Digital Signal Processing Reference
In-Depth Information
Applying the time-shifting property, frequency-shifting property, and time
reversal property on
u(n)
, we can derive the DTFT of a few more discrete-time
functions. For example
e
−
jωk
π
∞
1
u(n
−
k)
⇔
δ(ω
−
2
πk)
+
(3.46)
e
−
jω
)
(
1
−
k
=−∞
π
∞
1
e
jω
0
n
u(n)
⇔
δ(ω
−
ω
0
−
2
πk)
+
(3.47)
e
−
j(ω
−
ω
0
)
)
(
1
−
k
=−∞
π
∞
1
2
1
cos
(ω
0
n)u(n)
⇔
δ(ω
−
ω
0
−
2
πk)
+
(
1
−
e
−
j(ω
−
ω
0
)
)
k
=−∞
∞
1
+
π
δ(ω
+
ω
0
−
2
πk)
+
(3.48)
e
−
j(ω
+
ω
0
)
)
(
1
−
k
=−∞
It is worth comparing the DTFT of
e
jω
0
n
u(n)
given above with the DTFT of
e
−
an
u(n)
,where
|
a
|
<
1:
1
e
−
an
u(n)
⇔
(3.49)
e
−
a
e
−
jω
1
−
3.4.1 Differentiation Property
=
n
=−∞
j
[
dX(e
jω
)
]
/dω
, we start with
X(e
jω
)
To prove that
nx(n)
⇔
x(n)e
−
jωn
[
dX(e
jω
)
]
/dω
and
differentiate
both
sides
to
get
=
n
=−∞
x(n)(
jn)e
−
jωn
−
and multiplying
both
sides
by
j
, we t
=
n
=−∞
nx(n)e
−
jωn
. The proof
j
[
dX(e
jω
)
]
/dω
is
similar
to that used
in Chapter 2 to prove that the
z
transform of
nx(n)u(n)
is
−
z
[
dX(z)
]
/dz
.
a
n
u(n)
ae
−
jω
)
X(e
jω
)
, we can derive the follow-
Given
x(n)
=
⇔
1
/(
1
−
=
ing, using the differentiation property:
j
−
ae
−
jω
)
2
j
dX(e
jω
)
dω
jae
−
jω
(
1
−
ae
−
jω
(
1
−
=
=
ae
−
jω
)
2
ae
−
jω
(
1
−
na
n
u(n)
⇔
(3.50)
ae
−
jω
)
2
Since the DTFT of
a
n
u(n)
is 1
/(
1
ae
−
jω
)
, we add this DTFT to that of
na
n
u(n)
−
and get
1
+
1
)a
n
u(n)
(n
⇔
(3.51)
ae
−
jω
)
2
(
1
−
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