]=ℒ\left[t\right]\\ =\frac{1}{s^2}\end{array}$

For $f{\left(t\right)}_3$, compare it with equation (1) to get the Laplace transform of it.

$\begin{array}{c}ℒ\left[f{\left(t\right)}_3\right]=ℒ\left[-3\right]\\ =\frac{-3}{s}\end{array}$

For $f{\left(t\right)}_4$, simplify the function and compare it with equation (2) to get the Laplace transform of it. Here, $b=1$.

$\begin{array}{c}ℒ\left[f{\left(t\right)}_4\right]=ℒ\left[e^{-\left(t-3\right)}\right]\\ =ℒ\left[e^{-t+3}\right]\\ =ℒ\left[e^{-t}\cdot e^3\right]\\ =e^3ℒ\left[e^{-t}\right]\\ =\frac{1}{s+1}\cdot e^3\end{array}$

For $f{\left(t\right)}_5$, there is no Laplace transform available. Thus,

$ℒ\left[f{\left(t\right)}_5\right]=ℒ\left[5/t\right]$

Combine the Laplace transforms of $f{\left(t\right)}_1,f{\left(t\right)}_2,f{\left(t\right)}_3,f{\left(t\right)}_4,\text{ and }f{\left(t\right)}_5$ to get the Laplace transform of $f\left(t\right)$ as:

$\begin{array}{c}ℒ\left[f\left(t\right)\right]=ℒ\left[f{\left(t\right)}_1\right]+ℒ\left[f{\left(t\right)}_2\right]+ℒ\left[f{\left(t\right)}_3\right]+ℒ\left[f{\left(t\right)}_4\right]+ℒ\left[f{\left(t\right)}_5\right]\\ =\frac{16}{s+16}+\frac{1}{s^2}+\left(\frac{-3}{s}\right)+\left(\frac{1}{s+1}\cdot e^3\right)+ℒ\left[5/t\right]\\ =\frac{16}{s+16}+\frac{1}{s^2}-\frac{3}{s}+\left(\frac{1}{s+1}\cdot e^3\right)+ℒ\left[5/t\right]\end{array}$

Interpretation Introduction

(c)

Interpretation:

Laplace transform of $f\left(t\right)=t\cos \left(4t\right)+t/5$ is to be determined.

Concept introduction:

For a function $f\left(t\right)$, the Laplace transform is given by,

$F\left(s\right)=ℒ\left[f\left(t\right)\right]={\displaystyle {\int }_0^{\infty }f\left(f\right)e^{-st}dt}$

Here, $F\left(s\right)$ represents the Laplace transform, $s$ is a variable that is complex and independent, $f\left(t\right)$ is any function of time which is being transformed, and $ℒ$ is the operator which is defined by an integral.

Answer to Problem 3.1E

$ℒ\left[f\left(t\right)\right]=\frac{s^2-16}{{\left(s+16\right)}^2}+\frac{1}{5s^2}$

Explanation of Solution

The Laplace transform of the function of type $f\left(t\right)=t$ which is a ramp function is:

$ℒ\left[t\right]=\frac{1}{s^2}$ ...... (5)

The Laplace transform of the function of type $f\left(t\right)=t\cos \left(\omega t\right)$, where $\omega$ is a constant and $s>\left|\omega \right|$, is:

$ℒ\left[t\cos \left(\omega t\right)\right]=\frac{s^2-{\omega }^2}{{\left(s^2+{\omega }^2\right)}^2}$   ...... (6)

Now, for the given function $f\left(t\right)=t\cos \left(4t\right)+t/5$, break it into smaller functions as:

$\begin{array}{c}f\left(t\right)=f{\left(t\right)}_1+f{\left(t\right)}_2\\ f{\left(t\right)}_1=t\cos \left(4t\right)\\ f{\left(t\right)}_2=t/5\end{array}$

For $f{\left(t\right)}_1$, compare it with equation (6) to get the Laplace transform of it. Here, $\omega =4$

$\begin{array}{c}ℒ\left[f{\left(t\right)}_1\right]=ℒ\left[t\cos \left(4t\right)\right]\\ =\frac{s^2-4^2}{{\left(s^2+4^2\right)}^2}\\ =\frac{s^2-16}{{\left(s+16\right)}^2}\end{array}$

For $f{\left(t\right)}_2$, compare it with equation (5) to get the Laplace transform of it.

