MITS5003 Wireless Networks & Communication Assignment 1 Solution

This document provides a comprehensive solution to a wireless networks and communication assignment, covering several key concepts. The solution begins by differentiating between the IP and Network Access layers in the TCP/IP protocol, explaining the control information embedded in each. It then presents a scenario involving translation between French and Chinese prime ministers, illustrating the interaction at each level. The document includes calculations of amplitude, frequency, time period, and phase for given waveforms and equations. It also computes the isotropic free space loss at 6 GHz, determines the fundamental frequency, spectrum, and bandwidth of a signal, and calculates channel capacity using the Nyquist criteria. The solution further explains how to increase the data rate over a channel without increasing bandwidth, discussing the disadvantages of this approach. It then compares and contrasts packet switching and circuit switching, outlining their advantages and disadvantages. Finally, the document calculates the appropriate heights of two antennas in a LOS communication scenario. References are also provided.

Prime Minister
Translator
Telephone Telephone
Translator
Prime Minister
Chinese Prime Minister
Chinese Translator
English Translator English Translator
French Translator
French Prime Minister
1. Describe the difference between IP and Network Access layer in TCP protocol and also
explain the control information embedded in these two layers.
 Network access layer is a combination of data link and physical layer, whereas, IP
layer is a network layer.
 First layer of the four layer TCP/IP model is the Network access layer, whereas,
IP layer is the second layer of the four layer TCP/IP model.
 The protocols included in Network access layer are Ethernet, Token Ring, FDDI,
X.25, etc, whereas, the protocols included in IP layer are internet control message
protocol (ICMP), address resolution protocol (ARP), internet group management
(IGMP), and reverse address resolution protocol (RARP).
 Network access layer is responsible for placement of data on the medium through
host, whereas, IP layer is responsible for routing of IP data grams.
 Network access layer defines the details of how the data is physically sent through
the network with the help of cables like coaxial cable, optical fiber and twisted
pair copper wire, whereas, the IP layer directs the host to insert data packets into
required network and to deliver them independently to the destination.
2. The French and Chinese prime ministers need to come to an agreement by telephone, but
neither speaks the other’s language. Further, neither has on hand a translator that can
translate to the language of the other. However, both prime ministers have English
translators on their staffs. Draw a diagram similar to Figure 1 to depict the situation, and
describe the interaction and each level.
Telephone line
Telephone line

 Chinese prime minister will talk to the Chinese translator via Chinese, and the Chinese
translator will talk to English translator.
 French prime minister will talk to the French translator via French, and the French
translator will talk to English translator.
 English translators from the both the sides will talk with each other in English through
the telephone.
3. From the following figures, compute the maximum amplitude, frequency, time period
and phase for each of the wave. The x-axis represents the time in sec and y-axis
represents the amplitude.
Amplitude: the height from center line to the crust or trough is called the amplitude. In
other words, the magnitude of wave is called the amplitude.
Time period: the distance between 2 troughs or crusts is the time period. In other words,
time the wave takes to complete one cycle.
Frequency: it is inversely proportional to the time period. Total number of waves passing
through a specific point is called the frequency.
Phase: phase is nothing but the angle. Phase shift is how far the function or wave shifted
from the center point. Mainly there are 2 types of shifts. Horizontal and vertical shifts.
Horizontal shift is how far the function or wave shifted horizontally from the center
point.
Vertical shift is how far the function or wave shifted vertically from the center point.
a.

Amplitude 15
Frequency 333.33
Time period 3
Phase 0
b.
Amplitude 4
Frequency 142.85
Time period 7
Phase 0
c.
Amplitude 7.8
Frequency 500
Time period 2
Phase 90

