Problem 38.
Problem 40.
Problem 45.
The bit error rate or the probability of error, BER, of a coherent BPSK system is given by
BER =
where E = A2
E
1
erfc
N0
2
T
.
2
Thus, A = 2 N 0 R e rfinv (1 2 BER ) .
R=1kbps
R=10kbps
R=100kbps
A=0.0060
A=0.0191
A=0.0
Homework # 9
11.
Offered Traffic = 63.9; Carried Traffic = 63. 9x 0.98 =66.62 Erl
Consumption (Erl) by a single user = holding time(hour) x number of calls/hour
= (2/60) x 2 = 0.0667 Erl
Total Number of Users possible
= Carried Traffic/traffic generated b
Solutions to HW # 6
Ans 33)
(a) The BER of the QPSK scheme is the same as the BER of BPSK. The symbol
error rate of the QPSK is twice the BER, equal to 2 104 .
(b) BWQPSK = 0.5BWBPSK = 30 kHz .
(c) Spectral efficiency of QPSK is twice the spectral efficie
Ans 1)
Ans 12)
Ans 17)
=0.75
Ans 24)
Ans 26)
26. The phases are based on the following phase constellation
11
1 1
1 1 /4
1 1 3/4
1 1 5/4
1 1 7/4
1 1
1 1
1
I
0
Q
Phases 7/4
Phase shifts
0
I
1
Q
3/4
1
I
/2
1
Q
/4
/2
0
I
1
Q
3/4