Chapter 18, Problem 24P

(a)

To determine

Find the Fourier transform of $x\left(t\right)=f\left(t\right)+3$.

Answer to Problem 24P

The Fourier transform of $x\left(t\right)=f\left(t\right)+3$ is $\underset{\_}{\frac{j}{\omega }\left(e^{-j\omega }-1\right)+6\pi \delta \left(\omega \right)}$.

Explanation of Solution

Given data:

$F\left[f\left(t\right)\right]=\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)$.

$x\left(t\right)=f\left(t\right)+3$ (1)

Formula used:

Consider the general form of Fourier transform of $f\left(t\right)$ is represented as $F\left(\omega \right)$.

$F\left(f\left(t\right)\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(t\right)e^{-j\omega t}dt}$

Calculation:

Apply Fourier transform to equation (1) as follows.

$\begin{array}{l}F\left[x\left(t\right)\right]=F\left[f\left(t\right)+3\right]\\ X\left(\omega \right)=F\left[f\left(t\right)\right]+F\left(3\right)\end{array}$

Substitute $\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)$ for $F\left(f\left(t\right)\right)$ as follows.

$\begin{array}{c}X\left(\omega \right)=\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)+3\left[2\pi \delta \left(\omega \right)\right]\left\{\because F\left(1\right)=2\pi \delta \left(\omega \right)\right\}\\ =\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)+6\pi \delta \left(\omega \right)\end{array}$

Conclusion:

Thus, the Fourier transform of $x\left(t\right)=f\left(t\right)+3$ is $\underset{\_}{\frac{j}{\omega }\left(e^{-j\omega }-1\right)+6\pi \delta \left(\omega \right)}$.

(b)

To determine

Find the Fourier transform of $y\left(t\right)=f\left(t-2\right)$.

Answer to Problem 24P

The Fourier transform of $y\left(t\right)=f\left(t-2\right)$ is $\underset{\_}{\frac{j{e}^{-j2\omega }}{\omega }\left(e^{-j\omega }-1\right)}$.

Explanation of Solution

Given data:

$F\left(f\left(t\right)\right)=\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)$.

$y\left(t\right)=f\left(t-2\right)$ (2)

Calculation:

Apply Fourier transform to equation (2) as follows.

$\begin{array}{l}F\left[y\left(t\right)\right]=F\left[f\left(t-2\right)\right]\\ Y\left(\omega \right)=e^{-j\omega 2}F\left(f\left(t\right)\right)\left\{\because F\left[f\left(t-a\right)\right]=e^{-j\omega a}F\left(\omega \right)\right\}\end{array}$

Substitute $\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)$ for $F\left(f\left(t\right)\right)$ as follows.

$\begin{array}{c}Y\left(\omega \right)=e^{-j2\omega }\left[\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)\right]\\ =\left(\frac{j{e}^{-j2\omega }}{\omega }\right)\left(e^{-j\omega }-1\right)\end{array}$

Conclusion:

Thus, the Fourier transform of $y\left(t\right)=f\left(t-2\right)$ is $\underset{\_}{\frac{j{e}^{-j2\omega }}{\omega }\left(e^{-j\omega }-1\right)}$.

(c)

To determine

Find the Fourier transform of $h\left(t\right)=f^{\prime }\left(t\right)$.

Answer to Problem 24P

The Fourier transform of $h\left(t\right)=f^{\prime }\left(t\right)$ is $\underset{\_}{1-e^{-j\omega }}$.

Explanation of Solution

Given data:

$F\left(f\left(t\right)\right)=\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)$.

$h\left(t\right)=f^{\prime }\left(t\right)$ (3)

Calculation:

Apply Fourier transform to equation (3) as follows.

$\begin{array}{l}F\left[h\left(t\right)\right]=F\left[f^{\prime }\left(t\right)\right]\\ H\left(\omega \right)=j\omega F\left[f\left(t\right)\right]\left\{\because F\left[f^{\prime }\left(t\right)\right]=j\omega F\left(\omega \right)\right\}\end{array}$

Substitute $\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)$ for $F\left(f\left(t\right)\right)$ as follows.

