Electrical Engineering: Communication Systems Assignment 2 Solution

This document presents a comprehensive solution to a Communication Systems assignment, covering a range of topics including matched filters, QPSK modulation, binary communication systems, M-ary PSK, and communication system efficiency. The solution begins with an analysis of a matched filter, deriving its impulse response and Fourier transform, and determining the maximum signal-to-noise ratio. It then delves into QPSK modulation, including signal representation, bit error probability, and MATLAB implementation. The assignment further explores binary communication systems, analyzing a-posteriori probabilities, likelihood ratios, and error probabilities. Additional sections address the efficiency of communication systems, including the impact of data rates and the analysis of signal-to-noise ratio. The document also includes graphical representations and Matlab implementation to illustrate the concepts and results.

ASSIGNMENT 2
COMMUNICATION SYSTEMS
INSTITUTIONAL AFFILIATION
LOCATION
INSTRUCTOR (PROFESSOR)
STUDENT NAME
STUDENT REGISTRATION NUMBER
DATE OF SUBMISSION
1

QUESTION 1
Part A
(i) H(t) as a function of A and T
s ( t )= {A , for 0 ≤ t ≤T
0 , otherwise
Introducing the noise signal,
H ( t ) =s ( t ) +n ( t ) 0 ≤ t ≤ T
η= |s0 ( t )|2
E [ n2 ( t ) ]
h ( t )=F−1 [ H ( f ) ]
2 K
N0 [ ∫
−∞
∞
S¿ ( f ) e j w t0
e− j wt df ] ¿
¿ 2 K
N0
[ s ( t0 −t ) ] ¿
From the impulse response output, the constant value is given as,
Cs= 2 K
N0
Therefore,
h( t)=Cs (t 0−t )
(ii) H(f), the Fourier transform of h(t)
2

H ( f ) =2 K
N0
S¿ ( f ) e− j w t 0
(iii) Maximum signal to noise ratio at the output of h(t).
(iv) Show
The impulse response is demonstrated by plotting the images below,
Part B
Section 1
QPSK transmitted with a required probability of bit error, PB
3

0 5 10 15 20 25 30 35 40 45
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Sm ( t )= √ ( 2 Pr ) cos [ 2 π f 0 t +ϕ ( m ) ] 0 ≤ t< T
ϕ ( m ) ϵ { π
4 , 3 π
4 , 5 π
4 , 7 π
4 }
Pr −received average signal power
r ( t )=sm ( t )+ n(t)
Pr
N 0
, Pr= 1
T ∫
0
T
sm
2 ( t ) dt
SQPSK = {√ Es cos [ (i−1 ) π
2 ] ϕ1 ( t )− √ Es sin [ ( i−1 ) ] ϕ2 ( t ) }
The bit rate in either I or Q channel are ½ input data rate.
Section 2
Matlab Implementation
4

0 5 10 15
SNR in db
10-10
10-8
10-6
10-4
10-2
100
probability of error
Probability of Error Performance
QUESTION 2
Binary communication system
Part a
P0∧P1 =1−P0 be arbitrary
V →rv ( conditional probability density )
f v∨u { v|am }−finite∧non−zero ∀ v ∈ R
m∈ {0,1 }
To determine the a-posteriori probability for the input,
pu∨v ( am|v )= ( Pm f v∨u ( v|am ) )
f v ( u )
Part b
It is possible to obtain the likelihood ratio of the MAP decision,
5

A ( v ) = f v∨u ( v|a0 )
f v∨u ( v|a1 ) , p1
p0
=η
Part c
The overall probability of the error is given as,
Pr {e }= p0 Pr {e|U =a0 }+ p1 {e|U =a1 }
P ( error ) =0.5 Q
( √ Es
N 0
2 ) +0.5Q
( √ Es
N 0
2 )
P ( error ) =Q ( √ 2 Es
N 0 ) =0.5 erfc ( √ Es
N0 )
Part d
Replacing with real values,
f R ∨0 = √ 1
2 π e− ( 5 y2 )
f R ∨1 = √ 1
2 π e−¿¿
f Y /H i ( y / Hi ) =∏
k =1
m
1
√ 2 πσ exp [ − ( yk −si ,k )
2
2σ 2 ]
To obtain the error probability,
Perror = ( 0.5 ) ( 0.86675 ) + ( 0.5 ) ( 0.56703 ) =0.716578
QUESTION 3
Communication system: binary information transmission
6

