Chapter 18, Problem 27P

(a)

To determine

Find the Inverse Fourier transform of $F\left(\omega \right)=\frac{100}{j\omega \left(j\omega +10\right)}$.

Answer to Problem 27P

The Inverse Fourier transform of $F\left(\omega \right)=\frac{100}{j\omega \left(j\omega +10\right)}$ is $\underset{\_}{5\mathrm{sgn}\left(t\right)-10e^{-10t}u\left(t\right)}$

Explanation of Solution

Given data:

$F\left(\omega \right)=\frac{100}{j\omega \left(j\omega +10\right)}$ (1)

Calculation:

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

Substitute $s$ for $j\omega$ in equation (1) as follows.

$F\left(s\right)=\frac{100}{s\left(s+10\right)}$

Take partial fraction for the equation.

$F\left(s\right)=\frac{A}{s}+\frac{B}{\left(s+10\right)}$ (2)

Where

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

Substitute $\frac{100}{s\left(s+10\right)}$ for $F\left(s\right)$ as follows.

$\begin{array}{c}A=\left(s\right){\left[\frac{100}{s\left(s+10\right)}\right]}_{s=0}\\ ={\left[\frac{100}{\left(s+10\right)}\right]}_{s=0}\\ =\left[\frac{100}{10}\right]\\ =10\end{array}$

Similarly,

$B=\left(s+10\right){F\left(s\right)|}_{s=-10}$

Substitute $\frac{100}{s\left(s+10\right)}$ for $F\left(s\right)$ as follows.

$\begin{array}{c}B=\left(s+10\right){\left[\frac{100}{s\left(s+10\right)}\right]}_{s=-10}\\ ={\left[\frac{100}{s}\right]}_{s=-10}\\ =\left[\frac{100}{-10}\right]\\ =-10\end{array}$

Substitute $10$ for $A$ and $-10$ for $B$ in equation (2) as follows.

$F\left(s\right)=\frac{10}{s}+\frac{-10}{\left(s+10\right)}$

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

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

Apply inverse Fourier transform on both sides of equation.

$\begin{array}{l}{F}^{-1}\left[F\left(\omega \right)\right]=F^{-1}\left[\frac{10}{j\omega }-\frac{10}{\left(j\omega +10\right)}\right]\\ f\left(t\right)=F^{-1}\left[\frac{10}{j\omega }\right]-F^{-1}\left[\frac{10}{\left(j\omega +10\right)}\right]\\ f\left(t\right)=5F^{-1}\left[\frac{2}{j\omega }\right]-10F^{-1}\left[\frac{1}{j\omega +10}\right]\\ f\left(t\right)=5\mathrm{sgn}\left(t\right)-10e^{-10t}u\left(t\right)\left\{\begin{array}{l}\because F^{-1}\left[\frac{1}{a+j\omega }\right]=e^{-at}u\left(t\right)\\ \because F^{-1}\left[\frac{2}{j\omega }\right]=\mathrm{sgn}\left(t\right)\end{array}\right\}\end{array}$

Conclusion:

Thus, the Inverse Fourier transform of $F\left(\omega \right)=\frac{100}{j\omega \left(j\omega +10\right)}$ is $\underset{\_}{5\mathrm{sgn}\left(t\right)-10e^{-10t}u\left(t\right)}$.

(b)

To determine

Find the Inverse Fourier transform of $G\left(\omega \right)=\frac{10j\omega }{\left(-j\omega +2\right)\left(j\omega +3\right)}$.

Answer to Problem 27P

The Inverse Fourier transform of $G\left(\omega \right)=\frac{10j\omega }{\left(-j\omega +2\right)\left(j\omega +3\right)}$ is $\underset{\_}{4e^{2t}u\left(-t\right)-6e^{-3t}u\left(t\right)}$

Explanation of Solution

Given data:

$G\left(\omega \right)=\frac{10j\omega }{\left(-j\omega +2\right)\left(j\omega +3\right)}$ (3)

Calculation:

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

Substitute $s$ for $j\omega$ in equation (3) as follows.

$G\left(s\right)=\frac{10s}{\left(2-s\right)\left(3+s\right)}$

Take partial fraction for the equation.

$G\left(s\right)=\frac{A}{\left(2-s\right)}+\frac{B}{\left(3+s\right)}$ (4)

Where

$A=\left(2-s\right){G\left(s\right)|}_{s=2}$

Substitute $\frac{10s}{\left(2-s\right)\left(3+s\right)}$ for $G\left(s\right)$ as follows.

$\begin{array}{c}A=\left(2-s\right){\left[\frac{10s}{\left(2-s\right)\left(3+s\right)}\right]}_{s=2}\\ ={\left[\frac{10s}{\left(3+s\right)}\right]}_{s=2}\\ =\frac{20}{5}\\ =4\end{array}$

Similarly,

$B=\left(3+s\right){G\left(s\right)|}_{s=-3}$

Substitute $\frac{10s}{\left(2-s\right)\left(3+s\right)}$ for $G\left(s\right)$ as follows.

