Signals and Systems - Electrical Engineering

576 C H A P T E R 10: Fourier Analysis of Discrete-Time Signals and Systems

a sinusoidx[n]=cos(ω 0 n+θ), which is not absolutely summable. According to the above pairs,

we get

x[n]=

1

2

[

ej(ω^0 n+θ)+e−j(ω^0 n+θ)

]

⇔X(ejω)=π

[

ejθδ(ω−ω 0 )+e−jθδ(ω+ω 0 )

]

The DTFT of the cosine indicates that its power is concentrated at the frequencyω 0.

The “dual” pairs

δ[n−k], integerk⇔e−jωk (10.10)

e−jω^0 n, −π≤ω 0 < π⇔ 2 πδ(ω+ω 0 ) (10.11)

allow us to obtain the DTFT of signals that do not satisfy the absolutely summable condition. Thus, in general,

we have

∑

k

X[k]e−jωkn⇔

∑

k

2 πX[k]δ(ω+ωk) (10.12)

The linearity of the DTFT and the above result give that for a signal that is not absolutely summable

x[n]=

∑

`

A`cos(ω`n+θ`)

its DTFT is

X(ejω)=

∑

`

πA`

[

ejθ`δ(ω−ω`)+e−jθ`δ(ω+ω`)

]

−π≤ω < π

Ifx[n]is periodic, the discrete frequencies are harmonically related, i.e.,ω`=`ω 0 whereω 0 is the fundamental

frequency ofx[n].

nExample 10.2

The DTFT of a signalx[n] is

X(ejω)= 1 +δ(ω− 4 )+δ(ω+ 4 )+0.5δ(ω− 2 )+0.5δ(ω+ 2 )

The signalx[n]=A+Bcos(ω 0 n)cos(ω 1 n)is given as a possible signal that givesX(ejω). Determine

whether you can findA,B, andω 0 andω 1 to obtain the desired DTFT. If not, provide a betterx[n].

Solution

Using trigonometric identities or Euler’s identity, we have that

cos(ω 0 n)cos(ω 1 n)=0.5 cos((ω 0 +ω 1 )n)+0.5 cos((ω 1 −ω 0 )n)