Signals and Systems - Electrical Engineering

5.5 Linearity, Inverse Proportionality, and Duality 311

This can be shown by considering the inverse Fourier transform

x(t)=

1

2 π

∫∞

−∞

X(ρ)ejρtdρ

and replacingtby−and multiplying by 2πto get

2 πx(−)=

∫∞

−∞

X(ρ)e−jρdρ

=

∫∞

−∞

X(t)e−jtdt

=F[X(t)]

To understand the above equations you need to realize thatρandtare dummy variables inside the
integral, and as such they are not reflected outside the integral.

Remarks

n This duality property allows us to obtain the Fourier transform of signals for which we already have a
Fourier pair and that would be difficult to obtain directly. It is thus one more method to obtain the Fourier
transform, besides the Laplace transform and the integral definition of the Fourier transform.
n When computing the Fourier transform of a constant signal, x(t)=A, we indicated that it would be
X()= 2 πAδ(). Indeed, we have the dual pairs

Aδ(t) ⇔ A

A ⇔ 2 πAδ(−)= 2 πAδ() (5.9)

where in the second equation we use the fact thatδ()is even.

nExample 5.5

Use the duality property to find the Fourier transform of the sinc signal

x(t)=A

sin(0.5t)

0.5t

=Asinc(0.5t) −∞<t<∞

Solution

The Fourier transform of the sinc signal cannot be found using the Laplace transform or the integral

definition of the Fourier transform. The duality property provides a way to obtain it. We found

before, forτ=0.5, the following pair of Fourier transforms:

A[u(t+0.5)−u(t−0.5)] ⇔ A

sin(0.5)

0.5

=Asinc(0.5)