Solution of Midterm Examination
EE5131/6131 (Semester I, 12/13)
1 [10 marks]
(a) For BPSK, Pb =
1
.
4b
To guarantee the required BER performance, b =
1
4Pb
=
250.
[1 mark]
Therefore, the required received power is
Pr = N0 B b = 7.5 1010 mW.
[1 mark]
Since
Continuous Assessment 4
EE6131 Advanced Wireless Communications
Semester I, 12/13
CA 4 contains 6 questions with 50 marks in total. Please answer all the questions.
Please submit your answers on November 19, 2012 (Monday) in class.
1. [5 marks] Consider a
Continuous Assessment 3
EE6131 Advanced Wireless Communications
Semester I, 12/13
CA 3 contains 4 questions with 50 marks in total. Please answer all the questions.
Please submit your answers on October 22, 2012 (Monday) in class.
1. [14 marks] Consider t
Solution of Continuous Assessment 3
EE6131 (Semester I, 12/13)
1. [14 marks]
(a) Since h is obtained by a linear transformation of the CSCG random vector hw ,
h is CSCG distributed.
Furthermore, we have
1
E[h] = E[Rr /2 hw ] = R1/2 E[hw ] = R1/2 0 = 0
r
r
Solution of Continuous Assessment 1
EE5131/EE6131 (Semester I, 12/13)
2-7 [4 marks]
The delay spread is 0.022 106 sec. It thus follows that
x+x l
162 + d2 d2
=
= 0.022 106
c
c
d = 16.1m.
Moreover, =
c
fc
= 1 m. As a result,
3
Gl
1 =
=
= 0.0016,
4 l
4d
G
Continuous Assessment 1
EE6131 Advanced Wireless Communications
Semester I, 12/13
CA 1 contains 6 questions with 50 marks in total. Please answer all the questions.
Please submit your answers on September 10, 2012 (Monday) in class.
1. [4 marks] Textbook
Solution of Continuous Assessment 1
EE5131/EE6131 (Semester I, 12/13)
2-7 [4 marks]
The delay spread is 0.022 106 sec. It thus follows that
x+x l
162 + d2 d2
=
= 0.022 106
c
c
d = 16.1m.
Moreover, =
c
fc
= 1 m. As a result,
3
Gl
1 =
=
= 0.0016,
4 l
4d
G