Faculty of Information and
Engineering Technology
Dr. Tallal Elshabrawy
Bar Code
Wireless Communication (NETW 701)
Winter 2014
Final Examination
Jan. 5th 2015
Please read carefully before proceeding.
1. The duration of this exam is 3 hours
2. Only calculators are permitted for this exam
3. This exam booklet contains 22 pages including this one. Two extra sheets of
scratch papers are attached
Good Luck!
Marks:
Problem
Number
1
2
3
4
5
Total
Possible
Marks
Final
Marks
20
15
20
25
15
95
Problem 1:
a) In the following, choose the correct answer:
(10 Marks)
1- The main assumption in Ricean Fading Channels is
i.
ii.
iii.
iv.
Multiple waves arriving with NO line of sight component
Rayleigh Channel IN ADDITION TO a line of sight component
ONLY line of sight component
None of the above
2- The equation describing the instantaneous received power at a distance ‘d’ due to
large-scale path-loss (distance path-loss and log-normal shadowing) as well as flat
small-scale fading Rayleigh channel could be written as:
𝑋
𝑑0 𝑛
𝑃𝑟 (𝑑) = 𝑃𝑟 (𝑑0 ) ( ) × 1010 × 𝜌2
𝑑
The random variables 𝑋 and 𝜌 follow:
i.
ii.
iii.
iv.
𝑿
𝑋
𝑋
𝑋
Gaussian Distribution and 𝝆 Rayleigh Distribution
Gaussian Distribution and 𝜌 Log-Rayleigh Distribution
Log-Normal Distribution and 𝜌 Rayleigh Distribution
Log-Normal Distribution and 𝜌 Log-Rayleigh Distribution
3- Suppose a communication system where the signal BW is BS=1 MHz and the delay
spread is σΤ=1 μs. Given coherence bandwidth is defined as BC=1/5σΤ, which
access techniques is most suitable for this system
i.
ii.
iii.
iv.
CDMA
FDMA
OFDM
None of the above
4- Maximal length sequence codes at T=0 (i.e., when perfectly synchronized) have
i. The same cross-correlation properties as WALSH codes
ii. Better cross-correlation properties compared to WALSH codes
iii. Worse cross-correlation properties compared to GOLD codes
iv. Better cross-correlation properties compared to GOLD codes
5- For a primitive polynomial with degree 7, each pseudorandom sequence has
i.
ii.
iii.
iv.
63 zeros and 64 ones
64 zeros and 63 ones
3 zeros and 4 ones
4 zeros and 3 ones
2
6- After CDMA signal dispreading
i. The bandwidth of the desired signal is larger than the bandwidth of the
interference signal
ii. The bandwidth of the desired signal is smaller than the bandwidth of
the interference signal
iii. The bandwidth of the desired signal is equal to than the bandwidth of the
interference signal
iv. None of the above
7- The TOTAL system data rates of TDMA, FDMA, CDMA generally have the following
