Communication Systems: Solved Assignments and Essays

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 t 0
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 (t0−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