University of Illinois
Spring 2011
ECE 361:
Problem Set 8: Solutions
Equalizers in BandLimited Communications
1.
[Zeroforcing transversal filter]
The zeroforcing equalizer is a transversal filter designed to produce a signal that is as much like a unit
pulse as possible when the input is the channel pulse response. For a threetap equalizer, we need to
find
c
[

1],
c
[0] and
c
[1] such that
c
[

1]
h
[0] +
c
[0]
h
[

1] +
c
[1]
h
[

2] = +1
.
00
c
[

1] + 0
.
20
c
[0]

0
.
05
c
[1] = 0
c
[

1]
h
[1] +
c
[0]
h
[0]
+
c
[1]
h
[

1] = +0
.
30
c
[

1] + 1
.
00
c
[0] + 0
.
20
c
[1] = 1
c
[

1]
h
[2] +
c
[0]
h
[1]
+
c
[1]
h
[0]
=

0
.
07
c
[

1] + 0
.
30
c
[0] + 1
.
00
c
[1] = 0
which set of three linear equations in three unknowns can be solved to get the equalizer coefficients
c
[

1] =

0
.
215
0
.
8692
≈ 
0
.
247354,
c
[0] =
0
.
9965
0
.
8692
≈
1
.
146457, and
c
[1] =

0
.
314
0
.
8692
≈ 
0
.
361252. Thus,
ˆ
Y
[
m
]
≈ 
0
.
247354
Y
[
m
+ 1] + 1
.
146457
Y
[
m
]

0
.
361252
Y
[
m

1]
or, substituting the
X
[
m
+
i
] for
Y
[
m
+ 1]
,
Y
[
m
]
,
Y
[
m

1],

0
.
247354
×
(

0
.
05
X
[
m
+ 3] +0
.
2
X
[
m
+ 2]
+
X
[
m
+ 1]+0
.
3
X
[
m
]

0
.
07
X
[
m

1])
+1
.
146457
×
(

0
.
05
X
[
m
+ 2] +0
.
2
X
[
m
+ 1]
+
X
[
m
]
+0
.
3
X
[
m

1]

0
.
07
X
[
m

2])

0
.
361252
×
(

0
.
05
X
[
m
+ 1]+0
.
2
X
[
m
]
+
X
[
m

1]
+0
.
3
X
[
m

2]

0
.
07
X
[
m

3])
≈
0
.
02
X
[
m
+ 3]

0
.
11
X
[
m
+ 2]
+
X
[
m
]

0
.
19
X
[
m

2] +0
.
03
X
[
m

3]
showing that the contributions of
X
[
m
±
1] have been eliminated in the ISI affecting
ˆ
Y
[
m
] (though
contributions from
X
[
m
±
3] have been introduced). Note also that eliminating the ISI from
X
[
m
±
1]
has
increased
the ISI from
X
[
m
±
2] which previously had coefficients

0
.
07 and

0
.
05 but now have
coefficients

0
.
11 and

0
.
19. However, the maximum ISI is reduced from 0
.
62
√
E
T
in
Y
[
m
] to only
≈
0
.
33
√
E
T
in
ˆ
Y
[
m
].
Conversely,
X
[
m
] will not show up in
ˆ
Y
[
m
±
1] but will show up with larger
coefficients in
ˆ
Y
[
m
±
2] than it did in
Y
[
m
±
2], and
X
[
m
] will show up in
ˆ
Y
[
m
±
3] whereas it does not
affect
Y
[
m
±
3] at all.
2.
[Comparison of equalizers]
(a)
Y
[
m
] =
√
E
T
+ 0
.
5
X
[
m

1] +
N
[
m
]
∼ N
(
√
E
T
(1
±
0
.
5)
, σ
2
). Hence,
•
the maximum conditional error probability is
Q
(0
.
5
√
E
T
/σ
) =
Q
(1
.
5) = 6
.
681
×
10

2
.
•
the minimum conditional error probability is
Q
(1
.
5
√
E
T
/σ
) =
Q
(4
.
5) = 3
.
398
×
10

6
.
•
the average error probability
P
(
E
) is
1
2
Q
(1
.
5) +
Q
(4
.
5)
≈
1
2
Q
(1
.
5) = 3
.
340
×
10

2
.
In the absence of ISI, the error probability would be
Q
(
p
E
T
/σ
2
) =
Q
(3) = 1
.
350
×
10

3
, an
order of magnitude smaller than the maximum conditional error probability and the average error
probability.
Note that the
SNR
is 9 and
SINR
=
E
T
0
.
25
E
T
+
σ
2
=
SNR
1 + 0
.
25
SNR
=
9
3
.
25
= 2
.
769.
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 Spring '09
 Conditional Probability, Probability, Probability theory, error probability, The Equalizer, conditional error probability

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