relationship
i.
ii.
iii.
iv.
RCDMA> RTDMA> RFDMA
RCDMA=RTDMA= RFDMA
RFDMA> RTDMA> RCDMA
RCDMA> RFDMA> RTDMA
8- As the spreading factor of a CDMA code assigned to a given user increases
i. The user data rate increases
ii. The user data rate decreases
iii. The user data rate may increase and may decrease depending on the assigned
code
iv. The user data rate does not change
9- Coherent detection (i.e. with phase recovery) of BPSK over a Rayleigh channel with
instantaneous amplitude 𝛼 has a BER performance of:
i. 0.5
ii. 𝑄(√2𝐸𝑏 /𝑁0 )
iii. 𝑄(√2𝛼𝐸𝑏 /𝑁0 )
iv. 𝑸(𝜶√𝟐𝑬𝒃 /𝑵𝟎 )
10Which of the following frequency separations is considered to satisfy the criteria
for a valid OFDM system with DFT assuming a symbol duration of 𝑇𝑆
i. 1/2𝑇𝑠
ii. 3/2𝑇𝑠
iii. 𝟐/𝑻𝒔
iv. None of the above
3
b) If the average received power (Pr(d) = PT – PL(d)) is -59 dBm at d=1.5 Km taking
into account the shadowing effect having a Gaussian Distribution with zero mean
and standard deviation σ = 10dB. Find the percentage of area within a 1.5 Km
radius cell that receives signals strength greater than a value of -65dBm. (Assume
Path-loss exponent is 4)
(5 Marks)
𝑷𝒓 (𝒅) = ̅̅̅̅̅̅̅̅
𝑷𝒓(𝒅) + 𝑿𝝈
𝑷𝒓 (𝟏. 𝟓 𝒌𝒎) = ̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅̅
𝑷𝒓(𝟏. 𝟓 𝒌𝒎) + 𝑿𝝈
𝑷𝒓 (𝟏. 𝟓 𝒌𝒎) = −𝟓𝟗 𝒅𝑩𝒎 + 𝑿𝝈
𝑷𝒓[𝑷𝒓 (𝟏. 𝟓 𝒌𝒎) > −𝟔𝟓 𝒅𝑩𝒎] = 𝑷𝒓[−𝟓𝟗 𝒅𝑩𝒎 + 𝑿𝝈 > −𝟔𝟓 𝒅𝑩𝒎]
𝑷𝒓[𝑷𝒓 (𝟏. 𝟓 𝒌𝒎) > −𝟔𝟓 𝒅𝑩𝒎] = 𝑷𝒓[𝑿𝝈 > −𝟔 𝒅𝑩]
𝑷𝒓[𝑷𝒓 (𝟏. 𝟓 𝒌𝒎) > −𝟔𝟓 𝒅𝑩𝒎] = 𝑸(−𝟎. 𝟔)
𝑷𝒓[𝑷𝒓 (𝟏. 𝟓 𝒌𝒎) > −𝟔𝟓 𝒅𝑩𝒎] = 𝟏 − 𝑸(𝟎. 𝟔)
𝑭𝒓𝒐𝒎 𝑸 − 𝑻𝒂𝒃𝒍𝒆𝒔 𝑸(𝟎. 𝟔) = 𝟎. 𝟐𝟕𝟒𝟐𝟓
𝑷𝒓[𝑷𝒓 (𝟏. 𝟓 𝒌𝒎) > −𝟔𝟓 𝒅𝑩𝒎] = 𝟏 − 𝟎. 𝟐𝟕𝟒𝟐𝟓 = 𝟎. 𝟕𝟐𝟓𝟕𝟓
𝑭𝒓𝒐𝒎 𝑪𝒖𝒓𝒗𝒆𝒔
𝑷𝒆𝒓𝒄𝒆𝒏𝒕𝒂𝒈𝒆 𝒐𝒇 𝑪𝒐𝒗𝒆𝒓𝒂𝒈𝒆 = 𝟗𝟐 %
4
c) A hexagonal cell within a four-cell system has a radius of 1.387 km. A total of 60
channels are used within the entire system. If the load per user is 0.029 Erlangs,
and λ = 1 call/hour, compute the following for an Erlang C system that has a 5%
probability of a delayed call:
(5 Marks)
i. How many users per square kilometer will this system support
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑪𝒉𝒂𝒏𝒏𝒆𝒍𝒔 𝒑𝒆𝒓 𝑪𝒆𝒍𝒍 = 𝟏𝟓
𝑭𝒓𝒐𝒎 𝑬𝒓𝒍𝒂𝒏𝒈 𝑪 𝒄𝒖𝒓𝒗𝒆𝒔 𝒇𝒐𝒓 𝟓% 𝒅𝒆𝒍𝒂𝒚 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚
𝑻𝒉𝒆 𝒍𝒐𝒂𝒅 𝒑𝒆𝒓 𝒄𝒆𝒍𝒍 𝒊𝒔 𝑨 = 𝟗 𝑬𝒓𝒍𝒂𝒏𝒈
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒖𝒔𝒆𝒓𝒔 𝒑𝒆𝒓 𝒄𝒆𝒍𝒍 𝑼 𝒊𝒔
𝑼=
𝑨
𝟗
=
= 𝟑𝟏𝟎. 𝟑𝟒𝟒 = 𝟑𝟏𝟎
𝑨𝑼 𝟎. 𝟎𝟐𝟗
𝑪𝒆𝒍𝒍 𝑨𝒓𝒆𝒂 = 𝟐. 𝟓𝟗𝟖𝟏𝑹𝟐 = 𝟓 𝑲𝒎𝟐
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑼𝒔𝒆𝒓𝒔 𝒑𝒆𝒓 𝑲𝒎𝟐 = 𝟔𝟐
ii. What is the probability that a delayed call will have to wait for more than 10 s?
