Solved Communication Systems Assignment - Electrical Engineering

This document presents a comprehensive solution to a Communication Systems assignment, addressing key concepts such as Code Division Multiple Access (CDMA) and Orthogonal Frequency Division Multiplexing (OFDM). The solution includes detailed calculations and explanations for various aspects of CDMA, including the analysis of spreading codes, user bit sequences, and despread signals. It also covers the calculation of traffic load and the impact of sectoring. For OFDM, the solution addresses the Fast Fourier Transform (FFT) size, sub-carrier spacing, and symbol duration. The document also explores bit rate calculations for different modulation schemes, including BPSK, QPSK, and QAM, and analyzes the number of resource blocks and elements. This assignment provides a thorough understanding of the principles and applications of modern communication systems, making it a valuable resource for students studying electrical engineering.

Running head: COMMUNICATION SYSTEM 1
Communication Systems
Name
Institution

COMMUNICATION SYSTEMS 2
CDMA
Question 1
To test whether the spreading codes are orthogonal, we apply the dot product.
U1 U2= [ −1 1−11 1 ]∗ [−1−11−1−1 ] =−3
U2 U3= [ −1−1 1−1−1 ]∗[ 1−1 11−1 ] =1
U1 U3= [ −1 1−1 11 ]∗ [ 1−1 1 1−1 ] =−3
Therefore, the spreading codes are not orthogonal.
Question 2
User U1=[ −11−11 1 ]
Bit sequence b=[b0 , b1 ]=¿ [1 1]
The transmitted symbols will be= [ ~
b , b , ~
b , b , b ]=¿=[0,0,1,1,0,0,1,1,1,1]
User U2= [−1−11−1−1 ]
Bit sequence b=[b0 , b1 ]=¿ [1 1]
The transmitted symbols will be= [ ~
b , ~
b , b , ~
b , , ~
b ]=[ 0,0,0,0,1,1,0,0,0,0]
User U3= [ 1−11 1−1 ]
Bit sequence b=[b0 , b1 ]=¿ [1 -1]
The transmitted symbols will be= ¿

COMMUNICATION SYSTEMS 3
Question 3
The received signal ignoring noise is the same as the transmitted signal which will be
[0,0,1,1,0,0,1,1,1,1] for user U1
[0,0,0,0,1,1,0,0,0,0] for user U2
[1,−1 ,0 ,−1,1 ,−1,1 ,−1,0 ,−1] for user U3
Question 4
Despread signals for the three users are computed as follows:
User U1=[−11−11 1 ], Despread signal for user U1=[ 1,0 ,1,0,0]
User U2= [−1−11−1−1 ] Despread signal for user U2=[ 1,1,0,1,1]
User U3= [ 1−11 1−1 ] Despread signal for user U3=[0,1,0,0,1]
Question 5
The number of users, N=1+
W
R
Eb
Io
=1+ 128
6 =22.3333 ≅ 23 users
Question 6
Each subscriber generates 2 calls/day¿ 10× 2 min
day
The traffic load is
¿ 10× 2× 60
24 ×3600 = 1
72=0.013888889 Erlangs
subscriber

COMMUNICATION SYSTEMS 4
Number of subscribers is 0.013888889
0.01 ×23=31.9 ≅ 32 subscribers
Question 7
Sectoring ¿ 3
Including Voice activation of 60%=0.6 and 50% =0.5 of interference,
SIR is reduced by 0.6 × 0.5=0.3
New SIR=( 1−0.3 ) × 6+3=7.2 dB
The number of users, N=1+
W
R
Eb
Io
=1+ 128
7.2 =18.77778 ≅ 19 users
Given the traffic load of 1
72=0.013888889 Erlangs
subscriber
The number of subscribers is 0.013888889
0.01 ×19=26.38 ≅ 27 subscribers
OFDM
Question 1
a) The FFT size for a 5MHz bandwidth is 512
Effective bandwidth=Total bandwidth−2 ×Guard band=5 MHz−200 kHz=4.8 MHz
The size of a sub-carrier in OFDM is 15kHz
Therefore, the number of active subcarriers =
Effective Bandwidth
Subcarrier ¿ ¿ 48 MHz
15 kHz =320 ¿

COMMUNICATION SYSTEMS 5
b) The OFDM symbol duration
Useful OFDM symbol duration =1/sub carrier size= 1
15 kHz =66.67 μs
CP overhead for OFDM is 6.67% of OFDM symbol duration¿ 6.67
100 × 66.67 μs=4.44 μs
c) For a bit rate of12Mbps
Modulation Bit rate Baud
BPSK 1 12
QPSK 2 12
8-PSK 3 4
16-QAM 4 3
64-QAM 6 2
For bit rate of 15Mbps
Modulation Bit rate Baud
BPSK 1 15
QPSK 2 15
8-PSK 3 5
16-QAM 4 3.5
64-QAM 6 2.5
Question 2

COMMUNICATION SYSTEMS 6
When the bandwidth is 5MHz, there is 100kHz on each side.
The guard band will occupy approximately 10% of the total bandwidth.
Effective bandwidth=Total bandwidth−Guard band
Effective bandwidth=5− 10
100 × 5=5−0.2=4.8 MHz
Number of resources blocks= Effective bandwidth
100 KHz = 4.8 MHz
100 kHz =48
Download rate=1mbps, bit rate is 4bits per 16QAM
Number of resource elements(RE)= 1000 bps
4 bits per 16 QAM =1000
4 =250 REs
Assuming 2 slots on both sides, the Number of users can be calculated as:
Number of users¿ Number of resource elements
2 slots =250
2 =125