Part A
ωc= 2 π
T s
=N ( 2 π
T b )
ρ= A2 T b
2η
Part B
π0=Pr { transmits s0 ( t ) }= 1
2
π1=Pr {transmits s1 ( t ) }=1
2
H0 :r ( t ) =S0 ( t ) +n ( t ) , 0 ≤ t <T
H1 : r ( t )=S1 ( t )+ n ( t ) , 0 ≤t <T
Sn ( f )= N 0
2 ,−∞< f <∞
s0 ( t )= {A ,0 ≤ t ≤ 3
4 T
0 , otherwise
s1 ( t ) = {A , 1
4 T ≤ t ≤ T
0 , otherwise
ωc= 2 π
T s
=N ( 2 π
T b )
7

ρ= A2 T b
2η
SN Re=10 log10 [ A2 Tb
2 η ]
The maximum data rate is given as,
Eb =π0∫
0
T
s0
2 ( t ) dt+ π1∫
0
T
s1
2 ( t ) dt
To obtain the probability of error,
Pb=erfc ( √ Eb
N0 ) [ 1− 1
2 erfc ( √ Eb
N 0 ) ]
Part C
It is observed that an increased in the rb value causes a pb value decrease. The illustration below
expounds on the relationship,
QUESTION 4
Part A
For an M-ary PSK modulation
ϕi ( t ) =Acos ( ωc t+θi ) 0<t ≤T s
θi =0 , 2 π
M , … , 2 ( M−1 ) π
M
8

sϕ ( ω ) = A2 T s S a2
[ ( ω−ωc ) T s
2 ]
T b= T s
log2 M
Pϵ ≈2 erfc √ 2 Es
η sin2 π
M m>2
Pϵ ≈2 erfc √ 2 Es
η sin2 π
√ 2 M
Γ =
sin2 π
M
sin2 π
√2 M
-5 0 5 10 15 20
Es/No, dB
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Symbol Error Rate
Symbol error probability curve for QPSK
aplha=0
aplha=0.15
aplha=0.25
aplha=0.35
aplha=0.45
Part B
9

-5 0 5 10 15 20
Es/No, dB
0
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
0.18
0.2
Symbol Error Rate
Symbol error probability curve for QPSK
aplha=0
aplha=0.15
aplha=0.25
aplha=0.35
aplha=0.45
QUESTION 5
Efficiency of 0.55
EIR Ptransmit =10 log ( Ptransmitter ) −Transmitte rfeede rloss
+Transmitte r Antennagain
− Antenn apointin gloss
(dBW )
EIR Ptransmit =10 log 10−0.5 dB+45.21−0.70
EIR Ptransmit =54.01 dB
Gr
T s [ dB
K ] =−4 dB/ K
10

Pr =Pt (1−|Γt |2
. gt . ( λ
4 πd )2
. 1
αm . Aeff (1−|Γr|2
) )
Lbf =20 log10 ( 4 πd
λ )
Lb f uplink
=20 log10
( 4 π∗36000
c
f )=20 log10 ( 144e6 π
0.01 )
Lb f downlink
=20 log10
( 4 π∗36000
c
f )=20 log10 ( 144e6 π
0.015 )
Lb f downlink
=209.5884 dB
Part B
P1=Pr {1 bit error|s1 (t)}
¿ Pr {s2 ( t )|s1 ( t ) }+ Pr {s4 ( t )|s1 ( t ) }+ Pr { s8 ( t )|s1 ( t ) }
The autocorrelation is given as,
Rc ( τ ) = 1
T c
∫
−Tc
2
T c
2
c ( t ) c ( t−τ ) dt
11

0 5 10 15
SNR in db
10-10
10-8
10-6
10-4
10-2
100
probability of error
Probability of Error Performance
The receiver output is given as,
d ( t ) c2 (t)=d (t)
Sk ( t )= √ 2 Eb
T mk ( t ) pk ( t ) cos ( 2 π f c t+ φk )
12