𝑷𝒓[𝒅𝒆𝒍𝒂𝒚 > 𝒕|𝒅𝒆𝒍𝒂𝒚 > 𝟎] = 𝒆
−(
𝑪−𝑨
)𝒕
𝑯
−(
𝟏𝟓−𝟗
)×𝟏𝟎
𝟏𝟎𝟒.𝟒
𝑨𝑼 = 𝝀𝑯
𝑯 = 𝟎. 𝟎𝟐𝟗 ∗ 𝟑𝟔𝟎𝟎 = 𝟏𝟎𝟒. 𝟒 𝒔
𝑷𝒓[𝒅𝒆𝒍𝒂𝒚 > 𝒕|𝒅𝒆𝒍𝒂𝒚 > 𝟎] = 𝒆
= 𝟎. 𝟓𝟔𝟐𝟖
iii. What is the probability that a call will be delayed for more than 10 seconds?
𝑷𝒓[𝒅𝒆𝒍𝒂𝒚 > 𝒕] = 𝑷𝒓[𝒅𝒆𝒍𝒂𝒚 > 𝒕|𝒅𝒆𝒍𝒂𝒚 > 𝟎]𝑷𝒓[𝒅𝒆𝒍𝒂𝒚 > 𝟎]
𝑷𝒓[𝒅𝒆𝒍𝒂𝒚 > 𝟏𝟎] = 𝟎. 𝟓𝟔𝟐𝟖 × 𝟎. 𝟎𝟓 = 𝟎. 𝟎𝟐𝟖𝟏𝟒
5
Problem 2:
Given a corridor as shown below where the transmitter is at ht=3m and the receiver is
at hr=1.6m, the carrier frequency is 2 GHz. The walls and the ground behave as
perfect reflectors.
1m
1m
1m
ht =3m
1m
hr =1.6m
12m
Derive an exact expression for the total amount of received power at the receiver
while considering the line-of-sight ray, ray reflected from ground and rays that are
reflected once from walls before reaching the receiver. (Assume Normal Incidence and
Reflection Coefficient = -1 for ALL Reflections). Calculate the exact value of received
power if PT=1mW and GT = GR = 0dB.
(15 Marks)
6
7
8
Problem 3:
A vehicle receives a 2GHz transmission while traveling at a constant velocity of 60
Km/hr. The large-scale pathloss is adjusted to be at an SNR level of 18 dB. To cope
with the dynamic behavior of small-scale fading, adaptive modulation and coding is
adopted in such system as per the table below:
(20 Marks)
Modulation Modulation in Channel Coding b/s/Hz
Bits/Symbol
Rate (C)
(M*C)
(M)
QPSK
2
1/2
1
QPSK
2
3/4
1.5
16 QAM
4
1/2
2
16 QAM
4
3/4
3
SNR Range
6 dB -8.5 dB
8.5dB-11.5 dB
11.5dB-15 dB
>15
i. Calculate the average bit rate in b/s/Hz used by the transmitter. Assume the
transmitter is capable of perfectly estimating the received SNR level at the
receiver and that for SNR<6dB the transmitter becomes silent.
𝑭𝒐𝒓 𝟏 𝒃/𝒔/𝑯𝒛: − 𝟏𝟐𝒅𝑩 < 𝝆𝟐 < −𝟗. 𝟓 𝒅𝑩 ⇒ 𝟎. 𝟎𝟔𝟑 < 𝝆𝟐 < 𝟎. 𝟏𝟏𝟐
𝑭𝒐𝒓 𝟏. 𝟓 𝒃/𝒔/𝑯𝒛: − 𝟗. 𝟓𝒅𝑩 < 𝝆𝟐 < −𝟔. 𝟓 𝒅𝑩 ⇒ 𝟎. 𝟏𝟏𝟐 < 𝝆𝟐 < 𝟎. 𝟐𝟐𝟒
𝑭𝒐𝒓 𝟐 𝒃/𝒔/𝑯𝒛: − 𝟔. 𝟓𝒅𝑩 < 𝝆𝟐 < −𝟑 𝒅𝑩 ⇒ 𝟎. 𝟐𝟐𝟒 < 𝝆𝟐 < 𝟎. 𝟓𝟎𝟏
𝑭𝒐𝒓 𝟑 𝒃/𝒔/𝑯𝒛: 𝝆𝟐 > −𝟑 𝒅𝑩 ⇒ 𝝆𝟐 > 𝟎. 𝟓𝟎𝟏
𝟐
𝑷𝒓[𝒓𝒆𝒄𝒆𝒊𝒗𝒆𝒅 𝒍𝒆𝒗𝒆𝒍 < 𝝆𝟐 ] = 𝟏 − 𝒆−𝝆
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟎] = 𝟎. 𝟎𝟔𝟏
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟏 𝒃/𝒔/𝑯𝒛] = 𝟎. 𝟎𝟒𝟓
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟏. 𝟓 𝒃/𝒔/𝑯𝒛] = 𝟎. 𝟎𝟗𝟓
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟐 𝒃/𝒔/𝑯𝒛] = 𝟎. 𝟏𝟗𝟑
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟑 𝒃/𝒔/𝑯𝒛] = 𝟎. 𝟔𝟎𝟔
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑩𝒊𝒕 𝑹𝒂𝒕𝒆 = 𝟎 × 𝟎. 𝟎𝟔𝟏 + 𝟏 × 𝟎. 𝟎𝟒𝟓 + 𝟏. 𝟓 × 𝟎. 𝟎𝟗𝟓 + 𝟐 × 𝟎. 𝟏𝟗𝟑 + 𝟑 × 𝟎. 𝟔𝟎𝟔 = 𝟐. 𝟑𝟗𝟏𝟓 𝒃/𝒔/𝑯𝒛
9
ii. Calculate the mean duration over which the transmitter is able to transmit at a
rate that is equal to 3 b/s/Hz.