$\begin{array}{c}ℒ\left[f{\left(t\right)}_2\right]=ℒ\left[t/5\right]\\ =\frac{1}{5s^2}\end{array}$

Combine the Laplace transforms of $f{\left(t\right)}_1,\text{ and }f{\left(t\right)}_2$ to get the Laplace transform of $f\left(t\right)$ as:

$\begin{array}{c}ℒ\left[f\left(t\right)\right]=ℒ\left[f{\left(t\right)}_1\right]+ℒ\left[f{\left(t\right)}_2\right]\\ =\frac{s^2-16}{{\left(s+16\right)}^2}+\frac{1}{5s^2}\end{array}$

Interpretation Introduction

(d)

Interpretation:

Laplace transform of $f\left(t\right)=\left(t-1\right)\cos \left(4\left(t-1\right)\right)+t^2$ is to be determined.

Concept introduction:

For a function $f\left(t\right)$, the Laplace transform is given by,

$F\left(s\right)=ℒ\left[f\left(t\right)\right]={\displaystyle {\int }_0^{\infty }f\left(f\right)e^{-st}dt}$

Here, $F\left(s\right)$ represents the Laplace transform, $s$ is a variable that is complex and independent, $f\left(t\right)$ is any function of time which is being transformed, and $ℒ$ is the operator which is defined by an integral.

Answer to Problem 3.1E

$ℒ\left[f\left(t\right)\right]=\cos \left(4\right)\left(-\frac{16\left(s+1\right)}{{\left(s+16\right)}^2}\right)+\sin \left(4\right)\left(\frac{4\left(s-16\right)}{{\left(s+16\right)}^2}\right)+\frac{2}{s^3}$

Explanation of Solution

Simplify the given function and expand using the formula $\cos \left(A-B\right)=\cos A\cos B+\sin A\sin B$ as:

$\begin{array}{c}f\left(t\right)=\left(t-1\right)\cos \left(4\left(t-1\right)\right)+t^2\\ =t\left(\cos \left(4t-4\right)\right)-\left(\cos \left(4t-4\right)\right)+t^2\\ =t\left(\cos \left(4t\right)\cos \left(4\right)+\sin \left(4t\right)\sin \left(4\right)\right)-\left(\cos \left(4t\right)\cos \left(4\right)+\sin \left(4t\right)\sin \left(4\right)\right)+t^2\\ =\cos \left(4\right)t\cos \left(4t\right)+\sin \left(4\right)t\sin \left(4t\right)-\cos \left(4\right)\cos \left(4t\right)-\sin \left(4\right)\sin \left(4t\right)+t^2\end{array}$

Note that $\cos \left(4\right)\text{ and sin}\left(4\right)$ are constants.

The Laplace transform of the function of type $f\left(t\right)=\sin \left(\omega t\right)$, where $\omega$ is a constant, is:

$ℒ\left[\sin \left(\omega t\right)\right]=\frac{\omega }{s^2+{\omega }^2}$   ...... (4)

The Laplace transform of the function of type $f\left(t\right)=t\cos \left(\omega t\right)$, where $\omega$ is a constant and $s>\left|\omega \right|$, is:

$ℒ\left[t\cos \left(\omega t\right)\right]=\frac{s^2-{\omega }^2}{{\left(s^2+{\omega }^2\right)}^2}$   ...... (6)

The Laplace transform of the function of type $f\left(t\right)=t\sin \left(\omega t\right)$, where $\omega$ is a constant and $s>\left|\omega \right|$, is:

$ℒ\left[t\sin \left(\omega t\right)\right]=\frac{2\omega s}{{\left(s^2+{\omega }^2\right)}^2}$   ...... (7)

The Laplace transform of the function of type $f\left(t\right)=\cos \left(\omega t\right)$, where $\omega$ is a constant, is:

$ℒ\left[\cos \left(\omega t\right)\right]=\frac{s}{s^2+{\omega }^2}$   ...... (8)

The Laplace transform of the function of type $f\left(t\right)=t^n$, where $n$ is positive integer, is:

$ℒ\left[t^2\right]=\frac{n!}{s^{n+1}}$   ...... (9)

Now, for the simplified function $f\left(t\right)=\cos \left(4\right)t\cos \left(4t\right)+\sin \left(4\right)t\sin \left(4t\right)-\cos \left(4\right)\cos \left(4t\right)-\sin \left(4\right)\sin \left(4t\right)+t^2$, break it into smaller functions as:

$\begin{array}{c}f\left(t\right)=f{\left(t\right)}_1+f{\left(t\right)}_2+f{\left(t\right)}_3+f{\left(t\right)}_4+f{\left(t\right)}_5\\ f{\left(t\right)}_1=\cos \left(4\right)t\cos \left(4t\right)\\ f{\left(t\right)}_2=\sin \left(4\right)t\sin \left(4t\right)\\ f{\left(t\right)}_3=-\cos \left(4\right)\cos \left(4t\right)\\ f{\left(t\right)}_4=-\sin \left(4\right)\sin \left(4t\right)\\ f{\left(t\right)}_5=t^2\end{array}$

