University of Illinois
Spring 2011
ECE 361:
Problem Set 2: Solutions
Matched and Unmatched Filtering, and Signal Design
Throughout this Problem Set, assume that the bits being transmitted are equally likely to be 0 or 1.
1.
[Matched Filtering]
(a)
E
0
=
Z
T/
2

T/
2
A
2
dt
=
A
2
T
;
E
1
=
Z
T/
2

T/
2
2
A
2
cos
2
(
πt/T
)
dt
=
A
2
T
. (rms amplitude!)
h
s
0
, s
1
i
=
Z
T/
2

T/
2
√
2
A
2
cos(
πt/T
)
dt
=
√
2
A
2
T
π
sin(
πt/T
)
T/
2

T/
2
=
2
√
2
A
2
T
π
.
Hence,
P
(
E
) =
Q
r
E
0
+
E
1

2
h
s
0
, s
1
i
2
N
0
=
Q
s
A
2
T
(1

2
√
2
/π
)
N
0
.
(b) Now obviously
E
0
=
E
1
=
A
2
T
as before while
h
s
0
, s
1
i
=

2
√
2
A
2
T
π
giving
P
(
E
) =
Q
s
A
2
T
(1 + 2
√
2
/π
)
N
0
.
This is smaller than the answer to part (a) because the
argument of
Q
(
·
) is larger. Simplistically, the signals in part (a) “resemble” each other while the
signals in part (b) “look more like” antipodal signals which are known to give the smallest error
probability.
(c) Now
E
0
=
E
1
=
A
2
T
as before while
h
s
0
, s
1
i
= 0 giving
P
(
E
) =
Q
s
A
2
T
N
0
!
. The signals are
now orthogonal and the error probability is between the answers to parts (a) and (b). Intuitively,
orthogonal signals are better than signals that “resemble” each other but antipodal is even better.
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 Spring '09
 Trigraph, dt, −T /2

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