𝟐
𝝉̅ =
𝒆𝝆 − 𝟏
√𝟐𝝅𝒇𝒎 𝝆
𝒇𝒎 =
𝒗 𝟔𝟎 × 𝟏𝟎𝟎𝟎
𝟏
=
×
= 𝟏𝟏𝟏. 𝟏𝟏 𝑯𝒛
𝝀
𝟑𝟔𝟎𝟎
𝟎. 𝟏𝟓
𝝆𝟐 = 𝟎. 𝟓𝟎𝟏
𝝉̅ = 𝟑. 𝟑 𝒎𝒔
𝑭𝒐𝒓 𝒂 𝒓𝒂𝒕𝒆 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 𝟑 𝒃/𝒔/𝑯𝒛
𝟐
𝑵𝑹 = √𝟐𝝅𝒇𝒎 𝝆𝒆−𝝆 = 𝟏𝟏𝟗. 𝟒𝟓
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝒕𝒊𝒎𝒆 𝒓𝒂𝒕𝒆 𝒆𝒒𝒖𝒂𝒍 𝒕𝒐 𝟑 𝒃/𝒔/𝑯𝒛
𝑻=
𝟏
𝟏
− 𝝉̅ =
− 𝟎. 𝟎𝟎𝟑𝟑 = 𝟓. 𝟎𝟕 𝒎𝒔
𝑵𝑹
𝟏𝟏𝟗. 𝟒𝟓
10
iii. Assume each user could be allocated M independent frequency channels where
selection diversity is utilized. What is the value of minimum M required to
achieve an average rate of 2.7 b/s/Hz or better.
𝟐 𝑴
𝑷𝒓[𝒓𝒆𝒄𝒆𝒊𝒗𝒆𝒅 𝒍𝒆𝒗𝒆𝒍 < 𝝆𝟐 ] = [𝟏 − 𝒆−𝝆 ]
𝑴=𝟐
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟎] = 𝟎. 𝟎𝟎𝟑𝟕𝟓
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟏 𝒃/𝒔/𝑯𝒛] = 𝟎. 𝟎𝟎𝟕𝟓
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟏. 𝟓 𝒃/𝒔/𝑯𝒛] = 𝟎. 𝟎𝟐𝟗
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟐 𝒃/𝒔/𝑯𝒛] = 𝟎. 𝟏𝟏𝟓
𝑷𝒓[𝒓𝒂𝒕𝒆 = 𝟑 𝒃/𝒔/𝑯𝒛] = 𝟎. 𝟖𝟒𝟒𝟕𝟓
𝑨𝒗𝒆𝒓𝒂𝒈𝒆 𝑩𝒊𝒕 𝑹𝒂𝒕𝒆 = 𝟎 × 𝟎. 𝟎𝟎𝟑𝟕𝟓 + 𝟏 × 𝟎. 𝟎𝟎𝟕𝟓 + 𝟏. 𝟓 × 𝟎. 𝟎𝟐𝟗 + 𝟐 × 𝟎. 𝟏𝟏𝟓 + 𝟑 × 𝟎. 𝟖𝟒𝟒𝟕𝟓
= 𝟐. 𝟖𝟏𝟓𝟑 𝒃/𝒔/𝑯𝒛 > 𝟐. 𝟕
𝑻𝒉𝒆𝒓𝒆𝒇𝒐𝒓𝒆 𝑴 = 𝟐
11
Problem 4:
a) Consider WALSH codes in Orthogonal Variable Spreading Factor (OVSF). Let c41=[1
1 -1 -1], c42=[1 -1 1 -1] be two codes with a spreading factor of 4 that are
allocated to users within a certain cell.
