ECE 402: Communications Engineering
Homework 7 Solutions (Fall 2009)
1.
(a)
As noted in class, for the sinc pulse the null bandwidth is
W
null
= 1
/T
and the data rate is
R
=
k/T
, so
η
=
R/W
null
=
k
= 6 bps/Hz.
(b)
For
M
= 64, we want
10

8
=
P
b
≈
4
log
2
M
±
1

1
√
M
¶
Q
ˆs
3
PT
(
M

1)
N
0
!
=
⇒
Q
ˆ
s
3
PT
63
N
0
!
= 1
.
714
×
10

8
.
In MATLAB, we ﬁnd
qfuncinv(1.7140e008)=5.518
, so the required SNR is
s
3
PT
63
N
0
= 5
.
518 =
⇒
PT
N
0
=
63
×
(5
.
518)
2
3
= 639
.
4
.
We conclude that
E
b
/N
0
=
PT/kN
0
= 106
.
6 or in decibels, 20
.
28 dB.
(c)
The spectral eﬃciency of this system is
η
= 6 bps/Hz. By Shannon’s Formula, the best possible
tradeoﬀ between power and bandwidth eﬃciency is
E
b
/N
0
≥
(2
η

1)
/η
. The best power
eﬃciency is therefore
E
b
/N
0
≥
(2
6

1)
/
6 = 10
.
5 or in decibels, 10.21 dB, which is much smaller
than in part (b)!
2.
(a)
Since antenna absorption and scattering losses are ignored,
η
T
=
η
R
= 1. The wavelength is
λ
=
c/f
= 0
.
15 m and the diameter of the transmit and receive antennas is
D
T
=
D
R
= 1
.
22 m.
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 Fall '08
 Townsend
 Decibel, sampling rate, symbol rate, dBw, data amplitudes

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