EEET3028 Communication Systems Assignment 2 (2019 SP2) - UniSA

This document presents a comprehensive solution to a Communication Systems assignment (EEET3028) focusing on sampling and digital communication principles. The assignment explores the Fourier transform and magnitude spectrum of a sinc-squared signal, determining the Nyquist rate and analyzing the effects of different sampling periods. It utilizes MATLAB for simulations, including signal reconstruction, plotting, and comparison of original and reconstructed signals. The solution also delves into pulse shaping, constellation mapping, and modulation techniques, specifically addressing intersymbol interference (ISI) and quadrature amplitude modulation (QAM). Additionally, it covers receiver block diagrams, the impact of noise, and decision regions in a digital transmission system, providing a thorough understanding of communication system components and their behavior. The assignment is a practical application of theoretical concepts, with detailed MATLAB code and explanations.

STUDENTS NAME:
STUDENTS NUMBER:
COURSE TITLE:
SUBJECT TITLE: COMMUNICATION SYSTEMS
ASSIGNMENT
PROFESSERS NAME:

Assignment 2 (2019 SP2)
General information
This assignment covers sampling and principles of digital communication systems, and their
simulation in MATLAB.
Your report should include written answers, block diagrams, sketches of signals, plots produced by
MATLAB and the corresponding MATLAB code. Explain how you came to your answers and
conclusions, and provide concise answers.
Hand-written answers, mathematical derivations, sketches, and block diagrams are fine if done neatly and
clearly. You may scan them and include them in your document.
The report is to be submitted online as one single PDF file and on time. Late submissions may not
be considered. Make sure that you have added the cover page.
The deadline for submission is 16 June 2019.
Problems
Sampling (23 marks)
1. Consider the following signal
x(t) = sinc 2
(2t).
Write the Fourier transform X(f ) of x(t). Sketch the magnitude spectrum. (2 marks)
Solution:
From Fourier properties the Fourier transform of sinc squared function is a triangular pulse these can
be summarised as:
sinc2 t F . T
↔ tri ( f )
X ( f )=FT ( sinc2 2t )=tri (2 f )= {2 f +1 ,−0.5<f ≤ 0
−2 f +1 , 0<f ≤ 0.5
0 , otherwise
The transformation is as shown below:
Figure 1: Magnitude spectrum of the resulting triangular signal.

2. What is the Nyquist rate fN for this signal? Note that any sampling time larger than 1/fN
may prevent from perfect reconstruction of x(t) from its samples.
Nyquist rate for x(t) is given by:
Nyquist rate=2∗f max =2∗0.5=1 Hz
3.
Solution:
We will use convolution principle to obtained the sampled signal. We will first obtain the Fourier
transform of each signal. And then multiply to obtain the sampled signal.
x ( t )∗s (t) F . T
↔ X ( f ) S( f )
Therefore:
Y ( f ) =X ( f ) S ( f )
X ( f ) =0.5 ⋀ ( 0.5 f )
S ( f )=f s ∑
n=−∞
∞
σ ( f − ( n−0.2 ) f s )=¿ 2 f n ∑
n=−∞
∞
σ ( f −2 ( n−0.2 ) f n ) ¿
Y ( f )=0.5 ⋀ ( 0.5 f ) 2 f n ∑
n=−∞
∞
σ ( f −2 ( n−0.2 ) f n )
The spectral magnitude of the signal Y (f) of y(t) is as shown in figure below.
Figure 2: The spectral magnitude of the signal Y(f) of y(t) signal.

0.9
f
We can perfectly reconstruct the signal x(t) from y(t), this because the sampling frequency is
twice greater than the highest frequency of the signal. This fulfils the sampling theorem which
states that the sampling frequency should be equal or greater than twice the frequency of the
highest frequency signal.
4. Repeat Question 3 with sampling period Ts = 1 . (5 marks)
Solution:
The approach to the solution is as in the previous question. We will use convolution principle to
obtained the sampled signal. We will first obtain the Fourier transform of each signal that is the
sampled signal and the sampling signal. And then multiply to obtain the sampled signal as given
in the convolution principle below.
x (t )∗s (t) F . T
↔ X ( f ) S( f )
Therefore:
Y ( f )=X ( f ) S ( f )
X ( f )=0.5 ⋀ ( 0.5 f )
S ( f )=f s ∑
n=−∞
∞
σ ( f − ( n−0.2 ) f s )=¿ 0.9 f n ∑
n=−∞
∞
σ ( f −0.9 ( n−0.2 ) f n ) ¿
Y ( f )=0.5 ⋀ ( 0.5 f ) 0.9 f n ∑
n=−∞
∞
σ ( f −0.9 ( n−0.2 ) f n )
The spectral magnitude of Y (f) of y(t) is as shown in figure below.
Figure 3: The spectral magnitude of the signal Y(f) of y(t) signal.
We cannot perfectly reconstruct the signal x(t) from y(t), this because the sampling frequency is less
than two times the highest frequency of the signal. This does not fulfil the sampling theorem
which states that the sampling frequency should be greater or equal to two times the frequency of
the highest frequency signal.

T
s
5. MATLAB simulation: reconstruct x(t) from the samples y.
the same figure, stem plot the signal y in the range [-2:2].
Solution:
Matlab-code:
Results:
-2 -1.5 -1 -0.5 0 0.5 1 1.5 2
Time t
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Magnitude
Original
sampled signal
Figure 4: Plot of original signal and sampled signal.
(a) Let the reconstruction filter impulse response be h(t) = sinc( t ), t = [−6Ts : T s /5 : 6Ts].