(15 Marks)
i. How many codes with spreading factor 64 remain for allocation in such system
𝟑𝟐
ii. How many codes with spreading factor 2 remain for allocation in such system
𝒁𝒆𝒓𝒐
iii. Plot the cross correlation function between c41 and c42
Τ
12
iv. Plot the auto correlation function for c41 and c42
Τ
-4ΤC -3ΤC -2ΤC -ΤC
ΤC
2ΤC
3ΤC
Τ
4ΤC
-4ΤC -3ΤC -2ΤC -ΤC
ΤC
2ΤC
3ΤC
4ΤC
v. From the plot in part iv, determine the regions of shifts where c41 has better
auto-correlation properties (Explain your answer)
𝑻𝒉𝒆 𝒄𝒐𝒎𝒑𝒂𝒓𝒊𝒔𝒐𝒏 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝒕𝒉𝒆 𝒂𝒖𝒕𝒐 𝒄𝒐𝒓𝒓𝒆𝒍𝒂𝒕𝒊𝒐𝒏 𝒑𝒆𝒓𝒇𝒐𝒓𝒎𝒂𝒏𝒄𝒆 𝒔𝒉𝒐𝒖𝒍𝒅 𝒃𝒆 𝒃𝒂𝒔𝒆𝒅 𝒐𝒏 𝒕𝒉𝒆 𝒂𝒃𝒔𝒐𝒍𝒖𝒕𝒆 𝒗𝒂𝒍𝒖𝒆
𝟐
𝟑
ΤC
𝟒
𝟑
ΤC
Τ
-4ΤC
-3ΤC
-2ΤC
𝑪𝟒𝟐 𝒃𝒆𝒕𝒕𝒆𝒓 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝟎 𝒂𝒏𝒅
𝑪𝟒𝟏 𝒃𝒆𝒕𝒕𝒆𝒓 𝒃𝒆𝒕𝒘𝒆𝒆𝒏
-ΤC
ΤC
2ΤC
3ΤC
4ΤC
𝟐
𝑻
𝟑 𝑪
𝟐
𝟒
𝑻 𝒂𝒏𝒅 𝑻𝑪
𝟑 𝑪
𝟑
13
vi. Suppose the transmission of a binary-1 bit using c41 over a flat channel with
coherence time of 0.3255 microsec. What is the received signal value after the
de-spreading (i.e., multiplying by the code of c1 and integration) assuming that
the channel changes from amplitude of 1 to amplitude of 0.5 back to amplitude
of 1 every coherence time period. (Assume that channel starts at an amplitude
of 1 and the chip rate is 4.096 Mchips/sec.
𝑻𝑪 = 𝟎. 𝟑𝟐𝟓𝟓 𝝁𝒔
𝑻𝑪𝒉𝒊𝒑 = 𝟎. 𝟐𝟒𝟒𝟏𝟒 𝝁𝒔
𝑻𝑪 = 𝟏. 𝟑𝟑𝟑𝟑 𝑻𝑪𝒉𝒊𝒑
𝑹𝒙 𝑺𝒊𝒈𝒏𝒂𝒍 = 𝟏. 𝟑𝟑 ∗ 𝟏 + 𝟏. 𝟑𝟑 ∗ 𝟎. 𝟓 + 𝟏. 𝟑𝟑 ∗ 𝟏 = 𝟑. 𝟑𝟐𝟓
14
b) Calculate the SIR at the cell edge for a CDMA cellular network given the following
conditions:
(10 Marks)
-
The system bandwidth is 5 MHz
The chip rate is 4.096×106 chips/sec.
Each voice call requires 2 Kbps and is modulated by BPSK.
The cellular network should support a city of area = 20 Km2.
The cell radius is 250 m
The number of subscribers is 1000000.
The load per user is 0.005 Erlang
Probability of a new call being blocked should not exceed 2 % given that calls
are blocked when all available codes are assigned to active users
Reuse of 1 and first tier of interferers are only considered.
