# 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.
Solution:
(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
$$\operatorname{sgn}(t)=f(t)=u(t)-u(-t)$$
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,
$$\mathcal{F}\left[e^{-a|t|}\right]=Y(\omega)+Y(-\omega)=\left(\frac{1}{a+j \omega}+\frac{1}{a-j \omega}\right)=\frac{2 a}{a^{2}+\omega^{2}}$$
(c)
$$\mathcal{F}\left[u\left(t+\frac{\tau}{2}\right)-u\left(t-\frac{\tau}{2}\right)\right]=\tau \frac{\sin (\omega \tau / 2)}{\omega \tau / 2}=\tau \operatorname{sinc} \frac{\omega \tau}{2}$$
Setting $\tau / 2=1$ gives $$\mathcal{F}[u(t+1)-u(t-1)]=2 \frac{\sin \omega}{\omega}$$ Applying the duality property,
$$\mathcal{F}\left[2 \frac{\sin t}{t}\right]=2 \pi[U(\omega+1)-U(\omega-1)]$$
or
$$\mathcal{F}\left[\frac{\sin t}{t}\right]=\pi[U(\omega+1)-U(\omega-1)]$$
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,
$$(j \omega)^{2} F(\omega)=e^{j \omega}-2+e^{-j \omega}=-2+2 \cos \omega$$
or
$$F(\omega)=\frac{2(1-\cos \omega)}{\omega^{2}}$$
Example 3: Obtain the inverse Fourier transform of:
(a) $F(\omega)=\frac{10 j \omega+4}{(j \omega)^{2}+6 j \omega+8}$
(b) $G(\omega)=\frac{\omega^{2}+21}{\omega^{2}+9}$
Solution:
(a) To avoid complex algebra, we can replace $j \omega$ with $s$ for the moment. Using partial fraction expansion,
$$F(s)=\frac{10 s+4}{s^{2}+6 s+8}=\frac{10 s+4}{(s+4)(s+2)}=\frac{A}{s+4}+\frac{B}{s+2}$$
where
$$\begin{array}{l}A=\left.(s+4) F(s)\right|_{s=-4}=\left.\frac{10 s+4}{(s+2)}\right|_{s=-4}=\frac{-36}{-2}=18 \\B=\left.(s+2) F(s)\right|_{s=-2}=\left.\frac{10 s+4}{(s+4)}\right|_{s=-2}=\frac{-16}{2}=-8\end{array}$$
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
$$G(\omega)=\frac{\omega^{2}+21}{\omega^{2}+9}=1+\frac{12}{\omega^{2}+9}$$