MITS5003 Assignment 1: Networking and Data Communication

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MITS5003
Assignment-1

Table of Contents
Question 1........................................................................................................................................3
Question 2........................................................................................................................................4
Question 3........................................................................................................................................5
Question 4........................................................................................................................................8
Question 5......................................................................................................................................12
Question 6......................................................................................................................................13
Question 7......................................................................................................................................14
Question 8......................................................................................................................................15
Question 9......................................................................................................................................16
References......................................................................................................................................17

Question 1
Internet protocol (IP) provides the connectivity between two nodes and allows the user to share
the data within the same network. the id address is assigned to each system connected with the
network so that the system can be identified. The IP address is like an address of the system
which can be used for delivery of the data packets (Aksoy and Gunes, 2016). The IP is
responsible for managing the sharing of the data between the machines over the internet
connection. The address is used by different other machines for transmitting the data to correct
location and preventing form packet loss. These small packets contain the destination and source
address which is used for identification.
The Network Access layer provides the connectivity between the machine and nodes using
physical connectivity. The Network Access layer is liked with the two layers of the OSI models
which are Link Layer and Physical layer (Zhang, Lo, Lu, Shu and Wong, 2019). The hardware
working is defined by this layer and the using of the network for transmitting the data packets.
This layer is placed over the lowest side of the OSI models (Iskander, Yun, Qazi, Sasaki and
Das, 2016). The use of new hardware is possible by implementing Network Access layer.

Question 2
Figure 1: Communication Process between China and France using Telephone lines
The diagram shows the communication between two Prime Minister of different countries. As
both countries have different languages, translators are used. Both translators convert the
information form their language to English and from English to their own language. Telephones
are used for commutation while using the telephone network. The Prime Minister also have
direct connection between them.

Question 3
 The maximum amplitude is the maximum displacement of the wave. So, based on the
analysis of the given graph, the maximum amplitude is found to be 15.
 The formula for the frequency calculation based on time can be calculated as Frequency
= 1/Time.
The following graph shows the time for each complete cycle is 3.
So, the Frequency can be calculated for a given Wave as Frequency = 1/3 = 0.33 Hertz.
 As the above calculation Time period calculated is 3 seconds.
 For calculating phase angle of the wave, the formula can be used
x(t) = Amax * sin (θ) = 0 where Amax is maximum amplitude and θ is angle of the
phase.
This will give a phase angle = zero.

 The maximum amplitude is the maximum displacement of the wave. So, based on the
analysis of the given graph, the maximum amplitude is found to be 4.
 The formula for the frequency calculation based on time can be calculated as Frequency
= 1/Time.
The following graph shows the time for each complete cycle is 6.5.
So, the Frequency can be calculated for a given Wave as Frequency = 1/6.5 = 0.15 Hertz.
 As the above calculation Time period calculated is 6.5 seconds.
 For calculating phase angle of the wave, the formula can be used
x(t) = Amax * sin (θ) = 0 where Amax is maximum amplitude and θ is angle of the
phase.
This will give a phase angle = zero.

 The maximum amplitude is the maximum displacement of the wave. So, based on the
analysis of the given graph, the maximum amplitude is found to be 7.5.
 The formula for the frequency calculation based on time can be calculated as Frequency
= 1/Time.
The following graph shows the time for each complete cycle is 2.25.
So, the Frequency can be calculated for a given Wave as Frequency = 1/2.25 = 0.44
Hertz.
 As the above calculation Time period calculated is 2.25 seconds.
 For calculating phase angle of the wave, the formula can be used
x(t) = Amax * sin ( θ) = Amax where Amax is maximum amplitude and θ is the angle of
the phase.
This will give a phase angle = 90.

Question 4
a. 3 s𝑖𝑛 (2 𝜋 (200) 𝑡)
According to the given equation, we find
 Maximum Amplitude = 3
 Frequency of the wave from equation = 200 Hertz
 Time period for each cycle = 0.005 seconds
 The phase angle of the wave = 0
b. 14 s𝑖𝑛 (2 𝜋 (50) 𝑡 + 90)
According to the given equation, we find
 Maximum Amplitude = 14
 Frequency of the wave from equation = 50 Hertz
 Time period for each cycle = 0.02 seconds

 The phase angle of the wave = 90
c. 4 s𝑖𝑛 (650 𝜋 𝑡 + 180)
According to the given equation, we find
 Maximum Amplitude = 4
 Frequency of the wave from equation = 325 Hertz
 Time period for each cycle = 0.003 seconds
 The phase angle of the wave = 180

d. 6 s𝑖𝑛 (700 𝜋 𝑡 + 270)
According to the given equation, we find
 Maximum Amplitude = 6
 Frequency of the wave from equation = 350 Hertz
 Time period for each cycle = 0.002 seconds
 The phase angle of the wave = 270

Question 5