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 $.
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
But from Eq. (A),
Applying this and the reversal property, we obtain
Second, another way of writing the signum function in terms of the unit step function is
Taking the Fourier transform of each term gives
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,
as obtained previously.
(b) The double-sided exponential can be expressed as
where $ y(t)=e^{-a t} u(t) $ so that $ Y(\omega)=1 /(a+j \omega) $. Applying the reversal property,
(c)
Setting $ \tau / 2=1 $ gives
$$\mathcal{F}[u(t+1)-u(t-1)]=2 \frac{\sin \omega}{\omega}$$
Applying the duality property,
or
(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
$$\operatorname{sgn}(t)=f(t)=u(t)-u(-t)$$
$$\mathcal{F}[u(t)]=mathcal{F} \left[ \int_{-\infty}^t \delta(t) \pi \delta(t) dt \right]= \frac{1}{j \omega}+\pi \delta(\omega) \tag{A}$$
$$U(\omega)=\mathcal{F}[u(t)]=\pi \delta(\omega)+\frac{1}{j \omega}$$
$$\begin{aligned}\mathcal{F}[\operatorname{sgn}(t)] &=U(\omega)-U(-\omega) \\&=\left(\pi \delta(\omega)+\frac{1}{j \omega}\right)-\left(\pi \delta(-\omega)+\frac{1}{-j \omega}\right)=\frac{2}{j \omega}\end{aligned}$$
$$f(t)=\operatorname{sgn}(t)=-1+2 u(t)$$
$$F(\omega)=-2 \pi \delta(\omega)+2\left(\pi \delta(\omega)+\frac{1}{j \omega}\right)=\frac{2}{j \omega}$$
$$j \omega F(\omega)=2 \quad \Longrightarrow \quad F(\omega)=\frac{2}{j \omega}$$
$$f(t)=e^{-a|t|}=e^{-a t} u(t)+e^{a t} u(-t)=y(t)+y(-t)$$
$$\mathcal{F}\left[e^{-a|t|}\right]=Y(\omega)+Y(-\omega)=\left(\frac{1}{a+j \omega}+\frac{1}{a-j \omega}\right)=\frac{2 a}{a^{2}+\omega^{2}}$$
$$\mathcal{F}\left[u\left(t+\frac{\tau}{2}\right)-u\left(t-\frac{\tau}{2}\right)\right]=\tau \frac{\sin (\omega \tau / 2)}{\omega \tau / 2}=\tau \operatorname{sinc} \frac{\omega \tau}{2}$$
$$\mathcal{F}\left[2 \frac{\sin t}{t}\right]=2 \pi[U(\omega+1)-U(\omega-1)]$$
$$\mathcal{F}\left[\frac{\sin t}{t}\right]=\pi[U(\omega+1)-U(\omega-1)]$$
Example 2: Find the Fourier transform of the function in Fig. 2.
Solution: The Fourier transform can be found directly using Eq. (B),
but it is much easier to find it using the derivative property. We can express the function as
Its first derivative is shown in Fig. 3(a) and is given by

Its second derivative is in Fig. 3(b) and is given by
Taking the Fourier transform of both sides,
or
Fig. 2: For Example 2.
$$ F(\omega)= \mathcal{F} [f(t)] = \int_{-\infty}^\infty f (t) e^{-j\omega t} dt \tag{B}$$
$$f(t)=\left\{\begin{array}{lr}1+t, & -1 < t < 0 \\1-t, & 0 < t < 1\end{array}\right.$$
$$f^{\prime}(t)=\left\{\begin{array}{rr}1, & -1 < t < 0 \\-1, & 0 < t < 1\end{array}\right.$$

Fig. 3: First and second derivatives of f (t) in Fig. 2;
for Example 2.
$$f^{\prime \prime}(t)=\delta(t+1)-2 \delta(t)+\delta(t-1)$$
$$(j \omega)^{2} F(\omega)=e^{j \omega}-2+e^{-j \omega}=-2+2 \cos \omega$$
$$F(\omega)=\frac{2(1-\cos \omega)}{\omega^{2}}$$
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} $ Solution:
(a) To avoid complex algebra, we can replace $ j \omega $ with $ s $ for the moment. Using partial fraction expansion,
where
Substituting $ A=18 $ and $ B=-8 $ in $ F(s) $ and $ s $ with $ j \omega $ gives
With the aid of Table 17.2, we obtain the inverse transform as
(b) We simplify $ G(\omega) $ as
(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,
$$F(s)=\frac{10 s+4}{s^{2}+6 s+8}=\frac{10 s+4}{(s+4)(s+2)}=\frac{A}{s+4}+\frac{B}{s+2}$$
$$\begin{array}{l}A=\left.(s+4) F(s)\right|_{s=-4}=\left.\frac{10 s+4}{(s+2)}\right|_{s=-4}=\frac{-36}{-2}=18 \\B=\left.(s+2) F(s)\right|_{s=-2}=\left.\frac{10 s+4}{(s+4)}\right|_{s=-2}=\frac{-16}{2}=-8\end{array}$$
$$F(j \omega)=\frac{18}{j \omega+4}+\frac{-8}{j \omega+2}$$
$$f(t)=\left(18 e^{-4 t}-8 e^{-2 t}\right) u(t)$$
$$G(\omega)=\frac{\omega^{2}+21}{\omega^{2}+9}=1+\frac{12}{\omega^{2}+9}$$
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