Digital Signal Processing Reference
In-Depth Information
Proof
The magnitude of the complex function
X
(
−ω
)
=
Re
{
X
(
−ω
)
+
jIm
{
X
(
−ω
)
}
is given by
(Re
X
(
−ω
)
)
2
+
(Im
X
(
−ω
)
)
2
.
X
(
−ω
)
=
Substituting Re
X
(
−ω
)
=
Re
X
(
ω
)
}
and Im
{
X
(
−ω
)
=−
Im
{
X
(
ω
)
}
,
obtained from the alternative form I of the Hermitian symmetry property in
the above expression, yields
=
X
(
ω
)
,
X
(
−ω
)
=
(Re
X
(
ω
)
)
2
+
(
−
Im
X
(
ω
)
)
2
which proves that the magnitude spectrum
X
(
ω
)
of a real-valued signal is
even. Alternatively, consider the phase of the complex function
X
(
−ω
)
=
Re
X
(
−ω
)
+
jIm
{
X
(
−ω
)
}
as given by
Re
X
(
−
ω
)
Im
X
(
−ω
)
−
1
<
X
(
−ω
)
=
tan
.
Substituting Re
{
X
(
−ω
)
=
Re
X
(
ω
)
}
and Im
{
X
(
−ω
)
=−
Im
{
X
(
ω
)
}
yields
Re
X
(
−ω
)
−
Im
X
(
−ω
)
−
1
<
X
(
−ω
)
=
tan
=−<
X
(
ω
)
,
which proves that the phase spectrum
<
X
(
ω
) of a real-valued signal is odd.
Example 5.11
Consider a function
g
(
t
) whose CTFT is given by
G
(
ω
)
=
1
+
2
πδ
(
ω − ω
0
).
Determine if
g
(
t
) is a real-valued function.
Solution
Substituting
ω
by
−ω
in the CTFT
G
(
ω
) yields
=
1
+
2
πδ
(
ω + ω
0
)
.
G
(
−ω
)
=
1
+
2
πδ
(
−ω − ω
0
)
The complex conjugate of
G
(
ω
)isgivenby
∗
∗
=
1
+
2
πδ
(
ω − ω
0
)
.
G
(
ω
)
=
[1
+
2
πδ
(
ω − ω
0
)]
Comparing the two expressions, it is clear that
G
*(
ω
)
=
G
(
−ω
), and therefore
that
g
(
t
) is not a real-valued function. In order to verify the result, we calculate
the inverse CTFT of
G
(
ω
) as follows:
=ℑ
−
1
G
(
ω
)
=ℑ
−
1
1
+
2
πδ
(
ω − ω
0
)
=ℑ
−
1
1
+
2
π ℑ
−
1
δ
(
ω − ω
0
)
,
g
(
t
)
which results in
= δ
(
t
)
+
e
j
ω
0
t
,
g
(
t
)
or
= δ
(
t
)
+
cos(
ω
0
t
)
+
j sin(
ω
0
t
)
,
g
(
t
)
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