The pathloss exponent in the whole network is 3.5
-
𝑺𝑰𝑹 =
𝑮
(𝑲 − 𝟏 ) + 𝟐𝑲[(𝑸 − 𝟏)−𝒏 + (𝑸 + 𝟏)−𝒏 + 𝑸−𝒏 ]
𝑮 = 𝟐𝟎𝟒𝟖
𝑪𝒆𝒍𝒍 𝑨𝒓𝒆𝒂 = 𝟐. 𝟓𝟗𝟖𝟏 ∗ 𝑹𝟐 = 𝟐. 𝟓𝟗𝟖𝟏 ∗ (𝟎. 𝟐𝟓)𝟐 = 𝟎. 𝟏𝟔𝟐𝟒 𝑲𝒎𝟐
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑪𝒆𝒍𝒍𝒔 = 𝟐𝟎/𝟎. 𝟏𝟔𝟐𝟒 = 𝟏𝟐𝟑 𝑪𝒆𝒍𝒍𝒔
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑺𝒖𝒃𝒔𝒄𝒓𝒊𝒃𝒆𝒓𝒔 𝒑𝒆𝒓 𝑪𝒆𝒍𝒍 = 𝟏, 𝟎𝟎𝟎, 𝟎𝟎𝟎/𝟏𝟐𝟑 = 𝟖𝟏𝟑𝟎
𝑳𝒐𝒂𝒅 𝒑𝒆𝒓 𝑪𝒆𝒍𝒍 𝑨 = 𝟎. 𝟎𝟎𝟓 ∗ 𝟖𝟏𝟑𝟎 = 𝟒𝟎. 𝟔𝟓 𝑬𝒓𝒍𝒂𝒏𝒈
𝑭𝒓𝒐𝒎 𝑬𝒓𝒍𝒂𝒏𝒈 𝑩 𝑪𝒖𝒓𝒗𝒆𝒔 𝒂𝒏𝒅 𝒇𝒐𝒓 𝟐% 𝑩𝒍𝒐𝒄𝒌𝒊𝒏𝒈 𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚
𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝑻𝒓𝒖𝒏𝒌𝒔 𝑲 = 𝟓𝟎
𝑸 = √𝟑
𝟐𝟎𝟒𝟖
𝑺𝑰𝑹 =
𝟒𝟗 + 𝟏𝟎𝟎 [(√𝟑 − 𝟏)
−𝟑.𝟓
−𝟑.𝟓
+ (√𝟑 + 𝟏)
+ √𝟑
−𝟑.𝟓
= 𝟓. 𝟔𝟏𝟖𝟒 = 𝟕. 𝟒𝟗 𝒅𝑩
]
15
Problem 5:
Consider an FDMA indoor cellular network applied within an office building as shown
below where each floor is 3m high. The building is to be covered by multiple cells with
the base station installed in center of the ceiling and receivers are also located in the
center within a given floor at a height of 1m from the ground as shown. Each cell is
supposed to cover one floor or more. Frequency reuse is applied such that frequencies
are reused every N floors. If the reuse factor N is odd then the cell is composed of the
center cell (where the base station is located) and equal number of floors from above
and below the center cell (i.e., if N=3 then the cell is composed of central floor i along
with the floor above i+1 and floor below i-1). If the reuse factor is even then the cell
is composed of the center cell (where the base station is located) and N/2 floors from
