Chapter 18, Problem 23P

(a)

To determine

Find the Fourier transform of $f\left(-3t\right)$.

Answer to Problem 23P

The Fourier transform of $f\left(-3t\right)$ is $\underset{\_}{\frac{30}{\left(6-j\omega \right)\left(15-j\omega \right)}}$

Explanation of Solution

Given data:

$F\left(\omega \right)=\frac{10}{\left(2+j\omega \right)\left(5+j\omega \right)}$

Formula used:

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

$F\left(f\left(t\right)\right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}f\left(t\right)e^{-j\omega t}dt}$ (1)

Consider the scaling property of the Fourier transform.

$F\left[f\left(at\right)\right]=\frac{1}{\left|a\right|}F\left(\frac{\omega }{a}\right)$

Calculation:

Find $F\left[f\left(-3t\right)\right]$.

$\begin{array}{c}F\left[f\left(-3t\right)\right]=\frac{1}{\left|-3\right|}\frac{10}{\left(2+\frac{j\omega }{-3}\right)\left(5+\frac{j\omega }{-3}\right)}\left\{\because F\left(\omega \right)=\frac{10}{\left(2+j\omega \right)\left(5+j\omega \right)}\right\}\\ =\frac{1}{\left|-3\right|}\frac{10}{\left(2-\frac{j\omega }{3}\right)\left(5-\frac{j\omega }{3}\right)}\\ =\frac{1}{3}\frac{10}{\left(\frac{6-j\omega }{3}\right)\left(\frac{15-j\omega }{3}\right)}\\ =\frac{1}{3}\frac{10\times 3\times 3}{\left(6-j\omega \right)\left(15-j\omega \right)}\end{array}$

$F\left[f\left(-3t\right)\right]=\frac{30}{\left(6-j\omega \right)\left(15-j\omega \right)}$

Conclusion:

Thus, the Fourier transform of $f\left(-3t\right)$ is $\underset{\_}{\frac{30}{\left(6-j\omega \right)\left(15-j\omega \right)}}$.

(b)

To determine

Find the Fourier transform of $f\left(2t-1\right)$.

Answer to Problem 23P

The Fourier transform of $f\left(2t-1\right)$ is $\underset{\_}{\frac{20e^{-\frac{j\omega }{2}}}{\left(4+j\omega \right)\left(10+j\omega \right)}}$

Explanation of Solution

Formula used:

Consider the Time shift property of the Fourier transform.

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

Consider the scaling property of the Fourier transform.

$F\left[f\left(at\right)\right]=\frac{1}{\left|a\right|}F\left(\frac{\omega }{a}\right)$

Calculation:

Find $F\left[f\left(2t-1\right)\right]$, using scaling and time shift property

$\begin{array}{c}F\left[f\left(2t-1\right)\right]=\frac{1}{\left|2\right|}{e}^{-\frac{j\omega }{2}}\frac{10}{\left(2+\frac{j\omega }{2}\right)\left(5+\frac{j\omega }{2}\right)}\left\{\because F\left(\omega \right)=\frac{10}{\left(2+j\omega \right)\left(5+j\omega \right)}\right\}\\ =\frac{1}{2}{e}^{-\frac{j\omega }{2}}\frac{10}{\left(\frac{4+j\omega }{2}\right)\left(\frac{10+j\omega }{2}\right)}\\ =\frac{1}{2}{e}^{-\frac{j\omega }{2}}\frac{10\times 2\times 2}{\left(4+j\omega \right)\left(10+j\omega \right)}\\ =\frac{20e^{-\frac{j\omega }{2}}}{\left(4+j\omega \right)\left(10+j\omega \right)}\end{array}$

Conclusion:

Thus, the Fourier transform of $f\left(2t-1\right)$ is $\underset{\_}{\frac{20e^{-\frac{j\omega }{2}}}{\left(4+j\omega \right)\left(10+j\omega \right)}}$.

(c)

To determine

Find the Fourier transform of $f\left(t\right)\cos 2t$.

Answer to Problem 23P

The Fourier transform of $f\left(t\right)\cos 2t$ is $\underset{\_}{\frac{5}{\left[2+j\left(\omega +2\right)\right]\left[5+j\left(\omega +2\right)\right]}+\frac{5}{\left[2+j\left(\omega -2\right)\right]\left[5+j\left(\omega -2\right)\right]}}$.

Explanation of Solution

Formula used:

Consider the Modulation property of the Fourier transform.

$F\left[\cos \left({\omega }_{o}t\right)f\left(t\right)\right]=\frac{1}{2}\left[F\left(\omega +{\omega }_o\right)+F\left(\omega -{\omega }_o\right)\right]$

Calculation:

Find $F\left[f\left(t\right)\cos 2t\right]$, using modulation property.