For $f{\left(t\right)}_1$, compare it with equation (6) to get the Laplace transform of it. Here, $\omega =4$

$\begin{array}{c}ℒ\left[f{\left(t\right)}_1\right]=ℒ\left[\cos \left(4\right)t\cos \left(4t\right)\right]\\ =\cos \left(4\right)ℒ\left[t\cos \left(4t\right)\right]\\ =\cos \left(4\right)\cdot \frac{s^2-4^2}{{\left(s^2+4^2\right)}^2}\\ =\cos \left(4\right)\cdot \frac{s^2-16}{{\left(s+16\right)}^2}\end{array}$

For $f{\left(t\right)}_2$, compare it with equation (7) to get the Laplace transform of it. Here, $\omega =4$

$\begin{array}{c}ℒ\left[f{\left(t\right)}_2\right]=ℒ\left[\sin \left(4\right)t\sin \left(4t\right)\right]\\ =\sin \left(4\right)ℒ\left[t\sin \left(4t\right)\right]\\ =\sin \left(4\right)\cdot \frac{2\left(4\right)s}{{\left(s^2+4^2\right)}^2}\\ =\sin \left(4\right)\cdot \frac{8s}{{\left(s+16\right)}^2}\end{array}$

For $f{\left(t\right)}_3$, compare it with equation (8) to get the Laplace transform of it. Here, $\omega =4$

$\begin{array}{c}ℒ\left[f{\left(t\right)}_3\right]=ℒ\left[-\cos \left(4\right)\cos \left(4t\right)\right]\\ =-\cos \left(4\right)ℒ\left[\cos \left(4t\right)\right]\\ =-\cos \left(4\right)\cdot \frac{s}{\left(s^2+4^2\right)}\\ =-\cos \left(4\right)\cdot \frac{s}{\left(s+16\right)}\end{array}$

For $f{\left(t\right)}_4$, compare it with equation (4) to get the Laplace transform of it. Here, $\omega =4$

$\begin{array}{c}ℒ\left[f{\left(t\right)}_4\right]=ℒ\left[-\sin \left(4\right)\sin \left(4t\right)\right]\\ =-\sin \left(4\right)ℒ\left[\sin \left(4t\right)\right]\\ =-\sin \left(4\right)\cdot \frac{4}{\left(s^2+4^2\right)}\\ =-\sin \left(4\right)\cdot \frac{4}{\left(s+16\right)}\end{array}$

For $f{\left(t\right)}_5$, compare it with equation (9) to get the Laplace transform of it. Here, $n=2$

$\begin{array}{c}ℒ\left[f{\left(t\right)}_5\right]=ℒ\left[t^2\right]\\ =\frac{2!}{s^{2+1}}\\ =\frac{2}{s^3}\end{array}$

Combine the Laplace transforms of $f{\left(t\right)}_1,f{\left(t\right)}_2,f{\left(t\right)}_3,f{\left(t\right)}_4\text{, and }f{\left(t\right)}_5$ to get the Laplace transform of $f\left(t\right)$ as:

$\begin{array}{c}ℒ\left[f\left(t\right)\right]=ℒ\left[f{\left(t\right)}_1\right]+ℒ\left[f{\left(t\right)}_2\right]+ℒ\left[f{\left(t\right)}_3\right]+ℒ\left[f{\left(t\right)}_4\right]+ℒ\left[f{\left(t\right)}_5\right]\\ =\left(\cos \left(4\right)\cdot \frac{s^2-16}{{\left(s+16\right)}^2}\right)+\left(\sin \left(4\right)\cdot \frac{8s}{{\left(s+16\right)}^2}\right)+\left(-\cos \left(4\right)\cdot \frac{s}{\left(s+16\right)}\right)+\left(-\sin \left(4\right)\cdot \frac{4}{\left(s+16\right)}\right)+\frac{2}{s^3}\\ =\cos \left(4\right)\left(\frac{s^2-16}{{\left(s+16\right)}^2}-\frac{s}{\left(s+16\right)}\right)+\sin \left(4\right)\left(\frac{8s}{{\left(s+16\right)}^2}-\frac{4}{\left(s+16\right)}\right)+\frac{2}{s^3}\\ =\cos \left(4\right)\left(-\frac{16\left(s+1\right)}{{\left(s+16\right)}^2}\right)+\sin \left(4\right)\left(\frac{4\left(s-16\right)}{{\left(s+16\right)}^2}\right)+\frac{2}{s^3}\end{array}$