Chapter 14
1. B=10 MHz
B
total
= [(10 + 2)100]MHz = 1200MHz
2. B
total
= 25MHz
125 TDMA channels
8 user time slots per channel
R=270.833Kbps
(a)
Pre
1 2
3 4 5
6
7
8 Trail
3 58 data bits
58 data bits
26 training
bits
3 8.25
start bits
stop bits
data bits
as guard
time
Figure 1: Problem 2a
(b)
2(58)
3+2(58)+26+3+8.25
= 0.7424
Information rate = 201.066 Kbps
(c) Frame duration = 8 × timeslotduration
time slot duration =
2(3)+2(58)+26+8.25
270.833×10
3
= 576.92µs
Frame duration = 4.61 ms
Latency =
7
8
(frame duration) = 4.038 ms
(d) duration of guard band + stop bits =
3+8.25
270.833×10
3
= 41.54µs
∴ max delay spread T
m
< 41.54µs for guard band to be useful.
3. B=10MHz
G=100
cross-correlation = 1/G
Interference limited system (N=0)
(a) SIR =
P
r
1
G
(k−1)P
r
=
G
K−1
(b) P
b
= Q(
√
2γ
b
) = 10
−3
SIR = 3.0902
K = 1 +
G
SIR
= b33.36c = 33
(c) B=10MHz
Information signal Bandwidth =
10MHz
100
= 100KHz Total number of users = 100
(d) SIR =
G
αSIR
= 100 ⇒ α = 0.3269
No, this is not a reasonable voice activity factor. It is too low.
4. (a) Since the system is interference limited
SIR =
1
(m−1)
= γ
b
P
e
= Q(
p
(γ
b
)) (assuming binary signalling)
P
e
= Q(
1
√
m−1
)
