10.2 Discrete-Time Fourier Transform 573
or that x[n]be absolutely summable. This means that only for those signals we can use the above direct
(Eq. 10.1) and inverse (Eq. 10.2) DTFT definitions. We will see in the next section how to obtain the
DTFT of signals that do not satisfy the absolutely summable condition.
10.2.1 Sampling, Z-Transform, Eigenfunctions, and the DTFT
The connection of the DTFT with sampling, eigenfunctions, and the Z-transform can be shown as
follows:
n Sampling and the DTFT.When sampling an analog signalx(t), the sampled signalxs(t)can be
written as
xs(t)=
∑
n
x(nTs)δ(t−nTs)
Its Fourier transform is then
F[xs(t)]=
∑
n
x(nTs)F[δ(t−nTs)]
=
∑
n
x(nTs)e−jnTs
Lettingω=Ts, the discrete frequency in radians, the above equation can be written as
Xs(ejω)=F[xs(t)]=
∑
n
x(nTs)e−jnω (10.3)
coinciding with the DTFT of the discrete-time signalx(nTs)=x(t)|t=nTsorx[n].
At the same time, the spectrum of the sampled signal can be equally represented as
Xs(ejTs)=Xs(ejω)=
∑
k
1
Ts
X
(
ω
Ts
−
2 πk
Ts
)
(10.4)
which is a periodic repetition, with period 2π/Ts, of the spectrum of the analog signal being
sampled. Thus, sampling converts a continuous-time signal into a discrete-time signal with a
periodic spectrum varying continuously in frequency.
n Z-transform and the DTFT.If in the above we ignoreTsand considerx(nTs)a function ofn, we can
see that
Xs(ejω)=X(z)|z=ejω (10.5)
That is, it is the Z-transform computed on the unit circle. For the above to happen,X(z)must
have a region of convergence (ROC) that includes the unit circle. There are discrete-time sig-
nals for which we cannot find their DTFTs from the Z-transform because they are not absolutely
summable—that is, their ROCs do not include the unit circle. However, any discrete-time signal
x[n], of finite support in time, has a Z-transformX(z)with a region of convergence the whole
z-plane, excluding either the origin or infinity, and as such its DTFTX(ejω)is computed fromX(z)
by lettingz=ejω.
