TCP/IP Model and Network Communication: A Comprehensive Overview

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Table of Contents
Answer for 1st Question...............................................................................................................................2
Answer for 2nd Question..............................................................................................................................3
Answer for 3rd Question...............................................................................................................................4
Answer for 4th Question...............................................................................................................................7
Answer for 5th Question.............................................................................................................................11
Answer for 6th Question.............................................................................................................................12
Answer for 7th Question.............................................................................................................................13
Answer for 8th Question.............................................................................................................................14
Answer for 9th Question.............................................................................................................................15
References...................................................................................................................................................16

Answer for 1st Question
The lowest layer of the TCP/IP model is the network access layer. This layer uses various kinds
of protocols that are utilised by the computer for a smooth transmission of data to various
devices on other location but over a sale network. In this layer the properties regarding the frame
usage is defined. Over the network the data is shred and exchanged. The data transmitted uses
the physical address which is embedded in the network adapter card of the node on the network.
The internet protocol layer or the internet layer is responsible for successful transmission of the
data over the network using routing techniques. The data that are called the packets are assigned
a logical/IP address. This logical address is used for the determination of the device which leads
to an easy and control transmission. Every layer in the TCP/IP model uses the IP layer for data
transmission.

Answer for 2nd Question
The French prime minister cannot converse directly to the Chinese prime minister as there is a
communication gap as both prime ministers doesn’t have the knowledge of each other’s
respective native languages.
The prime minister of both the countries have English speaking staff. Both the prime ministers
uses their respective translators as a medium for conversing. The French translator hears the
message from the French prime minister and similarly the Chinese translator hears the message
from the Chinese prime minister. Both the translators translate their respective messages into
English language and converse in English. The received messages are then translated from the
English language to the French language for the French prime minister and to the Chinese
language for the Chinese prime minister. The conversation is done via telephonic medium.

Answer for 3rd Question
Note: For this question, the amplitude will be defined as A,
The frequency will be defined as f,
The time period will be defined is T and,
The phase angle will be defined as ɸ.
A = 15
T = 3 s
f = 1
T
f = 1
3
f = 0.33 Hz
ɸ = 0˚

A = 4
T = 6.5 s
f = 1
T
f = 1
6.5
f = 0.154 Hz
ɸ = 0˚
A = 7.5
T = 2.25 s

f = 1
T
f = ½.25
f = 0.44 Hz
ɸ = 90˚

Answer for 4th Question
Part a)
Equation: 3𝑆𝑖𝑛 (2𝜋 (200) 𝑡)
Amplitude = 3
Frequency = 200 Hz
Time = 0.005 s
Phase angle = 0˚

Part b)
Equation: 14𝑆𝑖𝑛 (2𝜋 (50) 𝑡+90)
Amplitude = 14
Frequency = 50 Hz
Time = 0.02 s
Phase angle = 90˚

Part c)
Equation: 4𝑆𝑖𝑛 (650𝜋𝑡+180)
Amplitude = 4
Frequency = 325 Hz
Time = 0.003 s
Phase angle = 180˚

Part d)
Equation: 6𝑆𝑖𝑛 (700𝜋𝑡+270)
Amplitude = 6
Frequency = 350 Hz
Time = 0.002 s
Phase angle = 270˚

Answer for 5th Question
Let the isotropic free loss be IFL,
According to the formula for calculating the isotropic free loss,
IFL= {20log10 (35.863x106) +20log10 (6x109)}–147.56
IFL= (151.092932324+195.563025008) –147.56
IFL= 346.655957332-147.56
IFL= 199.095957332 dB

Answer for 6th Question
Equation,
S (t) =5 sin (100𝜋𝑡) + sin (300𝜋𝑡) + sin (600𝜋𝑡)
All frequencies calculated are 50 Hz, 150 Hz and 300 Hz.
The fundamental frequency calculated is 50 Hz.
The equation defining the spectrum is written below:
S (f) = i/2{5(d(f-50)-d(f+50))+(d(f-150)-d(f+150))+d(f-300)-d(f+300))}
Let Bandwidth be H,
H= 50+300+150
H= 500 Hz
To find the channel capacity the Nyquist formula for noiseless channel is followed where the
bandwidth is fixed.
Channel Capacity= 2*H*log2 (N)
Let Channel Capacity be CC and number of signal level be N.
For N= 2,
CC= 1000 bits/sec
For N= 4,
CC= 2000 bits/sec
For N=8,
CC= 4000 bits/sec