$\begin{array}{c}F\left[f\left(t\right)\cos 2t\right]=\frac{1}{2}\left[\begin{array}{l}\frac{10}{\left[2+j\left(\omega +2\right)\right]\left[5+j\left(\omega +2\right)\right]}+\\ \frac{10}{\left[2+j\left(\omega -2\right)\right]\left[5+j\left(\omega -2\right)\right]}\end{array}\right]\left\{\because F\left(\omega \right)=\frac{10}{\left(2+j\omega \right)\left(5+j\omega \right)}\right\}\\ =\left\{\frac{1}{2}\frac{10}{\left[2+j\left(\omega +2\right)\right]\left[5+j\left(\omega +2\right)\right]}+\frac{1}{2}\frac{10}{\left[2+j\left(\omega -2\right)\right]\left[5+j\left(\omega -2\right)\right]}\right\}\\ =\frac{5}{\left[2+j\left(\omega +2\right)\right]\left[5+j\left(\omega +2\right)\right]}+\frac{5}{\left[2+j\left(\omega -2\right)\right]\left[5+j\left(\omega -2\right)\right]}\end{array}$

Conclusion:

Thus, the Fourier transform of $f\left(t\right)\cos 2t$ is $\underset{\_}{\frac{5}{\left[2+j\left(\omega +2\right)\right]\left[5+j\left(\omega +2\right)\right]}+\frac{5}{\left[2+j\left(\omega -2\right)\right]\left[5+j\left(\omega -2\right)\right]}}$.

(d)

To determine

Find the Fourier transform of $\frac{d}{dt}f\left(t\right)$.

Answer to Problem 23P

The Fourier transform of $\frac{d}{dt}f\left(t\right)$ is $\underset{\_}{\frac{j\omega \left(10\right)}{\left(2+j\omega \right)\left(5+j\omega \right)}}$.

Explanation of Solution

Formula used:

Consider the Time differentiation property of the Fourier transform.

$F\left[\frac{df}{dt}\right]=j\omega F\left(\omega \right)$

Calculation:

Find $F\left[\frac{d}{dt}f\left(t\right)\right]$, using time differentiation property.

$F\left[\frac{d}{dt}f\left(t\right)\right]=j\omega \left[\frac{10}{\left(2+j\omega \right)\left(5+j\omega \right)}\right]\left\{\because F\left(\omega \right)=\frac{10}{\left(2+j\omega \right)\left(5+j\omega \right)}\right\}$

Conclusion:

Thus, the Fourier transform of $\frac{d}{dt}f\left(t\right)$ is $\underset{\_}{\frac{j\omega \left(10\right)}{\left(2+j\omega \right)\left(5+j\omega \right)}}$.

(e)

To determine

Find the Fourier transform of ${\displaystyle \underset{-\infty }{\overset{t}{\int }}f\left(t\right)dt}$.

Answer to Problem 23P

The Fourier transform of ${\displaystyle \underset{-\infty }{\overset{t}{\int }}f\left(t\right)dt}$ is $\underset{\_}{\frac{10}{j\omega \left(2+j\omega \right)\left(5+j\omega \right)}+\pi \delta \left(\omega \right)}$.

Explanation of Solution

Formula used:

Consider the Time integration property of the Fourier transform.

$F\left[{\displaystyle \underset{-\infty }{\overset{t}{\int }}f\left(t\right)dt}\right]=\frac{F\left(\omega \right)}{j\omega }+\pi F\left(0\right)\delta \left(\omega \right)$

Calculation

Find $F\left[{\displaystyle \underset{-\infty }{\overset{t}{\int }}f\left(t\right)dt}\right]$, using time integration property.

$\begin{array}{c}F\left[{\displaystyle \underset{-\infty }{\overset{t}{\int }}f\left(t\right)dt}\right]=\left[\begin{array}{l}\frac{\frac{10}{\left(2+j\omega \right)\left(5+j\omega \right)}}{j\omega }+\\ \pi \frac{10}{\left(2+0\right)\left(5+0\right)}\delta \left(\omega \right)\end{array}\right]\left\{\because F\left(\omega \right)=\frac{10}{\left(2+j\omega \right)\left(5+j\omega \right)}\right\}\\ =\frac{10}{j\omega \left(2+j\omega \right)\left(5+j\omega \right)}+\pi \frac{10}{\left(2\right)\left(5\right)}\delta \left(\omega \right)\\ =\frac{10}{j\omega \left(2+j\omega \right)\left(5+j\omega \right)}+\pi \left(\frac{10}{10}\right)\delta \left(\omega \right)\\ =\frac{10}{j\omega \left(2+j\omega \right)\left(5+j\omega \right)}+\pi \delta \left(\omega \right)\end{array}$

Conclusion:

Thus, the Fourier transform of ${\displaystyle \underset{-\infty }{\overset{t}{\int }}f\left(t\right)dt}$ is $\underset{\_}{\frac{10}{j\omega \left(2+j\omega \right)\left(5+j\omega \right)}+\pi \delta \left(\omega \right)}$.