4. Compute the amplitude, frequency, time period and phase for each of the following
equations and also draw their respective waveforms.
a. 3𝑆𝑖𝑛(2𝜋(200)𝑡)
=3cos (2𝜋 (200) 𝑡-90)
Amplitude 3
Frequency 200
Time period 5
Phase -90
b. 14𝑆𝑖𝑛(2𝜋(50)𝑡 + 90)
=14cos (2𝜋 (50) 𝑡)
Amplitude 14
Frequency 50
Time period 20
Phase 0

c. 4𝑆𝑖𝑛(650𝜋𝑡 + 180)
=4cos (650𝜋𝑡 + 90)
Amplitude 4
Frequency 325
Time period 3.07
Phase 90
d. 6𝑆𝑖𝑛(700𝜋𝑡 + 270)
=6cos (700𝜋𝑡 +180)
Amplitude 6
Frequency 350

Time period 2.85
Phase 180
5. Determine the isotropic free space loss at 6 GHz for the shortest path to a synchronous
satellite from earth (35,863 km).
An isotropic radiator is an ideal antenna which radiates the power with unit gain. The loss
between 2 isotropic radiators in the free space is called as the isotropic free space loss.
We have formula for calculating the isotropic free space loss,
PL=10 log10 ( ( 4 ᴨ df
c )2
)
PL=20 log10 ( 4 ᴨ df
c )
PL=20 log10 ( f )+ 20 log10 ( d ) −20 log10 ( 4 ᴨ
c )
PL=20 log10 ( 6 ×109 ) +20 log10 ( 35.863× 106 )−147.56dB
PL=199 dB
Hence the isotropic free space loss between 2 isotropic radiators.

6. Given a signal as follows, compute the fundamental frequency, spectrum and bandwidth.
Also calculate the channel capacity using Nyquest criteria using M= 2, 4, 8, where M is
the number of levels. 𝑠(𝑡) = 5 sin(100𝜋𝑡) + sin(300𝜋𝑡) + sin(600𝜋𝑡)
Given,
𝑠 (𝑡) = 5 sin (100𝜋𝑡) + sin (300𝜋𝑡) + sin (600𝜋𝑡)
Upon converting the sine function into cos function, we get,
s ( t )=5 cos (100 π t− π
2 )+ cos (300 π t − π
2 )+ cos (600 π t− π
2 )
s ( t )=0+5 cos (2 π (50)t− π
2 )+ cos (2 π (150) t− π
2 )+cos (2 π (300)t − π
2 )
The fundamental frequency: f = 50Hz
We have,
Fourier series coefficients (note this solution is in terms of single sided Fourier series
representation):
A0 = 0,
A1=5, ɸ1= - π
2
A2 = 0,
A3=1, ɸ3= - π
2
A4 = 0,
A5 = 0,
A6=5, ɸ6= - π
2
To obtain the two – sided Fourier series representation, we can write, using Euler formula

s ( t ) =0+2. 5 e− j π
2 e j 2 ᴨ ( 50 ) t +2. 5 e j π
2 e− j 2ᴨ ( 50 ) t +0. 5 e− j π
2 e j 2ᴨ ( 150 ) t +0.5 e j π
2 e− j 2 ᴨ ( 150 ) t +0.5 e− j π
2 e j 2 ᴨ ( 300 ) t +0. 5 e j π
2 e−
Thus the two – sided Fourier series coefficients are:
S0 = 0,
S1 =2. 5 e− j π
2
S-1 =2. 5 e j π
2
S3 =e− j π
2
S-3 =e j π
2
S6 =e− j π
2
S-6 =e j π
2
The two – sided line spectrum of the signal is as follows (line height only indicates the
magnitude)

Bandwidth is given by,
BW =f × ( 70.0 % ) 2
BW =50 ×0.5=25 Hz
Bandwidth is constant in a noiseless channel.
Channel rate is given by,
C=2 B log2 M
C=2 ×25 × log2 2=50 bits / s / Hz
C=2 ×25 × log2 4=100 bits / s / Hz
C=2 ×25 × log2 8=150 bits / s / Hz
7. Explain how the data rate over a channel can be increased, without increasing the
bandwidth? What is the disadvantage of this approach? Hint: Nyquist Theorem
Nyquist gives the upper bound for the bit rate of a transmission system by calculating the
bit rate directly from the number of bits in a symbol (or signal levels) and the bandwidth
of the system (assuming 2 symbols/per cycle and first harmonic).
Nyquist theorem states that for a noiseless channel:
C=2 B log2 2n
C= capacity in bps
B = bandwidth in Hz
n=data rate
In baseband transmission, the bit rate is 2 times the bandwidth if we use only the first
harmonic in the worst case.
Nyquist’s theorem states that for properly representing a waveform, a bit rate of at least
twice the signal frequency (bandwidth) is required. Aliasing is a method using which the

data rate over a channel can be increased without increasing the bandwidth. Aliasing can
make 2 different signals look alike by limiting the bandwidth. In other words, aliasing
folds the higher frequency signals to appear as lower frequency signal. But, the
disadvantage of this approach is that, the signals are not properly reconstructed at the
receiver side, meaning low quality signals get reproduced.
8. What is the main difference between Packet switching and Circuit Switching? Also
discuss the advantages of Packet switching over Circuit Switching and vice versa.
Circuit switching Packet switching
Setting up of a call is required. Setting up of a call is not required.
Circuit switching is physical Packet switching is not physical
Connection will be lost, if any one of the
link connecting both the ends gets broken.
Connection could continue if any one of
the route breaks, as the host can choose
other routes.
Same route will be chosen for information
delivery between the hosts
Different routes could be chosen for
information delivery between the hosts
Always the information arrives the
destination in order.
Not always the information arrives the
destination in order.
Fixed bandwidth is available Variable bandwidth is available
Congestion is call based Congestion is packet based
Utilization of bandwidth is partial Utilization of bandwidth is full
Store and forward transmission technique
is not used
Store and forward transmission technique
is used
Circuit switching is transparent Packet switching is not transparent
Cost is based on time consumption Cost is based on number of packets utilized
End terminals are telephone and modem End terminals are computers
Information type being used are analog
voice or PCM digital voice
Information being used is binary
information (0’s and 1’s)
Transmission system being used is analog
and digital data over different transmission
media
Transmission system being used is digital
data over different transmission media
Advantages of packet switching over circuit switching:
As in case of packet switching the entire bandwidth is not reserved, at the end of
transmission, there won’t be any wastage of bandwidth, whereas in case of circuit
switching, the entire bandwidth is reserved in advance and there occurs wastage of
bandwidth, if not fully consumed.
As in case of packet switching, even if one of the links breaks, the packets choose
different route for transmission purpose, therefore the communication doesn’t get
disturbed, whereas in case of circuit switching, as the information moves through single
path, if it gets broken, then the whole communication gets disturbed.

Store and forward transmission in packet switching helps to store the information,
whereas in case of circuit switching, the store and forward transmission is not used.
Advantages of circuit switching over packet switching:
In case of circuit switching, the information arrives in order at destination, which makes
this switching more organized when compared to the packet switching, where the
information doesn’t arrive in order.
In case of circuit switching, the transmission / communication is transparent, whereas in
case of packet switching, there will be no transparency, which makes circuit switching
more reliable.
9. In a LOS communication, consider d = 60km, the requirement is to make two antennas
(transmitter and receiver) such that the height of one antenna should be four time of the
other. Considering this, find the appropriate heights of these two antennas.
We have formula for calculating the distance between 2 antennas in a Line Of Sight
communication,
d=3.57 √ K h
d=3.57 ¿
We have K= 4
3
d=4.12( √ hT + √ 4 hT ) As per given condition
d=4.12 ×3 √ hT
d=12 .36 √hT
60=12.36 √hT
hT =23.56 m
hR=4 × 23.56=94.25 m
References:

Giangrandi, I. (2019). Isotropic loss of free-space radiation. [online] Giangrandi.ch.
Available at: http://www.giangrandi.ch/electronics/anttool/iso-loss.shtml
System, O., Mohan, R., yousif, y., apake, A., Banda, A., Karr, W., yejashwi, k. and
Vaghani, N. (2019). Difference Between Circuit Switching and Packet Switching (with
Comparison Chart)- Tech Differences. [online] Tech Differences. Available at:
https://techdifferences.com/difference-between-circuit-switching-and-packet-
switching.html