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)