ECE 6325 Fall 2009: Exam 2
NAME:
• Solve each problem completely, showing each step. Each successful step shown can be awarded
partial credit. For example, write down the definition before doing the math.
• Feel free to use extra sheets to answer any question. Include these extra sheets with the exam
booklet and label each sheet with your name.
• For clarity, copy your final answer to the box after the question. (Explanations do not need to be
copied.)
• There is no partial credit on multiple choice questions – so there’s no need to show your work.
• You will have 60 minutes for this exam.
• You may have your portfolio (written homework / example problem descriptions and work).
• THIS IS THE EXAM FOR 6325 STUDENTS.
#
1
Score
Out of
25
2
20
3
40
4
15
Total
100
2
ECE 6325 Fall 2009
1 × 10−2 =
2.3263
1.5 × 10−2
= 2.1701
−2
2 × 10 = 2.0537
3 × 10−2 = 1.8808
4 × 10−2 = 1.7507
5 × 10−2 = 1.6449
6 × 10−2 = 1.5548
7 × 10−2 = 1.4758
8 × 10−2 = 1.4051
9 × 10−2 = 1.3408
1 × 10−1 =
1.2816
1.5 × 10−1
= 1.0364
−1
2 × 10 = 0.84162
3 × 10−1 = 0.5244
4 × 10−1 = 0.25335
5 × 10−1 = 0
5
4.5
−1
Inverse Q function, Q (x)
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
Q−1
TABLE OF THE Q−1
(·) FUNCTION:
−1
−6
−4
1 × 10
=
Q
1 × 10
=
Q−1
4.7534
3.719
−1
−6
−4
1.5 × 10 = 4.6708 Q
1.5 × 10 = 3.6153 Q−1
2 × 10−6 = 4.6114
Q−1 2 × 10−4 = 3.5401
Q−1
−1
−6
−4
3 × 10 = 4.5264
Q
3 × 10 = 3.4316
Q−1
−1
−6
−4
4 × 10 = 4.4652
Q
4 × 10 = 3.3528
Q−1
5 × 10−6 = 4.4172
Q−1 5 × 10−4 = 3.2905
Q−1
−1
−6
−4
Q
Q−1
6 × 10 = 4.3776
6 × 10 = 3.2389
Q−1 7 × 10−4 = 3.1947
Q−1
7 × 10−6 = 4.3439
−1
−6
−4
8 × 10 = 4.3145
Q
8 × 10 = 3.1559
Q−1
−1
−6
−4
9 × 10 = 4.2884
Q
9 × 10 = 3.1214
Q−1
Q−1 1 × 10−3 =
Q−1
1 × 10−5 =
4.2649
3.0902
−1
−5
−3
1.5 × 10 = 4.1735 Q
1.5 × 10 = 2.9677 Q−1
2 × 10−5 = 4.1075
Q−1 2 × 10−3 = 2.8782
Q−1
−1
−5
−3
3 × 10 = 4.0128
Q
3 × 10 = 2.7478
Q−1
−1
−5
−3
4 × 10 = 3.9444
Q
4 × 10 = 2.6521
Q−1
5 × 10−5 = 3.8906
Q−1 5 × 10−3 = 2.5758
Q−1
−1
−5
−3
6 × 10 = 3.8461
Q
6 × 10 = 2.5121
−1
−5
7 × 10 = 3.8082
Q
7 × 10−3 = 2.4573
−1
−5
8 × 10 = 3.775
Q
8 × 10−3 = 2.4089
−1
−5
9 × 10
= 3.7455
Q
9 × 10−3 = 2.3656
4
3.5
3
2.5
2
1.5
1
0.5
0 −7
10
−6
10
−5
10
−4
10
−3
10
Value, x
−2
10
−1
10
0
10
ECE 6325 Fall 2009
3
Problems
1. (25 pts total) Compute the received power in the diffracted path shown in the above figure. The
BS has height 30 m, but the MS has height 1 m and is on the opposite side of a 30 m tall building.
The signal, at fc = 450 MHz, arrives via diffraction over the building only. Assume TX power of
10 dBW and antenna gains of 9 dBi at the BS and 1 dBi at the mobile.
ECE 6325 Fall 2009
4
2. (20 pts total) 6325 Short answer: Compare Rician (K > 0) and Rayleigh fading:
(a) What assumptions are made about multipath in Rician channels? What assumptions are made
about multipath in Rayleigh channels? (Two sentences max)
(b) Which, the Rician or the Raleigh channel, requires a larger fade margin? (Two sentences max)
ECE 6325 Fall 2009
5
3. (40 pts total) Assuming that a 30 kHz is available for a downlink channel in a new cellular system,
and SRRC pulse shaping with α = 0.30 will be used.
(a) What data rate (bits per second) can be achieved for OQPSK, binary coherent FSK, and 64QAM?
(b) For these three modulations, what SNR is required to achieve P [bit error] = 3 × 10−4 ?
ECE 6325 Fall 2009
6
4. (15 pts) Short answer: What is the benefit of OQPSK compared to QPSK? (Two sentences max)