Chapter 18, Problem 35P

(a)

To determine

Find the Fourier transform of $x\left(t\right)$.

Answer to Problem 35P

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

Explanation of Solution

Given data:

$F\left(\omega \right)=\frac{1}{2+j\omega }$

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

Formula used:

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

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

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

$f\left(t\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}F\left(\omega \right)e^{j\omega t}d\omega }$

Consider scaling property.

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

Consider Time shift property.

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

Calculation:

Modify equation (1) as follows.

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

Substitute $f\left(3t-1\right)$ for $x\left(t\right)$ as follows.

$X\left(\omega \right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[f\left(3t-1\right)\right]e^{-j\omega t}dt}$

From scaling and time shift property, equation (1) can be write as follows.

$\begin{array}{c}X\left(\omega \right)=\frac{1}{\left|3\right|}{e}^{-\frac{j\omega }{3}}F\left(\frac{\omega }{3}\right)\\ =\frac{1}{3}{e}^{-\frac{j\omega }{3}}F\left(\frac{\omega }{3}\right)\end{array}$

Substitute $\frac{1}{2+j\omega }$ for $F\left(\omega \right)$ as follows.

$\begin{array}{c}x\left(\omega \right)=\frac{1}{3}{e}^{-\frac{j\omega }{3}}\left(\frac{1}{2+\frac{j\omega }{3}}\right)\left\{\begin{array}{l}\because x\left(t\right)=f\left(3t-1\right)\\ \because F\left(\frac{\omega }{3}\right)=\frac{1}{\left(2+\frac{j\omega }{3}\right)}\end{array}\right\}\\ =\frac{1}{3}{e}^{-\frac{j\omega }{3}}\left(\frac{3}{6+j\omega }\right)\\ =\frac{e^{-\frac{j\omega }{3}}}{6+j\omega }\end{array}$

Conclusion:

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

(b)

To determine

Find the Fourier transform of $y\left(t\right)$.

Answer to Problem 35P

The Fourier transform of $y\left(t\right)$ is $\underset{\_}{\frac{1}{2}\left[\frac{1}{2+j\left(\omega +5\right)}+\frac{1}{2+j\left(\omega -5\right)}\right]}$.

Explanation of Solution

Given data:

$F\left(\omega \right)=\frac{1}{2+j\omega }$

$y\left(t\right)=f\left(t\right)\cos 5t$

Formula used:

Consider Modulation property.

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

Calculation:

Modify equation (1) as follows.

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

Substitute $f\left(t\right)\cos 5t$ for $y\left(t\right)$ as follows.

$Y\left(\omega \right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[f\left(t\right)\cos 5t\right]e^{-j\omega t}dt}$

From modulation property, equation can be write as follows.

$\begin{array}{c}Y\left(\omega \right)=\frac{1}{2}\left[F\left(\omega +5\right)+F\left(\omega -5\right)\right]\\ =\frac{1}{2}\left[\frac{1}{2+j\left(\omega +5\right)}+\frac{1}{2+j\left(\omega -5\right)}\right]\end{array}$

Conclusion:

Thus, the Fourier transform of $y\left(t\right)$ is $\underset{\_}{\frac{1}{2}\left[\frac{1}{2+j\left(\omega +5\right)}+\frac{1}{2+j\left(\omega -5\right)}\right]}$.

(c)

To determine

Find the Fourier transform of $z\left(t\right)$.

Answer to Problem 35P

The Fourier transform of $z\left(t\right)$ is $\underset{\_}{\frac{j\omega }{2+j\omega }}$.

Explanation of Solution

Given data:

$F\left(\omega \right)=\frac{1}{2+j\omega }$

$z\left(t\right)=\frac{d}{dt}f\left(t\right)$

Formula used:

Consider Time differentiation property.

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

Calculation:

Modify equation (1) as follows:

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

Substitute $\frac{d}{dt}f\left(t\right)$ for $z\left(t\right)$ as follows:

$Z\left(\omega \right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[\frac{d}{dt}f\left(t\right)\right]e^{-j\omega t}dt}$

From modulation property, equation can be write as follows:

$\begin{array}{c}Z\left(\omega \right)=j\omega F\left(\omega \right)\\ =\frac{j\omega }{2+j\omega }\end{array}$

Conclusion:

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

(d)

To determine

Find the Fourier transform of $h\left(t\right)$.

Answer to Problem 35P

The Fourier transform of $h\left(t\right)$ is $\underset{\_}{\frac{1}{{\left(2+j\omega \right)}^2}}$.

Explanation of Solution

Given data:

$F\left(\omega \right)=\frac{1}{2+j\omega }$

$h\left(t\right)=f\left(t\right)\ast f\left(t\right)$

Formula used:

Consider Convolution in t property.

$f_1\left(t\right)\ast f_2\left(t\right)=F_1\left(\omega \right)F_2\left(\omega \right)$

Calculation:

Modify equation (1) as follows.

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

Substitute $f\left(t\right)\ast f\left(t\right)$ for $h\left(t\right)$ as follows.

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

From Convolution in t property, equation can be write as follows.

$\begin{array}{c}H\left(\omega \right)=F\left(\omega \right)F\left(\omega \right)\\ =\left(\frac{1}{2+j\omega }\right)\left(\frac{1}{2+j\omega }\right)\\ =\frac{1}{{\left(2+j\omega \right)}^2}\end{array}$

Conclusion:

Thus, the Fourier transform of $h\left(t\right)$ is $\underset{\_}{\frac{1}{{\left(2+j\omega \right)}^2}}$.

(e)

To determine

Find the Fourier transform of $i\left(t\right)$.

Answer to Problem 35P

The Fourier transform of $i\left(t\right)$ is $\underset{\_}{\frac{1}{{\left(2+j\omega \right)}^2}}$.

Explanation of Solution

Given data:

$F\left(\omega \right)=\frac{1}{2+j\omega }$

$i\left(t\right)=tf\left(t\right)$

Formula used:

Consider Frequency differentiation property.

$t^{2}f\left(t\right)={\left(j\right)}^n\frac{d^n}{d{\omega }^n}F\left(\omega \right)$

Calculation:

Modify equation (1) as follows.

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

Substitute $tf\left(t\right)$ for $i\left(t\right)$ as follows.

$I\left(\omega \right)={\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[tf\left(t\right)\right]e^{-j\omega t}dt}$

From Frequency differentiation property, equation can be write as follows.

$\begin{array}{c}I\left(\omega \right)={\left(j\right)}^0\frac{d^0}{d{\omega }^0}F\left(\omega \right)\\ =j\frac{d}{d\omega }F\left(\omega \right)\\ =j\frac{d}{d\omega }\frac{1}{\left(2+j\omega \right)}\\ =j\frac{\left(0-j\right)}{{\left(2+j\omega \right)}^2}\end{array}$

Simplify the equation as follows.

$\begin{array}{c}I\left(\omega \right)=\frac{-j^2}{{\left(2+j\omega \right)}^2}\\ =\frac{1}{{\left(2+j\omega \right)}^2}\end{array}$

Conclusion:

Thus, the Fourier transform of $i\left(t\right)$ is $\underset{\_}{\frac{1}{{\left(2+j\omega \right)}^2}}$.