Examples of the Fourier Transform

Example 1: Find the Fourier transforms of the following functions:
(a) signum function $ \operatorname{sgn}(t) $, shown in Fig. 1,
(b) the double-sided exponential $ e^{-a|t|} $, and
(c) the sinc function $ (\sin t) / t $.
Fig. 1: The signum function of Example 1.
(a) We can obtain the Fourier transform of the signum function in three ways. First, we can write the signum function in terms of the unit step function as
But from Eq. (A),
$$\mathcal{F}[u(t)]=mathcal{F} \left[ \int_{-\infty}^t \delta(t) \pi \delta(t) dt \right]= \frac{1}{j \omega}+\pi \delta(\omega) \tag{A}$$
$$U(\omega)=\mathcal{F}[u(t)]=\pi \delta(\omega)+\frac{1}{j \omega}$$
Applying this and the reversal property, we obtain
$$\begin{aligned}\mathcal{F}[\operatorname{sgn}(t)] &=U(\omega)-U(-\omega) \\&=\left(\pi \delta(\omega)+\frac{1}{j \omega}\right)-\left(\pi \delta(-\omega)+\frac{1}{-j \omega}\right)=\frac{2}{j \omega}\end{aligned}$$
Second, another way of writing the signum function in terms of the unit step function is
$$f(t)=\operatorname{sgn}(t)=-1+2 u(t)$$
Taking the Fourier transform of each term gives
$$F(\omega)=-2 \pi \delta(\omega)+2\left(\pi \delta(\omega)+\frac{1}{j \omega}\right)=\frac{2}{j \omega}$$
Third, we can take the derivative of the signum function in Fig. $ 1 $ and obtain $$f^{\prime}(t)=2 \delta(t)$$ Taking the transform of this,
$$j \omega F(\omega)=2 \quad \Longrightarrow \quad F(\omega)=\frac{2}{j \omega}$$
as obtained previously.
(b) The double-sided exponential can be expressed as
$$f(t)=e^{-a|t|}=e^{-a t} u(t)+e^{a t} u(-t)=y(t)+y(-t)$$
where $ y(t)=e^{-a t} u(t) $ so that $ Y(\omega)=1 /(a+j \omega) $. Applying the reversal property,
$$\mathcal{F}\left[e^{-a|t|}\right]=Y(\omega)+Y(-\omega)=\left(\frac{1}{a+j \omega}+\frac{1}{a-j \omega}\right)=\frac{2 a}{a^{2}+\omega^{2}}$$
$$\mathcal{F}\left[u\left(t+\frac{\tau}{2}\right)-u\left(t-\frac{\tau}{2}\right)\right]=\tau \frac{\sin (\omega \tau / 2)}{\omega \tau / 2}=\tau \operatorname{sinc} \frac{\omega \tau}{2}$$
Setting $ \tau / 2=1 $ gives $$\mathcal{F}[u(t+1)-u(t-1)]=2 \frac{\sin \omega}{\omega}$$ Applying the duality property,
$$\mathcal{F}\left[2 \frac{\sin t}{t}\right]=2 \pi[U(\omega+1)-U(\omega-1)]$$
$$\mathcal{F}\left[\frac{\sin t}{t}\right]=\pi[U(\omega+1)-U(\omega-1)]$$
Example 2: Find the Fourier transform of the function in Fig. 2.
Fig. 2: For Example 2.
Solution: The Fourier transform can be found directly using Eq. (B),
$$ F(\omega)= \mathcal{F} [f(t)] = \int_{-\infty}^\infty f (t) e^{-j\omega t} dt \tag{B}$$
but it is much easier to find it using the derivative property. We can express the function as
$$f(t)=\left\{\begin{array}{lr}1+t, & -1 < t < 0 \\1-t, & 0 < t < 1\end{array}\right.$$
Its first derivative is shown in Fig. 3(a) and is given by
$$f^{\prime}(t)=\left\{\begin{array}{rr}1, & -1 < t < 0 \\-1, & 0 < t < 1\end{array}\right.$$
Fig. 3: First and second derivatives of f (t) in Fig. 2; for Example 2.
Its second derivative is in Fig. 3(b) and is given by
$$f^{\prime \prime}(t)=\delta(t+1)-2 \delta(t)+\delta(t-1)$$
Taking the Fourier transform of both sides,
$$(j \omega)^{2} F(\omega)=e^{j \omega}-2+e^{-j \omega}=-2+2 \cos \omega$$
$$F(\omega)=\frac{2(1-\cos \omega)}{\omega^{2}}$$
Example 3: Obtain the inverse Fourier transform of:
(a) $ F(\omega)=\frac{10 j \omega+4}{(j \omega)^{2}+6 j \omega+8} $
(b) $ G(\omega)=\frac{\omega^{2}+21}{\omega^{2}+9} $
(a) To avoid complex algebra, we can replace $ j \omega $ with $ s $ for the moment. Using partial fraction expansion,
$$F(s)=\frac{10 s+4}{s^{2}+6 s+8}=\frac{10 s+4}{(s+4)(s+2)}=\frac{A}{s+4}+\frac{B}{s+2}$$
$$\begin{array}{l}A=\left.(s+4) F(s)\right|_{s=-4}=\left.\frac{10 s+4}{(s+2)}\right|_{s=-4}=\frac{-36}{-2}=18 \\B=\left.(s+2) F(s)\right|_{s=-2}=\left.\frac{10 s+4}{(s+4)}\right|_{s=-2}=\frac{-16}{2}=-8\end{array}$$
Substituting $ A=18 $ and $ B=-8 $ in $ F(s) $ and $ s $ with $ j \omega $ gives
$$F(j \omega)=\frac{18}{j \omega+4}+\frac{-8}{j \omega+2}$$
With the aid of Table 17.2, we obtain the inverse transform as
$$f(t)=\left(18 e^{-4 t}-8 e^{-2 t}\right) u(t)$$
(b) We simplify $ G(\omega) $ as