above and N/2-1 from below (i.e., if N=4 then the cell is composed of central floor i
along with the floors above i+1,i+2 and floor below i-1) Suppose that co-channel
interference is only perceived from the two nearest co-channel cells as shown in the
figure. The attenuation through a given floor constitutes a loss of L dB. The path loss
exponent is n
(15 Marks)
3m
N Floors
Frequency Reuse
floor i+1
3m
Rx
1m
floor i
Rx
3m
1m
floor i-1
N Floors
Frequency Reuse
3m
Rx
1m
3m
16
i. Derive a formula for the worst case SIR as a function of the reuse factor N and
floor loss L along with the path loss exponent n (Assume path loss follows the
𝑑
equation 𝑃𝐿(𝑑) = 𝑃𝐿(𝑑0 ) + 10𝑛𝑙𝑜𝑔 ( ) + ∑𝐾
𝑘=1 𝐿 where 𝐾 is the number of floors the
𝑑0
signal must penetrate between the transmitter and a given receiver)
𝑭𝒐𝒓 𝑶𝒅𝒅 𝑵
𝑾𝒐𝒓𝒔𝒕 𝑪𝒂𝒔𝒆 𝑺𝒊𝒈𝒏𝒂𝒍 𝒊𝒔 𝒂𝒕 𝒂 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝟐 +
𝑾𝒐𝒓𝒔𝒕 𝑪𝒂𝒔𝒆 𝑺𝒊𝒈𝒏𝒂𝒍 𝒑𝒆𝒏𝒆𝒕𝒓𝒂𝒕𝒆𝒔
(𝑵 − 𝟏)
×𝟑
𝟐
(𝑵 − 𝟏)
𝒘𝒂𝒍𝒍𝒔
𝟐
𝑵+𝟏
𝟑𝑵 − 𝟏
𝑰𝒏𝒕𝒆𝒓𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑨𝒃𝒐𝒗𝒆 𝑪𝒆𝒍𝒍 𝒊𝒔 𝒂𝒕 𝒂 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝟐 + (
+ 𝑵 − 𝟏) × 𝟑 = 𝟐 + (
)×𝟑
𝟐
𝟐
𝟑𝑵 − 𝟏
𝑰𝒏𝒕𝒆𝒓𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑨𝒃𝒐𝒗𝒆 𝑪𝒆𝒍𝒍 𝒑𝒆𝒏𝒆𝒕𝒓𝒂𝒕𝒆𝒔 (
) 𝒘𝒂𝒍𝒍𝒔
𝟐
𝑵−𝟏
𝑰𝒏𝒕𝒆𝒓𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑩𝒆𝒍𝒐𝒘 𝑪𝒆𝒍𝒍 𝒊𝒔 𝒂𝒕 𝒂 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝟏 + (
)×𝟑
𝟐
𝑰𝒏𝒕𝒆𝒓𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑩𝒆𝒍𝒐𝒘 𝑪𝒆𝒍𝒍 𝒑𝒆𝒏𝒆𝒕𝒓𝒂𝒕𝒆𝒔
𝑷𝒓(𝒅𝟎 ) (
𝑺𝑰𝑹 =
𝑷𝒓(𝒅𝟎 ) (
𝟐+(
𝟐+
(𝑵 + 𝟏)
𝒘𝒂𝒍𝒍𝒔
𝟐
−𝒏
(𝑵 − 𝟏)
(𝑵−𝟏)
×𝟑
𝟐
) × 𝑳− 𝟐
𝒅𝟎
−𝒏
−𝒏
𝟑𝑵 − 𝟏
𝑵−𝟏
)×𝟑
𝟏
+
(
)
×
𝟑
(𝑵+𝟏)
