ECE 402: Communications Engineering
Homework 6 Solutions
(Fall 2009)
1.
(a)
For polar signaling,
P
b
=
Q
(
p
2
E
b
/N
0
). From the reference sheet, we see that
Q
(3
.
719) = 10

4
and
Q
(4
.
264) = 10

5
. Hence to obtain
P
b
= 10

4
, we must have
s
2
E
b
N
0
= 3
.
719 =
⇒
E
b
N
0
¶
dB
= 10 log
10
ˆ
(3
.
719)
2
2
!
= 8
.
4 dB
.
To obtain
P
b
= 10

5
, we need
s
2
E
b
N
0
= 4
.
264 =
⇒
E
b
N
0
¶
dB
= 10 log
10
ˆ
(4
.
264)
2
2
!
= 9
.
6 dB
.
(b)
Unipolar signaling (
P
b
=
Q
(
p
E
b
/N
0
)) requires twice as much (3 dB more) energyperbit than
polar signaling for the same
P
b
, so
E
b
/N
0
= 11
.
4 dB for
P
b
= 10

4
and
E
b
/N
0
= 12
.
6 dB for
P
b
= 10

5
.
2. The error probability of BPSK is
P
b
=
Q
(
p
2
PT/N
0
).
From the reference sheet, we see that
Q
(4
.
753) = 10

6
so
q
2
PT/N
0
= 4
.
753 =
⇒
P
=
11
.
3
N
0
T
= 11
.
3
N
0
R
Since
N
0
=
kT
r
, the required noise temperature is
T
r
=
P
11
.
3
Rk
=
5
×
10

18
11
.
3
·
4740
·
1
.
38
×
10

23
= 6
.
76
o
K .
3. The baseband version of 16QAM has data amplitudes
a
n
=
a
+
jb
, where
a
and
b
are
±
A
√
T,
±
3
A
√
T
.
The pulse waveform is
g
(
t
) = (1
/
√
T
)sinc(
t/T
). For 16QAM, we have
μ
=
E{
a
n
}
=
1
16
X
a,b
=
±
A
√
T,
±
3
A
√
T
[
a
+
jb
] = 0
σ
2
=
E{
a
n

μ

2
}
=
1
16
X
a,b
=
±
A
√
T,
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 Fall '08
 Townsend
 Bandwidth, Baseband, Passband, Wnull

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