ELEC-E7120 Wireless Systems
Exercise Session 2
Problem 1: (Link level performance)
The throughput in LTE can be approximated using the formula
R = NRB*WRB* A *log2 (1 + B*SNR)
(1)
Where NRB is the number of assigned Resource Blocks (RBs), W = 180 kHz is the bandwidth
of a single RB. A = 0.88 is a parameter that accounts the bandwidth efficiency of LTE, and
B=1/1.25 is a parameter that models the SNR efficiency of LTE.
(a) If received SNR in the inner part of the cell is 10 dB, how many RBs are needed to
achieve a data rate of 2 Mbps for a given mobile user?
Solving (1) for NRB:
𝑅𝑅
�
𝑁𝑁𝑅𝑅𝑅𝑅 = �
𝑊𝑊𝑅𝑅𝑅𝑅 ∙ 𝐴𝐴 ∙ log 2 (1 + 𝐵𝐵 ∙ 𝑆𝑆𝑆𝑆𝑆𝑆)
𝑁𝑁𝑅𝑅𝑅𝑅 = �
180 ×
103
6
(2)
2 × 10
�=4
∙ 0.88 ∙ log 2 (1 + 0.8 ∙ 1010/10 )
(b) Aim is to provide a data rate of at least 700 kbps for each of the 20 mobile users that are
located on the cell edge. When LTE operates on a 20MHz bandwidth, there are 96 RBs
available for users in total (WRB = 180 kHz). What is the minimum SNR that users at the
cell edge should perceive to satisfy this requirement? Express the final result in dB.
Solving (1) for SNR:
𝑆𝑆𝑆𝑆𝑆𝑆 =
𝑆𝑆𝑆𝑆𝑆𝑆 =
20∙700×103
�
3 ∙0.88 �
96∙180×10
2
0.8
�
2
𝑅𝑅
�
𝑁𝑁𝑅𝑅𝑅𝑅 ∙𝑊𝑊𝑅𝑅𝑅𝑅 ∙𝐴𝐴
−1
𝐵𝐵
−1
= 1.1162 [linear] = 0.477 dB
(3)
Problem 2: (Channel capacity analysis)
What is the theoretical channel maximum data rate that can be supported in:
(a) In a 10 MHz channel for SNR = 13 dB and 4x4 MIMO configuration in a LTE System.
Using Shannon Shannon's channel capacity formula from the slides, the maximum
possible data rate is
R = B *log2 (1 + SNR)
(4)
and solving (4) for R
13
𝑅𝑅 = 10 × 106 ∙ log 2 �1 + 1010 � = 43.89 Mbps
(5)
𝑅𝑅𝑡𝑡𝑡𝑡𝑡𝑡 = 4 × 𝑅𝑅 = 175.5 Mbps
(6)
Since the system is using 4x4 MIMO configuration, the maximum channel data rate is:
(b) A WCDMA system that works in a 5 MHz channel has a received signal with SNR = 26
dB after despreading. The spreading factor of the spreading code is M = 16.
Using the formula for WCDMA:
R = B *log2 (1 + SNR/M)
(7)
Solving (7) for R:
26
1010
𝑅𝑅 = 5 × 10 ∙ log 2 �1 +
� = 23.47 Mbps
16
6
(8)
Problem 3: (Multiple Access Techniques)
Consider a TDMA/FDD communications system. The transmission is subdivided in TDMA time
slots. The information for a time slot has the following structure in bits (1 bit= 2 μs):
N. Bits
Information Element
100
Data & signaling bits
10
Guard Period
The guard period keeps the timeslot synchronized, it compensates for propagation delay due to
the unknown distance from the users to the base station.
The further away the user is from the base, the time slot will take more time to arrive at the base
station due to propagation delay. Eventually, there comes a certain point where the transmitted
slot will arrive so late that it would occur outside its designated timeslot and would interfere with
the next time slot allocated to another user.
a) Determine the maximum distance the user can be from the base station and still transmit
in the correct time slot.
User 1
Base
User 2
In order for the user to stay synchronized with the base station, it needs to receive the
synchronization signal during the guard period. So, the total distance is halved because of the
two way communication.
Using the equation for distance:
1
𝑑𝑑 = (3 × 108 m/s)(20 × 10−6 s) � � = 3000 m
2
(9)
Timing Advance is a technique where the base station can tell the users the time of advance
transmission to compensate for propagation delays. The base station examines the user
sequence and sees how long it arrived after the time that it expected. If the base station sees
that the time slot is late by a single bit, it knows the propagation delay is the bit time. This delay
in bits transforms into the value of Timing Advance.
b) For the same TDMA/FDD system, determine the Guard Period in bits for a maximum cell
radius of 25 km.
If the radius cell is 25 km, the total distance the information has to travel is twice the
radius. We also can figure out the propagation delay for 1 bit (3x108 m/s *2 μs/bit= 600
m/bit). Then
2(25 × 103 m)
� = 84 bits
Guard Period = �
(10)
600 m/bit