Chapter 18, Problem 32P

(a)

To determine

Find the inverse Fourier transform of $F_1\left(\omega \right)$.

Answer to Problem 32P

The inverse Fourier transform of $F_1\left(\omega \right)$ is $\underset{\_}{e^{\left(t+1\right)}u\left(-t-1\right)}$

Explanation of Solution

Given data:

$F_1\left(\omega \right)=\frac{e^{j\omega }}{-j\omega +1}$        (1)

Formula used:

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

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

Calculation:

Modify equation (1) as follows.

$F_1\left(\omega \right)=\frac{e^{-j\omega }}{j\omega +1}$

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

From Reversal property,

$f\left(-t\right)=F\left(-\omega \right)$

Therefore,

$F_1\left(\omega \right)=\frac{e^{j\omega }}{-j\omega +1}$

Apply inverse Fourier transform to equation (1) as follows.

$\begin{array}{c}{F}^{-1}\left[F_1\left(\omega \right)\right]=F^{-1}\left[\frac{e^{j\omega }}{-j\omega +1}\right]\\ =e^{-\left(-t-1\right)}u\left(-t-1\right)\\ =e^{\left(t+1\right)}u\left(-t-1\right)\end{array}$

Conclusion:

Thus, the inverse Fourier transform of $F_1\left(\omega \right)$ is $\underset{\_}{e^{\left(t+1\right)}u\left(-t-1\right)}$.

(b)

To determine

Find the inverse Fourier transform of $F_2\left(\omega \right)$.

Answer to Problem 32P

The inverse Fourier transform of $F_2\left(\omega \right)$ is $\underset{\_}{\frac{2}{\pi \left(t^2+1\right)}}$.

Explanation of Solution

Given data:

$F_2\left(\omega \right)=2e^{\left|\omega \right|}$        (2)

Calculation:

Since $F^{-1}\left[\frac{2}{t^2+1}\right]=2\pi e^{-\left|\omega \right|}$.

From Reversal property,

$f\left(-t\right)=F\left(-\omega \right)$

Therefore,

$F_2\left(\omega \right)=2e^{-\left|\omega \right|}$

Apply inverse Fourier transform to equation (2) as follows.

$\begin{array}{c}{F}^{-1}\left[F_2\left(\omega \right)\right]=F^{-1}\left[2e^{-\left|\omega \right|}\right]\\ =\frac{2}{\pi \left(t^2+1\right)}\end{array}$

Conclusion:

Thus, the inverse Fourier transform of $F_2\left(\omega \right)$ is $\underset{\_}{\frac{2}{\pi \left(t^2+1\right)}}$.

(c)

To determine

Find the inverse Fourier transform of $F_3\left(\omega \right)$.

Answer to Problem 32P

The inverse Fourier transform of $F_3\left(\omega \right)$ is $\underset{\_}{\frac{1}{4}\left(t+1\right)e^{-t}u\left(t\right)+\frac{1}{4}\left(t-1\right)e^{t}u\left(t\right)}$.

Explanation of Solution

Given data:

$F_3\left(\omega \right)=\frac{1}{{\left(1+{\omega }^2\right)}^2}$        (3)

Calculation:

Modify equation (3) as follows.

$F_3\left(\omega \right)=\frac{1}{{\left(1+j\omega \right)}^2{\left(1-j\omega \right)}^2}$

Substitute $s$ for $j\omega$ to reduce complex algebra.

$F_3\left(s\right)=\frac{1}{{\left(1+s\right)}^2{\left(1-s\right)}^2}$

Take partial fraction for the equation.

$\frac{1}{{\left(1+s\right)}^2{\left(1-s\right)}^2}=\frac{A}{\left(1+s\right)}+\frac{B}{{\left(1+s\right)}^2}+\frac{C}{\left(1-s\right)}+\frac{D}{{\left(1-s\right)}^2}$        (4)

Find the partial fraction coefficients.

