Digital Signal Processing Reference
In-Depth Information
b
o
(
t
)
b
o
(
t
) represents physical transmission of logic "0"
1
0
t
T
b
1
(
t
)
b
1
(
t
) represents physical transmission of logic "1"
1
T
t
− 1
T/2
s
(
t
) =
b
1
(
t
)
b
o
(
t
)
b
1
(
t
) ... = Tx data: 101 ...
0 1 2 3 4 0 k mod 5
1
T
s
=
T
/ 5
T
s
(
t
)
t
− 1
T/2
2
T
3
T
0
1 2 3 4
0
1 2 3 4
0
1 . . .
k mod 5 (k = sample index)
M
o
is matched
to
s
o
(
t
)
M
o
c
o
1
c
1
1
c
2
1
c
3
1
c
4
1
Decision at
k mod 5 =4:
M
o
>
M
1
?
y = {
"0" if
M
o
>
M
1
"1" if
M
1
>
M
o
z
−1
z
−1
z
−1
z
−1
r
(
t
)
h
o
−1
h
1
−1
h
2
1
h
3
1
h
4
1
kT
s
Clk from Sych. Ct.
M
1
is matched
to
s
1
(
t
)
M
1
5
÷
Fig. 2.30 Digital matched filter for orthogonal binary signaling with T
s
= T/5. Above wave-
forms for transmitting logic ''0'' and logic ''1''. The transmitted signal s(t) represents the data
string 1 0 1. Below digital matched filter receiver
The output of the ith matched filter at the kth sampling instant is given by the
convolution:
M
i
ð
n
Þ¼
X
N
1
r
ð
k
n
Þ
h
i
ð
n
Þ
n
¼
0
where r(t) is the received signal. Decisions (comparisons) are made only when
kN
¼
N
1 [every 5 samples for the scenario represented in Fig. (
2.30
)]. Note that
in Fig. (
2.30
) the simpler notation h
n
= h
1
(n) and c
n
= h
0
(n) is used-under noise-
free condition and full synchronization with the transmitter at k = 4 (after 5
samples) the shift register has the following contents: a = 1, b = 1, c = 1,
d =-1, e =-1. Hence, M
o
= 1, M
1
= 5 [ M
o
, and the decision is that ''1''
was transmitted. As an exercise try to find the decision of the circuit after 10 and
15 samples.
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