Spring 2011
ECE 361:
Problem Set 9: Solutions
MaximumLikelihood Sequence Estimation; Passband Signals
1.
[Maximumlikelihood sequence estimation]
(a) The squared distance is proportional to (0
.
1
x
[
m

2] + 0
.
2
x
[
m

1] +
x
[
m
]

y
[
m
])
2
giving the
trellis shown below.
2
.
25
0
.
01
0
.
00
3
.
61
3
.
61
4
.
00
0
.
01
4
.
84
0
.
09
0
.
04
2
.
89
2
.
25
++

+
+


00
++
++

+

+


+

0

+

0+
0
.
9

0
.
7

1
.
1
3
.
24
0
.
8
0
.
01
3
.
61
0
.
25
0
.
04
4
.
41
5
.
76
0
,
25
0
.
01
0
.
16
(b) Using the Viterbi algorithm, we progress through the trellis as shown in the ﬁgure below.
The best paths are 00
→
0+
→
+
 → 
+
→
++, 00
→
0+
→
+
 →  → 
+,
00
→
0+
→
+
 → 
+
→
+

, 00
→
0+
→
+
 →  → 
,
2.
[Trellis diagram structure]
(a) The ﬁrst statement is true: the two edges correspond to data bits 0 and 1 (or
±
1 or
±
√
E
T
). The
second statement is false: the two edges have the
data
bit since the data bit labels on the edges
become part of the label of the node being entered.
(b) The squareddistance labels on the two edges leaving a node at depth
m
in the trellis are of the
form
M
X
i
=1
h
[
i
]
x
[
m

i
] + 1

y
[
m
]
!
2
and
M
X
i
=1
h
[
i
]
x
[
m

i
]

1

y
[
m
]
!
2
and these are obviously diﬀerent
except
when the two quantities being squared are negatives of
each other. Thus, we have
M
X
i
=1
h
[
i
]
x
[
m

i
] + 1

y
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 Spring '09
 Baseband, Passband, FC, −T /2, passband signal, Passband Signals

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