𝟑𝑵−𝟏
𝟐
𝟐
) × 𝑳−( 𝟐 ) + 𝑷𝒓(𝒅𝟎 ) (
) × 𝑳− 𝟐
𝒅𝟎
𝒅𝟎
−𝒏
(𝑵−𝟏)
(𝑵 − 𝟏)
× 𝟑) × 𝑳− 𝟐
𝟐
=
−𝒏
−𝒏
(𝑵+𝟏)
𝟑𝑵−𝟏
𝟑𝑵 − 𝟏
𝑵−𝟏
−(
)
𝟐
(𝟐 + (
) × 𝟑) × 𝑳
+(𝟏+(
) × 𝟑) × 𝑳− 𝟐
𝟐
𝟐
(𝟐 +
𝑺𝑰𝑹𝒐𝒅𝒅
17
𝑭𝒐𝒓 𝑬𝒗𝒆𝒏 𝑵
𝑵
𝑾𝒐𝒓𝒔𝒕 𝑪𝒂𝒔𝒆 𝑺𝒊𝒈𝒏𝒂𝒍 𝒊𝒔 𝒂𝒕 𝒂 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝟐 + ( − 𝟏) × 𝟑
𝟐
𝑵
𝑾𝒐𝒓𝒔𝒕 𝑪𝒂𝒔𝒆 𝑺𝒊𝒈𝒏𝒂𝒍 𝒑𝒆𝒏𝒆𝒕𝒓𝒂𝒕𝒆𝒔 ( − 𝟏) 𝒘𝒂𝒍𝒍𝒔
𝟐
𝑵
𝟑𝑵 − 𝟐
𝑰𝒏𝒕𝒆𝒓𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑨𝒃𝒐𝒗𝒆 𝑪𝒆𝒍𝒍 𝒊𝒔 𝒂𝒕 𝒂 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝟐 + ( + 𝑵 − 𝟏) × 𝟑 = 𝟐 + (
)×𝟑
𝟐
𝟐
𝟑𝑵 − 𝟐
𝑰𝒏𝒕𝒆𝒓𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑨𝒃𝒐𝒗𝒆 𝑪𝒆𝒍𝒍 𝒑𝒆𝒏𝒆𝒕𝒓𝒂𝒕𝒆𝒔 (
) 𝒘𝒂𝒍𝒍𝒔
𝟐
𝑵
𝑰𝒏𝒕𝒆𝒓𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑩𝒆𝒍𝒐𝒘 𝑪𝒆𝒍𝒍 𝒊𝒔 𝒂𝒕 𝒂 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 = 𝟏 + ( ) × 𝟑
𝟐
𝑰𝒏𝒕𝒆𝒓𝒇𝒆𝒓𝒆𝒏𝒄𝒆 𝒇𝒓𝒐𝒎 𝑩𝒆𝒍𝒐𝒘 𝑪𝒆𝒍𝒍 𝒑𝒆𝒏𝒆𝒕𝒓𝒂𝒕𝒆𝒔
𝑵
+ 𝟏 𝒘𝒂𝒍𝒍𝒔
𝟐
−𝒏
𝑵
𝟐 + ( − 𝟏) × 𝟑
𝑵
𝟐
𝑷𝒓(𝒅𝟎 ) (
) × 𝑳− ( 𝟐 −𝟏)
𝒅𝟎
𝑺𝑰𝑹 =
𝑷𝒓(𝒅𝟎 )
𝑺𝑰𝑹𝒆𝒗𝒆𝒏
(𝟐 + (
−𝒏
−𝒏
𝟑𝑵 − 𝟐
𝑵
) × 𝟑)
𝟏+( )×𝟑
𝟑𝑵−𝟐
𝑵
−(
)
𝟐
𝟐
𝟐
×𝑳
+ 𝑷𝒓(𝒅𝟎 ) (
) × 𝑳−( 𝟐 +𝟏)
𝒅𝟎
𝒅𝟎
−𝒏
𝑵
𝑵
( 𝟐 + ( − 𝟏) × 𝟑) × 𝑳−( 𝟐 −𝟏)
𝟐
=
−𝒏
−𝒏
𝟑𝑵−𝟐
𝑵
𝟑𝑵 − 𝟐
𝑵
(𝟐 + (
) × 𝟑) × 𝑳− ( 𝟐 ) + ( 𝟏 + ( ) × 𝟑) × 𝑳−( 𝟐 +𝟏)
𝟐
𝟐
18
ii. Calculate the reuse factor required to guarantee a worst case SIR of 8 dB
(pathloss exponent 𝑛 = 2.5 and 𝐿 = 6 𝑑𝐵)
𝑳≈𝟒
𝐅𝐨𝐫 𝑵 = 𝟏
𝑺𝑰𝑹𝟏 =
(𝟐)−𝟐.𝟓
= 𝟎. 𝟔𝟗𝟓 = −𝟏. 𝟓𝟖 𝒅𝑩
(𝟐 + 𝟑)−𝟐.𝟓 × 𝟒−𝟏 + ( 𝟏)−𝟐.𝟓 × 𝟒−𝟏
𝐅𝐨𝐫 𝑵 = 𝟐
𝑺𝑰𝑹𝟐 =
(𝟐)−𝟐.𝟓
= 𝟕𝟔. 𝟗 = 𝟏𝟖. 𝟖𝟔 𝒅𝑩
(𝟐 + 𝟔)−𝟐.𝟓 × 𝟒−𝟐 + ( 𝟒)−𝟐.𝟓 × 𝟒−𝟐
𝐅𝐨𝐫 𝑵 = 𝟑
𝑺𝑰𝑹𝟑 =
(𝟓)−𝟐.𝟓 × 𝟒−𝟏
= 𝟐. 𝟐𝟖 = 𝟑. 𝟓𝟖 𝒅𝑩
(𝟐 + 𝟏𝟐)−𝟐.𝟓 × 𝟒−𝟒 + ( 𝟏 + 𝟑)−𝟐.𝟓 × 𝟒−𝟐
𝐅𝐨𝐫 𝑵 = 𝟒
𝑺𝑰𝑹𝟒 =
(𝟓)−𝟐.𝟓 × 𝟒−𝟏
= 𝟑𝟔. 𝟖𝟓 = 𝟏𝟓. 𝟔𝟔 𝒅𝑩
(𝟐 + 𝟏𝟓)−𝟐.𝟓 × 𝟒−𝟓 + ( 𝟏 + 𝟔)−𝟐.𝟓 × 𝟒−𝟑
19
20