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COMMUNICATION SYSTEMS
ASSIGNMENT 2
STUDENT NAME
STUDENT ID NUMBER
INSTRUCTOR
INSTITUTIONAL AFFILIATION
LOCATION
DATE OF SUBMISSION
1

QUESTION 1
SOLUTION
(1)
s ( t )= {A , for 0 ≤ t ≤T
0 , otherwise
N 0
2
watts
Hz ... power spectral density
Representing h(t) as a function of A and T, the following illustration is used to demonstrated the
function,
2

The unit impulse response is matched with the s(t) signal
h(t)=Cs (t0−t )
It gives the signal an impulse response and it is pushed backwards in the output. The Simulink
plot shows the output,
(2)
Introducing the zero-mean white Gaussian noise,
H ( f ) =2 K
N0
S¿ ( f ) e− j w t 0
Obtaining the Fourier transform of the impulse response,
h ( t )=F−1 [ H ( f ) ]= 2 K
N0
∫
−∞
∞
S¿ ( f ) e j w t0
e− j wt df
2 K
N0 [ ∫
−∞
∞
S¿ ( f ) e j 2 πf ( t0−t ) df ]
¿
= 2 K
N0
[ s ( t0 −t ) ] ¿
x ( t ) =s ( t ) +n ( t ) 0 ≤ t ≤ T
(3)
The maximum SNR at the impulse output is obtained as,
3

2 K
N0
=Cs
(4)
Further, it is possible to show that,
y ( T ) =∫
0
T
x ( τ ) h (T −τ )dτ= [ x ( t ) , s ( t ) ] L2
Using convolution, the resulting illustration shows the sifting property of the continuous time
impulse at infinite number of impulses as h(T−τ )
x ( τ ) h ( T −τ ) =0 t ≠ τ
∫
0
T
x ( τ ) h(T −τ)dτ
¿ x (t )∗∫
0
T
h(T −τ )dτ
y ( T ) =x ( t )∗h ( t )=x ( t )
4

Solution
The received signal is given as,
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
To determine the maximum bit rate , BER, in bps
φ ( m ) ε { π
4 , 3 π
4 , 5 π
4 , 7 π
4 }
5

Pr −received average signal power
The theoretical BER is given by,
Q
( Eb
√Eb N 0
2 )=Q ( √ 2 Eb
N0 )
With a standard matched filter, the signal output is given for 2 BPSK receivers using phase-
orthogonal carriers.
σ 2= Eb N 0
2
The simulated BER is close to Q ( √ 2 Eb
N0 )=Q ( A
√12 )
Solution
6