On MATLAB, the reconstruction can be implemented by
• upsampling y by 5 to create yup.
• convolving yup with h(t).
Plot the reconstructed signal. Compare with the original signal and explain any
discrepency. (10 marks)
Solution:
Matlab-code:
Results:
0 20 40 60 80 100 120 140
Time t
-1
0
1
2
3
4
Magnitude
Reconstructed signal
0 10 20 30 40 50 60 70
Time t
0
0.2
0.4
0.6
0.8
1
Magnitude
Original signal
Figure 5: Plot of the reconstructed signal and original signal for comparison.
Discussion:

Comparison: it can be observed from the plots that the signals are quite similar in shape in general.
However, the reconstructed signal goes beyond the x axis that is has a negative magnitude while the
original signal doesn’t also, the magnitudes are not equal.
Reason: during sampling we may some important information of the signal and therefore when we
reconstructed the sampled signal will be similar to the original signal. Also, the signal we use to
reconstruct the signal may introduce some amplitude which through its gain.
6. The pulse p(τ ) has unit energy. Find c. (3 marks).
Solution:
E∞ =∫
−∞
∞
| p(τ )|
2
dτ= ∫
0
2∗10−3
|c cos πτ
4 |
2
dτ=1
c2 (sin πτ
4 cos πτ
4 ) 0.002
0 =1
c=19
7. Explain the roles of the blocks in Figure 1 (3 marks).
Mapping and constellation: this block represent the step in which the data is represented or
mapped into a point in the complex plane.
Pulse shape: this block is used to map digital information into a discrete sequence of voltage
signal this is also known as code set up.
Modulation: this block interpolates between these voltage waveforms to produce a discrete
sequence voltage that is transmitted.
8. Calculate the average energy ( Es) of the constellation. (2 marks)
Solution:
Eavg = 2
3 ( M −1 )= 2
3 ( 4−1 )=2

9. What are the criteria (in both time and frequency domain) for intersymbol-interference (ISI)
free transmission? Use Matlab to plot the function g(t) and explain whether ISI-free is
satisfied by the system in Figure 1.
(8 marks)
Solution:
According Nyquist criterion we represent the impulse response of the channel as q(t), equation below
shows the ISI free response condition as:
q ( t ) .∑
−∞
∞
σ ( t−kTs ) =σ (t)
h ( nTs ) ={ 1 n=0
0 ;n ≠ 0
Given Ts is the symbol period. In all integers n, the Nyquist theorem can be shown as:
1
Ts ∑
−∞
∞
H ( f − k
Ts )=1 ∀ f ,
Where Q(f) is the Fourier transform of q(t).
10. Assume the data bit sequence is 011110001001.
(a) Compute the transmit symbols ak, k = 1, 2, . . ..
Solution:
We first begin by grouping the signals into groups of two:
01 11 10 00 10 01
a1=Ac cos(2 π f c t+ π
2 )
a2=Ac cos(2 π f c t+ π )
a3=Ac cos(2 π f c t)
a4= Ac cos (2 π f c t + π
2 )
a5=Ac cos(2 π f c t+ 3 π
2 )
a6= Ac cos(2 π f c t+ π
2 )

(b) Plot the transmitted waveform x(t) using MATLAB.
Solution:
Matlab-code:
Plots:
0
0.5
1
1.5
amplitude
plot of digital data signal
0 1 2 3 4 5 6 7 8 9 10 11
time
Figure 6: Plot of transmitted data

0 100 200 300 400 500 600 700
time(sec)
-200
-150
-100
-50
0
50
100
150
200
amplitude
QPSK modulated signal (sum of inphase and Quadrature phase signal)
Figure 7: Plot of the transmitted waveform.
11. Draw a diagram illustrating how you modify Figure 1 to realize quadrature amplitude
modulation (QAM) transmission. (7 marks)
Solution:
To realize quadrature amplitude modulation transmitter from the quadrature phase shift keying
in figure one we will replace mapping in figure 1 with QAM symbol mapper and pilot
symbol mapper. We will also introduce two pulse shaping and I/Q mux. The MQAM
operates as follows: the data stream in the transmitter is first fed into an M-QAM mapper.
At the M-QAM mapper, the data bit stream is split into in-phase (I) and quadrature (Q) bit
streams, which are separately Gray coded as -AM signals and mapped to complex symbols.
The I and Q parts of the complex M-QAM symbol are from the set, and in order to guarantee equal
average energy per bit for fair performance comparison with a different modulation order. Figure below
shows and QAM block diagram.
Figure 8: Quadrature amplitude modulation (QAM) transmission block diagram.

Digital receiver (15 marks)
12. Draw a
block diagram of the receiver. Explain how noise on the communication channel affects the
signals in the receiver, and how it affects the symbol estimates and the bit estimates in the
receiver. (5 marks)
Solution:
Block diagram of the receiver
The figure below shows a generalized block diagram of a receiver.
Figure 9: Generalised block diagram of a receiver
The receiver is comprised of an antenna a demodulator, matched filter, pilot symbol aided channel
estimation, coherent demodulator with MRC and scaling and final symbol de-mapper.
How noise on the communication channel affects the signals in the receiver:
Noise has the following effects:
i. It degrades system performance for both digital and analogue systems.
ii. The receiver is not able to capture the information correctly.
iii. Reduce the efficiency of the system.
iv. The signal received at the receiver losses a lot of its power and its information, the
information is degraded and therefore not meaningful.
How noise affects the symbol estimates and the bit estimates in the receiver:
Noise affects the BER (bit error rate) of the receiver. This is the ratio of the number of decoded bits that
remain incorrect after error correction to the total number of decoded bits.

13. Consider now transmission of the bit sequence in question 10 over a noise-free
communication channel. Using the pulse p(t) described in Figure 1. The receiver filter is
assumed to be the match filter such that
q(t) = p∗ (−t) = p(t).
(a) Plot in MATLAB the output of the receive filter.
Solution:
Matlab-code: