10.2 Discrete-Time Fourier Transform 587
(a) (b)
024681012141618
−0.4
−0.2
0
s[
n]−
x[
n]
0 2 4 6 8 10 12 14 16 18
0
0.2
0.4
0.6
0.8
1
n
s[
n],
x[
n]
Interpolated
Original
0 2 4 6 8 10 12 14 16
0 2 4 6 8 10121416
− 1
−0.5
0
0.5
1
n
s[
n],
x[
n]
−0.15
−0.1
−0.05
0
0.05
0.1
nn
s[
n]−
x[
n]
Interpolated
Original
FIGURE 10.5
Interpolation of (a) non-band-limited and (b) bandlimited discrete-time signals. The interpolated signal is
compared to the original signal, and the interpolation error is shown. The errors signals show that the original
signal can be recovered almost exactly when the signal satisfies the bandlimiting condition, not otherwise.
If the DTFT of a finite-energy signalx[n] isX(ejω), we have that the energy of the signal is given by
Ex=
∑∞
n=−∞
|x[n]|^2 =
1
2 π
∫π
−π
|X(ejω)|^2 dω (10.18)
The magnitude square|X(ejω)|^2 has the units of energy per radian, and so it is called anenergy density.
When|X(ejω)|^2 is plotted against frequencyω, the plot is called theenergy spectrumof the signal, or
how the energy of the signal is distributed over frequencies.
10.2.6 Time and Frequency Shifts........................................................
Shifting in time does not change the frequency content of a signal. Thus, the magnitude of the signal
DTFT is not affected, only the phase is. Indeed, ifx[n] has a DTFTX(ejω), then the DTFT ofx[n−N]
for some integerNis
F(x[n−N])=
∑
n
x[n−N]e−jωn
=
∑
m
x[m]e−jω(m+N)=e−jωNX(ejω)
Ifx[n]has a DTFT
X(ejω)=|X(ejω)|ejθ(ω)
whereθ(ω)is the phase, the shifted signalx 1 [n]=x[n−N]has a DTFT of
X 1 (ejω)=X(ejω)e−jωN
=|X(ejω)|e−j(ωN−θ(ω))
