MITS5003 Wireless Networks & Communication Assignment 1

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MITS5003
WIRELESS NETWORKS &
COMMUNICATION
ASSIGNMENT NO 1

Contents
Question 1........................................................................................................................................3
Question 2........................................................................................................................................4
Question 3........................................................................................................................................6
Question 4........................................................................................................................................9
Question 5......................................................................................................................................12
Question 6......................................................................................................................................12
Question 7......................................................................................................................................14
Question 8......................................................................................................................................15
Question 9......................................................................................................................................16
References......................................................................................................................................17

Question 1
Solution:
IP: It is also referred as Internet Protocol which consists of rules and regulations developed for
the communication, sharing of data and transmission of different files from one system to
another over different physical locations and connected with a network. It controls the entire
transmission and sharing of information from one device to another along with their management
for the safety and security of data. It consists of a unique set of address or programs developed
for the transmission of entire information or data over the multiple systems connected within the
network. It doesn’t restrict the total number of devices or systems connected within the network
and provide complete path for the transmission of information of important data over the Internet
in order to utilize by different receivers located at various geographical locations. It is located in
3rd layer of the OSI model developed for the communication system. It has two different versions
(i.e. IPv4 and IPv6) used by different systems connected within the network (Goralski, 2017).
Network Access Layer: It is one of the last layer present in OSI model developed for the
communication system which consists of different requirements of connected devices and
hardware in order to perform the transmission or establish the communication between two or
more connected devices. It also includes different rules and regulation to manage the IP used for
the transmission or the communication in-between multiple systems. It can perform different
action which is managed by other lower layers available in OSI model of the communication
system. The main working of this layer is to provide encapsulation and security to the data in
order to develop a frame to be communicated or shared over the internet along with
identification and establishment of communication path based on IP address of different
connected devices within the network (Goralski, 2017).

Question 2
Solution:
Step 1st:
1. Direct communication between Prime Minister speaking French and the Prime Minister
speaking Chinese with the help of translator to translate the French language into the
Chinese language.
2. The message or word used by the Prime Minister in the French language transferred to
the translator in order to translate that message.
3. Translated message by French translator communicated and shared with another
translator through the internet or the telephone line.
4. The received message is again translated by the Chinese translator so that the Prime
Minister speaking Chinese language can under the message used by another Prime
minister.
Step 2nd:
1. Direct communication between Prime Minister speaking Chinese and the Prime Minister
speaking French with the help of translator to translate the Chinese language into the
English language.
2. The message or word used by the Prime Minister in Chinese language transferred to the
translator in order to translate that message and works as a mediator in-between English
& French.

3. Translated message by the mediator communicated and shared with another mediator
through the internet or the telephone line.
4. The received message is again translated by the English translator so that the Prime
Minister speaking French language can under the message used by another Prime
minister with the help of the French translator.

Question 3
Solution:
i) As per the image provided in the question,
From the provided image (Light-measurement, 2019);
Am as max. Amplitude = 15 (calculated from the image)
Value of time-period = 3 in seconds (i.e. T)
Now, taking the equation for frequency, we have;
Frequency denoted as f = 3rd part of a unit
So, f = 1 divided by 3
i.e. f = 0.3333 Hz
Taking the equation for (), we have;
(Value of Am) multiplied by Sin = X(t)
So, X(t) = 0 (Zero)
ii) As per the image provided in question,

From the provide image;
Am as max. Amplitude = 4 (calculated from the image)
Value of time-period = 6.5 in seconds (i.e. T)
Now, taking the equation for frequency, we have;
Frequency denoted as f = 6.5th part of a unit
So, f = 1 divided by 6.5
i.e. f = 0.153846 Hz
Taking the equation for (), we have;
(Value of Am) multiplied by Sin = X(t)
So, X(t) = 0 (Zero)
iii) As per the image provided in question,

From the provide image;
Am as max. Amplitude = 7.5 (calculated from the image)
Value of time-period = 2.25 in seconds (i.e. T)
Now, taking the equation for frequency, we have;
Frequency denoted as f = 2.25th part of the unit
So, f = 1 divided by 2.25
i.e. f = 0.444444 Hz
Taking the equation for (), we have;
(Value of Am) multiplied by Sin = X(t) (which is Am)
So, X(t) = 90 (Ninety degree)

Question 4
Solution:
a. 3𝑆𝑖𝑛(2𝜋(200)𝑡)
Here, A is 3,
Value of f is 200 in Hz
And, the value of T is (1 divided by 200) (DeCross and Williams, 2019).
i.e. T = 0.05 second
Also, the value of  is Zero.
Thus using the above results, we get;
b. 14𝑆𝑖𝑛(2𝜋(50)𝑡 + 90)
Here, A is 14,
Value of f is 50 in Hz
And, the value of T is (1 divided by 50)
i.e. T = 0.02 second
Also, the value of  is 90.
Thus using the above results, we get;

c. 4𝑆𝑖𝑛(650𝜋𝑡 + 180)
Here, A is 4,
Value of Fis (650 divided by 2) in Hz which is 325 Hz.
And, the value of T is (1 divided by 325)
i.e. T = 0.003 second
Also, the value of  is 180.
Thus using the above results, we get;
d. 6𝑆𝑖𝑛(700𝜋𝑡 + 270)
Here, A is 6,
Value of f is (700 divided by 2) in Hz which is 350 Hz.

And, the value of T is (1 divided by 350)
i.e. T = 0.002 second
Also, the value of  is 270.
Thus using the above results, we get;

Question 5
Solution:
As per the data provided in the question,
Value of f is 6GHz
Value of R is 35,863 in Km
Now, calculating the value of λ, we have (Tsai, 2017);
i.e. λ = (3 multiplied by 108) / (6 multiplied by 109) = 0.05
Thus, the value of loss in dB is given as;
LdB = 20log10 ((6*3.14*35863*1000)/(0.05)
i.e. LdB is 139.04 dB which is the final result.
Question 6
Solution:
As per the equation provided in question;
(𝑡) = 5 sin(100𝜋𝑡) + sin(300𝜋𝑡) + sin(600𝜋𝑡)
Now, the value of frequency as per the above equation,
f = 50Hz, 150Hz and 300Hz
Putting Fourier transform on the provided equation (Fund, 2017);
i.e. {S(f)}= (1/2i) multiplied by {(d(f-300)-d(f+300))+(d multiplied by (f-150)-d(f+150))+5(d(f-
50)-d(f+50))}
In which; the maximum value of F is value of Bandwidth addition with minimum value of F.
i.e. B = 300 + 50
i.e. B = 350 MHz
Now, implementing equation of Nyquist, we have;