Digital Signal Processing Reference
In-Depth Information
P
(
j
ω
)
+ j
P
(
j
ω
)
P
(
j
ω
)
−
j
P
(
j
ω
)
(a)
(b)
2
- -
2
P
+
ω
ω
0
0
Figure 2.21
. (a), (b) The right half and left half of the Fourier transform
P
(
jω
),
separated using the Hilbert transform.
cos
ω
t
c
x
(
t
)
SSB-1
x
s
(
n
)
P
(
j
ω)
D/C
T
x
(
t
)
x
H
(
j
ω)
hi
−
1
SSB-2
Hilbert
transformer
sin ω
t
c
Figure 2.22
. Block diagram for generating the two types of SSB signals.
2.4.2 Extracting baseband PAM from modulated versions
Given the modulated PAM signal in DSB form, or in SSB form of either type,
we can extract the baseband signal
x
(
t
) at the receiver by using the structure
shown in Fig. 2.23. The input to this system is the modulated signal
x
mod
(
t
)=
x
(
t
)cos
ω
c
t
+
α
x
(
t
)sin
ω
c
t,
where
α
is0forDSB,1forType1SSB,and
1 for Type 2 SSB. This signal is
multiplied by the locally generated carrier 2 cos
ω
c
t
to yield the output
6
−
=
x
(
t
)cos
2
ω
c
t
+2
α
2
x
mod
(
t
)cos
ω
c
t
x
(
t
)sin
ω
c
t
cos
ω
c
t
=
x
(
t
)(1 + cos 2
ω
c
t
)+
α
x
(
t
)sin2
ω
c
t
=
x
(
t
)+
x
(
t
) cos 2
ω
c
t
+
α
x
(
t
)sin2
ω
c
t.
The second and third terms in the last line above are high frequency components,
and their energy is contained mostly around the frequency 2
ω
c
. The first term
6
This assumes that the receiver knows the carrier frequency
ω
c
and furthermore that there
is no unknown phase offset. (If cos
ω
c
t
is received as cos(
ω
c
t
+
φ
)
,
where
φ
is
unknown
,wesay
there is an unknown phase offset.)
Search WWH ::
Custom Search