Signals and Systems - Electrical Engineering

312 CHAPTER 5: Frequency Analysis: The Fourier Transform

− 1 −0.5 0 0.5 1

0

2

4

6

8

10

t

− 100 − 50 0 50 100

0

5

10

Ω

− 1 −0.5 0 0.5 1

0

20

40

60

Ω

x^1

(t

)

− 100 − 50 0 50 100

0

5

10

t

x(

t)

=

X^1

(t

)

X^1

(Ω

)

X

(Ω

)=

2

πx

( 1

−Ω

)

FIGURE 5.4

Application of duality to find the Fourier transform ofx(t)=10 sinc(0.5t). Notice that

X()= 2 πx 1 ()≈6.28x 1 ()=62.8[u(+0.5)−u(−0.5)].

Then according to the duality property, the Fourier transform ofx(t)is

X()= 2 πA[u(−+0.5)−u(−−0.5)]= 2 πA[u(+0.5)−u(−0.5)]

given that the function is even with respect to. So the Fourier transform of the sinc is a rectangular

pulse in frequency, in the same way that the Fourier transform of a pulse in time is a sinc function

in frequency. Figure 5.4 shows the dual pairs forA=10. n

nExample 5.6

Find the Fourier transform ofx(t)=Acos( 0 t)using duality.

Solution

The Fourier transform ofx(t)cannot be computed using the integral definition since this signal

is not absolutely integrable, or using the Laplace transform sincex(t)does not have a Laplace

transform. As a periodic signal,x(t)has a Fourier series representation and we will use it later to

find its Fourier transform. For now, let us consider the Fourier pair

δ(t−ρ 0 )+δ(t+ρ 0 ) ⇔ e−jρ^0 +ejρ^0 =2 cos(ρ 0 )