$\begin{array}{c}B=\left(3+s\right){\left[\frac{10s}{\left(2-s\right)\left(3+s\right)}\right]}_{s=-3}\\ ={\left[\frac{10s}{2-s}\right]}_{s=-3}\\ =\frac{-30}{5}\\ =-6\end{array}$

Substitute $4$ for $A$ and $-6$ for $B$ in equation (4) as follows.

$G\left(s\right)=\frac{4}{\left(2-s\right)}+\frac{-6}{\left(3+s\right)}$

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

$G\left(\omega \right)=\frac{4}{\left(2-j\omega \right)}-\frac{6}{\left(3+j\omega \right)}$

Apply inverse Fourier transform on both sides of equation.

$\begin{array}{l}{F}^{-1}\left[G\left(\omega \right)\right]=F^{-1}\left[\frac{4}{\left(2-j\omega \right)}-\frac{6}{\left(3+j\omega \right)}\right]\\ g\left(t\right)=F^{-1}\left[\frac{4}{\left(2-j\omega \right)}\right]-F^{-1}\left[\frac{6}{\left(3+j\omega \right)}\right]\\ g\left(t\right)=4F^{-1}\left[\frac{1}{2-j\omega }\right]-6F^{-1}\left[\frac{1}{3+j\omega }\right]\\ f\left(t\right)=4e^{2t}u\left(-t\right)-6e^{-3t}u\left(t\right)\left\{\begin{array}{l}\because F^{-1}\left[\frac{1}{a+j\omega }\right]=e^{-at}u\left(t\right)\\ \because F^{-1}\left[\frac{1}{a-j\omega }\right]=e^{at}u\left(-t\right)\end{array}\right\}\end{array}$

Conclusion:

Thus, the Inverse Fourier transform of $G\left(\omega \right)=\frac{10j\omega }{\left(-j\omega +2\right)\left(j\omega +3\right)}$ is $\underset{\_}{4e^{2t}u\left(-t\right)-6e^{-3t}u\left(t\right)}$.

(c)

To determine

Find the Inverse Fourier transform of $H\left(\omega \right)=\frac{60}{-{\omega }^2+j40\omega +1300}$.

Answer to Problem 27P

The Inverse Fourier transform of $H\left(\omega \right)=\frac{60}{-{\omega }^2+j40\omega +1300}$ is $\underset{\_}{2e^{-20t}\sin \left(30t\right)u\left(t\right)}$

Explanation of Solution

Given data:

$H\left(\omega \right)=\frac{60}{-{\omega }^2+j40\omega +1300}$ (5)

Calculation:

Modify equation (5) as follows.

$\begin{array}{c}H\left(\omega \right)=\frac{60}{{\left(j\omega \right)}^2+j40\omega +1300}\\ =\frac{60}{{\left(j\omega +20\right)}^2+900}\\ =2\left(\frac{60}{{\left(j\omega +20\right)}^2+{30}^2}\right)\end{array}$

Apply inverse transform on both sides of the equation.

$\begin{array}{c}{F}^{-1}\left[H\left(\omega \right)\right]=2F^{-1}\left[\frac{30}{{\left(j\omega +20\right)}^2+{30}^2}\right]\\ =2e^{-20t}\sin \left(30t\right)u\left(t\right)\left\{\because F^{-1}\left[\frac{{\omega }_o}{{\left(a+j\omega \right)}^2+{\omega }_o^2}\right]=e^{-at}\sin {\omega }_{o}tu\left(t\right)\right\}\end{array}$

Conclusion:

Thus, the Inverse Fourier transform of $H\left(\omega \right)=\frac{60}{-{\omega }^2+j40\omega +1300}$ is $\underset{\_}{2e^{-20t}\sin \left(30t\right)u\left(t\right)}$

(d)

To determine

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

Answer to Problem 27P

The Inverse Fourier transform of $Y\left(\omega \right)=\frac{\delta \left(\omega \right)}{\left(j\omega +1\right)\left(j\omega +2\right)}$ is $\underset{\_}{\frac{1}{4\pi }}$.

Explanation of Solution

Given data:

$Y\left(\omega \right)=\frac{\delta \left(\omega \right)}{\left(j\omega +1\right)\left(j\omega +2\right)}$.

Formula used:

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

$f\left(t\right)=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}F\left(\omega \right)e^{j\omega t}d\omega }$ (6)

Calculation:

Modify equation (6) as follows.

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

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

$\begin{array}{l}{F}^{-1}\left[Y\left(\omega \right)\right]=\frac{1}{2\pi }{\displaystyle \underset{-\infty }{\overset{\infty }{\int }}\frac{\delta \left(\omega \right)}{\left(j\omega +1\right)\left(j\omega +2\right)}{e}^{j\omega t}d\omega }\\ y\left(t\right)=\frac{1}{2\pi }\left(\frac{1}{2}\right)\\ y\left(t\right)=\frac{1}{4\pi }\end{array}$

Conclusion:

Thus, the Inverse Fourier transform of $Y\left(\omega \right)=\frac{\delta \left(\omega \right)}{\left(j\omega +1\right)\left(j\omega +2\right)}$ is $\underset{\_}{\frac{1}{4\pi }}$.