EE161 - Spring 2020 San Jose´ State University
Solution of Midterm Exam # 1
Problem 1 (25 points) A binary communication system uses polar mapping and NRZ pulses.
The bit rate is 100 Mbps.
(a) Three bits are sent through an AWGN channel. The correlator outputs are {−0.8, 1.2,−0.5}.
Determine the estimated bit sequence {Bˆk}.
Solution: There are two possible polar mappings. Consequently, either one of the following
two answers is considered correct: {Bˆk} = {0, 1, 0}, {Bˆk} = {1, 0, 1}.
(b) At the receiver the average AWGN power is σ2W = N0/2 = 2.5 × 10−10. Find the minimum
average received pulse amplitude such that the average probability of a bit error Pb < 10
−6.
(Note: You may use the tables on the last page.)
Solution: Using the table of the inverse Q-function:(
Eb
N0
)
min
=
1
2
[
Q−1
(
10−6
)]2
=
(4.76)2
2
= 11.3
It follows that Eb,min = (11.3) · 5× 10−10 = 5.7× 10−9. For NRZ pulses and polar mapping,
the pulse amplitude is given by
amin =
√
Eb,min
T
=
√
5.7× 10−9
10−8
= 0.75 V.
Problem 2 (35 points) Consider a 4-PAM wireless communication system using RZ pulses.
(a) Sketch carefully the signal associated to the bit sequence {0, 1, 0, 0, 1, 1}. You must show the
mapping of bits to symbols that you used in producing the sketch.
Solution: With the Gray mapping used in the class lectures, the signal is a sequence of three
RZ pulses (not shown) of amplitudes −a,−3a and +a.
(b) Determine the minimum signal energy-to-noise ratio Es/N0 that is required to obtain an
average probability of a bit error of at most Pb = 10
−5. You may use the following formula:
Pb = Q
(√
4Eb
5N0
)
.
Solution: In terms of the signal energy, the average bit error probability is approximately
10−5 = Q
(√
2Es
5N0
)
−→ Es
N0
=
5
2
[
Q−1
(
10−5
)]2
= 2.5(4.27)2 = 45.6 or 16.6 dB.
(c) If QPSK mapping is used then what is the minimum value of Es/N0? Find the difference
in dB between the minimum Es/N0 values in parts (b) and (c). What mapping gives the
smallest value?
Solution: For QPSK,
10−5 = Q
(√
Es
N0
)
−→ Es
N0
=
[
Q−1
(
10−5
)]2
= (4.27)2 = 18.2 or 12.6 dB.
This is 4 dB smaller compared to 4-PAM. QPSK mapping results in the smallest required
signal energy.
Problem 3 (40 points) Binary OOK modulation is used in a wireless link.
(a) Sketch the signal constellation used by the mapper and a detailed block diagram of the
receiver.
Solution: The signal constellation (sketch not shown) consists of two points along the line
ψ(t) of amplitudes s1 = 0 for B = 0 and s2 = a =
√
2Es for B = 1. A block diagram of
the receiver must include a down-conversion stage, a matched filter or correlator for an NRZ
pulse, and a decision device with rule
Bˆk =
{
0, Yk ≤ a/2,
1, Yk > a/2,
with a =
√
2Es.
(b) Sketch the waveform corresponding to the bit sequence {Bk} = {1, 0, 1, 0, 1}. You may assume
any carrier frequency value in your sketch.
Solution: In OOK modulation, baseband NRZ pulses are amplitude modulated by a sinu-
soidal signal. A possible sketch is shown in the figure below.
!"
#$!%"
&" '"&"'" &"
(c) The average bit energy-to-noise ratio is Eb/N0 = 15 dB. Evaluate the average probability of
a bit error Pb.
Solution: Eb/N0 = 10
15/10 = 31.6 and Pb = Q
(√
Eb
N0
)
= Q(5.6) = 10−8.
学霸联盟