$\begin{array}{c}H\left(\omega \right)=j\omega \left[\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)\right]\\ =\left(\frac{j^2\omega }{\omega }\right)\left(e^{-j\omega }-1\right)\\ =\left(-1\right)\left(e^{-j\omega }-1\right)\\ =1-e^{-j\omega }\end{array}$

Conclusion:

Thus, the Fourier transform of $h\left(t\right)=f^{\prime }\left(t\right)$ is $\underset{\_}{1-e^{-j\omega }}$.

(d)

To determine

Find the Fourier transform of $g\left(t\right)=4f\left(\frac{2}{3}t\right)+10f\left(\frac{5}{3}t\right)$.

Answer to Problem 24P

The Fourier transform of $g\left(t\right)=4f\left(\frac{2}{3}t\right)+10f\left(\frac{5}{3}t\right)$ is $\underset{\_}{\frac{j4}{\omega }\left(e^{-\frac{j3\omega }{2}}-1\right)+\frac{j10}{\omega }\left(e^{-\frac{j3\omega }{5}}-1\right)}$.

Explanation of Solution

Given data:

$F\left(f\left(t\right)\right)=\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)$.

$g\left(t\right)=4f\left(\frac{2}{3}t\right)+10f\left(\frac{5}{3}t\right)$ (4)

Calculation:

Apply Fourier transform to equation (4) as follows.

$\begin{array}{l}F\left[g\left(t\right)\right]=F\left[4f\left(\frac{2}{3}t\right)+10f\left(\frac{5}{3}t\right)\right]\\ G\left(\omega \right)=F\left[4f\left(\frac{2}{3}t\right)+10f\left(\frac{5}{3}t\right)\right]\\ G\left(\omega \right)=4F\left[f\left(\frac{2}{3}t\right)\right]+10F\left[f\left(\frac{5}{3}t\right)\right]\\ G\left(\omega \right)=4\frac{1}{\left(\frac{2}{3}\right)}F\left[f\left(\frac{t}{\left(\frac{2}{3}\right)}\right)\right]+10\frac{1}{\left(\frac{5}{3}\right)}F\left[f\left(\frac{t}{\left(\frac{5}{3}\right)}\right)\right]\left\{\because F\left[f\left(at\right)=\frac{1}{\left|a\right|}F\left(\frac{\omega }{a}\right)\right]\right\}\end{array}$

Simplify the equation as follows.

$G\left(\omega \right)=4\left(\frac{3}{2}\right)F\left[f\left(\frac{3}{2}t\right)\right]+10\left(\frac{3}{5}\right)F\left[f\left(\frac{3}{5}t\right)\right]$

Substitute $\left(\frac{j}{\omega }\right)\left(e^{-j\omega }-1\right)$ for $F\left(f\left(t\right)\right)$ as follows.

$\begin{array}{c}G\left(\omega \right)=4\left(\frac{3}{2}\right)\left[\left(\frac{j}{\frac{3}{2}\omega }\right)\left(e^{-\frac{j3\omega }{2}}-1\right)\right]+10\left(\frac{3}{5}\right)\left[\left(\frac{j}{\frac{3}{5}\omega }\right)\left(e^{-\frac{j3\omega }{5}}-1\right)\right]\\ =4\left[\left(\frac{j}{\omega }\right)\left(e^{-\frac{j3\omega }{2}}-1\right)\right]+10\left[\left(\frac{j}{\omega }\right)\left(e^{-\frac{j3\omega }{5}}-1\right)\right]\\ =\left(\frac{j4}{\omega }\right)\left(e^{-\frac{j3\omega }{2}}-1\right)+\left(\frac{j10}{\omega }\right)\left(e^{-\frac{j3\omega }{5}}-1\right)\end{array}$

Conclusion:

Thus, the Fourier transform of $g\left(t\right)=4f\left(\frac{2}{3}t\right)+10f\left(\frac{5}{3}t\right)$ is $\underset{\_}{\frac{j4}{\omega }\left(e^{-\frac{j3\omega }{2}}-1\right)+\frac{j10}{\omega }\left(e^{-\frac{j3\omega }{5}}-1\right)}$.