Chapter 18, Problem 28P

(a)

To determine

Find the Inverse Fourier transform of $\frac{\pi \delta \left(\omega \right)}{\left(5+j\omega \right)\left(2+j\omega \right)}$.

Answer to Problem 28P

The Inverse Fourier transform of $\frac{\pi \delta \left(\omega \right)}{\left(5+j\omega \right)\left(2+j\omega \right)}$ is $\underset{\_}{0.05}$.

Explanation of Solution

Given data:

$\frac{\pi \delta \left(\omega \right)}{\left(5+j\omega \right)\left(2+j\omega \right)}$

Formula used:

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

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

Calculation:

Substitute $\frac{\pi \delta \left(\omega \right)}{\left(5+j\omega \right)\left(2+j\omega \right)}$ for $F\left(\omega \right)$ in equation (1) as follows.

$\begin{array}{c}f\left(t\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[\frac{\pi \delta \left(\omega \right)}{\left(5+j\omega \right)\left(2+j\omega \right)}\right]e^{j\omega t}d\omega }\\ =\frac{\pi }{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{\delta \left(\omega \right)}{\left(5+j\omega \right)\left(2+j\omega \right)}{e}^{j\omega t}d\omega }\\ =\frac{1}{2}\left(\frac{1}{\left(2\right)\left(5\right)}\right)\\ =\frac{1}{20}\end{array}$

$f\left(t\right)=0.05$

Conclusion:

Thus, the Inverse Fourier transform of $\frac{\pi \delta \left(\omega \right)}{\left(5+j\omega \right)\left(2+j\omega \right)}$ is $\underset{\_}{0.05}$

(b)

To determine

Find the Inverse Fourier transform of $\frac{10\delta \left(\omega +2\right)}{j\omega \left(j\omega +1\right)}$.

Answer to Problem 28P

The Inverse Fourier transform of $\frac{10\delta \left(\omega +2\right)}{j\omega \left(j\omega +1\right)}$ is $\underset{\_}{\frac{\left(-2+j\right)e^{-j2t}}{2\pi }}$.

Explanation of Solution

Given data:

$\frac{10\delta \left(\omega +2\right)}{j\omega \left(j\omega +1\right)}$

Calculation:

Substitute $\frac{10\delta \left(\omega +2\right)}{j\omega \left(j\omega +1\right)}$ for $F\left(\omega \right)$ in equation (1) as follows.

$\begin{array}{c}f\left(t\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[\frac{10\delta \left(\omega +2\right)}{j\omega \left(j\omega +1\right)}\right]e^{j\omega t}d\omega }\\ =\frac{10}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{\delta \left(\omega +2\right)}{j\omega \left(j\omega +1\right)}{e}^{j\omega t}d\omega }\\ =\frac{10}{2\pi }\left(\frac{e^{-j2t}}{\left(-j2\right)\left(-j2+1\right)}\right)\\ =\frac{j5}{2\pi }\left(\frac{e^{-j2t}}{1-j2}\right)\end{array}$

$f\left(t\right)=\frac{\left(-2+j\right)e^{-j2t}}{2\pi }$

Conclusion:

Thus, the Inverse Fourier transform of $\frac{10\delta \left(\omega +2\right)}{j\omega \left(j\omega +1\right)}$ is $\underset{\_}{\frac{\left(-2+j\right)e^{-j2t}}{2\pi }}$.

(c)

To determine

Find the Inverse Fourier transform of $\frac{20\delta \left(\omega -1\right)}{\left(2+j\omega \right)\left(3+j\omega \right)}$.

Answer to Problem 28P

The Inverse Fourier transform of $\frac{20\delta \left(\omega -1\right)}{\left(2+j\omega \right)\left(3+j\omega \right)}$ is $\underset{\_}{\frac{\left(1-j\right)e^{jt}}{\pi }}$.

Explanation of Solution

Given data:

$\frac{20\delta \left(\omega -1\right)}{\left(2+j\omega \right)\left(3+j\omega \right)}$.

Calculation:

Substitute $\frac{20\delta \left(\omega -1\right)}{\left(2+j\omega \right)\left(3+j\omega \right)}$ for $F\left(\omega \right)$ in equation (1) as follows.

$\begin{array}{c}f\left(t\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[\frac{20\delta \left(\omega -1\right)}{\left(2+j\omega \right)\left(3+j\omega \right)}\right]e^{j\omega t}d\omega }\\ =\frac{20}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{\delta \left(\omega -1\right)}{\left(2+j\omega \right)\left(3+j\omega \right)}{e}^{j\omega t}d\omega }\\ =\frac{20}{2\pi }\left(\frac{e^{jt}}{\left(2+j\right)\left(3+j\right)}\right)\\ =\frac{20e^{jt}}{2\pi \left(5+5j\right)}\end{array}$

Simplify the equation as follows.

$\begin{array}{c}f\left(t\right)=\frac{20e^{jt}}{10\pi \left(1+j\right)}\\ =\frac{2e^{jt}}{\pi \left(1+j\right)}\\ =\frac{\left(1-j\right)e^{jt}}{\pi }\end{array}$

Conclusion:

Thus, the Inverse Fourier transform of $\frac{20\delta \left(\omega -1\right)}{\left(2+j\omega \right)\left(3+j\omega \right)}$ is $\underset{\_}{\frac{\left(1-j\right)e^{jt}}{\pi }}$.

(d)

To determine

Find the Inverse Fourier transform of $\frac{5\pi \delta \left(\omega \right)}{5+j\omega }+\frac{5}{j\omega \left(5+j\omega \right)}$.

