8.2 Discrete-Time Signals 461
FIGURE 8.1
(a) Signalx 2 [n]
(nonperiodic forn≥ 0 )
and (b) signalx 1 [n]
(periodic forn≥ 0 ).
The arrows show that
the values are not
equal forx 2 [n]and
equal forx 1 [n]. (c) The
square of the signals
differ slightly,
suggesting that if
x 1 [n]has finite power
so doesx 2 [n].
− 10 0 10 20 30 40
− 2
− 1
0
1
2
n
x^2
[n
]
− 10 0 102030 40
− 2
− 1
0
1
2
n
x^1
[n
]
(a) (b)
− 10 0 10 20 30 40
0
1
2
3
4
n
x
2 [n 1
],
x
2 [ 2
n]
x^22 [n]
x^21 [n]
(c)
Solution
The energy is given by
εx=
∑∞
n= 0
4 (0.5)^2 n= 4
∑∞
n= 0
(0.25)n=
4
1 −0.25
=
16
3
thus,x[n] is a finite-energy signal. Just as with continuous-time signals, a finite-energy signal is a
finite-power (actually zero power) signal. n
8.2.3 Even and Odd Signals
Time shifting and scaling of discrete-time signals are very similar to the continuous-time cases, the
only difference being that the operations are now done using integers.
A discrete-time signalx[n]is said to be
n DelayedbyN(an integer) samples ifx[n−N]isx[n]shifted to the rightNsamples.
n AdvancedbyM(an integer) samples ifx[n+M]isx[n]shifted to the leftMsamples.
n Reflectedif the variableninx[n]is negated (i.e.,x[−n]).
The shifting to the right or the left can be readily seen by considering wherex[0] is attained. Forx[n−
N], this is whenn=N(i.e.,Nsamples to the right of the origin), orx[n] is delayed byNsamples.