$\begin{array}{c}A={\frac{d}{ds}\left[{\left(1+s\right)}^2\frac{1}{{\left(1+s\right)}^2{\left(1-s\right)}^2}\right]|}_{s=-1}\\ ={\frac{d}{ds}\left[\frac{1}{{\left(1-s\right)}^2}\right]|}_{s=-1}\\ ={\frac{2}{{\left(1-s\right)}^3}|}_{s=-1}\\ =\frac{2}{8}\end{array}$

$A=\frac{1}{4}$

$\begin{array}{c}B={{\left(1+s\right)}^2\frac{1}{{\left(1+s\right)}^2{\left(1-s\right)}^2}|}_{s=-1}\\ ={\frac{1}{{\left(1-s\right)}^2}|}_{s=-1}\\ =\frac{1}{4}\end{array}$

$\begin{array}{c}C={\frac{d}{ds}\left[{\left(1-s\right)}^2\frac{1}{{\left(1+s\right)}^2{\left(1-s\right)}^2}\right]|}_{s=1}\\ ={\frac{d}{ds}\left[\frac{1}{{\left(1+s\right)}^2}\right]|}_{s=1}\\ ={-\frac{2}{{\left(1+s\right)}^3}|}_{s=1}\\ =-\frac{1}{4}\end{array}$

$\begin{array}{c}D={{\left(1-s\right)}^2\frac{1}{{\left(1+s\right)}^2{\left(1-s\right)}^2}|}_{s=1}\\ ={\frac{1}{{\left(1+s\right)}^2}|}_{s=1}\\ =\frac{1}{4}\end{array}$

Substitute $\frac{1}{4}$ for A, $\frac{1}{4}$ for B, $-\frac{1}{4}$ for C and  $\frac{1}{4}$ for D in equation (4).

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

Substitute $j\omega$ for $s$ as follows.

$\frac{1}{{\left(1+j\omega \right)}^2{\left(1-j\omega \right)}^2}=\frac{1}{4}\left[\frac{1}{\left(1+j\omega \right)}+\frac{1}{{\left(1+j\omega \right)}^2}-\frac{1}{\left(1-j\omega \right)}+\frac{1}{{\left(1-j\omega \right)}^2}\right]$

Apply inverse transform on both sides of equation.

$\begin{array}{l}{F}^{-1}\left[\frac{1}{{\left(1+j\omega \right)}^2{\left(1-j\omega \right)}^2}\right]=\frac{1}{4}{F}^{-1}\left[\frac{1}{\left(1+j\omega \right)}+\frac{1}{{\left(1+j\omega \right)}^2}-\frac{1}{\left(1-j\omega \right)}+\frac{1}{{\left(1-j\omega \right)}^2}\right]\\ f_3\left(t\right)=\frac{1}{4}\left(e^{-t}+t{e}^{-t}-e^t+t{e}^t\right)u\left(t\right)\\ f_3\left(t\right)=\frac{1}{4}\left(t+1\right)e^{-t}u\left(t\right)+\frac{1}{4}\left(t-1\right)e^{t}u\left(t\right)\end{array}$

Conclusion:

Thus, the inverse Fourier transform of $F_3\left(\omega \right)$ is $\underset{\_}{\frac{1}{4}\left(t+1\right)e^{-t}u\left(t\right)+\frac{1}{4}\left(t-1\right)e^{t}u\left(t\right)}$.

(d)

To determine

Find the inverse Fourier transform of $F_4\left(\omega \right)$.

Answer to Problem 32P

The inverse Fourier transform of $F_4\left(\omega \right)$ is $\underset{\_}{\frac{1}{2\pi }}$.

Explanation of Solution

Given data:

$F_4\left(\omega \right)=\frac{\delta \left(\omega \right)}{1+j2\omega }$        (5)

Calculation:

Apply inverse Fourier transform to equation (5) as follows.

$\begin{array}{c}{F}^{-1}\left[F_4\left(\omega \right)\right]=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{\delta \left(\omega \right)}{1+j2\omega }{e}^{j\omega t}}dt\\ =\frac{1}{2\pi }\left(1\right)\\ =\frac{1}{2\pi }\end{array}$

Conclusion:

Thus, the inverse Fourier transform of $F_4\left(\omega \right)$ is $\underset{\_}{\frac{1}{2\pi }}$.