Answer to Problem 28P

The Inverse Fourier transform of $\frac{5\pi \delta \left(\omega \right)}{5+j\omega }+\frac{5}{j\omega \left(5+j\omega \right)}$ is $\underset{\_}{u\left(t\right)-e^{-5t}u\left(t\right)}$.

Explanation of Solution

Given data:

$\frac{5\pi \delta \left(\omega \right)}{5+j\omega }+\frac{5}{j\omega \left(5+j\omega \right)}$

Here,

$F_1\left(\omega \right)=\frac{5\pi \delta \left(\omega \right)}{5+j\omega }$.

$F_2\left(\omega \right)=\frac{5}{j\omega \left(5+j\omega \right)}$

Calculation:

Modify equation (1) as follows.

$f_1\left(t\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}{F}_1\left(\omega \right)e^{j\omega t}d\omega }$

Substitute $\frac{5\pi \delta \left(\omega \right)}{5+j\omega }$ for $F_1\left(\omega \right)$ as follows.

$\begin{array}{c}{f}_1\left(t\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[\frac{5\pi \delta \left(\omega \right)}{5+j\omega }\right]e^{j\omega t}d\omega }\\ =\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\left[\frac{5\pi \delta \left(\omega \right)}{5+j\omega }\right]e^{j\omega t}d\omega }\\ =\frac{5\pi }{2\pi }\left(\frac{1}{5}\right)\\ =0.5\end{array}$

Let $F_2\left(\omega \right)=\frac{5}{j\omega \left(5+j\omega \right)}$ (2)

Consider $s=j\omega$ to reduce complex algebra.

Substitute $s$ for $j\omega$ in equation (2) as follows.

$F_2\left(s\right)=\frac{5}{s\left(5+s\right)}$

Take partial fraction for the equation.

$F_2\left(s\right)=\frac{A}{s}+\frac{B}{s+5}$ (3)

Where

$A=\left(s\right){F_2\left(s\right)|}_{s=0}$

Substitute $\frac{5}{s\left(5+s\right)}$ for $F_2\left(s\right)$ as follows.

$\begin{array}{c}A=\left(s\right){\left[\frac{5}{s\left(5+s\right)}\right]}_{s=0}\\ ={\left[\frac{5}{\left(5+s\right)}\right]}_{s=0}\\ =\left[\frac{5}{5}\right]\\ =1\end{array}$

Similarly,

$B=\left(5+s\right){F\left(s\right)|}_{s=-5}$

Substitute $\frac{5}{s\left(5+s\right)}$ for $F_2\left(s\right)$ as follows.

$\begin{array}{c}B=\left(5+s\right){\left[\frac{5}{s\left(5+s\right)}\right]}_{s=-5}\\ ={\left[\frac{5}{s}\right]}_{s=-5}\\ =\left[\frac{5}{-5}\right]\\ =-1\end{array}$

Substitute $1$ for $A$ and $-1$ for $B$ in equation (3) as follows.

$F_2\left(s\right)=\frac{1}{s}-\frac{1}{s+5}$

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

$F_2\left(\omega \right)=\frac{1}{j\omega }-\frac{1}{j\omega +5}$

Apply inverse Fourier transform on both sides of equation.

$\begin{array}{l}{F}^{-1}\left[F_2\left(\omega \right)\right]=F^{-1}\left[\frac{1}{j\omega }-\frac{1}{j\omega +5}\right]\\ f_2\left(t\right)=F^{-1}\left[\frac{1}{j\omega }\right]-F^{-1}\left[\frac{1}{j\omega +5}\right]\\ f_2\left(t\right)=\frac{1}{2}{F}^{-1}\left[\frac{2}{j\omega }\right]-F^{-1}\left[\frac{1}{j\omega +5}\right]\\ f_2\left(t\right)=\frac{1}{2}\mathrm{sgn}\left(t\right)-e^{-5t}u\left(t\right)\left\{\begin{array}{l}\because F^{-1}\left[\frac{1}{a+j\omega }\right]=e^{-at}u\left(t\right)\\ \because F^{-1}\left[\frac{2}{j\omega }\right]=\mathrm{sgn}\left(t\right)\end{array}\right\}\end{array}$

$f_2\left(t\right)=-\frac{1}{2}+u\left(t\right)-e^{-5t}u\left(t\right)$

As $F\left(\omega \right)=F_1\left(\omega \right)+F_2\left(\omega \right)$

Apply inverse Fourier transform on both sides of equation.

$\begin{array}{l}{F}^{-1}\left[F\left(\omega \right)\right]=F^{-1}\left[F_1\left(\omega \right)+F_2\left(\omega \right)\right]\\ f\left(t\right)=f_1\left(t\right)+f_2\left(t\right)\end{array}$

Substitute $\frac{1}{2}-\frac{1}{2}+u\left(t\right)$ for $f_1\left(t\right)$ and $e^{-5t}u\left(t\right)$ for $f_2\left(t\right)$ as follows.

$\begin{array}{c}f\left(t\right)=\frac{1}{2}-\frac{1}{2}+u\left(t\right)-e^{-5t}u\left(t\right)\\ =u\left(t\right)-e^{-5t}u\left(t\right)\end{array}$

Conclusion:

Thus, the Inverse Fourier transform of $\frac{5\pi \delta \left(\omega \right)}{5+j\omega }+\frac{5}{j\omega \left(5+j\omega \right)}$ is $\underset{\_}{u\left(t\right)-e^{-5t}u\left(t\right)}$.