Electrical Engineering: Networking and Communication Assignment

This document presents a comprehensive solution to a networking assignment, covering a range of topics within the field of electrical engineering and computer science. The assignment addresses key concepts such as the TCP/IP model, contrasting the internet protocol (IP) and network access layers, and exploring the communication processes between networks and translators. It includes detailed calculations and analysis of waveform characteristics, including amplitude, frequency, time period, and phase difference. Furthermore, the solution delves into frequency calculations and free space loss, channel capacity, and the Nyquist hypothesis, providing insights into data rate optimization. The assignment also compares and contrasts packet switching and circuit switching, highlighting their respective strengths and weaknesses. Finally, it includes calculations related to antenna height and distance, providing a well-rounded overview of essential networking principles and their practical applications. The assignment is contributed by a student and is available on Desklib, a platform offering AI-based study tools.

NETWORKING 0
Wireless network and communication

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Question 1
TCP/IP is defined as a networking model which contains four communication layers
including internet, application, and transport and network access. IP is also called internet
protocol which is the third layer of the TCP model which is used for determining the
addresses of the signals and delivering packets to the servers (Oracle, 2016). Network
access is the fourth layer of the TCP model which is mainly used for controlling physical
network and interconnections in the system.
Mainly, the internet protocol layer contains addresses and destination of the receiver but
the network access layer contains networks related information and performs data
communication in the systems. It is argued that the IP layer included major two protocols
such as IPv6 and IPv4 but network access point to point protocol which is enough capable
for data transmission over the networks (Jave Point, 2018). It is evaluated that in IP layer
addresses and location of the sources are embedded in an effective manner but in-network
access layer network protocols and communication channels are embedded in order to
control information transmitted from the source.
Question 2
The below figure shows the connection between networks and translators along with the
configuration of telephonic systems:
For performing data communication both PMs uses translators and a telephonic system
where the translator plays a significant role in the data transmission process. Presently,
assume the French PM speak with Chinese leader then the transmitted information will go
through the interpreter and phone to the interpreter utilized by Chinese PM. The translator

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used by French PM convert signals into English language and move to the goal which
additionally contains translator for changing signal into its original form. In this way,
performing correspondence between the two PMs translator helps for converting signals
into another form and telephone line provide a platform where both PMs can easily
exchange data over the translators.
Question 3
For above waveform:
Amplitude= 15
Frequency= 1/3 Hz
Time period= 3 seconds
Phase difference= 0 degree

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For above waveform:
Amplitude= 4
Frequency= 1/6.44 Hz
Time period= 6.44 seconds
Phase difference= 0 degree
For above waveform:
Amplitude= 7.86
Frequency= 1/2.34 Hz

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Time period= 2.34 seconds
Phase difference= 90 degree
Question 4
Part A
1 130 259 388 517 646 775 904 1033116212911420
-4.00000
-3.00000
-2.00000
-1.00000
0.00000
1.00000
2.00000
3.00000
4.00000
time y
Part B

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0.00020.00210.00400.00590.00780.00970.01160.01350.01540.01730.0192
-20.00000
-15.00000
-10.00000
-5.00000
0.00000
5.00000
10.00000
15.00000
20.00000
Y
Part C
1 19 37 55 73 91 109 127 145 163 181 199 217 235 253 271 289
-5
-4
-3
-2
-1
0
1
2
3
4
5
time y
Part D

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1 13 25 37 49 61 73 85 97 109 121 133 145 157 169 181 193
-8
-6
-4
-2
0
2
4
6
8
time y
Question 5
Frequency= 6 GHz
Total distance= 35863 kilometer
Free space loss= 20log10 (6*109) +20log10 (35.863*106) - 147.56 dB
After calculating the above problem it is examined that at 6 GHz frequency, around 199.09
dB free space loss will be occurred in the system.
Question 6
S(𝑡) = 5*sin(100𝜋 𝑡 ) + sin(600𝜋𝑡) + sin(300𝜋 )𝑡

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From the above problem, it is identified that fundamental frequency is 50 Hz because there
are three frequency signals presented in the equation including 50 Hz, 15o Hz and 300.
Total bandwidth= 300- 50= 250 Hz
Spectrum is shown in the below chart:
-300 -250 -200 -150 -100 -50 0 50 100 150 200 250 300
0.5
0 0
0.5
0
2.5
0
2.5
0
0.5
0 0
0.5
Freq. Spectrum
Frequency
Channel capacity:
C= 2*B*log2(M)
Put, B= 250 Hz and M= 2, 4 and 8
Therefore, at different numbers of levels the value of channel capacity will be different
which is given below:
At M= 2
C= 500 bits/s/Hz
At M= 4:
C= 1000 bits/s/Hz
At M= 8:

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C= 1500 bits/s/Hz
Question 7
It is seen that the Nyquist hypothesis give the connection between information rate and
transfer speed by which the shoppers can improve information rate over the channels
(Tech target, 2018). For expanding information rate without changing transmission
capacity customers can build quantities of levels utilized in the correspondence channels
however it isn't much successful procedure that can influence the execution of transmitted
sign which is a key disadvantage of this procedure.
Question 8:
Packet switching Circuit switching
Support forward transmission Does not support the forward transmission
Does not waste bandwidth Waste bandwidth while performing data
communication
Data packets move independently (Johnson,
2016)
Packets require a path
It is stated that the key benefit of packet switching is larger security due to which the users
can exchange data effectively. In the context of circuit switching the users may suffer from
security-related issues. However, the major strength of circuit switching is that it provides
quality of service which is a limitation of packet switching (Apposite technology, 2018).
Question 9:
D= 3.57*sqrt(kh1+kh2)
Here, D= 60 KM
H1= 4*H2
K= 1.3

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So, D= 3.57*sqrt(1.3(5h2))
Or, by solving the above problem it is found that H1= 169.476 meter and H2= 42.369 meter

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References
Apposite technology (2018) Packet Switching vs. Circuit Switching. [online]. Available from:
https://www.apposite-tech.com/blog/packet-switching-vs-circuit-switching/ [Accessed
5/9/2019].
Jave Point (2018) TCP/IP model. [online]. Available from:
https://www.javatpoint.com/computer-network-tcp-ip-model [Accessed 5/9/2019].
Johnson A. (2016) Packet Switching and Circuit Switching. [online]. Available from:
http://www.pearsonitcertification.com/articles/article.aspx?p=2481850&seqNum=3
[Accessed 5/9/2019].
Oracle (2016) TCP/IP Protocol Architecture Model. [online]. Available from:
https://docs.oracle.com/cd/E19455-01/806-0916/ipov-10/index.html [Accessed
5/9/2019].